R.D. Sharma Solutions Class 9th: Ch 24 Measure of Central Tendency MCQ's

December 19, 2018, 10:00 pm

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Chapter 24 Measure of Central Tendency R.D. Sharma Solutions for Class 9th MCQ's

Exercise 24.4

1. If the ratio of mean and median of a certain data is 2:3, then find the ratio of its mode and mean.

Solution

Given that the ratio of mean and median of a certain data is 2:3. That is,

2. If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.

Solution

Given that the ratio of mode and median of a certain data is 6:5. That is,

3. If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x .

Solution

The given data is x+2, 2x+3, 3x+4, 4x+5. They are four in numbers.
The mean is

4. The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.

Solution

If the difference of mode and median of a data is 24, then find the difference of median and mean.
Given that the arithmetic mean and mode of a data are 24 and 12 respectively. That is,
Mean = 24
Mode = 12
We have to find median.
We know that,
Mode = 3 × Median - 2 × Mean
⇒ 12 = 3 × Median - 2 × 24
⇒ 3 × Median = 12 + (2 × 24)
⇒ 3 × Median = 12 + 48
⇒ 3 × Median = 60
⇒ Median = 60/3
⇒ Median = 20

5. If the difference of mode and median of a data is 24, then find the difference of median and mean.

Solution

Given that the difference of mode and median of a data is 24. That is,
Mode - Median = 24
⇒ Mode = Median + 24
We have to find the difference between median and mean
We know that,
Mode = 3 × Median - 2 × Mean
⇒ Median + 24 = × Median - 2 × Mean
⇒ 24 = 3 × Median - Median - 2 × Mean
⇒ 24 = 2 × Median - 2 × Mean
⇒ 2 × Median - 2 × Mean = 24
⇒ 2(Median - Mean) = 24
⇒ Median - Mean = 24/2
⇒ Median - Mean = 12

6. If the median of scores x/2, x/3, x/4, x/5 and x/6 (where x > 0) is 6, then find the value of x/6.

Solution

7. If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.

Solution

The given data is 2,4,6,8,x,y. They are 6 in numbers.
The mean is
2+4+6+8+x+y/6 = 20+x+y/6
But, it is given that the mean is 5. Hence, we have
20+x+y/6 = 5
⇒ 20+x+y = 30
⇒ x+y = 30 - 20
⇒ x+y = 10

8. If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.

Solution

The given data is 3,4,5,4,6,6,x.

The mode is the value which occur maximum number of times, that is, the mode has maximum frequency. If the maximum frequency occurs for more than 1 value, then the number of mode is more than 1 and is not unique.

Here it is given that the mode is 4. So, x must be 4, otherwise it contradicts that the mode is 4. Hence x = 4

9. If the median of 33, 28, 20, 25, 34, x is 29, find the maximum possible value of x.

Solution

10. If the median of the scores 1, 2, x, 4, 5 (where 1 < 2 < x < 4 < 5) is 3, then find the mean of the scores.

Solution

The given data is 1, 2, x, 4 and 5. Since 1<2<x<4<5, the given data is already in ascending order.
Here, the number of observation n = 5, which is an odd number.
Hence, the median is

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Notes of Ch 9 Vital Villages, Thriving Towns| Class 6th History

December 20, 2018, 9:31 pm

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Notes of Chapter 9 Vital Villages, Thriving Towns Class 6th History

Iron tools and agriculture

• Around 2500 years ago, there is evidence for the growing use of iron tools.

• These included axes for clearing forests and the iron ploughshare.

Other steps to increase production: irrigation

• Irrigation was also used to increased production other than use of new tools and the system of transplantation.

• Irrigation works that were built during this time included canals, wells, tanks, and artificial lakes.

Who lived in the villages?

• There were at least three different kinds of people living in most villages in the southern and northern parts of the subcontinent.

• In the Tamil region,
→ Large landowners were known as vellalar
→ Ordinary ploughmen were known as uzhavar
→ Landless labourers, including slaves, were known as kadaisiyar and adimai.

