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पाठ 2- इंडो-चाइना में राष्ट्रवादी आंदोलन इतिहास के नोट्स| Class 10th

January 7, 2019, 10:33 pm
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पठन सामग्री और नोट्स (Notes)| पाठ 2-  इंडो-चाइना में राष्ट्रवादी आंदोलन (Indo-China me Rashtravaadi Aandolan) Itihas Class 10th

चीन के साये से आजादी

• इंडो-चाइना में आधुनिक देश वियतनाम, लाओस और कंबोडिया शामिल हैं।

• 1945 में वियतनाम ने फ्रांस से औपचारिक स्वतंत्रता प्राप्त की। स्वतंत्रता से पहले, यह चीनी साम्राज्य के अधीन था।

• स्वतंत्रता के बाद भी, वियतनाम ने चीनी संस्कृति और चीनी शासन व्यवस्था को अपनाए रखा था।

औपनिवेशिक वर्चस्व और उसका प्रतिरोध

• 1858 में फ्रांसीसी सैनिक वियतनाम की धरती पर डेरा डाला।
→ 1880 के दशक के मध्य तक उन्होंने उत्तरी क्षेत्र पर अपनी पकड़ मजबूत बना ली थी।

फ्रांसीसी ने कॉलोनियों को क्यों जरूरी समझा?

• प्राकृतिक संसाधनों और अन्य आवश्यक वस्तुओं की आपूर्ति करने के लिए।

• पिछड़े समाजों तक सभ्यता की रोशनी पहुँचना 'विकसित' यूरोपीय राष्ट्र का दायित्व है।

क्या उपनिवेशों का विकास करना जरूरी है?

• वियतनाम के आर्थिक विकास की बाधाऐें  
→ ज्यादा आबादी
→ कृषि की कम उत्पादकता।
→ किसानों पर भारी कर्ज़

वियतनाम एक कॉलोनी के रूप में कैसे विकसित हुआ?

• वियतनाम की अर्थव्यवस्था मुख्य रूप से चावल की खेती और रबर के बागानों पर आधारित था। इन पर मुख्य रूप से फ्रेंच और वियतनामी अभिजात वर्ग का स्वामित्व था।

• इस क्षेत्र की जरुरतों को पूरा करने के लिए रेल और बंदरगाह की सुविधाएं विकसित की गईं थी।

• रबर के बागानों में काम करने के लिये वियतनामी मजदूरों से एकतरफा अनुबंध किया गया था।

• फ्रांस ने अर्थव्यवस्था के औद्योगिकीकरण के लिए कोई खास काम नहीं किया।

औपनिवेशिक शिक्षा की दुविधा

• फ्रांसीसी ने देश के स्थानीय लोगों को सभ्य बनाने के लिए आधुनिक शिक्षा को ही एक रास्ते के रूप में देखते थे।

आधुनिक सोच : शिक्षा की भाषा

• चीनी भाषा वियतनामी कुलीन लोगों द्वारा इस्तेमाल की जाने वाली भाषा थी।

• चीनी भाषा वियतनामी धनी और अभिजात्य लोगों द्वारा इस्तेमाल की जाने वाली भाषा थी।

• कुछ नीति-निर्माताओं ने फ्रेंच भाषा को शिक्षा के माध्यम के रूप में उपयोग करने की आवश्यकता पर जोर दिया ताकि वे फ्रांसीसी संस्कृति की श्रेष्ठता के कायल हो जाए।

• अन्य लोगों का सुझाव था कि छोटी कक्षाओं में वियतनामी भाषा में और उच्च कक्षाओं में फ्रेंच में भाषा में शिक्षा दिया जाए।

• फ्रेंच सीखने और फ्रांसीसी संस्कृति अपनाने वाले लोगों को फ्रेंच नागरिकता से पुरस्कृत किया जाता था।

• वियतनाम के कवेल धनी वर्ग ही स्कूलों में दाखिला ले सकता था।

• स्कूल की पाठ्यपुस्तकों में फ्रांसीसीयों और औपनिवेशिक शासन को सही ठहराया था।

आधुनिक दिखने की चाह

• पश्चिमी शैली की शिक्षा प्रदान करने के लिए 1907 में टोनकिन फ्री स्कूल शुरू किया गया था।
• स्कूल अपने छात्रों को पश्चिमी शैलियों को अपनाने के लिए प्रोत्साहित किया जैसे बच्चों को छोटे -छोटे बाल रखने की सलाह दी जाती थी।

स्कूलों में विरोध के स्वर

• शिक्षकों और छात्रों ने पुस्तकों एवं पाठ्यक्रमों का विरोध किया।

• 1920 के दशक आते -आते छात्र- छात्राओं राजनीतिक दलों का गठन करने लगे थे, जैसे कि युवा अन्नान की पार्टी।

• स्कूल राजनीतिक और सांस्कृतिक लड़ाई के लिए एक महत्वपूर्ण स्थान बन गए।

• वियतनामी बुद्धिजीवियों ने वियतनामी भूभाग और संस्कृति दोनों के नुकसान होने की आशंका जताई।

साफ़- सफ़ाई, बीमारी और रोज़मर्रा प्रतिरोध

हनोई में प्लेग और हड़ताल

• फ्रांस का हनोई शहर एक सुंदर और स्वच्छ शहर के रूप में बनाया गया था।

• 1903 में, हनोई के आधुनिक हिस्से में ब्यूबॉनिक प्लेग फ़ैल गया था।  

• सीवरों ने एक बड़े परिवहन प्रणाली के रूप में भी काम किया, जिससे चूहों को शहर के चारों ओर निर्बाध घूमने के लिए आदर्श साबित हुआ।

चूहों की पकड़- धकड़

• चूहों की घुसपैठ रोकने के लिए 1902 में चूहों को पकड़ने की मुहिम शुरू किया गया था।

• फ्रांसीसी ने वियतनामी श्रमिकों को काम पर रखा था और उन्हें प्रत्येक चूहे पकड़ने पर भुगतान किया जाता था।

• वियतनामी श्रमिक चूहों को पकड़ कर उसकी पूंछ काट लेते थे और उसे जिंदा जाने देते थे।

• प्रतिरोध और असहयोग से तंग आकर फ्रांसीसियों ने चूहों के बदले पैसे देने की कार्यक्रम बंद कर दी।

धर्म और उपनिवेशवाद- विरोध

• वियतनामिओ के धार्मिक विश्वास बौद्ध धर्म, कन्फ्यूशीवाद और स्थानीय प्रथाओं के रिवाज़ पर आधारित था।

• वियतनाम में फ्रांसीसिओं ने ईसाई धर्म की शुरुआत की।

1868 में विद्वानों का विद्रोह

• इस विद्रोह का नेतृत्व इंपीरियल कोर्ट के अधिकारियों ने किया था।
• न्गू अन और हा तिएन ने विद्रोह का नेतृत्व किया और एक हजार कैथोलिकों को मार डाला।
• 18 वीं शताब्दी के मध्य तक 3,00,000 लोगों को ईसाई धर्म में परिवर्तित कर दिए गए।
• विद्रोह को फ्रांसीसिओं द्वारा कुचल दिया गया।

होआ हाओ आंदोलन

• होआ हाओ आंदोलन की शुरुआत 1939 में इसके संस्थापक हुइन्ह फू सो के नेतृव में हुई थी।

• उन्होंने फ़िजूल खर्च की आलोचना की और बालिका वधुओं की बिक्री, जुआ और शराब और अफीम के इस्तेमाल का भी विरोध किया।

• फ़्रांसिसी ने उसे पागल घोषित कर दिया और उसे पागलखाने में भेज दिया।

• 1941 में उन्हें मुक्त कर वितयनाम से लाओस भेज दिया और उनके अनुयायियों को यातना शिविर में डाल दिया गया।

आधुनिकीकरण की संकल्पना

• आधुनिकीकरण और राष्ट्रवाद के संबंध के दो राय
→ एक राय यह थी की पश्चिमी वर्चस्व का मुकाबला करने के लिए वियतनामी परंपराओं को मजबूत करना चाहिए।
→ दूसरी राय यह थी की विदेशी वर्चस्व का विरोध करते हुए वियतनामी को पश्चिमी सभ्यता से कुछ सीखना चाहिए।

• फान बोई चाऊ एक कन्फ्यूशियस विद्वान और एक राष्ट्रवादी थे।
→ उन्होंने 1903 में रिवोल्यूशनरी सोसाइटी नामक पार्टी का गठन किया।

• फान चू त्रिंह राजशाही के खिलाफ था और एक लोकतांत्रिक गणराज्य की स्थापना करना चाहता था।
→ वह पश्चिमी सभ्यता को पूरी तरह खारिज करने के खिलाफ थे।

आधुनिक बनने के अन्य तरीके: जापान और चीन

पूरब की ओर चलो

• 1907-08 में आधुनिक शिक्षा प्राप्त करने के लिए लगभग 300 वियतनामी छात्र जापान गए थे।
• उनका उद्देश्य फ्रांसीसी को बाहर निकालना और न्गूयेन वंश को फिर से गद्दी पर बैठना था।
• वे जापान से मदद चाहते थे और टोक्यो में एक पुनर्स्थापना सोसायटी की स्थापना की।
• लेकिन 1908 के बाद जापान सरकार ने पुनर्स्थापना सोसायटी जैसी गतिविधियों का दमन शुरू कर दिया, और फ़ान बोई चाऊ ने उनमें से बहुतों  को देश से बाहर निकल दिया। वियतनामी छात्रों को  चीन और थाईलैंड में शरण लेना पड़ा।

वियतनाम पर चीनी प्रभाव

• जब 1911 में सन यात सेन ने चीन में राजशाही को उखाड़ फेंका, इससे प्रेरणा लेते हुए वियतनामी छात्रों ने वियतनाम मुक्ति एसोसिएशन का गठन कर डाला।

• इसका उद्देश्य वियतनाम में एक लोकतांत्रिक गणराज्य और एक संवैधानिक राजतंत्र का स्थापना करना था।

कम्युनिस्ट आंदोलन और वियतनामी राष्ट्रवाद

• 1930 के महामंदी ने वियतनाम में बेरोजगारी, गरीबी और ग्रामीण विद्रोह को जन्म दिया।

• हो ची मिन्ह ने फरवरी 1930 में वियतनामी कम्युनिस्ट पार्टी की स्थापना की।

• जापान ने 1940 में वियतनाम पर कब्जा कर लिया।

• वियतनाम की आज़ादी के लिए वियतनाम स्वंत्रता लीग (जिसे विएट मिन्ह के रूप में जाना जाता है) ने जापानियों को मुँहतोड़ जबाब दिया और हनोई को आज़ाद करा लिया। हो ची मिन्ह सितंबर 1943 में वियतनाम के लोकतांत्रिक गणराज्य के अध्यक्ष बने।

वियतनाम गणराज्य

• फ़्रांसिसी शासक, सम्राट बाओ दाई को एक कठपुतली की तरह इस्तेमाल करते हुए देश पर कब्ज़ा जमाए रखना चाहता था।

• सालों की लड़ाई के बाद, आखिरकार 1954 में दीएन बिएन फू ने फ़्रांसिसीयों को हराया।

• फ़्रांसिसीयों के हार के बाद जिनेवा में आयोजित शांति वार्ता में वियतनाम को उत्तरी वियतनाम और दक्षिण वियतनाम बाँट दिया गया।
→ हो ची मिन्ह और कम्युनिस्टों ने उत्तर में सत्ता स्थापित की।
→ बाओ दाई ने दक्षिण में सत्ता स्थापित की।

• बाओ दाई शासन जल्द ही नगो दीन्ह डायम के नेतृत्व में तख्तापलट द्वारा समाप्त कर दिया गया था।

• नागो दीन्ह दीम के नेतृत्व में तानाशाह सरकार बनी।  
→ इसका विरोध नेशनल लिबरेशन फ्रंट (एनएलएफ) नाम के एक व्यापक मोर्चा द्वारा किया गया था।

• उत्तरी वियतनाम की मदद से एनएलएफ ने देश के एकीकरण के लिए लड़ाई लड़ी।

युद्ध में अमेरिका का प्रवेश

• साम्यवाद के डर से वियतनाम में अमेरिका ने हस्तक्षेप किया।

• हजारों अमेरिकी सैनिक भारी हथियारों और टैंकों से लैस होकर पहुंचे।

• घातक रासायनिक हथियार जैसे नापाम, एजेंट ऑरेंज और फॉस्फोरस बमों का उपयोग कर असंख्य गाँव को नष्ट कर दिया गया।

• युद्ध का असर अमेरिका में भी साफ महसूस किया जा सकता था।
→ अमेरिकी लोग भी युद्ध में शामिल होने के लिए सरकार की आलोचना कर रहे थे।

हो ची मिन्ह भूलभुलैया मार्ग

• फुटपाथों और सड़कों का एक विशाल नेटवर्क जिसका उपयोग उत्तर से दक्षिण तक सैनिकों और रसद सामग्रियों को ले जाने के लिए किया जाता था।

• अमेरिका ने नियमित आपूर्ति को बाधित करने के लिये इस मार्ग पर भारी बमबारी की, लेकिन गहन बमबारी के बाद भी इस महत्वपूर्ण आपूर्ति लाइन को बाधित नहीं कर पाए क्योंकि उन्हें फ़ौरन मरम्मत कर लिया जाता था।

युद्ध का अंत

• अमेरिका अपने लक्ष्य को हासिल करने में विफल रहा था।

• यह पहला युद्ध था जिसे टेलीविजन युद्ध कहा गया है क्योंकि इस युद्ध के दृश्य दैनिक समाचार कार्यक्रमों में टेलीविजन पर्दो पर दिखाए जाते थे।

• जनवरी 1974 में पेरिस शांति समझौते के साथ ही अमेरिका के साथ संघर्ष को समाप्त कर दिया।

• नेशनल लिबरेशन फ्रंट (एनएलएफ) ने 30 अप्रैल 1975 को साइगॉन के राष्ट्रपति भवन पर कब्जा किया और वियतनाम के दोनों हिस्सों को मिला कर एक राष्ट्र की स्थापना कर दी गई।  




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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.6

January 7, 2019, 10:56 pm
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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.6

Exercise 3.6

In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

1. 5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of1 pen and 1 pencil.

Solution


2. 7 audio cassettes and 3 video cassettes cost Rs 1110, while 5 audio cassettes and 4 video cassettes cost Rs 1350. Find the cost of an audio cassette and a video cassette.

Solution

Let the cost of a audio cassette be Rs x and that of a video cassette be . Rs y Then,
Hence, cost of one audio cassette = Rs 30 and cost of one video cassette = Rs 300 .

3. Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, then number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.

Solution 

Let the number of pens be x and that of pencil be y. then,

4. 4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950. Find the cost of 2 chairs and 1 table.

Solution

Let the cost of a table be Rs x and that of a chairs be Rs y Then,
Cost of 2 chairs = 300 Rs and cost of 1 table = 450
∴ The cost of 2 chairs and 1 table 300 + 450 + 750

5. 3 bags and 4 pens together cost Rs 257 whereas 4 bags and 3 pens together cost R 324. Find the total cost of 1 bag and 10 pens.

Solution


Hence, the total cost of 1 bag and 10 pens 75+ 80 = 155.

6. 5 books and 7 pens together cost Rs 79 whereas 7 books and 5 pens together cost Rs 77. Find the total cost of 1 book and 2 pens.

Solution

Let the cost of a book be Rs x and that of a pen be Rs y. Then,

Hence, the total cost of 1 book and 2 pens 6 + 14 = 20

7. Jamila sold a table and a chair for 1050 , thereby making a profit of 10% on a table and 25% on the chair . If she had taken a profit of 25% on the table and 10% on the chair she would have got Rs 1065. Find the cost price of each .

Solution

Let the CP of the table be Rs x and that of the chair be Rs y.

8. Susan nvested certain amount of money in two shemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively . She received Rs 1860 as annual interest . However, had she interchanged the amount of investment in the two schemes, she would have received Rs 20 more as annual interest . How much money did she invest in each scheme ?

Solution

Let the money invested in Scheme A be Rs x and that in Scheme B be Rs y.
So, the money invested in scheme A = Rs 12,000 and in scheme B = Rs 10,000.

9.The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution

x = 50
Hence, cost of 1 bat = x = 500
Hence, cost of 1 ball = x = 50

10. A lending library has a fixed charge for the first three days and additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution

To find:
(1) the fixed charge
(2) The charge for each day

Let the fixed charge be Rs x
And the extra charge per day be Rs y.
According to the given conditions,
x + 4y = 27
x + 4y - 27 = 0 ...(i)
x + 2y = 21
x + 2y - 21 = 0 ...(ii)
Subtracting equation 1 and 2 we get
2y = 6
y = 3
Substituting the value of y in equation 1 we get
x + 4(3) - 27 = 0
x + 12 - 27 = 0
x - 15 = 0
x = 15
Hence the fixed charge is x = Rs 15 and the charge of each day y = Rs 3

11. The cost of 4 pens and 4 pencils boxes is 100 . Three times the cost of a pen is 15 more than the cost of a pencil box . Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box .

Solution

Let the cost of 1 pen be x and that of 1 pencil box be y.
Now,
Cost of 4 pens + Cost of 4 pencil boxes = Rs 100 (given)
⇒ 4x + 4y = 100
⇒ x + y = 25 ...(i)
Also,
3 × Cost of a pen = Cost of a pencil box + 15
3x = y + 15
⇒ 3x - y = 15 ...(ii)
Adding (i) and (ii), we get
4x = 40
⇒ x = 10
Putting x = 10 in (i), we get
10 + y = 25
⇒ y = 15
Thus, the cost of a pen = 10 and that of a pencil box = 15.

12. One says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their respective capital ?

Solution

To find:
(1) Total amount of A.
(2) Total amount of B.
Suppose A has Rs x and B has Rs y
According to the given conditions,
x + 100 = 2(y − 100)
x + 100 = 2y − 200
x − 2y = −300 ....(1)
and
y + 10 = 6(x − 10)
y + 10 = 6x − 60
6x − y = 70 ....(2)
Multiplying equation (2) by 2 we get
12x − 2y = 140 ....(3)
Subtracting (1) from (3), we get
11x = 440
x = 40
Substituting the value of x in equation (1), we get
40 - 2y = -300
-2y = -340
y = 170
Hence A has x = Rs40 and B has y = Rs170

13. A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you." B replies, "if you give me 10, I will have thrice as many as left with you." How many mangoes does each have?