• In the northern part of the country,
→ The village headman was known as the grama bhojaka (often the largest landowner).
→ Independent farmers were known as grihapatis
→ Landless labourers were known as the dasa karmakara (had to earn a living working on the fields owned by others.)

• In most villages, there were also some crafts persons such as the blacksmith, potter, carpenter and weaver.

Finding out about cities: stories, travellers, sculpture and archaeology

• Jatakas stories that were probably composed by ordinary people, and then written down and preserved by Buddhist monks.

• Sculptors carved scenes depicting peoples’ lives in towns and villages, as well as in the forest which were used to decorate railings, pillars and gateways of buildings that were visited by people.

• Rows of pots, or ceramic rings arranged one on top of the other, known as ring wells, were used as toilets in some cases, and as drains and garbage dumps.

• Accounts of sailors and travellers who visited the early cities also provided information about the past.

Coins

• Archaeologists have found several thousands of punch-marked coins belonging to this period made up of silver or copper have been found belonging to this period.

Cities with many functions

• A city performed a variety of functions.

• Mathura was an important settlement for more than 2500 years as it was located at the cross roads of two major routes of travel and trade.
→ It was also a centre for arts, crafts, religion and administration according to inscriptions that have been found.

Crafts and crafts persons

• Archaeological evidence for crafts which include extremely fine pottery, known as the Northern Black Polished Ware (NBPW).

• The famous centres for cloth manufacturing were Varanasi and Madurai, where both men and women worked.

• Many crafts persons and merchants formed associations called shrenis.

A closer look — Arikamedu

• Between 2200 and 1900 years ago, Arikamedu was a coastal settlement where ships unloaded goods from distant lands.

• A massive brick structure, which may have been a warehouse, was found at the site.

• Finds also include pottery from the Mediterranean region, such as amphorae (tall double-handled jars that contained liquids such as wine or oil) and stamped red-glazed pottery, known as Arretine Ware, which was named after a city in Italy.

• Roman lamps, glassware and gems have also been found at the site.

NCERT Solutions of Chapter 9 Vital Villges, Thriving Towns

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R.D. Sharma Solutions Class 9th: Ch 25 Probability Exercise 25.1

December 21, 2018, 4:18 am

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Chapter 25 Probability R.D. Sharma Solutions for Class 9th Exercise 25.1

Exercise 25.1

1. A coin is tossed 1000 times with the following frequencies:
Head: 455, Tail: 545
Compute the probability for each event.

Solution

The coin is tossed 1000 times. So, the total number of trials is 1000.
Let A be the event of getting a head and B be the event of getting a tail.
The number of times A happens is 455 and the number of times B happens is 545.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by

P(A) = m/n

Therefore, we have

P(A) = 455/1000 = 0.455

P(B) = 545/1000 = 0.545

2. Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:
Two heads: 95 times
One tail: 290 times
No head: 115 times
Find the probability of occurrence of each of these events.

Solution

The total number of trials is 500.
Let A be the event of getting two heads, B be the event of getting one tail and C be the event of getting no head.
The number of times A happens is 95, the number of times B happens is 290 and the number of times C happens is 115.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by

P(A) = m/n

Therefore, we have

P(A) = 95/500 = 0.19

P(B) = 290/500 = 0.58

P(C) = 115/500 = 0.23

3. Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

Outcome	No head	One head	Two Head	Three Head
Frequency	14	38	36	12

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.

Solution

The total number of trials is 100.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) m/n

(i) Let A be the event of getting two heads.
The number of times A happens is 36.
Therefore,we have
P(A) = 36/100 = 0.36

(ii) Let B be the event of getting three heads
The number of times B happens is 12.
Therefore, we have.
P(B) = 12/100 = 0.12

(iii) Let C be the event of getting at least one head.
The number of times C happens is 38+36+12 = 86
Therefore, we have
P(C) = 86/100 = 0.86

(iv) Let D be the event of getting more heads than tails.
The number of times D happens is 36+12=48.
Therefore, we have
P(D) = 48/100 = 0.48

(v) Let E be the event of getting more tails than heads .
The number of times E happens is 14 + 38 = 52.
Therefore, We have
P(E) = 52/100 = 0.52.

4. 1500 families with 2 children were selected randomly and the following data were recorded:

Number of girls in a family	0	1	2
Number of families	211	814	475

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys

Solution

The total number of trials is 1500.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) m/n

(i) Let A be the event of getting two heads.
The number of times A happens is 211.
Therefore,we have
P(A) = 211/1500 = 0.1406

(ii) Let B be the event of getting three heads
The number of times B happens is 814.
Therefor, we have.
P(B) = 814/1500 = 0.5426

(iii) Let C be the event of having two girls.
The number of times C happens is 475.
Therefore, we have
P(C) = 475/1500 = 0.3166

(iv) Let D be the event of having at most one girl.
The number of times D happens is 211+814 = 1025.
Therefore, we have
P(D) = 1025/1500 = 0.6833

(v) Let E be the event of having more girls than boys.
The number of times E happens is 475.
Therefore, we have
P(E) = 475/1500 = 0.3166

5. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays.
(i) he hits boundary
(ii) he does not hit a boundary.

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by

and is given by
P(A) = m/n

(i) Let A be the event of hitting boundary.
The number of times A happens is 6.
Therefore, we have
P(A) = 6/30 = 0.2

(ii) Let B be the event of does not hitting boundary.
The number of times B happens is 30-6=24.
Therefore, we have
P(B) = 24/30 = 0.8

6. The percentage of marks obtained by a student in monthly unit tests are given below:

Unit Test	I	II	III	IV	V
Percentage of marks obtained	69	71	73	68	76

Find the probability that the student gets:
(i) More than 70% marks
(ii) Less than 70% marks
(iii) A distinction

Solution

The total number of trials is 5.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A be the event of getting more than 70% marks.
The number of times A happens is 3.
Therefore, we have
P(A) = 3/5 = 0.6

(ii) Let B be the event of getting less than 70% marks.
The number of times B happens is 2.
Therefore, we have
P(B) = 2/5 = 0.4

(iii) Let C be the event of getting a distinction.
The number of times C happens is 1.
Therefore, we have
P(C) = 1/5 = 0.2

7. To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	Like	Dislike
Number of student	135	65

Find the probability that a student chosen at random (i) likes Mathematics (ii) does not like it.

Solution

The total number of trials is 200.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A be the event of liking mathematics.
The number of times A happens is 135.
Therefore, we have
P(A) = 135/200 = 0.675

(ii) Let B be the event of disliking mathematics .
The number of times A happens is 65.
Therefore, we have
P(B) 65/200 = 0.325.

8. The blood groups of 30 students of class IX are recorded as follows:
A B O O AB O A O B A O B A O O A AB O A A O O AB B A O B A B O
A student is selected at random from the class from blood donation, Fin the probability that the blood group of the student chosen is:
(i) A
(ii) B
(iii) AB
(iv) O

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the blood group of a chosen student is A.
The number of times A1 happens is 9.
Therefore, we have
P(A1) = 9/30 = 0.3

(ii) Let A2 be the event that the blood group of a chosen student is B.
The number of times A2 happens is 6.
Therefore, we have
P(A2) = 6/30 = 0.2

(iii) Let A3 be the event that the blood group of a chosen student is AB.
The number of times A3 happens is 3.
Therefore, we have
P(A3) = 3/30 = 0.1

(iv) Let A4 be the event that the blood group of a chosen student is O.
The number of times A4 happens is 12.
Therefore, we have

9. Eleven bags of wheat flour, each marked 5 Kg, actually contained the following weights of flour (in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution

The total number of trials is 11.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the actual weight of a chosen bag contain more than 5 Kg of flour.
The number of times A1 happens is 7.
Therefore, we have P(A1) = 7/11

10. Following table shows the birth month of 40 students of class IX.

Jan.	Feb	March	April	May	June	July	Aug.	Sept.	Oct.	Nov.	Dec.
3	4	2	2	5	1	2	5	3	4	4	4

Find the probability that a student was born in August.