Solution

To find:
(1) Total mangoes of A.
(2) Total mangoes of B.

Suppose A has x mangoes and B has y mangoes,
According to the given conditions,
x + 30 = 2(y-30)
x + 30 = 2y - 60
x - 2y + 30 + 60 = 0
x - 2y + 90 = 0
y + 10 = 3(x-10)
y + 0 = 3x - 30
y - 3x + 10 + 30 = 0
y - 3x + 40 = 0
Multiplying eq. 1 by (3),
3x + 6y + 270 = 0 .... (3) and
Now adding eq.2 and eq.3
5y = 310
y = 310/5
y = 62
x - 2 × 62 + 90 = 0
x - 124 + 90 = 0
x - 34 = 0
x = 34
Hence A has 34 mangoes and B has 62 mangoes .

14. Vijay had some bananas, and he divided them into two lots A and B . He solds first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Rs 1 per banana and got a total of 400 . If he had sold the first lot at the rate of 1 per banana and the second lot at the rate of Rs 4 per five bananas, his total collection would have been Rs 460 . Find the total number of bananas he had.

Solution

Let the bananas in lot A be x and that in lot B be y.
Vijay sold 3 bananas for Rs 2 in lot A.
So, the cost of x bananas in lot A = 2/3 x
∴ 2/3 x + y = 400
⇒ 2x + 32x + 3y = 1200n ...(i)
Now, if he sells the first lot at the rate of Rs 1 per banana and second for Rs 4 for 5 bananas, then
x + 4/5 y = Rs 460
⇒ 5x + 4y = 2300 ...(ii)
Solving (i) and (ii), we get
x = 300 and y = 200
So, the total number ob bananas = x+y = 300 + 200 = 500

15. On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain the fridge at 5% loss. He gains Rs 1500 on the transaction. Find the actual prices of T.V. and fridge

Solution

Given:
(i) On selling of a T.V. at 5% gain and a fridge at 10% gain, shopkeeper gain Rs.2000.
(ii) Selling T.V. at 10% gain and fridge at 5% loss. He gains Rs. 1500.
To find: Actual price of T.V. and fridge.
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Quick Revision Notes of Chapter 1 Rise of Nationalism in Europe for Board Exams| Class 10th

January 9, 2019, 7:57 am
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Quick Revision Notes of Chapter 1 Rise of Nationalism in Europe for Board Exams| Class 10th - Download PDF 

• In 19th century Europe, nationalism brought drastic changes in the political and mental world of Europe.
→ This led to the emergence of different nation-State in place of the multi-national dynastic empires of Europe.

• French Revolution (1789) was the first expression of nationalism.
→ It ended monarchy in France and gave power to the citizens.
→ Napoleonic code (Civil Code of 1804) made administrative machinery more efficient.

• The word 'liberalism' stood for freedom in the political, social and economic spheres.

• In 1815, Napoleon was defeated in the 'Battle of Waterloo' by the collective European forces of Britain, Russia, Prussia and Austria.
→ The European forces drew up the Treaty of Vienna of 1815 with the object of undoing most of the changes happened during the Napoleonic wars.
→ Congress of Vienna did not tolerate criticism and dissent and resulted in several liberal-nationalist groups going underground.
→ Now, the revolutionary groups started seeing the formation of nation-state as a necessity to the freedom struggle.

• In January 1871, the Prussian king, William I, was proclaimed German Emperor when three war over seven years ended in Prussian victory and completed the process of unification under the leadership of Otto Van Bismarck.

• Chief Minister Cavour who led the movement to unify the regions of Italy
→ In 1861 Victor Emmanuel II was proclaimed king of united Italy.

• The Act of Union (1707) between England and Scotland that resulted in the formation of the ‘United Kingdom of Great Britain’.
→ In 1801, Ireland was forcibly taken by the British after the failed revolution.
→ A new ‘British Nation’ was founded through the propagation of a dominant English culture.

• Nations were portrayed as female figure (Allegory).
→ Marianne was the female allegory of France while Germania became the allegory of Germany.

• Nationalism along with imperialism led Europe to a disaster which ultimately resulted in the First World War in 1914.

Important terms 

1. Absolutist - A government or system of rule that has no restraints on the power exercised. in history, the term refers to a form of monarchical government that was centralised, militarised and
repressive.

2. Utopian - A vision of a society that is so ideal that it is unlikely to actually exist.

3. Plebiscite - A direct vote by which all the people of a region are asked to accept or reject a proposal.

4. Conservatism - A political philosophy that stressed the importance of tradition, established institutions and customs and preferred gradual development to quick change.

5. Zollverein - It was formed at the initiative of Prussia and joined by most of the German States. The union abolished tariff barriers and reduce the number of currencies from over thirty to two.
6. Romanticism - A cultural movement which sought to develop a particular form of Nationalist sentiment.

7. Liberalism - This word is derived from the Latin root liber meaning free.

8. Junkers - Important personalities and Large landowners of Prussia.

Important Personalities 

1. Mazzini - An Italian revolutionary who is known for his noble efforts to achieve the unification of Italy. He is regarded as the spiritual force behind the Italian unification. He founded Young Italy in Marseilles and Young Europe in Berne.

2. Garibaldi - He is known as the physical force or the Sword of Italy. He with Mazzini launched the young Italy movement for the unification of Italy. He involved The sardinian sailors to Revolt in 1835 A.D.

3. Cavour - Cavour was the chief minister of Sardinia-Piedmont. He was neither a revolutionary nor a democrat. He led the movement to unify the regions of Italy.

3. Giuseppe Garibaldi  - He helped Chief Minister, Cavour by leading a large number of armed volunteers and marching into South Italy and the Kingdom of the Two Sicilies and succeeded in winning the support of the local peasants in order to drive out the Spanish rulers.

4. Bismarck - He played the most important role in the unification of Germany, his policy of blood and iron was mainly responsible. Meaning of symbols.




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NCERT Solutions for Class 12th: Ch 8 Controlling (MCQ and Short Questions)

January 9, 2019, 8:30 am
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NCERT Solutions for Class 12th: Ch 8 Controlling (MCQ and Short Questions)

Exercises

Page No. 233

Multiple Choice

For the following, choose the right answer.
1. An efficient control system helps to
(a) Accomplishes organisational objectives
(b) Boosts employee morale
(c) Judges accuracy of standards
(d) All of the above
► (d) All of the above

2. Controlling function of an organisation is
(a) Forward looking
(b) Backward looking
(c) Forward as well as backward looking
(d) None of the above
► (c) Forward as well as backward looking

3. Management audit is a technique to keep a check on the
performance of
(a) Company
(b) Management of the company
(c) Shareholders
(d) Customers
► (b) Management of the company
4. Budgetary control requires the preparation of
(a) Training schedule
(b) Budgets
(c) Network diagram
(d) Responsibility centres
► (b) Budgets
5. Which of the following is not applicable to responsibility
accounting
(a) Investment centre
(b) Accounting centre
(c) Profit centre
(d) Cost centre
► (b) Accounting centre

Page No. 234

Short Answer Type

1. Explain the meaning of controlling.

Answer

Controlling is one of the important functions of a manager. It means ensuring that activities in an organisation are performed as per the plans. It also ensures that an organisation’s resources are being used effectively and efficiently for the achievement of predetermined goals. Controlling is, thus, a goal-oriented function.

2. ‘Planning is looking ahead and controlling is looking back.’ Comment.

Answer

Planning and controlling are inseparable twins of management. It is the planning, which decides the controlling process on the one hand. On the other hand, it is controlling which provides sound basis for planning. In this way, planning and controlling both of them are interdependent. The controlling process and technique is decided by planning. It is the controlling, which directs and diverts the course of planning. Controlling invites our attention to those areas, where planning is necessary. Controlling provides statistical information for planning.

3. 'An effort to control everything may end up in controlling nothing.' Explain.

Answer

It is not possible to look over all thing at the same time thus when we want to control everything it may end up in controlling nothing. It is neither economical nor easy to keep a check on each and every activity in an organisation. Control should, therefore, focus on key result areas (KRAs) which are critical to the success of an organisation. These KRAs are set as the critical points. If anything
goes wrong at the critical points, the entire organisation suffers.

4. Write a short note on budgetary control as a technique of managerial control.

Answer

Budgetary control is a technique of managerial control in which all operations are planned in advance
in the form of budgets and actual results are compared with budgetary standards. This comparison reveals the necessary actions to be taken so that organisational objectives are accomplished. Budgeting offers the following advantages:
• It focuses on specific and time-bound targets and thus, helps in attainment of organisational objectives.
• It is a source of motivation to the employees who know the standards against which their
performance will be appraised and thus, enables them to perform better.
• It helps in optimum utilisation of resources by allocating them according to the requirements of different departments.
• Budgeting is also used for achieving coordination among different departments of an organisation and highlights the interdependence between them.
• It facilitates management by exception by stressing on those operations which deviate from budgeted standards in a significant way.

5. Explain how management audit serves as an effective technique of controlling.

Answer

Management audit serves as an effective technique of controlling as it refers to systematic appraisal of the overall performance of the management of an organisation. It is helpful in identifying the deficiences in the performance of management function. The benefit of management audit are:
• It helps to locate present and potential deficiencies in the performance of management functions.
• It helps to improve the control system of an organisation by continuously monitoring the performance of management.
• It ensures updating of existing managerial policies and strategies in the light of environmental changes.

NCERT Solutions of Chapter 8 Controlling (Long Anwer Questions)

Go Back To NCERT Solutions of Business Studies Class 12th

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Science Sample Question Paper 2018-19| Class 10th

January 10, 2019, 4:14 am
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Download Sample Question Paper of Science Class 10th Term II 2018-19

Time: 3 Hours
MM: 80 Marks

General Instructions:
(i) The question paper comprises of five sections – A, B, C, D and E. You are to attempt all the sections.
(ii) All questions are compulsory.
(iii) Internal choice is given in sections B, C, D and E.
(iv) Question numbers 1 and 2 in Section-A are one mark questions. They are to be answered in one word or in one sentence.
(v) Question numbers 3 to 5 in Section- B are two marks questions. These are to be answered in about 30 words each.
(vi) Question numbers 6 to 15 in Section-C are three marks questions. These are to be answered in about 50 words each.
(vii) Question numbers 16 to 21 in Section-D are 5 marks questions. These are to be answered in about 70 words each.
(viii) Question numbers 22 to 27 in Section- E are based on practical skills. Each question is a two marks question. These are to be answered in brief. 


Section A

1. Name a common nutrient that is absorbed in the small intestine and reabsorbed by the kidney tubules .

2. The presence of a particular group of bacteria in water bodies indicates contamination. Identify the group .
Section B 

3. How is Magnesium Chloride formed by the transfer of electrons ? Why does the solution of Magnesium chloride conduct electricity ?

4. In a flowering plant, summarize the events that take place after fertilization.

5. A ray of light enters into benzene from air. If the refractive index of benzene is 1.50, by what percent does the speed of light reduce on entering the benzene ?
 OR 
For the same angle of incidence in media A,B and C, the angles of refraction are 20° ,30° and 40° respectively. In which medium will the velocity of light be maximum ? Give reason in support of your answer.
Section C  

6. What happens when aqueous solutions of Sodium sulphate and Barium chloride are mixed ? Give a balanced equation for the reaction with state symbols. Name and define the type of chemical reaction involved in the above change.


7. Identify the compound of calcium which is used for plastering of fractured bones. With the help of chemical equation show how is it prepared and what special precautions should be taken during the preparation of this compound.
OR
‘Sweet tooth may lead to tooth decay’. Explain why ? What is the role of tooth paste in preventing cavities ?

8. The electronic configuration of an element ‘X’ is 2,8,6. To which group and period of the modern periodic table does ‘X’ belong .State it valency and justify your answer in each case.

9. Pertaining to endocrine system, what will you interpret if
i) You observe swollen neck in people living in the hills
ii) Over secretion of Growth Hormone takes place during childhood
iii) Facial hair develops in boys aged 13.

10. A variegated leaf with green and yellow patches in used for an experiment to prove that chlorophyll is required for photosynthesis. Before the experiment the green portions (A), and the pale yellow portions (B), are observed. What will be the colour of ‘A’ just before and after the starch test? Also write the equation of photosynthesis and mark, as well as validate from which molecule the by-product is obtained.

11. The image of an object formed by a mirror is real, inverted and is of magnification -1. If the image is at the distance of 30 cm from the mirror, where is the object placed ? Find the position of the image if the object is now moved 20 cm towards the mirror. What is the nature of the image obtained ? Justify your answer with the help of ray diagram.
OR
What is meant by power of a lens ? You have three lenses L1, L2 and L3 of powers +10D, +5D and -10D respectively. State the nature and focal length of each lens. Explain which of the three lenses will form a virtual and magnified image of an object placed at 15 cm from the lens. Draw the ray diagram in support of your answer.

12. Two lamps, one rated 100 W at 220V and the other 200 W at 220V are connected (i) in series and (ii) in parallel to electric main supply of 220V. Find the current drawn in each case.

13. The figure below shows three cylindrical copper conductors along with their face areas and lengths. Compare the resistance and the resistivity of the three conductors. Justify your answer.

14. What is biogas ? Describe the steps involved in obtaining biogas.  

15. How is ozone both beneficial and damaging ? How can we prevent the damaging effect of ozone ?
OR 
The flow of energy between various components of the environment has been extensively studied. Give an outline of the findings. 

Section D

16. a) How will you show experimentally that metals are good conductors of heat.
b) Describe the extraction of Mercury metal from its ore Cinnabar (HgS).

17. A compound A (C2H4O2) reacts with Na metal to form a compound ‘B’ and evolves a gas which burns with a popsound. Compound ‘A’ on treatment with an alcohol ‘C’ in presence of an acid forms a Sweet smelling compound ‘D’(C4H8O2 ). On addition of NaOH to ‘D’ gives back B and C. Identify A, B, C and D write the reactions involved.   
OR
a) Explain why carbon forms covalent bond ? Give two reasons for carbon forming a large number of compounds.

b) Explain the formation of ammonia molecule.

18. a) Draw the diagram of female reproductive system and match and mark the part(s): 5
i) Where block is created surgically to prevent fertilization.
ii) Where CuT is inserted?
iii) Inside which condom can be placed.
b) Why do more and more people prefer to use condoms? What is the principle behind use of condoms ?

19. Name the phenomenon that governs the following:
i) Green beetles living in green bushes are not eaten by the crows.
ii) Number of blue beetles in green bushes increases, only because the red beetles living there were trampled by a herd of elephants.
iii) No ‘medium height plants’ are obtained in F1 generation, upon crossing pure tall and dwarf pea plants.
iv) Tails of mice were surgically removed for several generations; still mice had tails in the following generations.
v) A migrant beetle reproduces with the local population; as a result genes of migrant beetle enter the new population.
OR
a) What are fossils and how is age of fossils determined ?
b) During artificial selection, which features of wild cabbage were selected to give rise to i) Cabbage ii) Cauliflower

20. (a)What is meant by the term ‘power of accommodation’? Name the component of eye that is responsible for the power of accommodation.
(b) A student sitting at the back bench in a class has difficulty in reading. What could be his defect of vision ? Draw ray diagrams to illustrate the image formation of the blackboard when he is seated at the (i) back seat (ii) front seat. State two possible causes of this defect. Explain the method of correcting this defect with the help of a ray diagram.

21. (i) With the help of an activity, explain the method of inducing electric current in a coil with a moving magnet. State the rule used to find the direction of electric current thus generated in the coil. 

(ii) Two circular coils-1 and coil-2 are kept close to each other as shown in the diagram.Coil-1 is connected to a battery and key and coil-2 with a galvanometer. State your observation in the galvanometer:


(a) When key k closed ; (b) when key k is opened;
Give reason for your observations.

OR 
Name a device which converts mechanical energy into electrical energy. Explain the underlying principle and working of this device with the help of a labelled diagram.

Section E 


22. When few drops of phenolphthalein are added to a dilute solution of sodium hydroxide a pink colour is produced.
What will be the colour of the final mixture when excess of HCl is added to it ? (justify your answer)
OR  
Arrange the metals iron, magnesium, zinc and copper in the increasing order of their reactivity.
What will be the two observations made by the student when iron filings are added to copper sulphate solution ?

23. from an experiment to study the properties of acetic acid. Answer the following questions:
a) Name the substances which on addition to acetic acid produce carbon dioxide gas. Give relevant chemical equation for the above?
b) How is CO2 gas tested in the laboratory?

24. When observed under high power of the microscope, ‘chain of buds’ is visible in the microscopic view. In which organism can it be observed ? Explain the process.
OR 
In the experimental set up on ‘CO2 is released during respiration,’ if one forgets to keep the vial with KOH in the conical flask, how will the result vary? Give details.

25. You soak seeds of bean and observe them after 2-3 days. What will be your observations ? 

26. The current flowing through a resistor connected in an electrical circuit and the potential difference developed across its ends are shown in the given ammeter and voltmeter. Find the least count of the voltmeter and ammeter .What is the voltage and the current across the given resistor ?  

 27. Consider the path of a ray of light passing through a rectangular glass slab for different angles of incidence. (i) Which one is greater: angle of incidence or angle of refraction ? (ii) What happens to the emergent angle on increasing the incident angle at air-glass interface? (iii) State the conditions when no refraction occurs. 


OR


Sunita takes a mirror which is depressed at the centre and mounts it on a mirror stand. An erect and enlarged image of her face is formed. She places the mirror on a stand along a meter scale at 15 cm mark. In front of this mirror, she mounts a white screen and moves it back and forth along the meter scale till a sharp, well-defined inverted image of a distant tree is formed on the screen at 35 cm mark.

(i) Name the mirror and find its focal length.
(ii) Why does Sunita get sharp image of the distant building at 35 cm mark?

View Solutions of Science Sample Question Paper 2018-19

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Solutions of Science Sample Question Paper 2018-19| Class 10th

January 10, 2019, 4:40 am
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Solutions of Science Sample Question Paper 2018-19| Class 10th

View Questions of Science Sample Question Paper 2018-19| Class 10th

Section A 

1. Glucose/Amino acids
2. Coliform bacteria 
Section B
3. Mg - 12  2, 8, 2
    Cl -   17  2, 8, 7 
MgCl2

Solution of Magnesium Chloride conduct electricity because an ionic compound dissociates into ions when dissolved in water and hence conduct electricity.