Solution

The total number of trials is 40.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the birth month of a chosen student is august.
The number of times A1 happens is 5.
Therefore, we have
P(A1) = 5/40 = 1/8

11. Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Conc. of SO2	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24
No. of Days	4	8	9	2	4	3

Find the probability of concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.

Solution

The total number of trials is 30.
Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Let A1 be the event that the concentration of sulphur dioxide in a day is 0.12-0.16 parts per million. The number of times A1 happens is 2.
Therefore, we have
P(A1) = 2/30 = 1/15

12. A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

If a family is chosen, find the probability that family is:
(i) earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v) owning not more than 1 vehicle
(vi) owning at least one vehicle.

Solution

The total number of trials is 2400.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that a chosen family earns Rs 10000-13000 per month and owns exactly 2 vehicles.
The number of times A1 happens is 29.
Therefore, we have P(A1) = 29/2400

(ii) Let A2 be the event that a chosen family earns Rs 16000 or more per month and owns exactly 1 vehicle.
The number of times A2 happens is 579.
Therefore, we have P(A2) = 579/2400.

(iii) Let A3 be the event that a chosen family earns less than Rs 7000 per month and does not owns any vehicles.
The number of times A3 happens is 10.
Therefore, we have
P(A3) = 10/2400 = 1/240

(iv) Let A4 be the event that a chosen family earns Rs 13000-16000 per month and owns more than 2 vehicles.
The number of times A4 happens is 25.
Therefore, we have
P(A4) = 25/2400 = 1/96

(v) Let A5 be the event that a chosen family owns not more than 1 vehicle (may be 0 or 1). In this case the number of vehicles is independent of the income of the family.
The number of times A5 happens is
(10+0+1+2+1) + (160+305+535+469+579) = 2062.
Therefore, we have
P(A5) = 2062/2400 = 1031/1200

(vi) Let A6 be the event that a chosen family owns atleast 1 vehicle (may be 1 or 2 or above 2). In this case the number of vehicles is independent of the income of the family.
The number of times A6 happens is
(160+305+535+469+579) + (25+27+29+29+82) + (0+2+1+25+88) = 2356
Therefore, we have
P(A6) = 2356/2400 = 589/600

13. The following table gives the life time of 400 neon lamps:

Life time(in hours)	300-400	400-500	500-600	600-700	700-800	800-900	900-1000
Number of lamps:	14	56	60	86	74	62	48

A bulb is selected of random, Find the probability that the the life time of the selected bulb is:
(i) less than 400
(ii) between 300 to 800 hours
(iii) at least 700 hours.

Solution

The total number of trials is 400.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the lifetime of a chosen bulb is less than 400 hours.
The number of times A1 happens is 14.
Therefore, we have
P(A1) = 14/400 = 7/200

(ii) Let A2 be the event that the lifetime of a chosen bulb is in between 300 to 800 hours.
The number of times A2 happens is 14+56+60+86+74 = 290.
Therefore, we have
P(A2) = 290/400 = 29/40

(iii) Let A3 be the event that the lifetime of a chosen bulb is atleast 700 hours.
The number of times A3 happens is 74+62+48=184.
Therefore, we have
P(A3) = 184/400 = 23/50

14. Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

Wages (in Rs)	110-130	130-150	150-170	170-190	190-210	210-230	230-250
Number of workers	3	4	5	6	5	4	3

A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.

Solution

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n

(i) Let A1 be the event that the wages of a worker is less than Rs 150. The number of times A1 happens is 3 + 4 = 7
Therefore, we have P(A1) = 7/30.