4. (Fertilization results in formation of zygote). 
Zygote divides several times, to form an embryo. The ovule develops a thick coat and is into seed. The ovary grows rapidly and ripens to form the fruit.

5.

OR 

Section C

6. A white precipitate is formed. 
Na2SO4(aq) + BaCl2(ag) → 2NaCl(aq) + BaSO4(↓) 
Doubled Displacement reaction 
It is a reaction in which there is an exchange of ions between the reactants .  

7. Plaster of Paris CaSO4 . 1/2 H2O 
Calcium Sulphate Hemihydrate
Preparation 
Precaution
Gypsum should not be heated above 373 K otherwise it will form CaSO4 .
OR
Sweet tooth leads to tooth decay. Which is caused by the action of Bacteria on food particles remaining in the mouth and acid is formed. The pH of the mouth falls below 5.5 and the tooth enamel dissolves resulting in cavities Toothpastes are generally basic, they neutralise the excess acid produced in the mouth and prevent tooth decay. 

8. X–2, 8, 6
a) Since ‘X’ has three energy shells and period number of an element is equal to the number of energy
shells, X belongs to 3rd period. 
b) X has 6 valence electrons it belongs to group 16. 
c) Valency will be 2. To acquire noble gas configuration it will gain 2 electrons.

9. i) less intake of Iodine (in the diet)
ii) will lead to gigantism
iii) timely secretion of testosterone

10. 10. Just before Starch test – Pale yellow 3
Just after Starch test – Blue black
Chlorophyll
O2 is obtained from water (H2O), as splitting of water results in formation of Hydrogen (used for
making glucose) and oxygen (by-product). 



OR
Power of a lens is the degree of convergence of divergence of light rays achieved by a lens. 
 

Lens L2 will form a virtual and magnified image of an object placed at 15 cm from the convex lens
because concave lens can never form virtual and magnified image of an object and convex lens form such image only when the object is placed between the optical centre and principle focus of the convex lens. 


12. 

13. 
  

14. Biogas is a mixture of methane, carbon dioxide, hydrogen, hydrogen sulphide. Following steps are involved in obtaining biogas:

i) Mixing (Slurry of cattle-dung and water)
ii) Digesting (decomposition of cattle-dung by anaerobic bacteria)
iii) Formation of biogas
iv) Residue left after the formation of biogas.

15. Damaging as it is a deadly poison.
Beneficial as it shields the surface of the earth from UV radiations of the Sun.
By not using synthetic chemicals like CFCs, that deplete O3 layer.

OR 

• Flow of energy is unidirectional.
• Terrestrial plants take about 1% of the Sun’s energy and change it to chemical energy.
• A great deal of energy is -lost as heat/ used for digestion/doing work/growth and reproduction.
• Only 10% of organic matter present at each trophic level (and available to next trophic level).
• Food chains are mainly of 3-4 trophic levels (because of 10% law) .
• The number of producers are maximum (the number reduces in subsequent trophic levels).
• Food webs are more common (as compared to isolated food chains).
• Biological magnification can be observed.

Section D 

a) Procedure
Observation - Heat is transferred from one end of metal wire to the free end of wire which melts
the wax and pin falls. Shows metals conduct heat.

b)

17. 

A - CH3COOH 
B - CH3COONa
C - C2H5OH
D - CH3COOC2H5 

OR 
a) Carbon has electronic configuration 2, 4. It could gain four electrons forming C-4 anion or lose 4
electrons to form C+4 cation . Both are not possible due to energy considerations. Carbon overcomes this problem by sharing electrons and forming covalent compounds.


Two reasons for forming large number of compounds:
1) Catenation
2) Tetra valency

b) Formation of NH3 molecule
N – 2, 5
H – 1
Three hydrgen atoms each share their 1 electron with nitrogen to form three covalent bonds
and make an ammonia molecule (NH3) ammonia molecule .
18 a)
Correct diagram with correct labelling, correctly matched with the following parts:-
i) Fallopian Tube/Oviduct
ii) Uterus
iii) Vagina
b) People prefer use of condoms as it prevents STDs/gives privacy to the user. Condoms help
create a mechanical barrier preventing meeting of sperms and ovum.

19. i) Natural selection
ii) Genetic drift
iii) Law of Dominance
iv) Acquired characters are not inherited
v) Gene flow

OR


a) Body or its parts that are not decomposed/preserved traces of organisms. (to begin with new line)

• Upon digging the earth, the fossils that are found closer to the surface are more recent
than the fossils in deeper layers.

• By detecting the ratios of different isotopes of the same element in the fossil material.

b) i) By selecting very short distances between leaves.
ii) By selecting sterile flowers.

20. (a) Power of accommodation: It is the ability of the eye lens to adjust its focal length. Ciliary
muscles of eye are responsible for change in its focal length. 

(b) Myopia
Causes : i) excessive curvature of the eye lens 3
ii) elongation of eyeball
This defect can be corrected by using a concave lens of suitable power.


21.(i)

(ii) a) The galvanometer needle deflects momentary in one direction because when the key is closed
,magnetic field lines around coil-2 increases momentary that causes induced current in coil-2. 
b) The galvanometer needle deflects momentary but in opposite direction because when the key is
opened, magnetic field lines around coil-2 decreases momentary that causes induced current in coil-2.  

OR

Section E

22. The colour of dilute solution of Sodium hydroxide turns pink or adding Phenolphthalein as NaOH is a base .When excess of HCl is added the final mixture becomes colourless due to neutralisation of base with an acid.
OR

Metals in increasing order of reactivity - Copper, iron ,zinc and magnesium
1. Color of the solution changes from blue to green
2. Reddish brown deposits on iron filings

23. a)NaHCO3

CH3COOH  + NaHCO3 → CH3COONa + H2O + CO2 
b) CO2 on passing through lime water turns milky.

24. Yeast
Budding – The parent (yeast) cell produces a small protuberance that grows to form a bud. The nucleus of the parent (yeast) cell divides, such that the daughter nucleus moves into the daughter cell and the process continues to form a chain of buds.
OR 
In absence of KOH- CO2 released by germinating seeds is not absorbed, partial vacuum is not created in the conical flask, air pressure in the flask is not reduced, water level does not rise in the delivery tube.

25. Soft/ruptured seed coat, radicle which emerges first (add comma) , leafy plumule, between the two cotyledons. 

26.

27. (i) ∠i = ∠e. 2
(ii) Angle of emergence also increases.
(iii) The light ray falls along the normal or the refractive index of the two optical media are equal.

OR 
(i) Concave mirror ; f = 35 - 15 = 20 cm
(ii) Because the incident rays parallel to each other after reflection from concave mirror meets at
focus and produce sharp image at focus.

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Notes of Ch 1 Nature and Significance of Management| Class 12th Business Studies

January 10, 2019, 9:28 pm
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Summary and Notes of Ch 1 Nature and Significance of Management| Class 12th Business Studies

Concept

Definition of Management(According to Harold Koontz and Heinz Weihrich)

"Management is the process of designing and maintaining an environment in which individuals, working together in groups, efficiently accomplish selected aims."

Effectiveness Versus Efficiency

Effectiveness and efficiency are two sides of the same coin. Effectiveness refers to complete the job on time, no matter whatever is the cost and on the other hand, efficiency refers to do the job in a cost-effective manner.

Characteristics of Management

• Management is goal-oriented process.
• Management is all-pervasive.
• Management is Multidimensional .
• Management is a continuous process.
• Management is a group activity.
• Management is a dynamic function.
• Management is an intangible force.

Objective of management

1. Organisational objectives: While fixing these objectives, management keeps in to consideration the benefit of all the related parties like (shareholder, employees, customers and the government). This also fulfils organisational economic objectives which are survival, profit and growth.

2. Social objectives: It refers to the consideration of the interest of the society during managerial activities. Organisation runs through the resources made available by the society. That is why it becomes the responsibility of every organisation to account for social benefits.
Main social objectives of management are included in the following list:
• To make available employment opportunities.
• To save the environment from getting polluted.
• To contribute in improving living standard.

3. Personal objectives: Organisations are made up of people who have different personalities, background, experiences and objectives. They all become part of the organisation to satisfy their diverse needs. These vary from financial needs such as competitive salaries and perks, social needs such as peer recognition and higher level needs such as personal growth and development.

Importance of Management 

• Management helps in achieving group goals.
• Management increases efficiency.
• Management creates a dynamic organisation.
• Management helps in achieving personal objectives.
• Management helps in the development of society.

Management as Science, Art and Profession.

Management as an art

• Existence of theoretical knowledge: Management is an art as a lot of literature is available in various areas of management.

• Personalised application: The use of this basic knowledge varies from individual to individual for example two dancers, two actors, two speakers or two writers will always differ in demonstrating their art.

• Based on practice and creativity: Just as art can be embellished with the help of practice, in the same way managerial skill also improves with practice.

Management as a science

Science refers to that systematic body of knowledge which is acquired on the basis of observation and experiments and verification of this knowledge is possible

• Systematised Body of knowledge: Science is a systematic body of knowledge because its principles are based on a cause and effect relationship.

• Principle based on experimentation: Scientific principles are first developed through observation and then tested through repeated experimentation under controlled conditions.

• Universal validity: Scientific principles have universal validity and application.

The management cannot be treated as a perfect science, but as its principles are subject to change with time, situation and human nature, it is better to call it applied science or inexact science.

Management as a Profession

Well- defined Body of knowledge: All profession are based on a well- defined body of knowledge that can be acquired through instruction.

• Restricted Entry: The entry to a profession is restricted through an examination or educational degree. For example charted accountant in India a candidate has to clear a specified examination. But as for management is concerned there is no such condition for being a manager. Hence, on this basis management cannot be accepted as a profession.

• Professional association: All professions are affiliated to a professional association which regulates entry, grants certificate of practice and formulates and enforces a code of conduct for Example Bar council of India for lawyers, medical council of India for doctors etc. on this basis management cannot be accepted as a profession.

• Ethical code of conduct: All professions are bound by a code of conduct which guides the behaviour of its members for example when doctors take a oath of ethical practice at the time they enter the profession. On this basis management cannot be accepted as a profession.

• Service motive: The main motive of profession is to serve the society and this feature is same as management.

Levels of Management

• Top-level Management: In top level management, Board of Directors, Chief Executive officer, etc are included. Management has all the authorities, and because of these authorities officers of these levels are accountable to owners or shareholder of the company.
• They formulate overall organisation goals and strategies for their achievement.
• They are responsible for all the activities of the business and for its impact on society.
Determining objectives, determining policies, determining activities, assembling resources, controlling the work performance, approving budget etc are the function of the top level management.

• Middle-level Management:  It lies between top-level and lower –level management. Under this divisional head, departmental head, deputy departmental heads, plant superintendents, and operations managers are included.
The main work of middle level management:
• Interpret the policies framed by top management.
• Ensure that their department has the necessary personnel.
• Assign necessary duties and responsibilities to them.
• Co-operate with other departments for smooth functioning.

• Lower-level Management: It is also known as supervisory management or operational management. Supervisors directly oversee the efforts of the workforce. Their authority and responsibility is limited according to the plans drawn by the top management. Supervisory management interact with their actual work force and pass on instructions of the middle management to the workers.

Function of Management

1. Planning: It means setting a goal in advance and developing a way of achieving them efficiently and effectively. Planning cannot prevent problems, but it can predict them and prepare contingency plans to deal with them if and when they order.

2. Organising: It refers to harmonious adjustment of various parts to achieve common objectives. It is the function of assigning duties, grouping tasks, establishing authority and allocating resources.
Staffing: it means appointing competent persons according to the importance of the post in the organisation. This is also known as the human resource function and it involves activities such as recruitment, selection, placement, and training personnel.

3. Directing: It refers to instructing, guiding, communicating, and inspiring people in the organisation. Under directing following four activities are included:
• Supervision
• Communication
• Leadership
• Motivation

4. Controlling: It is the management function of monitoring organisational performance towards the attainment of organisational goal. The task involved is:
• It involved establishing standards of performance.
• Comparing this with established standards and taking corrective action where any deviation found.

5. Coordination: It is a process to establish harmony among the different activities of an organisation, so that the desired objectives can be achieved.
Characteristics of coordination
• Coordination Integrates Group effort.
• Coordination Ensures unity of action.
• Coordination is a continuous process.
• Coordination is an all- pervasive function.
• Coordination is the responsibility of All Managers.
• Coordination is a Deliberate Function.

Importance of coordination

Coordination is important as it integrates the efforts of individuals, departments and specialists. The primary reason for coordination is that departments and individuals in the organisation are interdependent.

• Growth in size: As organisations grow in size, the number of people employed by the organisation also increases. Therefore, for organisation efficiency it is important to harmonise individual goals and organisational goal through coordination.

• Functional Differentiation: organisation divided in to separate departments for each function. All these departments may have their own objectives, policies. And each unit and department is performing activities in isolation from others and barriers between departments are becoming more rigid. So, the activity of each department needs to be focused on attainment of common organisational goals.

• Specialisation: Specialists usually think that they only are qualified to evaluate, judge, and decide according to their professional criteria. They do not take advice from others in serious matter. This often leads to conflict amongst different specialists as well as others in the organisation.

NCERT Solutions of Chapter 1 Nature and Significance of Management

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NCERT Solutions for Class 12th: Ch 8 Controlling (Long Answer Questions)

January 11, 2019, 12:25 am
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NCERT Solutions for Class 12th: Ch 8 Controlling (Long Answer Questions)

Long Answer Type

1. Explain the various steps involved in the process of control.

Answer

Controlling is a systematic process involving the following steps:

(i) Setting Performance Standards: The first step in the controlling process is setting up of performance standards. Standards are the criteria against which actual performance would be measured. Standards can be set in both quantitative as well as qualitative terms. Some of the qualitative standards are cost to be incurred, product units to be produced, time to be spent in performing a task etc. Improving goodwill and motivation level of employees are examples of qualitative standards.

(ii) Measurement of Actual Performance: Once performance standards are set, the next step is measurement of actual performance. Performance should be measured in an objective and reliable manner. Some of the techniques used for measuring the performance are personal observation, sample checking performance reports etc.

(iii) Comparing Actual Performance with Standards: This step involves comparison of actual performance with the standards. Such comparison will reveal the deviation between actual and desired results. Comparison becomes easier when standards are set in quantitative terms. For instance, performance of a worker in terms of units produced in a week can be easily measured against the standard output for the week.

(iv) Analysing Deviations: Some deviations in performance can be expected in all activities. It is therefore, important to determine the acceptable range of deviations. Also, deviations in key areas of business need to be attended more urgently as compared to deviations in certain insignificant areas. Critical point control and management by exception should be used by a manager in this regard.

(v) Taking Corrective Action: The final step in the controlling process is taking corrective action. No corrective action is required when the deviations are within acceptable limits. However, when the deviations go beyond the acceptable range, especially in the important areas, it demands immediate managerial attention so that deviations do not occur again and standards are accomplished. In case the deviations cannot be corrected through managerial action, the standards may have to be revised.

2. Explain the techniques of managerial control.

Answer

The various techniques of managerial control may be classified into two broad categories: traditional techniques, and modern techniques.

(i) Traditional Techniques: Those techniques which have been used by the companies for a long time now. However, these techniques have not become obsolete and are still being used by companies.
These include:

(a) Personal Observation
Personal observation enables the manager to collect first hand information. It also creates a psychological pressure on the employees to perform well as they are aware that they are being observed personally in their job.

(b) Statistical Reports
Statistical analysis in the form of averages, percentages, ratios, correlation etc. Present useful information to the managers regarding performance of the organisation in various areas. Such information when presented in the form of
charts, graphs, tables etc enables the managers to read them more easily and allow a comparison to be made with performance in previous periods and also with the benchmarks.

(c) Break-even Analysis
It is a technique used by managers to study the relationship between costs, volume and profits. It determines the probable profits and losses at different levels of activity. The sales volume at which there is no profit, no loss is known as break-even point. It is a useful 

(d) Budgetary Control: It is a technique of managerial control in which all operations are planned in advance in the form of budgets and actual results are compared with budgetary standards. This comparison reveals the necessary actions to be taken so that organisational goals are accomplished. A budget is a quantative statement for a definite future period of time for the purpose of obtaining a given objective. It is also a statement which reflects the policy of that particular period. It will contain figures of forecasts both in terms of time and quantities.

(ii) Modern Techniques: Modern techniques of controlling are those which are of recent origin and are comparatively new in management literature. These techniques provide a refreshingly new thinking on the ways in which various aspects of an organisation can be controlled. These
include:

(a) Return on Investment: Return on Investment (ROI) is a useful technique which provides the basic yardstick for measuring whether or not invested capital has been used effectively for generating reasonable amount of return. It can be calculated as under

(b) Ratio Analysis It refers to analysis of financial statements through computation of ratios. The most commonly used ratios are:
• Liquidity Ratios: Calculated to determinedly short term solvency of business.
• Solvency Ratios: Calculated to determine the long term solvency of business are known as Solvency ratios.
• Profitability Ratios: Calculated to analyse the profitability position of a business.
• Turnover Ratios: Calculated to determine the efficiency of operations based on effective utilisation of resources.

(c) Responsibility Accounting: It is a system of accounting in which different sections, divisions and departments of an organisation are set up as ‘responsibility centres’. The head of the centre is responsible for achieving the target set for his centre. Responsibility centres may be of the following types
• Cost Centre: It is a segment of an organisation in which managers are held responsible for the cost incurred in the centre but not for the revenues e.g., production depart
• Revenue Centre: It is held responsible for generating revenue, e.g., marketing department.
• Profit Centre: It is responsible for both cost and revenue e.g., repair and maintenance department.
• Investment Centre: It is responsible not only for profits but also for investments made in the centre in the form of assets.

(d) Management Audit: It refers to systematic appraisal of the overall performance of the management of an organisation. The purpose is to review the efficiency and effectiveness of management and to improve its performance in future periods. It is helpful in identifying the deficiencies in the performance of management functions. The main advantages are:
• It helps to locate present and potential deficiencies in the performance of management functions.
• It helps to improve the control system of an organisation by continuously monitoring the performance of management.
• It improves coordination in the functioning of various departments so that they work together
effectively towards the achievement of organisational objectives.
• It ensures updating of existing managerial policies and strategies in the light of environmental
changes.