(ii) Let A2 be the event that the wages of a worker is atleast Rs 210.
The number of times A2 happens is 4+3 = 7 .
Therefore, we have P(A2) = 7/30.

(iii) Let A3 be the event that the wages of a worker is more than or equal to Rs 150 but less than Rs 210.
The number of times A3 happens is 5+6+5 = 16.
Therefore, we have
P(A3) = 16/30 = 8/15

↧

R.D. Sharma Solutions Class 9th: Ch 25 Probability MCQ's

December 23, 2018, 9:20 am

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Chapter 25 Probability R.D. Sharma Solutions for Class 9th MCQ's

Multiple Choice Questions

1. Mark the correct alternative in each of the following:
The probability of an impossible event is
(a) 1
(b) 0
(c) less than 0
(d) greater than 1

Solution

We have to find the probability of an impossible event.
Note that the number of occurrence of an impossible event is 0. This is the reason that’s why it is called impossible event.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Note that n is a positive integer, it can’t be zero. So, whatever may be the value of n, the probability of an impossible event is 0/n = 0.
Hence, the correct option is (b).

2. The probability of a certain event is
(a) 0
(b) 1
(c) greater than 1
(d) less than 0

Solution

We have to find the probability of a certain event.
Note that, the number of occurrence of an impossible event is same as the total number of trials. When we repeat the experiment, every times it occurs. This is the reason that’s why it is called certain event. Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Hence the correct option is (b).

3. The probability an event of a trial is
(a) 1
(b) 0
(c) less than 1
(d) more than 1

Solution

Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Note that m is always less than or equal to n and n is a positive integers, it can’t be zero. But, m is a non negative integer. So, the maximum value of probability of an event is n/n = 1 , which is the probability of a certain event and the minimum value of it is 0, which is the probability of an impossible event. For any other events the value is in between 0 and 1.
Hence the correct option is (c).

4. Which of the following cannot be the probability of an event ?
(a) 1/3
(b) 3/5
(c) 5/3
(d) 1

Solution

Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Note that m is always less than or equal to n and n is a positive integers, it can’t be zero. But, m is a non negative integer. So, the maximum value of probability of an event is n/n = 1, which is the probability of a certain event and the minimum value of it is 0, which is the probability of an impossible event. For any other events the value is in between 0 and 1.
All the options except (c) satisfy the above criteria’s.
Hence, the correct option is (c).

5. Two coins are tossed simultaneously. The probability of getting atmost one head is
(a) 1/4
(b) 3/4
(c) 1/2
(d) 1/4

Solution

The random experiment is tossing two coins simultaneously.
All the possible outcomes are HH, HT, TH, and TT.
Let A be the event of getting at most one head.
The number of times A happens is 3.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have
P(A) = 3/4
So, the correct choice is b .

6. A coin is tossed 1000 times, if the probability of getting a tail is 3/8, how many times head is obtained?
(a) 525
(b) 375
(c) 625
(d) 725

Solution

The total number of trials is 1000. Let x be the number of times a tail occurs.
Let A be the event of getting a tail.
The number of times A happens is x.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have P(A) = x/1000
But, it is given that P(A) = 3/8. So, we have
x/1000 = 3/8
⇒ 8x = 3000
⇒ x = 3000/8
⇒ x = 375
Hence, a tail is obtanined 375 times.
Consequently, a head is obtained 1000-375 = 625 times.
So, the correct choice is (c).

7. A dice is rolled 600 times and the occurrence of the outcomes 1, 2, 3, 4, 5 and 6 are given below:

The probability of getting a prime number is
(a) 1/3
(b) 2/3
(c) 49/60
(d) 39/125

Solution

The total number of trials is 600.
Let A be the event of getting a prime number (2, 3 and5).
The number of times A happens is 30+120+50 = 200.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
therefore, we have
P(A) = 200/600 =1/3
So. the correct choice is (A)

8. The percentage of attendance of different classes in a year in a school is given below:

Class	X	IX	VIII	VII	VI	V
Attendance	30	62	85	92	76	55

What is the probability that the class attendance is more than 75% ?
(a) 1/6
(b) 1/3
(c) 5/6
(d) 1/2

Solution

The total number of trials is 6.
Let A be the event that the attendance of a class is more than 75%.
The number of times A happens is 3 (for classes’ VIII, VII and VI).
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore , we have
P(A) = 3/6 = 1/2
So, the correct choice is (d).