(v) PERT and CPM: PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are important network techniques useful in planning and controlling. These techniques are especially useful for planning, scheduling and implementing time bound projects involving performance of a variety of complex, diverse and interrelated activities. These techniques deals with time scheduling and resource allocation for these activities and aims at effective execution of projects within given time schedule and structure of costs.

(vi)  Management Information System: It is a computer-based information system that provides information and support for effective managerial decision-making. A decision-maker requires up-to-date, accurate and timely information. MIS provides the required information to the managers by systematically processing a massive data generated in an organisation. Thus, MIS is an important communication tool for managers.

3. Explain the importance of controlling in an organisation. What are the problems faced by the organisation in implementing an effective control system?

Answer

Control is an indispensable function of management. A good control system helps an organisation
in the following ways:

(i) Accomplishing Organisational Goals: The controlling function measures progress towards the organisational goals and brings to light the deviations. If any, and indicates corrective action. It thus, guides the organisation and keeps it on the right track so that organisational goals might be achieved.

(ii) Judging Accuracy of Standards: A good control system enables management to verify whether the standards set are accurate and objective an efficient control system keeps a careful check on the changes taking place in the organisation and in the environment and helps to review and revise the standards in light of such changes.

(iii) Making Efficient Use of Resources: By exercising control, a manager seeks to reduce wastage and spoilage of resources. Each activity is performed in accordance with pre-determined standards and norms. This ensures that resources are used in the most efficient and effective manner.

(iv) Improving Employee Motivation: A good control system ensures that employees know well in advance what they are expected to do and what are the standards of performance on the basis of which they will be appraised. Thus, it motivates them and helps them to give better performancer.

(v) Ensuring Order and Discipline: Controlling creates an atmosphere of order and discipline in the organisation. It helps to minimise dishonest behaviour on the part of the employees by keeping a close check on their activities.

(vi) Facilitating Co-ordination in Action: Controlling provides direction Jo al! activities and efforts for achieving organisational goals. Each department and employee is governed by pre-determined standards which are well co-ordination with one another. This ensures that overall organisational objectives are accomplished.

Controlling suffers from the following limitations:

(i) Difficulty in Setting Quantitative Standards: Control system loses some of its effectiveness when standards cannot be defined in quantitative terms. This makes measurement of performance and their comparison with standards a difficult task. Employee morale, job satisfaction and human behaviour are such areas where this problem might arise.

(ii) Little Control on External Factors: Generally an enterprise cannot control external factors such as government policies, technological changes competition etc.

(iii) Resistance from Employees: Control is offer resisted by employees. They see it as a restriction on their freedom. For instance, employees might object when they are kept under a strict watch with the help of Closed Circuit Televisions (CCTVs).

(iv) Costly Affair: Control is a costly affair as it involves a lot of expenditure, time and effort. A small enterprise cannot afford to install an expensive control system. It cannot justify the expenses involved. Managers must ensure that the costs of installing and operating a control system should not exceed the benefits derived from it.

4. Discuss the relationship between planning and controlling.

Answer

Planning and controlling are inseparable, they are twins of management. A system of control pre-supposes the existence of certain standards. These standards of performance which serve as the basis of controlling are provided by planning. Once a plan becomes operational controlling is necessary to monitor the progress, measure it, discover deviations and initiate corrective measures to ensure that events conform to plans.
Planning is clearly a pre-requisite for controlling. Controlling cannot be accomplished with planning. With planning there is no pre-determined understanding of the desired performance, planning seeks consistent, integrated and articulated programmes while controlling seeks to compel events to conform to plans.

Application type

Following are some behaviours that you and others might engage in on the job. For each item, choose the behaviour that management must keep a check to ensure an efficient control system.
1. Biased performance appraisals.
2. Using company’s supplies for personal use.
3. Asking a person to violate company’s rules.
4. Calling office to take a day off when one is sick.
5. Overlooking boss’s error to prove loyalty
6.  Claiming credit for someone else’s morn.
7. Reporting a violation on noticing it.
8. Falsifying quality reports.
9. Taking longer than necessary to do the job.
10. Setting standards in consultation with workers.
You are also required to suggest the management how the undesirable behaviour can be controlled.

Answer

1. To avoid this, performance appraisal should be considered.
2. The statements are not so expensive, so it can be ignored.
3. Strict action should be taken.
4. There should be a written e-mail regarding leave, Mass bunking should not be allowed.
5. Secret suggestion box can be used to” collect feedback about the boss for appraisal.
6. Performance records of all employees should be maintained.
7. If minor can be over looked.
8. Quality control department take strict action regarding this.
9. Time should be fixed for each work.
10. The use of scientific techniques can help in fixing the most feasible and optimum standards.

Page No. 235

Case Problem

A company M limited is manufacturing mobile phones both for domestic Indian market as well as for export. It had enjoyed a substantial market share and also had a loyal customer following. But lately it has been experiencing problems because its targets have not been met with regard to sales and customer satisfaction. Also mobile market in India has grown tremendously and new players have come with better technology and pricing. This is causing problems for the company. It is planning to revamp its controlling system and take other steps necessary to rectify the problems it is facing.

1. Identify the benefits the company will define from a good control system.

Answer

• A good control system enables management to verify whether the standards set are accurate and objective.
• A good control system ensures that employees know well in advance what they are expected to do and what are the standards of performance on the basis of which they will be appraised.
• A good control system creates an atmosphere of order and discipline.

2. How can the company relate its planning with control in this line of business to ensure that its plans are actually implemented and targets attained?

Answer

A system of control presupposes the existence of certain standards. These standards of performance which serve as the basis of controlling are provided by planning. Once a plan becomes operational, controlling is necessary to monitor the progress, discover deviations and initiate corrective measures to ensure that events conform to plans.

3. Give the steps in the control process that the company should follow to remove the problems it is facing.

Answer

The company should follow these steps in a systematic manner:
(i) Setting performance standards.
(ii) Measurement of actual performance.
(iii) Comparison of actual performance with standards
(iv) Analysing deviations.
(v) Taking corrective actions.

4. What techniques of control can the company use?

Answer

The company should follow the modern techniques to control the systems:
(i) ROI (Return on Investment): ROI is used to measure the overall performance of organisation and evaluate the efficiency of investment.
(ii) Responsibility Accounting: Responsibility Accounting is a system of accounting in which different sections, divisions and departments of an organisation are set up as Responsibility centres. The head of this centres is responsible for achieving the target set for this sections.
Responsibility centres in the organisation are:-
(a) Cost Centre.
(b) Revenue Centre.
(d) Profit centre.
(iii) Investment centre.



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Quick Revision Notes of Chapter 2 The Nationalist Movement in Indo-China for Board Exams| Class 10th

January 11, 2019, 10:39 pm
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Quick Revision Notes of Chapter 2 The Nationalist Movement in Indo-China for Board Exams| Class 10th

• Indo-China region comprises the modern countries of Vietnam, Laos and Cambodia.

• Vietnam gained formal independence in 1945.

• French troops landed in Vietnam in 1858 and soon gained control of the area.

• In 1887 French Indo China was formed.

• Nationalism in Vietnam emerged through the efforts of different sections of society to fight against the French.

• Primarily, Vietnam was based on rice cultivation and rubber plantations owned by the French and small Vietnamese elite.
→ The French did very little to industrialise the economy of Vietnam. 

• The French started providing Western education in Vietnam as they want to their superiority to Vietnamese and also they needed an educated local labour force. 
→ In 1907, the Tonkin Free School was started for providing Western-style education.
→ Teachers and students opposed the curriculum.

• Religious movements like Scholars Revolt (1868) and Hoa Hao Movement started against the colonialism.

• In February 1930, Ho Chi Minh established the Vietnamese Communist Party inspired by the European Communist Parties.

• In 1940, Japan occupied Vietnam.

• In Septemeber 1943, Ho Chi Minh became the chairman of the Democratic Republic of Vietnam in September 1943.

• With the help of Ho Chi Minh Gwemment in the North, the National Liberation Front (NLF) fought
for the unification of the country.

• Fear of communism made the US intervene in Vietnam.
→ The war proved too costly to the Vietnamese as well as to the Americans.

• The Vand women played a major role in the war against the US.

• The US had failed to achieve its objectives.

• In January 1974: A peace settlement was signed in Paris which ended the conflict with the US.

• The National Liberation Front (NLF) occupied the presidential palace in Saigon on 30 April 1975 and unified Vietnam.

Important Terms

• Colonization: A system where one country subjugates another for exploitation of its resources.

• Syncretic Characterized by syncretism, aims to bring together different beliefs and practices seeing their essential Unity rather than their differences.

• Concentration camp: A prison where people are detained without due process of law. They are subjected to torture.

• Republic: A form of government based on popular consent and popular representation.

• Napalm: An organic compound used to thicken gasoline for fire bombs, developed in the US, was used in Vietnam Scholars revolt This revolt was led by Officials at the Imperial Court Angered by the spread of catholicism and French power.

Important Personalities

• Phan Bai Chau: He was a Confucian scholar activist. He Advocated that French should be driven out of Vietnam he formed the Revolutionary society.

• Phan Chau Trinh: He was intensely hostile to the monarchy and opposed to the idea of resisting the French with the help of the court.

• Huynh Phu So: He was founder of Hoa Hao Movement, He opposed the practice of sale of child and performed miracles

• Confucius: A Chinese thinker developed a system of philosophy based on good conduct, wisdom and proper social relationships.

NCERT Solutions of Chapter 2 The Nationalist Movement in Indo-China

Notes of Chapter 2 The Nationalist Movement in Indo-China
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Extra Questions for Class 10th Science

January 12, 2019, 10:33 pm
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Extra Questions for Class 10th Science with Important Questions and Answers

Science textbook for Class 10th Science is full of concepts from which questions can be asked in the examination. You need to understand every basic and advance concepts in order to pass the exam with flying colours. Thus, we have prepared the Extra Questions for Class 10th Science textbook form every chapter. We looked down upon every important thing which will help you in getting marks. Now you don't need to be panic if examinations are near and you have not covered the whole chapters. You can check your understanding by trying to solve the extra questions in the first step and then match your answer sheet with us. It will help you in clearing all your doubts. You only need to select your needy chapter from the list.

Extra Questions for Class 10th Science

Extra Questions for Class 10th Science




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Extra Questions for Class 9th Science

January 14, 2019, 10:08 am
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Extra Questions for Class 9th Science with Important Questions and Answers

There are many questions provided at the end of each chapter in the Class 9th Science textbook. These questions will check your basic understanding of the chapters. However, there are various concepts which are of importance left behind. In order to cover these parts, you need to practice more questions as these can come in the examination. Therefore, we have prepared the list of Extra Questions for Class 9th Science textbook. We are also providing the answers to these questions. While preparing these questions we have tried to cover every concept present in the chapter. This will save your precious time and will help you a lot in the exam preparation. You only need to select your favourite chapter from the list and get started.

Extra Questions for Class 9th Science

Extra Questions for Class 9th Science




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Quick Revision Notes of Chapter 3 Nationalism in India for Board Exams| Class 10th

January 15, 2019, 3:27 am
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Quick Revision Notes of Chapter 3 Nationalism in India for Board Exams| Class 10th 

• In India, the growth of modern nationalism is connected to the anti-colonical movement.

• The Indian National Congress under the leadership of Mahatma Gandhi tried to unite the different group within one movement.

• Gandhiji's idea of Satyagraha highlighted the power of truth and the need to search for truth.
→ Gandhiji decided to oppose Satyagraha when British government imposed Rowlatt Act  in 1919.

• On 13th April 1919, Jalianwala massacre happened.

• The Non-Cooperation-Khilafat Movement began in January 1921 to bring the Hindus and the Muslims closer to each other.

• In February 1922, Mahatma Gandhi decided to withdraw the Non-Cooperation Movement.
→ After the incident of Chauri Chaura in 1922, Gandhiji decided to call off the movement.

• On 6th April, 1930 Gandhiji ceremonially violated the Salt 'Law and the Civil Disobedience Movement started.
→ Civil diobedience was a widespread movement however Dailts and large sections of Muslims did not pariticpate in the movement.

• Nationalist Movement spread a sense of collective belongingness among different regions and communities.

• In the 20th century, the identity of India symbolised in the image of Bharat Mata.
→ History and fiction, folklore and songs, popular prints and symbols, all played a part in the making of nationalism.

Important Terms 

• Satyagrah - The idea of Satyagraha emphasised the power of Truth. It suggested that if the cause was true, if the struggle was against injustice, then physical force was not necessary.

• Boycott- Withdraw from commercial or social relations with a country organization or person as a punishment or protest. 

• Swadeshi- One's own country. 

• Civil disobedience movement- It was started by Mahatma Gandhi People were called upon to disobey the British laws. 

• Khalifa- Spiritual Heads of Islamic world.

• Begar- Labour that villagers were forced to contribute without any payment Harijan-'Children of the God' term given by Gandhiji to the ill trodden of the society. 

• Harijan - This term means 'Children of the god' which was given by Gandhiji to the ill trodden of the society.

• Picketing-A form of demonstration or protest by which people block the entrance to a shop, factory or office.




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ध्वनि का सार NCERT Class 8th Hindi

January 15, 2019, 4:43 am
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ध्वनि का सार (Dhwani ka summary) वसंत भाग - 3 हिंदी

सार

इस विडियो में हम पाठ ध्वनि के बारे में जानेगें| यह पाठ सूर्यकांत त्रिपाठी निराला द्वारा लिखा गया है जिसमें कवि ने मानव को अपने जीवन में कभी निराश नहीं होने की प्रेरणा दी है| वे मानव जाति को कठिनाइयों में भी आगे बढ़ते रहने को कह रहे हैं| वे इस कविता द्वारा मानव में जोश का संचार करने का प्रयास कर रहे हैं|

कवि का मानना है कि अभी उसके जीवन का अन्त नहीं होगा। अभी-अभी तो उसके जीवन में सुकुमार शिशु रूपी वसन्त का आगमन हुआ है। जिस प्रकार वसन्त के आने से प्रकृति में चारों ओर हरियाली छा जाती है। उसी तरह कवि भी अपने अच्छे कर्मों के माध्यम से अपनी ख्याति फैलाना चाहते हैं। कवि अपने सपनों से भरे कोमल हाथों को अलसाई कलियों पर फेरकर उन्हें सुबह दिखाना चाहते हैं यानी वह अपनी कविता द्वारा वह आलस्य में डूबे और निराशा से भरे युवाओं को प्रेरित कर उन्हें उत्साह से भर देना चाहते हैं ताकि वह नया सृजन कर सकें|

कवि सोए रहने वाले प्रत्येक पुष्प यानी युवा की नींद भरी आँखों से आलस्य हटाकर उन्हें जागरूक बनाना चाहते हैं। कवि उन पुष्पों को हरा-भरा बनाए रखने के लिए उन्हें अपने नवजीवन के अमृत से सींचना चाहते हैं। कवि के जीवन में अभी वसन्त का आगमन हुआ है| उनका अन्त बहुत दूर है। अभी उन्हें बहुत सारे काम करने हैं|

कठिन शब्दों के अर्थ

• मृदुल - कोमल
• पात - पत्ता
• गात - शरीर
• निद्रित - सोया हुआ
• प्रत्यूष - प्रात:काल
• तंद्रालस - नींद से अलसाया हुआ
• लालसा - कुछ पाने की चाह, अभिलाषा, इच्छा


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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.7

January 15, 2019, 11:13 am
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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.7

Exercise 3.7

1. The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Solution

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 8. Thus, we have x+y =8
The sum of the two numbers is four times their difference. Thus, we have
x +y=4(x-y)
⇒ x+y = 4x-4y
⇒ 4x-4y-x-y = 0
⇒ 3x-5y = 0
So, we have two equations
x + y = 8
3x – 5y = 0
Here, x and y are unknowns. We have to solve the above equations for x and y
Multiplying the first equation by 5 and then adding with the second equation, we have
5(x+y) + (3x-5y) = 5 × 8 + 0
⇒ 5x + 5y + 3x – 5y = 40
⇒ 8x = 40
⇒ x = 40/8
⇒ x = 5
Substituting the value of x in the first equation, we have
5 + y = 8
⇒ y = 8 – 5
⇒ y = 3
Hence, the number are 5 and 3 .

2. The sum of digits of a two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number ?

Solution

The sum of the digits of the number is 13. Thus, we have x + y = 13
After interchanging the digits, the number becomes 10x + y .
The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have
(10x + y) – (10y+x) = 45
⇒ 10x + y – 10y – x = 45
⇒ 9x – 9y = 45
⇒ 9(x-y) = 45
⇒ x-y = 45/9
⇒ x-y = 5
So, we have two equations
x+y = 13
x-y = 5
Here, x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have
(x+y) + (x-y) = 13+5
⇒ x + y + x – y = 18
⇒ 2x = 18
⇒ x = 9
Substituting the value of x in the first equation, we have
9 + y = 13
⇒ y = 13 – 9
⇒ y = 4
Hence, the number is 10 × 4 + 9 = 49 .

3. A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The sum of the digits of the number is 5. Thus, we have x+y = 5
After interchanging the digits, the number becomes 10x + y .
The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have
10x + y = 10y + x + 9
⇒ 10x + y – 10y – x = 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x - y = 9/9
⇒ x - y = 1
So, we have two equations
x + y = 5
x – y = 1
Here x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have
(x+y) + (x-y) = 5 + 1
⇒ x + y + x – y = 6
⇒ 2x = 6
⇒ x = 6/2
⇒ x = 3
Substituting the value of x in the first equation, we have
3 + y = 5
⇒ y = 5 - 3
⇒ y = 2
Hence, the number is 10 × 2 + 3 = 23 .

4. The sum of digits of a two number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The sum of the digits of the number is 15. Thus, we have x + y = 15
After interchanging the digits , the number becomes 10x + y .
The number obtained by interchanging the digits is exceeding by 9 from the original number.
Thus, we have
10x + y = 10y + x + 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x – y = 9/9
So, we have two equations
x+y = 15
x-y = 1
Here, x and y are unknowns . We have to solve the above equations for x and y .
Adding the two equations, we have
(x+y) + (x-y) = 15 + 1
⇒ x + y + x – y = 16
⇒ 2x = 16
⇒ x = 16/2
⇒ x = 8
Substituting the value of x in the first equation, we have
8 + y = 15
⇒ y = 15 – 8
⇒ y = 7
Hence, the number is 10 × 7 + 8 = 78 .

5. The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there ?

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The two digits of the number are differing by 2. Thus, we have x-y = ± 2 .
After interchanging the digits, the number becomes 10x + y .
The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have
(10x + y) + (10y + x) = 66
⇒ 10x + y + 10y + x = 66
⇒ 11x + 11y = 66
⇒ 11(x+y) = 66
⇒ x + y = 66/11
⇒ x + y = 6
So, we have two systems of simultaneous equations
x - y = 2 ,
x + y = 6
x – y = -2
x + y = 6
Here, x and y are unknows. We have to solve the above systems of equations for x and y .

(i) First, we solve the system
x – y =2 ,
x + y = 6
Adding the two equations, we have
(x-y) + (x+y) = 2 + 6
⇒ x – y + x + y = 8
⇒ 2x = 8
⇒ x = 8/2
⇒ x = 4
Substituting the value of x in the first equation, we have
4 – y = 2
⇒ y = 4 – 2
⇒ y = 2
Hence, the number is 10 × 2 + 4 = 24.

(ii) Now, we solve the system
x – y = -2
x + y = 6
Adding the two equations, we have
(x-y) + (x+y) = -2+6
⇒ x – y + x + y = 4
⇒ 2x = 4
⇒ x = 2
Substituting the value of x in the first equation, we have
2 – y = -2
⇒ y = 2 + 2
⇒ y = 4
Hence, the number is 10 × 4 + 2 = 42.
There are two such numbers .

6. The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers.

Solution

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 1000. Thus, we have x + y = 1000.
The difference between the squares of the two numbers is 256000. Thus, we have


7. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Solution

Let the digits at units and tens plac e of the given number be x and y respectively. Thus, the number is 10y + x.
The two digits of the number are differing by 3. Thus, we have x – y ± 3
After interchanging the digits, the number becomes 10x + y.
The sum of the numbers obtained by interchanging the digits and the original number is 99 .
Thus, we have
(10x + y) + (10y + x) = 99
⇒ 10x + y + 10y + x = 99
⇒ 11x + 11y = 99
⇒ 11(x+y) = 99
⇒ x+y = 99/11
⇒ x + y = 9
So, we have two systems of simultaneous equations
x-y=3,
x+y=9
x-y = -3
x+y=9
Here, x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system
x-y = 3,
x+y = 9
Adding the two equations, we have
(x-y) + (x+y) = 3 + 9
⇒ x – y + x + y = 12
⇒ 2x = 12
⇒ x = 12/2
⇒ 6
Substituting the value of x in the first equation, we have
6 – y = 3
⇒ y = 6 – 3
⇒ y = 3
Hence, the number is 10 × 3 + 6 = 36.
(ii) Now, we solve the system
x– y = -3
x + y = 9
Adding the two equations, we have
(x-y) + (x+y) = -3+9
⇒ x – y + x + y = 6
⇒ x = 6/2
Substituting the value of x in the first equation, we have
3 – y = -3
⇒ y = 3 + 3
⇒ y = 6
Hence, the number is 10×6+3 = 63.
Note that there are two such numbers .

8. A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The number is 4 times the sum of the two digits. Thus, we have
10y + x = 4(x+y)
⇒ 10y + x = 4x + 4y
⇒ 4x + 4y – 10y – x = 0
⇒ 3x – 6y = 0
⇒ 3(x-2y) = 0
⇒ x – 2y = 0
After interchanging the digits, the number becomes 10x + y.
If 18 is added to the number, the digits are reversed. Thus, we have
(10y + x) + 18 = 10x + y
⇒ 10x + y – 10y – x = 18
⇒ 9x – 9y = 18
⇒ 9(x-y) = 18
⇒ x – y = 18/9
⇒ x – y = 2
So, we have the systems of equations
x – 2y = 0 ,
x – y = 2
Here, x and y are unknowns . We have to solve the above systems of equations for x and y . Subtracting the first equation from the second , we have
(x – y) – (x-2y) = 2-0
⇒ x – y – x + 2y = 2
⇒ y = 2
Substituting the value of y in the first equation, we have
x – 2×2 = 0
⇒ x – 4 = 0
⇒ x = 4
Hence, the number is 10 × 2 + 4 = 24 .

9. A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The number is 3 more than 4 times the sum of the two digits . Thus, we have
10y + x = 4(x+y) + 3
⇒ 10y + x = 4x + 4y + 3
⇒ 4x + 4y – 10y – x = -3
⇒ 3x – 6y = -3
⇒ 3(x-2y) = -3
⇒ x - 2y = - 3/3
⇒ x – 2y = -1
After interchanging the digits, the number becomes 10x + y .
If 8 is added to the number, the digits are reversed . Thus, we have
(10y + x) + 18 = 10x + y
⇒ 10x + y – 10y – x = 18
⇒ 9x – 9y = 18
⇒ 9(x-y) = 18
⇒ x – y = 18/9
⇒ x – y = 2
So, we have systems of equations
x – y = -1,
x – y = 2
Here,  x and y are unknowns . We have to solve the above systems of equations for x ad y . Substracting the first equation from the second, we have
(x-y) – (x-2y) = 2 – (-1)
⇒ x – y – x + 2y = 3
⇒ y = 3
Substituting the value of y in the first equation, we have
x – 2×3 = -1
⇒ x – 6 = -1
⇒ x = -1 + 6
⇒ x = 5
Hence, the number is 10×3+5=35 .

10. A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively . Thus, the number is 10y + x. The number is 4 more than 6 times the sum of the two digits. Thus, we have
10y + x = 6(x+y) + 4
⇒ 10y + x = 6x + 6y + 4
⇒ 6x + 6y – 10y – x = -4
⇒ 5x – 4y = - 4
After interchanging the digits, the number becomes 10x + y.
If 18 is subtracted from the number, the digits are reversed. Thus, we have
(10y + x) – 18 = 10x + y
⇒ 10x + y – 10y – x = -18
⇒ 9x – 9y = -18
⇒ 9(x – y) = -18
⇒ x – y = - 18/9
⇒ x – y = -2
So, we have the systems of equations
5x – 4y = -4
x – y = -2
Here, x and y are unknowns. We have to solve the above systems of equations for x and y.
Multiplying the second equation by 5 and then subtracting from the first, we have
(5x – 4y) – (5x – 5y) = - 4 – (-2×5)
⇒ 5x – 4y – 5x + 5y = -4 × 10
Substituting the value of y in the second equation, we have
x = 6 = -2
⇒ x = 6 – 2
⇒ x = 4
Hence, the number is 10 × 6 + 4 = 64 .

11. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x. The number is 4 times the sum of the two digits. Thus, we have
10y + x = 4(x+y)
⇒ 10y + x = 4x + 4y
⇒ 4x + 4y – 10y – x = 0
⇒ 3x – 6y = 0
⇒ 3(x – 2y) = 0
⇒ x – 2y = 0
⇒ x = 2y
After interchanging the digits, the number becomes 10x + y .
The number is twice the product of the digits . Thus, we have 10y + x = 2xy
So, we have the systems of equations
x = 2y ,
10y + x = 2xy
Here, x and y are unknowns . We have to solve the above systems of equations for x and y.
Substituting x = 2y in the second equation, we get
10y + 2y = 2 × 2y × y
⇒ 12y = 4y2
⇒ 4y2 – 12y = 0
⇒ 4y (y – 3) = 0
⇒ y(y – 3) = 0
⇒ y = 0 or y = 3
Substituting the value of y in the first equation, we have

Hence , the number is 10 × 3 + 6 = 36 .
Note that the first pair of solution does not give a two digit number .

12. A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively . Thus, the number is 10y + x .
The product of the two digits of the number is 20 . Thus , we have xy = 20
After interchanging the digits, the number becomes 10x + y .
If 9 is added to the number, the digits interchange their places . Thus, we have
(10y + x) + 9 = 10x + y
⇒ 10y + x + 9 = 10x + y
⇒ 10x + y – 10y – x = 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x – y = 9/9
⇒ x – y = 1
So, we have the system of equations
xy = 20
x-y = 1
Here, x and y are unknows . We have to solve the above systems of equations for x and y. Substituting x = 1 + y from the second equation to the first equation, we get
(1+y)y = 20
⇒ y + y2 = 20
⇒ y2  + y – 20 = 0
⇒ y2+ 5y – 4y – 20 = 0
⇒ y(y+5) – 4(y+5) = 0
⇒ (y+5)(y-4) = 0
⇒ y = -5 or y = 4
Substituting the value of y in the second equation, we have

Hence, the number is 10 × 4 + 5 = 45
Note that in the first pair of solution the values of x and y are both negative . But, the digits of the number cant be negative . So, we must remove this pair .

13. The difference between two numbers is 26 and one number is three times the other. Find them.

Solution

Let the numbers are x and y . One of them must be greater than or equal to the other . Let us assume that x is greater than or equal to y .
The difference between the two numbers is 26. Thus, we have x - y = 26
One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have x = 3y
So, we have two equations
x– y = 26
x = 3y
Here, x and y are unknows . We have to solve the above equations for x and y .
Substituting x = 3y from the second equation in the first equation, we get
3y – y = 26
⇒ 2y = 26
⇒ y = 26/2
⇒ y = 13
Substituting the value of y in the first equation, we have
x – 13 = 26
⇒ x = 13 + 26
⇒ x = 39
Hence, the numbers are 39 and 13 .

14. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The sum of the two digits of the number is 9. Thus, we have x+y = 9
After interchanging the digits, the number becomes 10x + y .
Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have
9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 20x + 2y – 90y – 9x = 0
⇒ 11x – 88y = 0
⇒ 11(x-8y) = 0
⇒ x – 8y = 0
So, we have the systems of equations
x+y = 9
x – 8y = 0
Here, x and y are unknows . We have to solve the abpve systems of equations for x and y substituting x = 8y from the second equation, we get
8y + y = 9
⇒ 9y = 9
⇒ y = 9/9
⇒ y = 1
Substituting the value of y in the second equation, we have
x – 8 × 1 = 0
⇒ x – 8 = 0
⇒ x = 8
Hence, the number is 10 × 1 + 8 = 18 .

15. Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The difference between the two digits of the number is 3. Thus, we have x – y = ± 3
After interchanging the digits, the number becomes 10x + y .
Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have
7(10y + x) = 4 (10x + y)
⇒ 70y + 7x = 40x + 4y
⇒ 40x + 4y – 70y – 7x = 0
⇒ 33x – 66y = 0
⇒ 33(x-2y) = 0
⇒ x – 2y = 0
So, we have two systems of simultaneous equations
x – y = 3,
x – 2y = 0
x – y = -3,
x – 2y = 0
Here, x and y are unknowns . We have to solve the above systems of equations for x and y .

(i) First, we solve the system
x - y = 3,
x – 2y = 0
Multiplying the first equation by 2 and then subtracting from the second equation, we have
(x-2y) – 2 (x-y) = 0 -2 × 3
⇒ x – 2y – 2x + 2y = -6
⇒ -x = -6
⇒ x = 6
Substituting the value of x in the first equation , we have
6 – y = 3
⇒ y = 6 – 3
⇒ y = 3
Hence, the number is 10 × 3 + 6 = 36 .

(ii) Now, we solve the system
x-y = -3
⇒ x – 2y = 0
Multiplying the first equation by 2 and then subtracting from the second equation, we have
(x-2y) -2 (x-y) = 0 – (-3 ×2)
⇒ x – 2y – 2x + 2y = 6
⇒ -x = 6
⇒ x = -6
Substituting the value of x in the first equation , we have
-6 – y = -3
⇒ y = - 6 + 3
⇒ y = -3
But, the digits of the number can’t be negative. Hence, the second case must be removed.

16. Two numbers are in the ration 5:6 . If 8 is subtracted from each of the numbers , the ration becomes 4 : 5 . Find the numbers .

Solution

Let the two numbers be x and y
So,
x/y = 5/6
⇒ 6x = 5y
⇒ 6x – 5y = 0 ….(i)
Now, when 8 is subtracted from each of the numbers, then
x-8/y-8 = 4/5
⇒ 5x – 40 = 4y – 32
⇒ 5x – 4y = 8 …. (ii)
Multiplying (i) with 4 and (ii) with 5, we get
24x − 20y = 0 .....(iii)
25x − 20y = 40 .....(iv)
Subtracting (iii) from (iv), we get
x = 40
Putting x = 40 in (i), we get
240 − 5y = 0
⇒ y = 48
Thus, the two numbers are 40 and 48 .

17. A two – digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3 . Find the number .

Solution

Let the digits of the two digit number be x and y. So, the two digit number will be 10x + y.
Now, according to the given condition the number is obtained in two ways.
Case I: 8(x + y) − 5 = 10x + y
⇒ 8x + 8y − 5 = 10x + y
⇒ 2x −7y = −5 .....(1)

Case II: 16(x − y) + 3 = 10x + y
⇒ 16x − 16y + 3 = 10x + y
⇒ 6x − 17y = −3 .....(2)
Multiplying (1) by 3, we get
6x − 21y = −15 .....(3)
Subtracting (2) from (3), we get
− 4y = −12
⇒ y = 3
Putting y = 3 in (1), we get
2x − 21 = −5
⇒ 2x = 16
⇒ x = 8
So, the required number is 10x + y = 10 × 8 + 3 = 83.
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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.8

January 16, 2019, 3:35 am
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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.8

Exercise 3.7

1. The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Solution

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 8. Thus, we have x+y =8
The sum of the two numbers is four times their difference. Thus, we have
x +y=4(x-y)
⇒ x+y = 4x-4y
⇒ 4x-4y-x-y = 0
⇒ 3x-5y = 0
So, we have two equations 
x + y = 8
3x – 5y = 0
Here, x and y are unknowns. We have to solve the above equations for x and y
Multiplying the first equation by 5 and then adding with the second equation, we have
5(x+y) + (3x-5y) = 5 × 8 + 0
⇒ 5x + 5y + 3x – 5y = 40
⇒ 8x = 40
⇒ x = 40/8
⇒ x = 5
Substituting the value of x in the first equation, we have
5 + y = 8
⇒ y = 8 – 5
⇒ y = 3
Hence, the number are 5 and 3 .

2. The sum of digits of a two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number ?

Solution

The sum of the digits of the number is 13. Thus, we have x + y = 13
After interchanging the digits, the number becomes 10x + y .
The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have
(10x + y) – (10y+x) = 45
⇒ 10x + y – 10y – x = 45
⇒ 9x – 9y = 45
⇒ 9(x-y) = 45
⇒ x-y = 45/9
⇒ x-y = 5
So, we have two equations
x+y = 13
x-y = 5
Here, x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have
(x+y) + (x-y) = 13+5
⇒ x + y + x – y = 18
⇒ 2x = 18
⇒ x = 9
Substituting the value of x in the first equation, we have
9 + y = 13
⇒ y = 13 – 9
⇒ y = 4
Hence, the number is 10 × 4 + 9 = 49 .

3. A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The sum of the digits of the number is 5. Thus, we have x+y = 5
After interchanging the digits, the number becomes 10x + y .
The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have
10x + y = 10y + x + 9
⇒ 10x + y – 10y – x = 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x - y = 9/9
⇒ x - y = 1
So, we have two equations
x + y = 5
x – y = 1
Here, x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have
(x+y) + (x-y) = 5 + 1
⇒ x + y + x – y = 6
⇒ 2x = 6
⇒ x = 6/2
⇒ x = 3
Substituting the value of x in the first equation, we have
3 + y = 5
⇒ y = 5 - 3
⇒ y = 2
Hence, the number is 10 × 2 + 3 = 23 .

4. The sum of digits of a two number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The sum of the digits of the number is 15. Thus, we have x + y = 15
After interchanging the digits , the number becomes 10x + y .
The number obtained by interchanging the digits is exceeding by 9 from the original number.
Thus, we have
10x + y = 10y + x + 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x – y = 9/9
So, we have two equations
x+y = 15
x-y = 1
Here, x and y are unknowns . We have to solve the above equations for x and y .
Adding the two equations, we have
(x+y) + (x-y) = 15 + 1
⇒ x + y + x – y = 16
⇒ 2x = 16
⇒ x = 16/2
⇒ x = 8
Substituting the value of x in the first equation, we have
8 + y = 15
⇒ y = 15 – 8
⇒ y = 7
Hence, the number is 10 × 7 + 8 = 78 .

5. The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there ?

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The two digits of the number are differing by 2. Thus, we have x-y = ± 2 .
After interchanging the digits, the number becomes 10x + y .
The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have
(10x + y) + (10y + x) = 66
⇒ 10x + y + 10y + x = 66
⇒ 11x + 11y = 66
⇒ 11(x+y) = 66
⇒ x + y = 66/11
⇒ x + y = 6
So, we have two systems of simultaneous equations
x - y = 2 ,
x + y = 6
x – y = -2
x + y = 6
Here, x and y are unknows. We have to solve the above systems of equations for x and y .
(i) First, we solve the system
x – y =2 ,
x + y = 6
Adding the two equations, we have
(x-y) + (x+y) = 2 + 6
⇒ x – y + x + y = 8
⇒ 2x = 8
⇒ x = 8/2
⇒ x = 4
Substituting the value of x in the first equation, we have
4 – y = 2
⇒ y = 4 – 2
⇒ y = 2
Hence, the number is 10 × 2 + 4 = 24.

(ii) Now, we solve the system
x – y = -2
x + y = 6
Adding the two equations, we have
(x-y) + (x+y) = -2+6
⇒ x – y + x + y = 4
⇒ 2x = 4
⇒ x = 2
Substituting the value of x in the first equation, we have
2 – y = -2
⇒ y = 2 + 2
⇒ y = 4
Hence, the number is 10 × 4 + 2 = 42.
There are two such numbers .

6. The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers.

Solution

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 1000. Thus, we have x + y = 1000.
The difference between the squares of the two numbers is 256000. Thus, we have
 

7. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Solution

Let the digits at units and tens plac e of the given number be x and y respectively. Thus, the number is 10y + x.
The two digits of the number are differing by 3. Thus, we have x – y ± 3
After interchanging the digits, the number becomes 10x + y.
The sum of the numbers obtained by interchanging the digits and the original number is 99 .
Thus, we have
(10x + y) + (10y + x) = 99
⇒ 10x + y + 10y + x = 99
⇒ 11x + 11y = 99
⇒ 11(x+y) = 99
⇒ x+y = 99/11
⇒ x + y = 9
So, we have two systems of simultaneous equations
x-y=3,
x+y=9
x-y = -3
x+y=9
Here, x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system
x-y = 3,
x+y = 9
Adding the two equations, we have
(x-y) + (x+y) = 3 + 9
⇒ x – y + x + y = 12
⇒ 2x = 12
⇒ x = 12/2
⇒ 6
Substituting the value of x in the first equation, we have
6 – y = 3
⇒ y = 6 – 3
⇒ y = 3
Hence, the number is 10 × 3 + 6 = 36.