9. A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at random. The probability that the number on the coin is not a prime number, is
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5

Solution

The total number of trials is 50.
Let A be the event that the number on the picked coin is not a prime.
The prime’s lies in between 51 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. They are 10 in numbers. Therefore the numbers lies between 51 and 100 and which are not primes are 50-10 = 40 in numbers.
So, the number of times A happens is 40.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have
P(A) = 40/50 = 4/5
So, the correct choice is (d).

10. In a football match, Ronaldo makes 4 goals from 10 penalty kicks. The probability of converting a penalty kick into a goal by Ronaldo, is
(a) 1/4
(b) 1/6
(c) 1/3
(d) 2/5

Solution

The total number of trials is 10.
Let A be the event that Ronaldo makes a goal in a penalty kick.
The number of times A happens is 4.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) = m/n
Therefore, we have
P(A) = 4/10 = 2/5
So, the correct choice is (d) .

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CBSE 10th Datesheet 2019| Board Exam

December 23, 2018, 8:52 pm

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CBSE Class 10th Datesheet 2019| Board Exam Datesheet

CBSE has released Class 10th datesheet 2019 board examination. The Board will begin the exams from 22nd February 2018 and will continue till 4 April 2018. The exam will be conducted in the morning session, from 10.30 am to 1:30 pm, the answer books will be distributed at 10 am and question papers at 10:15 am.

Last year, the CBSE board exam datesheet for Class 10 was released on January 10.

The examinations would begin on 2nd March with Information Technology while the main subject exam, Math is on 7th March, 2019. This year Board has also decided to conduct the examination of skill subjects in latter half of February 2019. It is expected that about 16 lakh students of class 10 would appear in the CBSE Board Examination 2018. This year many of the chapters had been removed from textbooks such as English and Hindi.

CBSE date sheet 2019 is available at the official website cbse.nic.in.

Date	Subjects
7th March, Thursday	Mathematics (041)
13th March, Wednesday	Science (086)
19th March, Tuesday	Hindi Course-A (002) Hindi Course-B (085)
23rd March, Saturday	English Comm. (101) English LNG & LIT (184)
29th March, Friday	Social Science (087)

You can download Class 10th datesheet 2019 from the official website.

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CBSE 12th Datesheet 2019| Board Exam

December 23, 2018, 10:47 pm

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CBSE Class 12th Datesheet 2019| Board Exam Datesheet

CBSE has released Class 12th datesheet 2019 board examination. The Board will begin the exams from 15th February 2019 and will continue till 3rd April 2019.

The exam will be conducted in the morning session, from 10.30 am to 1:30 pm, the answer books will be distributed at 10 am and question papers at 10:15 am.

Last year, the CBSE board exam datesheet for Class 12 was released on January 10.

The examinations would begin on 2nd March with English. This year Board has also decided to conduct the examination of skill subjects in latter half of February 2019. It is expected that about 12 lakh students of class 10 would appear in the CBSE Board Examination 2018. This year many of the chapters had been removed from textbooks such as English and Hindi.

CBSE date sheet 2019 is available at the official website cbse.nic.in.