(ii) Now, we solve the system
x– y = -3
x + y = 9
Adding the two equations, we have
(x-y) + (x+y) = -3+9
⇒ x – y + x + y = 6
⇒ x = 6/2
Substituting the value of x in the first equation, we have
3 – y = -3
⇒ y = 3 + 3
⇒ y = 6
Hence, the number is 10×6+3 = 63.
Note that there are two such numbers .

8. A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The number is 4 times the sum of the two digits. Thus, we have
10y + x = 4(x+y)
⇒ 10y + x = 4x + 4y
⇒ 4x + 4y – 10y – x = 0
⇒ 3x – 6y = 0
⇒ 3(x-2y) = 0
⇒ x – 2y = 0
After interchanging the digits, the number becomes 10x + y.
If 18 is added to the number, the digits are reversed. Thus, we have
(10y + x) + 18 = 10x + y
⇒ 10x + y – 10y – x = 18
⇒ 9x – 9y = 18
⇒ 9(x-y) = 18
⇒ x – y = 18/9
⇒ x – y = 2
So, we have the systems of equations
x – 2y = 0 ,
x – y = 2
Here, x and y are unknowns . We have to solve the above systems of equations for x and y . Subtracting the first equation from the second , we have
(x – y) – (x-2y) = 2-0
⇒ x – y – x + 2y = 2
⇒ y = 2
Substituting the value of y in the first equation, we have
x – 2 × 2 = 0
⇒ x – 4 = 0
⇒ x = 4
Hence, the number is 10 × 2 + 4 = 24.

9. A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The number is 3 more than 4 times the sum of the two digits . Thus, we have
10y + x = 4(x+y) + 3
⇒ 10y + x = 4x + 4y + 3
⇒ 4x + 4y – 10y – x = -3
⇒ 3x – 6y = -3
⇒ 3(x-2y) = -3
⇒ x - 2y = - 3/3
⇒ x – 2y = -1
After interchanging the digits, the number becomes 10x + y .
If 8 is added to the number, the digits are reversed . Thus, we have
(10y + x) + 18 = 10x + y
⇒ 10x + y – 10y – x = 18
⇒ 9x – 9y = 18
⇒ 9(x-y) = 18
⇒ x – y = 18/9
⇒ x – y = 2
So, we have systems of equations
x – y = -1,
x – y = 2
Here,  x and y are unknowns . We have to solve the above systems of equations for x ad y . Substracting the first equation from the second, we have
(x-y) – (x-2y) = 2 – (-1)
⇒ x – y – x + 2y = 3
⇒ y = 3
Substituting the value of y in the first equation, we have
x – 2 × 3 = -1
⇒ x – 6 = -1
⇒ x = -1 + 6
⇒ x = 5
Hence, the number is 10×3+5=35.

10. A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively . Thus, the number is 10y + x. The number is 4 more than 6 times the sum of the two digits. Thus, we have
10y + x = 6(x+y) + 4
⇒ 10y + x = 6x + 6y + 4
⇒ 6x + 6y – 10y – x = -4
⇒ 5x – 4y = - 4
After interchanging the digits, the number becomes 10x + y.
If 18 is subtracted from the number, the digits are reversed. Thus, we have
(10y + x) – 18 = 10x + y
⇒ 10x + y – 10y – x = -18
⇒ 9x – 9y = -18
⇒ 9(x – y) = -18
⇒ x – y = - 18/9
⇒ x – y = -2
So, we have the systems of equations
5x – 4y = -4
x – y = -2
Here, x and y are unknowns. We have to solve the above systems of equations for x and y.
Multiplying the second equation by 5 and then subtracting from the first, we have
(5x – 4y) – (5x – 5y) = - 4 – (-2×5)
⇒ 5x – 4y – 5x + 5y = -4 × 10
Substituting the value of y in the second equation, we have
x = 6 = -2
⇒ x = 6 – 2
⇒ x = 4
Hence, the number is 10 × 6 + 4 = 64 .

11. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x. The number is 4 times the sum of the two digits. Thus, we have
10y + x = 4(x+y)
⇒ 10y + x = 4x + 4y
⇒ 4x + 4y – 10y – x = 0
⇒ 3x – 6y = 0
⇒ 3(x – 2y) = 0
⇒ x – 2y = 0
⇒ x = 2y 
After interchanging the digits, the number becomes 10x + y .
The number is twice the product of the digits . Thus, we have 10y + x = 2xy
So, we have the systems of equations
x = 2y ,
10y + x = 2xy
Here, x and y are unknowns . We have to solve the above systems of equations for x and y.
Substituting x = 2y in the second equation, we get
10y + 2y = 2 × 2y × y
⇒ 12y = 4y2
⇒ 4y2 – 12y = 0
⇒ 4y (y – 3) = 0
⇒ y(y – 3) = 0
⇒ y = 0 or y = 3
Substituting the value of y in the first equation, we have

Hence , the number is 10 × 3 + 6 = 36 .
Note that the first pair of solution does not give a two digit number .

12. A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively . Thus, the number is 10y + x .
The product of the two digits of the number is 20 . Thus , we have xy = 20
After interchanging the digits, the number becomes 10x + y .
If 9 is added to the number, the digits interchange their places . Thus, we have
(10y + x) + 9 = 10x + y
⇒ 10y + x + 9 = 10x + y
⇒ 10x + y – 10y – x = 9
⇒ 9x – 9y = 9
⇒ 9(x-y) = 9
⇒ x – y = 9/9
⇒ x – y = 1
So, we have the system of equations
xy = 20
x-y = 1
Here, x and y are unknows . We have to solve the above systems of equations for x and y. Substituting x = 1 + y from the second equation to the first equation, we get
(1+y)y = 20
⇒ y + y2 = 20
⇒ y2 + y – 20 = 0
⇒ y2+ 5y – 4y – 20 = 0
⇒ y(y+5) – 4(y+5) = 0
⇒ (y+5)(y-4) = 0
⇒ y = -5 or y = 4
Substituting the value of y in the second equation, we have

Hence, the number is 10 × 4 + 5 = 45
Note that in the first pair of solution the values of x and y are both negative . But, the digits of the number cant be negative . So, we must remove this pair .

13. The difference between two numbers is 26 and one number is three times the other. Find them.

Solution

Let the numbers are x and y . One of them must be greater than or equal to the other . Let us assume that x is greater than or equal to y .
The difference between the two numbers is 26. Thus, we have x - y = 26
One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have x = 3y
So, we have two equations
x– y = 26
x = 3y
Here, x and y are unknows . We have to solve the above equations for x and y .
Substituting x = 3y from the second equation in the first equation, we get
3y – y = 26
⇒ 2y = 26
⇒ y = 26/2
⇒ y = 13
Substituting the value of y in the first equation, we have
x – 13 = 26
⇒ x = 13 + 26
⇒ x = 39
Hence, the numbers are 39 and 13 .

14. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x .
The sum of the two digits of the number is 9. Thus, we have x+y = 9
After interchanging the digits, the number becomes 10x + y .
Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have
9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 20x + 2y – 90y – 9x = 0
⇒ 11x – 88y = 0
⇒ 11(x-8y) = 0
⇒ x – 8y = 0
So, we have the systems of equations
x+y = 9
x – 8y = 0
Here x and y are unknows . We have to solve the abpve systems of equations for x and y substituting x = 8y from the second equation, we get
8y + y = 9
⇒ 9y = 9
⇒ y = 9/9
⇒ y = 1
Substituting the value of y in the second equation, we have
x – 8 × 1 = 0
⇒ x – 8 = 0
⇒ x = 8
Hence, the number is 10 × 1 + 8 = 18 .

15. Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.

Solution

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.
The difference between the two digits of the number is 3. Thus, we have x – y = ± 3
After interchanging the digits, the number becomes 10x + y .
Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have
7(10y + x) = 4 (10x + y)
⇒ 70y + 7x = 40x + 4y
⇒ 40x + 4y – 70y – 7x = 0
⇒ 33x – 66y = 0
⇒ 33(x-2y) = 0
⇒ x – 2y = 0
So, we have two systems of simultaneous equations
x – y = 3,
x – 2y = 0
x – y = -3,
x – 2y = 0
Here x and y are unknowns . We have to solve the above systems of equations for x and y .

(i) First, we solve the system
x - y = 3,
x – 2y = 0
Multiplying the first equation by 2 and then subtracting from the second equation, we have
(x-2y) – 2 (x-y) = 0 -2 × 3
⇒ x – 2y – 2x + 2y = -6
⇒ -x = -6
⇒ x = 6
Substituting the value of x in the first equation , we have
6 – y = 3
⇒ y = 6 – 3
⇒ y = 3
Hence, the number is 10 × 3 + 6 = 36 .

(ii) Now, we solve the system
x-y = -3
⇒ x – 2y = 0
Multiplying the first equation by 2 and then subtracting from the second equation, we have
(x-2y) -2 (x-y) = 0 – (-3 ×2)
⇒ x – 2y – 2x + 2y = 6
⇒ -x = 6
⇒ x = -6
Substituting the value of x in the first equation , we have
-6 – y = -3
⇒ y = - 6 + 3
⇒ y = -3
But, the digits of the number can’t be negative. Hence, the second case must be removed.

16. Two numbers are in the ration 5:6 . If 8 is subtracted from each of the numbers , the ration becomes 4 : 5 . Find the numbers .

Solution

Let the two numbers be x and y
So,
x/y = 5/6
⇒ 6x = 5y
⇒ 6x – 5y = 0 ….(i)
Now, when 8 is subtracted from each of the numbers, then
x-8/y-8 = 4/5
⇒ 5x – 40 = 4y – 32
⇒ 5x – 4y = 8 …. (ii)
Multiplying (i) with 4 and (ii) with 5, we get
24x − 20y = 0 .....(iii)
25x − 20y = 40 .....(iv)
Subtracting (iii) from (iv), we get
x = 40
Putting x = 40 in (i), we get
240 − 5y = 0
⇒ y = 48
Thus, the two numbers are 40 and 48 .

17. A two – digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3 . Find the number

Solution

Let the digits of the two digit number be x and y. So, the two digit number will be 10x + y.
Now, according to the given condition the number is obtained in two ways.
Case I: 8(x + y) − 5 = 10x + y
⇒ 8x + 8y − 5 = 10x + y
⇒ 2x −7y = −5 .....(1)

Case II: 16(x − y) + 3 = 10x + y
⇒ 16x − 16y + 3 = 10x + y
⇒ 6x − 17y = −3 .....(2)

Multiplying (1) by 3, we get
6x − 21y = −15 .....(3)
Subtracting (2) from (3), we get
−4y = −12
⇒ y = 3
Putting y = 3 in (1), we get
2x − 21 = −5
⇒ 2x = 16
⇒ x = 8
So, the required number is 10x + y = 10 × 8 + 3 = 83
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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.9

January 16, 2019, 4:55 am
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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.9

Exercise 3.9

1. A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find the their present ages .

Solution

Let the present age of father be x years and the present age of son be y years. Father is three times as old as his son. Thus, we have
x = 3y
⇒ x-3y = 0
After 12 years, father’s age will be (x+12) years and son’s age will be (y+12) years . Thus using the given information, we have
x+12=2(y+12)
⇒ x+12=2y+24
⇒ x-2y-12=0
So, we have two equations
x-3y=0
x-2y-12=0
Here, x and y are unknows . We have to solve the above equations for x and y .
By using cross-multiplication, we have

2. Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B ?

Solution

Let the present age of A be x years and the present age of B be y years.
After 10 years, A’s age will be (x+10) years and B’s age will be (y+10) years. Thus using the given information, we have
x+10=2(y+10)
⇒ x+10 = 2y+20
⇒ x – 2y – 10 = 0
Before 5 years, the age of A was (x-5) years and the age of B was (y-5) years . Thus using the given information, we have
x-5=3(y-5)
⇒ x – 5 = 3y – 15
⇒ x – 3y + 10 =0
So, we have two equations
x-2y-10=0
x-3y+10=0
Here, x and y are unknowns. We have to solve the above equations for x and y .
By using cross – multiplication, we have

3. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?

Solution

Let the present age of Nuri be x years and the present age of Sonu be y years. After 10 years, Nuri’s age will be(x + 10) years and the age of Sonu will be(y + 10) years. Thus using the given information, we have
x + 10 = 2(y+10)
⇒ x + 10 = 2y + 20
⇒ x – 2y – 10 = 0
Before 5 years, the age of Nuri was(x – 5)years and the age of Sonu was(y – 5)years. Thus using the given information, we have
x - 5 = 3 (y-5)
⇒ x - 5 = 3y – 15
⇒ x – 3y + 10 = 0
So, we have two equations
x - 2y -10 = 0
x - 3y + 10 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have

4. Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Solution

Let the present age of the man be x years and the present age of his son be y years. After 6 years, the man’s age will be (x+6) years and son’s age will be (y+6) years . Thus using the given information, we have
x + 6 = 3(y+6)
⇒ x + 6 = 3y + 18
⇒ x – 3y – 12 = 0
Before 3 years, the age of the man was (x-3) years and the age of son’s was (y-3) years. Thus using the given information, we have
x – 3 = 9(y-3)
⇒ x -3 = 9y – 27
⇒ x – 9y + 24 = 0
So, we have two equations
x - 3y – 12 = 0
x - 9y + 24 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have

5. The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages .

Solution

Let the present age of father be x years and the present age of his son be y years. After 10 years, father’s age will be (x+10) years and son’s age will be (y+10) years . Thus using the given information, we have
x + 10 = 2(y+10)
⇒ x + 10 = 2y + 20
⇒ x – 2y – 10 = 0
Before 10 years, the age of father was (x-10)years and the age of son was (y-10) years . Thus using the given information, we have
x – 10 = 12(y-10)
⇒ x – 10 = 12y – 120
⇒ x – 12y + 110 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have

6. The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.

Solution

Let the present age of father be x years and the present age of his son be y years. The present age of father is three years more than three times the age of the son. Thus, we have
x = 3y + 3
⇒ x – 3y – 3 = 0
After 3 years, father’s age will be (x+3) years and son’s age will be (y+3) years.
Thus using the given information, we have
x + 3 = 2(y+3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x – 2y -13 = 0
So, we have two equations
x – 3y - 3 = 0
x - 2y - 13 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have

7. A father is three times as old as his son. In 12 years time, he will be twice as old as his son. Find the present ages of father and the son .

Solution

Let the present age of father be x years and the present age of his son be y years. The present age of father is three times the age of the son. Thus, we have
x = 3y
⇒ x – 3y = 0
After 12 years , father’s age will be (x+12) years and son’s age will be (y+12) years . Thus using the given information , we have
x + 12 = 2(y+2)
⇒ x + 12 = 2y + 24
⇒ x – 2y – 12 = 0
Here, x and y are unknowns. We have to solve the above equations for x and y. By using cross-multiplication, we have

8. Father's age is three times the sum of age of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.

Solution

Let the present age of father be x years and the present ages of his two children’s be y and z years.
The present age of father is three times the sum of the ages of the two children’s. Thus, we have
x = 3(y+z)
⇒ y+z = x/3
After 5 years, father’s age will be (x+5) years and the children’s age will be (y+5) and (z+5) years .
Thus using the given information,we have
x + 5 = 2{(y+5) + (z+5)}
⇒ x + 5 = 2(y+5+z+5)
⇒ x = 2(y+z) + 20 – 5
⇒ x = 2(y+z) + 15
So, we have two equations
y+z = x/3
x = 2(y+z) + 15
Here x,y and z are unknowns . We have to find the value of x .
Substituting the value of (y+z) from the first equation in the second equation, we have
By using cross – multiplication , we have

9. Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Solution

Let the present age of father be x years and the present age of his son be y years. After 2 years, father’s age will be (x+2) years and the age of son will be (y+2) years. Thus using the given information, we have
x+2 = 3(y+2)+8
⇒ x+2=3y+6+8
⇒ x-3y-12=0
Before 2 years, the age of father was (x-2) years and the age of son was (y-2) years . Thus using the given information , we have
x-2 =5(y-2)
⇒ x – 2 = 5y – 10
⇒ x – 5y + 8 = 0
So, we have two equations
x – 3y – 12 = 0
x – 5y + 8 = 0
Here, x and y are unknowns . We have to solve the above equations for x and y .
By using cross – multiplication, we have

10. A is elder to B by 2 years. A's father F is twice as old as A and B is twice as old as his sister S. If the ages of the father and sister differ by 40 years, find the age of A.

Solution

Let the present ages of A, B, F and S be x, y, z and t years respectively.
A is elder to B by 2 years. Thus, we have x = y + 2
F is twice as old as A. Thus, we have z = 2x
B is twice as old as S. Thus, we have y = 2t
The ages of F and S is differing by 40 years. Thus, we have z – t = 40
So, we have four equations
x = y+2 …..(1)
z = 2x, ……(2)
y = 2t, ……(3)
z – t = 40 …(4)
Here x, y, z and t are unknowns. We have to find the value of x.
By using the third equation, the first equation becomes x = 2t + 2
From the fourth equation, we have t = z – 40
Hence, we have
x = 2(z – 40) + 2
= 2z – 80 + 2
= 2z – 78
Using the second equation, we have
x = 2 × 2x – 78
⇒ x = 4x – 78
⇒ 4x – x = 78
⇒ 3x = 78
⇒ x = 78/3
⇒ x = 26
Hence , the age of A is 26 years .

11. The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharma is twice as old as Ani and Biju as twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution

Let the present ages of Ani, Biju, Dharam and Cathy be x, y, z and t years respectively.
The ages of Ani and Biju differ by 3 years. Thus, we have
x – y = ± 3
⇒ x = y ± 3
Dharam is twice as old as Ani. Thus, we have z = 2x
Biju is twice as old as Cathy. Thus, we have y = 2t
The ages of Cathy and Dharam differ by 30 years. Clearly, Dharam is older than Cathy. Thus, we have z-t = 30
So, we have two systems of simultaneous equations

(i) x = y + 3,
z = 2x,
y = 2t,
z-t = 30

(ii) x = y – 3 ,
z = 2x,
y = 2t,
z-t = 30
Here x, y, z and t are unknowns. We have to find the value of x and y.

(i) By using the third equation, the first equation becomes x = 2t + 3
From the fourth equation, we have
t = z -30
Hence, we have
x = 2(z-30)+3
= 2z – 60 + 3
= 2z – 57
Using the second equation, we have
x = 2 × 2x – 57
⇒ x = 4x – 57
⇒ 4x – x = 57
⇒ 3x = 57
⇒ x = 57/3
⇒ x = 19
From the first equation, we have
x = y+3
⇒ y = x – 3
⇒ y = 19 – 3
⇒ y = 16
Hence, the age of Ani is 19 years and the age of Biju is 16 years.

(ii) By using the third equation becomes x = 2t – 3
From the fourth equation, we have
t = z – 30
Hence, we have
x = 2(z-30) – 3
= 2z – 60 – 3
= 2z – 63
Using the second equation, we have
x = 2× 2x – 63
⇒ x = 4x – 63
⇒ 4x – x = 63
⇒ 3x = 63
⇒ x = 63/3
⇒ x = 21
From the first equation, we have
x = y-3
⇒ y = x + 3
⇒ y = 21+3
⇒ y = 24
Hence, the age of Ani is 21 years and the age of Biju is 24 years .
Note that there are two possibilities .

12. Two years ago , Salim was thrice as old as his daughter and six years later, he will be four years older than thrice her age . How old are they now ?

Solution

Let the present ages of Salim be x years and that of her daughter be y years.
Two years ago, the age of Salim was (x - 2) years and that of her daughter was (y -2).
It is given that Salim was thrice as old as her daughter two years ago. So,
x - 2 = 3(y - 2)
⇒ x - 2 = 3y - 6
⇒ x - 3y = -4 .....(i)
Six years later, the age of Salim will be (x + 6) and that of her daughter will be (y + 6).
∴ x + 6 = 2(y + 6) + 4
⇒ x - 2y = 10 .....(ii)
Subtracting (ii) from (i), we get
-y = -14
⇒ y = 14
Putting y = 14 in (ii), we get
x - 28 = 10
⇒ x = 38
Hence, the present age of Salim is 38 years and that of her daughter is 14 year .

13. The age of the father is twice the sum of the age of his two children . After 20 years, his age will be equal to the sum of the ages of his children . Find the age of the father .

Solution

Let the present age of the father be x years and the sum of the present ages of his two children be y years.
Now according to the given conditions,

Case I: x = 2y
⇒ x - 2y = 0 .....(i)

Case II: After 20 years, the age of the father will be (x + 20) years and the sum of the ages of the two children will be y + 20 + 20 = (y + 40) years.
So, x + 20 = y + 40
⇒ x - y = 20 .....(ii)
Subtracting (ii) from (i), we get
- y = -20
⇒ y = 20
Putting y = 20 in (i), we get
x - 40 = 0
⇒ x = 40
Hence, the present age of the father is 40 years.
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Notes of Ch 2 Principles of Management| Class 12th Business Studies

January 16, 2019, 7:01 am
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Summary and Notes of Ch 2 Principles of Management| Class 12th Business Studies 

Principles of management: The Concept

A managerial principle is a broad and general guideline for decision making and behaviour. Management principles are not as rigid as principles of pure science.

Difference between Principle of management and Technique of management

Principles of management
Techniques of management
Principles are guidelines.Techniques are procedures or methods.
It helps to take a decision or action while Practicing techniques.It involved a series of steps to be taken to accomplish desired goals.

Nature of Principles of Management

•Universal applicability: The principles of management are universal in nature that means they can be applied to all types of organisations, business as well as non-business, small as well large, public sector as well as private sector irrespective of their size and nature.

•General guidelines: The principles of management are guidelines to action and solve the problems, but these principles do not provide a readymade solution for all the problems because real business situations are very complex and dynamic.

•Formed by practice and experimentation: The principles of management are formed by experience and deep research work.

•Flexible:  The principles of management are not rigid they are flexible and can be modified by the manager when the situation so demands.

•Mainly behavioural: Management principles are formed to guide and influence the behaviour of employees.

• Cause and Effect Relationship: Management principles are based on cause and effect that means these principles are used in a similar situation in a large number of cases.

•Contingent: The application of principles of management is contingent or dependent upon the prevailing situation at a particular point of time. They may be changed according to the situation.

Significance of Principles of Management

These principles guide managers in taking and implementing decisions.
•Providing managers with useful insights into reality.
•Optimum utilisation of resources and effective administration.
•Scientific decision.
•Meeting changing environment requirements.
•Fulfilling social responsibility.
•Management training, education and research.

Principles of Scientific Management

According to Taylor "Scientific management means knowing exactly what you want men to do and seeing that they do it in the best and cheapest way".

Scientific principles of management
•Science not Rule of Thumb.
•Harmony, Not Discord.
•Cooperation, Not individualism.
•Development of Each and Every Person to His or Her Greatest Efficiency and Prosperity.

Techniques of Scientific Management

• Functional foremanship: In this technique, Taylor suggested that revolves the entire production in planning, implementation and control. Taylor advocated separation of planning and execution functions. This concept was extended to the lowest level of the shop floor.
(a)Planning Department
•Card clerk
•Route clerk
•Time and cost clerk
•Disciplinarian

(b) Operational Department
•Speed boss
•Gang boss
•Repair boss
•Inspector

• Standardisation and Simplification of work:  Setting standards for every business activity like for process, raw material, time, product, machinery, methods or working condition.
Method study:  The objective of method study is to find out one best way of doing the job. Taylor advised the concept of assembly line by using method study. The objective of the whole exercise is to minimise the cost of production and maximise the quality and satisfaction of the customer.

• Motion study: Motion study refers to the study of movements like lifting, putting objects, sitting and changing positions etc. which are undertaken by doing typical job. Unnecessary movements are ignored so, it takes less time to complete a job. Through motion study, Taylor was able to design equipment and tools to educate workers on their use and the result was fantastic.

• Time study: The standard time is fixed for the whole of the task by taking several readings. The method of time study will depend upon volume and frequency of the task, the cycle time of the operation and time measurement costs. The objective of the time study is to determine the number of workers to be employed, frame suitable incentive schemes and determine labour costs.

• Fatigue Study: It refers to determine the duration and frequency of rest intervals to complete a particular job. The main objective of this study is to maintain the efficiency level of workers.

• Wage System: It means wages are paid on the basis of work done and not on the basis of time spent on doing the work. In this system, two different wage rates are used
(a) High wage rate
(b) Low wage rate.

• Mental revolution: It refers to the change in the attitude of management and workers towards one another from competition to cooperation.

Fayol’s Principles of Management

Henry Fayol, a famous industrialist of France, has described fourteen principles of management in his book "General and Industrialist Management".

The fourteen principles given by Fayol are as under:

• Division of work: The whole work is divided into smaller parts and each individual should be assigned only one parts of the work according to his ability and taste.

• Authority and responsibility: According to this principle, authority and responsibility should go hand in hand.

• Discipline:  The organisational rules and employment agreement should be obeyed by both the superiors and subordinate which are necessary for the successful working of the organisation.

• Unity of command:  An individual employee should receive orders from only one superior at a time and that employee should be answerable only to that superior.

• Unity of direction: It means that there should be one head for one plan for a group of activities having the same objective.

• Subordination of individual interest to general interest: The interests of an organisation should take priority over the interests of any one individual employee.

• Remuneration of employees:  The employees should be paid fair remuneration which should give them at least a reasonable standard of living.

• Centralisation and Decentralisation:  The concentration of decision - making authority is called centralisation whereas its dispersal among more than one person is known as decentralisation.

• Scalar chain: The formal lines of authority from highest to lowest ranks are known as scalar chain.

• Order: A right person should be placed at the right job and a right thing should be placed at the right place.

• Equity: The managers should treat their subordinates as fairly as possible so that they develop a feeling of dedication for their work.

• Stability of personnel: There should be a stability of tenure of the employees so that the work continues efficiently.

• Initiative: Employees in the organisation must be given an opportunity in making and executing plan.

• Espirit De Corps: As per this principle, a manager should continuously make efforts to develop a team spirit among the subordinates.

Dissimilarities in the ideas of Taylor and Fayol

Basis of difference
Henry Fayol
F.W. Taylor
PerspectiveTop level of managementShop floor level of a factory
Unity of CommandStaunch proponentDid not feel that it is important as under functional foremanship a worker received orders from eight specialists.
Applicability Applicable universallyApplicable to specialised situations.
Basis of formationPersonal experienceObservations and experimentation.
FocusImproving overall administration.Increasing productivity.
PersonalityPractitionerScientist 

NCERT Solutions of Principles of Management

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पाठ 3- भारत में राष्ट्रवाद इतिहास के नोट्स| Class 10th

January 16, 2019, 7:06 pm
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पठन सामग्री और नोट्स (Notes)| पाठ 3-  भारत में राष्ट्रवाद  (Bharat me Rashtravaad) Itihas Class 10th

परिचय

• आधुनिक राष्ट्रवाद के साथ ही राष्ट्र-राज्यों का उदय हुआ था।

• भारत में आधुनिक राष्ट्रवाद का विकास कई अन्य उपनिवेशों की तरह उपनिवेश विरोधी आंदोलन से जुड़ा हुआ था?

पहला विश्व युद्ध, खिलाफत और असहयोग

• प्रथम विश्व युद्ध (1914-1918) ने एक नयी राजनीतिक और आर्थिक स्थिति पैदा कर दी थी।

• भारत को युद्ध के दौरान विभिन्न समस्याओं का सामना करना पड़ा था।
→ रक्षा खर्चे में वृद्धि।
→ युद्ध के दौरान कीमतों में वृद्धि हुई।
→ ग्रामीण क्षेत्रों में सिपाहियों को जबरन भर्ती किया गया।

• 1918-19 और 1920-21 के दौरान भारत के कई हिस्सों में फसल खराब हो गई।

• युद्ध खत्म होने के बाद भी मुश्किलें कम नहीं हुईं।

सत्याग्रह का विचार

• सत्याग्रह भारत में औपनिवेशिक शासन से लड़ने का एक नया तरीका है।
→ यह अत्याचार और अन्याय के खिलाफ एक गैर-आक्रामक और शांतिपूर्ण जन आंदोलन है।

• सत्याग्रह का अर्थ है सत्य को स्वीकार करने के लिए आग्रह करना।

• यह एक नैतिक चेतना है न कि निष्क्रिय प्रतिरोध।

• महात्मा गांधी जनवरी 1915 में भारत लौट आए।

• गांधीजी ने बिहार के चंपारण ज़िले में (1916), गुजरात के खेड़ा जिले में (1917) और अहमदाबाद में कपास मिल श्रमिकों (1918) के बीच सत्याग्रह आंदोलन किए।

रौलट एक्ट (1919)

• इस अधिनियम से सरकार को राजनीतिक गतिविधियों को कुचलने और राजनीतिक कैदियों को दो साल तक बिना मुकदमा चलाऐ जेल में बंद रखने का अधिकार मिल गया था।

जलियांवाला बाग हत्याकांड

• 13 अप्रैल 1919 को जलियांवाला बाग के मैदान में भारी भीड़ एकत्रित हुई थी।

• जनरल डायर हथियारबंद सैनिकों के साथ वहाँ पहुँचा और बाहर निकलने के सारे रास्ते बंद कर दिए। इसके बाद डायर ने भीड़ पर गोलियाँ चलाने के आदेश दे दिया जिससे सैकड़ों लोग मारे गए।

• खबर फैलते ही उत्तर भारत में हड़तालें होने लगी, लोग पुलिस से मोर्चा लेने लगे, और सरकारी इमारतों पर हमला करने लगे।

• सरकार ने इसका जबाब क्रूरता के साथ दिया।

खिलाफत आंदोलन

• खिलाफत आंदोलन का नेतृत्व दो भाइयों शौकत अली और मुहम्मद अली ने किया था।

• खलीफा की तात्कालिक शक्तियों की रक्षा के लिए मार्च 1919 में बॉम्बे में खिलाफत समिति का गठन किया गया था।

• गांधीजी ने कांग्रेस को खिलाफत आंदोलन का समर्थन और स्वराज के लिए एक असहयोग अभियान शुरू करने के लिए राजी कर लिया।

• दिसंबर 1920 में कांग्रेस के नागपुर अधिवेशन में असहयोग कार्यक्रम पर स्वीकृति की मुहर लगा दी गई।

आंदोलन के भीतर अलग-अलग धाराएँ

• असहयोग-खिलाफत आंदोलन जनवरी 1921 में शुरू हुआ।

शहरों में आंदोलन

• आंदोलन की शुरुआत शहरी मध्यवर्ग की हिस्सेदारी के साथ हुई।

• विद्यार्थियों ने स्कूल -कॉलेज , शिक्षकों ने अपनी नौकरी तथा वकीलों ने मुकदमा लड़ने का वहिष्कार किया और आंदोलन में शामिल हो गए।

• प्रांतो में परिषद चुनावों का बहिष्कार किया गया।

• विदेशी सामानों का बहिष्कार किया गया।

• शराब की दुकानों के सामने प्रदर्शन किया गया।

ग्रामीण इलाकों में विद्रोह

• किसानों और आदिवासियों का संघर्ष धीरे-धीरे हिंसक हो गया।

अवध में किसान आंदोलन

• किसानों का नेतृत्व बाबा रामचंद्र ने जमींदारों और तालुकेदारों के खिलाफ किया था।

• अवध किसान सभा की स्थापना 1920 में जवाहरलाल नेहरू, बाबा रामचंद्र और कुछ अन्य लोगों के नेतृत्व में की गई थी।

आंध्र प्रदेश में आदिवासियों का आंदोलन

• आंध्र प्रदेश के गुडेम पहाड़िओ में अल्लूरी सीताराम राजू के नेतृत्व में उग्र गुरिल्ला आंदोलन फैल गया।

• विद्रोहियों ने पुलिस थानों पर हमलें कियें।

• राजू को 1924 में पकड़ लिया गया और फांसी दे दी गई।

बागानों में स्वराज

• बागान श्रमिकों के लिए स्वराज का अर्थ था चारदीवारियों से बाहर स्वतंत्र रूप से आवाजाही।

• उन्होंने 1859 के अंतर्देशीय आप्रवासन अधिनियम का विरोध किया, जिसमें उन्हें बिना अनुमति के बागवानों से बाहर जाने से रोका जाता था।

• प्रत्येक समूह ने स्वराज शब्द की व्याख्या अपने तरीके से की।

सविनय अवज्ञा की ओर

• महात्मा गांधी ने फरवरी 1922 में असहयोग आंदोलन वापस लेने का फैसला किया।

• सी आर दास और मोतीलाल नेहरू ने परिषद राजनीति में वापस लौटने के लिए कांग्रेस के भीतर ही स्वराज पार्टी का गठन कर डाला।

• जवाहरलाल नेहरू और सुभाष चंद्र बोस जैसे युवा नेताओं ने ज्यादा उग्र जन जनांदोलन और पूर्ण स्वतंत्रता के लिए दबाव डाला।

1920 के बाद भारतीय राजनीति को दिशा निर्धारित वाले कारण

• विश्वव्यापी आर्थिक मंदी
→ 1930 के बाद कृषि उत्पादों की माँग गिरी और निर्यात कम होने लगा जिससे कृषि उत्पादों की कीमतों में भारी गिरावट आयी।

• साइमन कमीशन

- इस आयोग का गठन ब्रिटेन की टोरी सरकार द्वारा राष्ट्रवादियों की मांगों पर विचार करने और भारत के संवैधानिक व्यवस्था का अध्ययन करने और सुझाव देने के लिए किया था।
→ यह आयोग 1928 में भारत पहुंचा।
→ कांग्रेस ने इस आयोग का विरोध किया।

• जवाहरलाल नेहरू की अध्यक्षता में दिसंबर 1929 में कांग्रेस के लाहौर अधिवेशन में "पूर्ण स्वराज" की माँग को औपचारिक रूप दे दिया गया।

नमक यात्रा और सविनय अवज्ञा आंदोलन

• गांधीजी ने नमक को ऐसे माध्यम के रूप में चुना जो राष्ट्र को एकजुट कर सके क्योंकि नमक समाज के सभी वर्गों द्वारा प्रयोग में लाया जाता है।

नमक यात्रा

• नमक या दांडी यात्रा 12 मार्च 1930 को शुरू किया गया।
→ गांधीजी 6 अप्रैल 1930 को गुजरात के एक गाँव दांडी पहुँचे और वहाँ समुंद्र का पानी उबालकर नमक बनाना शुरू कर दिया। इस तरह उसने नमक का कानून तोड़ा।
→ इस प्रकार उसने सविनय अवज्ञा आंदोलन शुरू किया।

• यह असहयोग आंदोलन से अलग था क्योंकि लोगों को न केवल अंग्रेजों का सहयोग न करने के लिए बल्कि औपनिवेशिक कानूनों उल्लंघन करने के लिए आह्वान किया गया।

• विदेशी वस्तुओं का बहिष्कार, करों का भुगतान न करना और वन कानूनों को तोड़ना इसकी प्रमुख विशेषताएं थीं।

• ब्रिटिश सरकार आंदोलन को कुचलने के लिए क्रूर दमन की नीति अपनाया।

• ब्रिटिश सरकार ने गांधीजी और नेहरू सहित सभी बड़े नेताओं को गिरफ्तार कर लिया।

• गांधीजी ने आंदोलन वापस ले लिया।

गांधी-इरविन समझौता

• गांधीजी ने 5 मार्च 1931 को वायसराय लॉर्ड इरविन के साथ एक समझौते पर हस्ताक्षर किए।

• गांधीजी दिसंबर 1931 में दूसरे गोलमेज सम्मेलन में भाग लेने लंदन गए लेकिन वार्ता विफल होने के कारण उन्हें निराश होकर लौटना पड़ा।

• गांधी ने सविनय अवज्ञा आंदोलन दोबारा शुरू किया लेकिन 1934 तक आते -आते इसकी गति मन्द पड़ने लगी।

लोगों ने आंदोलन को कैसे लिया

संपन्न किसान

• अंग्रेजों ने जब राजस्व कर कम से इनकार कर दिया तब अमीर किसान समुदाय भी आंदोलन में शामिल हो गए।