CBSE Class 12 Science stream exam dates

Date	Subject
12th March, Saturday, March 2	English Elective-N (001) English Elective-C (101) English Core (301)
15th March, Tuesday, March 5	Physics (042)
12th March, Tuesday, March 12	Chemistry (043)
15th March, Friday, March 15	Biology (044)
18th March, Monday, March 18	Mathematics (041)

CBSE Class 12 Commerce stream exam dates

Date	Subject
16th February, Saturday, February 16	Cost Accounting (781) Fin. Accounting (780)
2nd March, Saturday, March 2	English Elective-N (001) English Elective-C (101) English Core (301)
14th March, Thursday, March 14	Business Studies (054)
18th March, Monday, March 18	Mathematics (041)
27th March, Wednesday, March 27	Economics (030)
28th March, Thursday, March 28	Informatics Prac. (065) Computer Science (083)

CBSE Class 12 Arts Stream exam dates

Date	Subject
20th February, Wednesday	Fashion Studies (053)
7th March, Thursday	Geography (029)
11th March, Monday	Sociology (039)
19th March, Tuesday	Political Science (028)
25th March, Monday	History (027)
27th March, Wednesday	Economics (030)
29th March, Friday	Psychology (0370
1st April, Monday	Home Science (064)
2nd April, Tuesday	Philosophy (0400

You can download Class 12th datesheet 2019 from the official website.

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Notes of Kshitiz Part I Class 9th

December 26, 2018, 2:08 am

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NCERT Notes of Kshitiz Part I Class 9th (पठन सामग्री और सार)

पाठ 1 - दो बैलों की कथा

इस कथा के माध्यम से प्रेमचंद ने कृषक समाज और पशुओं के भावात्मक संबंध को बताया है। चूँकि ये कहानी आजादी से पहले की है इसीलिए ये भी सन्देश देती है कि स्वतंत्रता आसानी से नहीं मिलती, उसके लिए बार-बार संघर्ष करना पड़ता है।

पाठ 2 - ल्हासा की ओर

यह अंश राहुल जी की प्रथम तिब्बत यात्रा से लिया गया है जो उन्होंने सन् 1929-30 में नेपाल के रास्ते की थी। उस समय भारतीयों को तिब्बत यात्रा की अनुमति नहीं थी, इसलिए उन्होंने यह यात्रा एक भिखमंग के छद्म वेश में की थी।

पाठ 3 - उपभोक्तावाद की संस्कृति

इस निबंध में लेखक बाजार की गिरफ्त में आ रहे समाज की वास्तविकता को प्रस्तुत करता है। लेखक का मानना है कि हम विज्ञापन की चमक-दमक के कारण वस्तुओं के पीछे भाग रहे हैं, हमारी निगाह गुणवत्ता पर नहीं है।

पाठ 4 - साँवले सपनों की याद

जून 1987 में प्रसिद्ध पक्षी विज्ञानी सालिम अली की मृत्यु के तुरंत बाद डायरी शैली में लिखा गया संस्मरण है। सालिम अली की मृत्यु से उत्पन्न दुख और अवसाद को लेखक ने साँवले सपनों की याद के रूप में व्यक्त किया है।

पाठ 5 - नाना साहब की पुत्री देवी मैना को भस्म कर दिया गया

यह एक ऐतिहासिक सच्चाई है कि सन् 1857 की क्रांति के विद्रोही नेता धुंधूपंत नाना साहब की पुत्री बालिका मैना आज़ादी की नन्हीं सिपाही थी जिसे अंग्रेजों ने जलाकर मार डाला। बालिका मैना के बलिदान की कहानी को चपला देवी ने इस गद्य रचना में प्रस्तुत किया है।

पाठ 6 - प्रेमचंद के फटे जूते

प्रेमचंद के फटे जूते शीर्षक निबंध में परसाई जी ने प्रेमचंद के व्यक्तित्व की सादगी के साथ एक रचनाकार की अंतर्भेदी सामाजिक दृष्टि का विवेचन करते हुए आज की दिखावे की प्रवृत्ति एवं अवसरवादिता पर व्यंग्य किया है।