• राजस्व दर घटाए बिना आंदोलन वापस ले लिया गया तो इन्हें निराशा हुई। जब आंदोलन दुबारा शुरू हुआ तो बहुतों ने शामिल होने से इनकार कर दिया।

गरीब किसान

• गरीब किसान चाहते थे उन्हें जमीन का भाड़ा माफ कर दिया जाए।

• अमीर किसानों और जमींदारों की नाराजगी के भय से कांग्रेस “भाड़ा विरोधी” आन्दोलनों का समर्थन करने के लिए तैयार नहीं थी।

व्यवसायी वर्ग

• प्रथम विश्व युद्ध के बाद इनके मुनाफें भारी कमी आयी ये विदेशी वस्तुओं के आयात के खिलाफ सुरक्षा चाहते थे।

• उग्रवादी गतिविधियों का प्रसार, लंबे समय तक व्यापारिक व्यवधानों की चिंता और युवा कांग्रेस के बीच समाजवाद के बढ़ते प्रभाव ने इन्हें आंदोलन में शामिल नहीं होने के लिए मजबूर किया।

महिलाएँ

• महिलाएँ भी जुलूस का हिस्सा बनी ,नमक बनाई, और शराब की दुकानों के सामने धरने में शामिल हुई।

• कांग्रेस महिलाओं को संगठन के भीतर किसी भी प्रकार के महत्वपूर्ण पद देने के पक्ष में नहीं थी।

सविनय अवज्ञा की सीमाएँ

• दलितों या अछूतों ने आंदोलन में सक्रिय रूप से भाग नहीं लिया उन्होंने सीटों के आरक्षण और अलग निर्वाचन क्षेत्रों की मांग की।

• दलितों के नेता डॉ. बी.आर. अंबेडकर ने 1930 में एक संस्था बनाई जिसका नाम दमित वर्ग एसोसिएशन रखा।

• गांधीजी का डॉ.बी.आर.अंबेडकर से काफी विवाद हुआ।

• गांधीजी और डॉ. बी.आर.अंबेडकर के बीच 1932 में पूना पैक्ट पर सहमति बनी। प्रांतीय और केंद्रीय परिषदों में आरक्षित सीटें दे दीं गई लेकिन मतदान सामान्य निर्वाचन क्षेत्रों में ही हुआ।

• मुस्लिम लीग के नेता एम ए जिन्ना केंद्रीय सभा में मुसलमानों के लिए आरक्षित सीटें चाहते थे।

• मुसलमानों के एक बड़े हिस्से ने सविनय अवज्ञा आंदोलन में हिस्सा नहीं लिया।

सामूहिक अपनेपन का भाव

• सामूहिक अपनेपन की भावना आंशिक रूप से संयुक्त संघर्षो के चलते पैदा हुई।

• इतिहास और साहित्य, लोककथाएँ व गीत , चित्र और प्रतीक सभी ने राष्ट्रवाद को साकार करने में अपना योगदान दिया।

• 1921 तक गांधीजी ने स्वराज का झंडा तैयार कर लिया था। यह एक तिरंगा (लाल, हरा और सफेद) था और जिसके मध्य में गांधीवादी चरखा था।


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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.10

January 16, 2019, 11:24 pm
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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.10

Exercise 3.10

1. Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars.

Solution


We have to find the speed of car
Let X and Y be two cars starting from points A and B respctively . Let the speed of car X be x km/hr ad that of car Y be y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point Q , Then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 7 hours.
Therefore, Distance travelled by car X in 7 hours = 7x km
AQ = 7x  
Distance traveled by car y in hours = 7 y km
BQ = 7Y
Clearly AQ – BQ = AB
7x – 7y = 70
Dividing both sides by common factor 7 we get,
x-y = 10….(i)
Case II : When two cars move in opposite direction
Suppose two cars meet at point. Then,
Distance travelled by car X = AP ,
Distance travelled by car Y = BP .
In this case, two cars meet in 1 hour
Therefore Distane travelled by car X in 1 Hour  = 1xKm
AP  = 1x  
Distance travelled by car Y in 1 hour = 1y km
BP = 1y
From the above clearly,
AP + BP = AB
AP + BP = AB
By solving equation (i) and (ii), we get
x-y = 10
x+y = 70
2x = 80
x = 80 / 2
x = 40
Substituting x = 40 in equation (ii) we get
x + y = 70
40 + y = 70
y = 70 – 40
y = 30
Hence, the speed of car starting from point A is 40 km/hr.
The speed of car starting from point B is 30/hr . 

2. A sailor goes 8 km downstream in 40 minutes and returns in 1 hours. Determine the speed of the sailor in still water and the speed of the current.

Solution

Let the speed of the sailor in still water x km/hr and the speed of the current be y km/hr
Speed upstream = (x-y) km/hr
Speed downstream = (x+y) km/hr
Now, Time taken to cover 8 km down stream = 8/x+y hhrs
Time taken to cover 8 km upstream = 8/x-y hrs
But, time taken to cover 8 km downstream in 40 minutes or 40/60 hours that is 2/3 hours
8/x+y = 2/3
8×3 = 2(x+y)
24 = 2x + 2y
Dividing both sides by common factor 2 we get
12 = x + y …(i)
Time taken to cover 8 km upstream in 1 hour
8/x-y = 1
8 = 1(x-y)
8 = x – y …(ii)
By solving  these equation (i) and (ii) we get
x+y = 12
x-y = 8
2x = 20
x = 20/2
x = 10
Substitute x = 10 in equation (i) we get
x+y = 12
10+y = 12
y = 12 – 10
y = 2
Hence the speed of sailor is 10km/hr
The speed of current is 2 km/hr .

3. The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.

Solution

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr
Speed upstream = (x-y) km/hr
Speed down stream = (x+y) km/hr
Time taken to cover 30 km upstream = 30/x-y hrs
Time taken to cover 44 km downstream = 44/x+y hrs 
But total time of journey is 10 hours
30/x-y + 44/x+y = 10 …(i)
Time taken to cover 40 km upstream = 40/x-y hrs
Time taken to cover 55 km downstream= 55/x+y hrs
In this case total time of journey is given to be 13 hours
Therefore, 40/x-y + 55/x+y = 13 …(ii)
Putting  1/x-y = u and 1/ x+y = v in equation (i) and (ii) we get
30u + 44v = 10
40u + 55v = 10
30u + 44v – 10 = 0 …(iii)
40u + 55v – 13 = 0 …(iv)
Solving these equations by cross multiplication we get 

4. A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in  6 fraction 1/2 hours . Find the speed of the boat in still water and also speed of the stream .

Solution


We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr then
Speed upstream = (x-y) km/hr
Speed downstream = (x+y) km/hr 

5. A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking. 

Solution

Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,
Distance covered = speed × distance
= x × y
= xy …(i)
If the speed is increased by 1/2 km/hr , then time of journey is reduced by 1 hour i.e., when speed is (x + 1/2) km/hr , time of journey is (y-1) hours
Distance covered = xy km   
-2x + y – 1 = 0 …(ii)  
When the speed is reduced by 1 km/hr, then the  time of journey is increased by 3 hours i.e., when speed is (x-1) km/hr, time of journey is (y+3) hours
Distance covered = xy
xy = (x-1)(y+3)
xy = (x-1)(y+3)
xy = xy – 1y + 3x – 3
xy = xy + 3x – 1y – 3
3x – 1y – 3 = 0 …(iii)
Thus we obtain the following equations
-2x + 1y – 1 = 0
3x – 1y – 3 = 0
By using elimination method, we have
-2x + 1y – 1 = 0
3x – 1y – 3 = 0
1x – 4 = 0
x = 4
Putting the value x = 4 in equation (iii) we get
3x – 1y – 3 = 0
3 × 4 – 1y – 3 = 0
12 – 1y – 3 = 0
9 – 1y = 0
-1y = -9
-1y = -9
 y = 9
Putting the value of x and y in equation (i) we get  
Distance covered = xy
= 4 × 9
= 36 km
Hence, the distance is 36 km ,
The speed of walking is 4 km/hr .  

6. A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km downstream. Find the speed of the stream.

Solution

Speed of the boat in still water = 5 km/h
Let the speed of stream = x km/h
∴  Speed of boat upstream = (5 – x) km/h
Speed of boat downstream = (5 + x) km/h
Time taken to row 40 km upstream = 40/5 – x
Time taken to row 40 km downstream = 40/5 + x
According to the given condition,
40/5 – x = 3(40/5+x)
⇒ 1/5 – x = 3/5 + x
⇒ 5 + x = 15 – 3x
⇒ 4x = 10
⇒ x = 10/4 = 2.5 km/h
Therefore, the speed of the stream is 2.5 km/h .

7. Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km. by train and the rest by car. He takes 12 minutes more if the travels 240 km by train and the rest by car. Find the speed of the train and car respectively.   

Solution

Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases
Case I: When Ramesh travels 760 Km by train and the rest by car

8. A man travels 600 km partly by train and partly by car. If the covers 400 km by train and the rest by car, it takes him 6 hours 30 minutes. But, if the travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.  

Solution

Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases: 
Case I: When a man travels 600Km by train and the rest by car

9. Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars.

Solution 

Let x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 8 hours.
Distance travelled by car X in 8 hours = 8x km
AQ = 8x
Distance travelled by car Y in 8 hours = 8y km
BQ = 8y
Clearly AQ – BQ = AB
8x – 8y = 80
Both sides divided by 8, we get
x – y = 10 …(i)

Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then ,
Distance travelled by X car X = AP
Distance travelled by Y car Y = BP 

10. A boat goes 12 km upstream and 40 km downstream in 8 hours. I can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream. 

Solution 

We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr
Speed upstream = (x-y) km/hr
Speed downstream = (x+y) km/hr

11. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. 

Solution

Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases

Case I: When Roohi travels 300 Km by train and the rest by bus 


























































































12. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution

Let the speed of rowing in still water be x km/hr and the speed of the current be y km/hr 
Speed upstream = (x-y) km/hr 
Speed downstream = (x+y) km/hr 
Now,

13. A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Solution

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Speed of boat upstream = x − y
Speed of boat downstream = x + y
It is given that, the boat travels 30 km upstream and 28 km downstream in 7 hours.
∴ 30/x-y + 28/x+y = 7
Also, the boat travels 21 km upstream and return in 5 hours .
∴ 21/x-y + 21/x+y = 5
Let 1/x-y = u and 1/x+y = v.
So, the equation becomes
30u + 28v = 7 …(i)
21u + 21v = 5 …(ii)
Multiplying (i) by 21 and (ii) by 30, we get
630u + 588v = 147 …(iii)
630u + 630v = 150 …(iv)
Solving (iii) and (iv), we get
v = 1/14 and u = 1/6
But, 1/x-y = u and 1/x+y = v
So,
1/x-y = 1/6 and 1/x+y = 1/14
⇒ x – y = 6 and x + y = 14
 Solving these two equations , we get
x = 10 and y = 4
So, the speed of boat in still water = 10 km/h and speed of stream = 4 km/h.


14. Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi.

Solution

Let the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases 

15. A train covered a certain distance at a uniform speed. If the train could have been 10km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10km/hr; it would have taken 3 hours more than the scheduled time. Find the distance coverec by train.

Solution

Let the actual speed of the train 1 km/hr and the actual time taken by y hours . Then,
Distance =  speed × Time
Distance covered = (xy)km …(i)
If the speed is increased by 10 km/hr , then time of journey is reduced by 2 hours
When speed is (x+10) km/hr, time of journey is (y-2)hours
Distance covered = (x+10)(y-2)
xy = (x+10)(y-2)
xy = xy + 10y – 2x – 20
-2x + 10y – 20 = 0
-2x + 3y – 12 = 0 – (ii)
When the speed is reduced  by 10 km/hr , then the time of journey is increased by 3 hours when speed is (x-10) km/hr, time of journey is (y+3) hours
Distance overed = (x – 10)(y+3)
xy = (x-10)(y+3)
xy = xy – 10y + 3x – 30
0 = -10y + 3x – 30
3x – 10y – 30 = 0 …(iii)
Thus, we obtain the following system of equations   :
-x + 5y – 10 = 0
3x – 10y – 30 = 0
By using cross multilication, we have
Hence, the length pf the journey is 600 km .

16. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars?

Solution

Let x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X=AQ
Distance travelled by car Y=BQ
It is given that two cars meet in 5 hours. 
Distance travelled by car X in 5 hours  =  5x km AQ = 5x
Distance travelled by car y  in 5 hours  = 5y km BQ = 5y
Clearly AQ – BQ = AB 5x – 5y = 100
Both sides divided by 5, we get x – y = 20 …(i)

Case II : When two cars move in opposite direction
Suppose two cars meet at point P, then ,
Distance travelled by X car X = AP
Distance travelled by Y car Y = BP
In this case, two cars meet in 1 hour
Therefore ,
Distance travelled by car x in 1 hours = 1x km
Distance travelled by car y in 1 hours = 1y km 
AP + BP = AB
1x + 1y = 100
x + y = 100
x + y = 100 – (ii)
By solving (i) and (ii) we get ,
x-y = 20
x+y = 100
2x = 120
x = 120 / 2
x = 60
By substituting x = 60  in equation (ii), we get
x + y = 100
60 + y = 100
y = 100 – 60
y = 40
Hence, speed of car X is 60 km/hr , speed of car Y is 40 km/hr .

17. While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.

Solution

Let the speed of Ajeet and Amit be x Km/hr respectively. Then, 
Time taken by Ajeet to cover 30 km = 3 /x hrs
Time taken by Amit to cover 30 km = 30/y hrs
By this given condition , we have
30/x – 30/y = 2 …(i)
If Ajeet doubles his speed , then speed of Ajeet is 2x km/hr
Time taken by Ajeet to cover 30 km = 30/2x hrs
Time taken by Amit to cover 30 km = 30/y hrs
According to the given condition, we have
-15x + 30/y = 1 …(ii)
Putting 1/x = u and 1/y = v , in equation (i) and (ii), we get
30u – 30v = 2 …(iii)
-15u + 30v = 1 …(iv)
Adding equations (iii) and (iv), we get
30u – 30v = 2
-15u + 30v = 1
15u = 3
u = 15/3
u = 1/5
Putting u = 1/5 in equation (iii) , we get
30u – 30v = 2
30 × 1/5 – 30v = 2
6 – 30v = 2
-30v = 2 – 6
 -30v = -4
v = - 4/-30
v = 2/15
Now, u = 1/5
1/x = 1/5
x = 5
and v = 2/15
1/y = 2/15
y = 15/2
y = 7.5
Hence the speed of Ajeet is 5 km/hr
The speed of Amit is 7.5 km/hr 

18. A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace(speed) he is ahead of B by 1½ hours. Find the speeds of A and B.

Solution

Let the speed of A and B be x Km/hr and y Km/hr respectively. Then,
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Notes of Ch 2 Business Environment| Class 12th Business Studies

January 17, 2019, 3:09 am
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Summary and Notes of Ch 3 Business Environment| Class 12th Business Studies 

Meaning of business environment

The term business environment means the sum total of all individuals, institutions and other forces that are outside the control of a business enterprise but that may affect its performance.

Characteristics of Business Environment

• Totality of external forces: It is sum total of all those factors/forces which are available outside the business and over which the business has no control.

• Specific and General Forces- specific forces affect the firms of an industry separately for example suppliers, competitive firms, investors etc and general forces affect all the firms of an industry equally for example social, political and technical situation.

• Interrelatedness- The different factors of business environment are co- related. A change in one factor affects the other factor.

• Dynamic Nature- Business environment is dynamic in that it keeps on changing whether in terms of technological improvement, shifts in consumer preferences or entry of new competition in the market.

• Uncertainty- Business environment is largely uncertain as it is very difficult to predict future happenings.

• Complexity- Environment comprises of many factors and individual effort cannot be recognise easily.
Relatively- Business environment is a relative concept since it differs from country to country and even region to region.

Importance of Business Environment

• It enables the firms to identify opportunities and getting the first mover advantage.
• It helps the firm to identify threats and early warning signals.
• It helps in tapping useful resources.
• It helps in coping with rapid changes.
• It helps in assisting in planning and policy formulation.
• It helps in improving performance.

Dimensions of Business Environment

Dimensions of business environment have the following important factors:

• Economic Environment
• Political  Environment
• Social Environment
• Legal Regulatory Environment
• Technological Environment

• Economic Environment:  It refers to the combination of economic systems, economic policies, and economic conditions of a country.

• Political Environment: It refers to political conditions in the country and attitude of the government towards business.

• Social Environment: It refers to the combination of all the characteristics of the society in which the organisation exists.

• Technological Environment: Technological Environment consists of innovations of new techniques to produce goods and services and operating the new business.

• Legal Regulatory Environment: It refers the some total of all the act passed by the government judgements of the courts, and decisions rendered by various commissions and agencies in the country.

Economic Environment in India

Liberalisation

Liberalisation means to unshackle the economy from bureaucratic cobweb to make it more competitive.
Following are its chief features:
• Abolishing licensing requirement in most of the industries except some.
• Freedom in deciding the scale of business activities.
• Removal of restrictions on the movement of goods and services.
• Freedom in fixing the prices of goods and services.
• Reduction in tax rate.
• Simplifying import-export procedure.
• Simplifying the process of attracting foreign capital and technology.

Privatisation

It refers to such an economic process through which some public sector undertakings are bought either partially or completely under private ownership.
Following are its chief features:
• Reducing the role of public sector and increasing the role of private sector.
• Reducing fiscal burden of the government.
• Reducing the size of the government machinery.
• Speeding up economic development.
• Improving management of enterprises.
• Increase in government treasury.

Globalisation

Globalisation means integrating the economy with the rest of the world.
Following are its chief features:
• Free flow of goods and services, capital, information and technology, movement of people in all the countries.
• The same conflict solving technique in all countries.

Demonetisation

Demonetisation means restrictions were placed on the convertibility of domestic money and bank deposits.
Following are its chief features:
• Demonetisation is viewed as a tax administration measure.
• Tax evasion will no longer be tolerated or accepted.
• Create less-cash or cash- lite economy.

Impact of government policy changes on business and industry

The Indian Corporate sector has come face to face with several challenges due to government policy changes. Some important points in this respect of the following:

• Increasing competition.
• More Demanding Customers.
• Rapidly Changing Technological environment.
• Necessity for change.
• Need for Developing Human Resources.
• Market orientation.
• Loss of Budgetary Support to the public sector.


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