पाठ 7 - मेरे बचपन के दिन

मेरे बचपन के दिन में महादेवी जी ने अपने बचपन के उन दिनों को स्मृति के सहारे लिखा है जब वे विद्यालय में पढ़ रही थीं। इस अंश में लड़कियों के प्रति सामाजिक रवैये, विद्यालय की सहपाठिनों, छात्रावास के जीवन और स्वतंत्रता आंदोलन के प्रसंगों का बहुत ही सजीव वर्णन है।

पाठ 8 - एक कुत्ता और एक मैना

एक कुत्ता और एक मैना निबंध में पशु-पक्षियों के प्रति मानवीय प्रेम प्रदर्शित है| इसमें रवींद्रनाथ की कविताओं और उनसे जुड़ी स्मृतियों के सहारे गुरुदेव की संवेदनशीलता, आंतरिक विराटता और सहजता के चित्र

तो उकेरे ही गए हैं, पशु-पक्षियों के संवेदनशील जीवन का भी बहुत सूक्ष्म निरीक्षण | है। यह निबंध सभी जीवों से प्रेम की प्रेरणा देता है।

पाठ 9 - साखियाँ एवं सबद

यहाँ संकलित साखियों में प्रेम का महत्व, संत के लक्षण, ज्ञान की महिमा, बाह्याडंबरों का विरोध आदि भावों का उल्लेख हुआ है। पहले सबद में बाह्याडंबरों का विरोध एवं अपने भीतर ही ईश्वर की व्याप्ति का संकेत है तो दूसरे सबद में ज्ञान की आँधी के रूपक के सहारे ज्ञान के महत्व का वर्णन है।

पाठ 10 - वाख

पहले वाख में ललद्यद ने ईश्वर-प्राप्ति के लिए किए जाने वाले | अपने प्रयासों की व्यर्थता की चर्चा की है। दूसरे में बाह्याडंबरों का विरोध करते हुए यह कहा गया है कि अंत:करण से समभावी होने पर ही मनुष्य की चेतना व्यापक हो सकती है। तीसरे वाख में कवयित्री के आत्मालोचन की अभिव्यक्ति है। वे अनुभव करती हैं कि भवसागर से पार जाने के लिए सद्कर्म ही सहायक होते हैं।

पाठ 11 - सवैये

पहले और दूसरे सवैये में कृष्ण और कृष्ण-भूमि के प्रति कवि का अनन्य समर्पण भाव व्यक्त हुआ है। तीसरे छंद में कृष्ण के रूप-सौंदर्य के प्रति गोपियों की उस मुग्धता का चित्रण है जिसमें वे स्वयं कृष्ण का रूप धारण कर लेना चाहती हैं। चौथे छंद में कृष्ण की मुरली की धुन और उनकी मुसकान के अचूक प्रभाव तथा गोपियों की विवशता का वर्णन है|

पाठ 12 - कैदी और कोकिला

यह कविता भारतीय स्वाधीनता सेनानियों के साथ जेल में किए गए दुर्व्यवहारों और यातनाओं का मार्मिक साक्ष्य प्रस्तुत करती है। कवि जेल में एकाकी और उदास है। कोकिल से अपने मन का दुख, असंतोष और ब्रितानी शासन के प्रति अपने आक्रोश को व्यक्त करते हुए वह कहता है कि यह समय मधुर गीत गाने का नहीं बल्कि मुक्ति का गीत सुनाने का है।

पाठ 13 - ग्राम श्री

ग्राम श्री कविता में पंत ने गाँव की प्राकृतिक सुषमा और समृद्धि का मनोहारी वर्णन किया है। खेतों में दूर तक फैली लहलहाती फसलें, फल-फूलों से लदी पेड़ों की डालियाँ और गंगा की सुंदर रेती कवि को रोमांचित करती है।

पाठ 14 - चंद्र गहना से लौटती बेर

प्रस्तुत कविता में कवि का प्रकृति के प्रति गहन अनुराग व्यक्त हुआ है। वह चंद्र गहना नामक स्थान से लौट रहा है। लौटते हुए उसके किसान मन को खेत-खलिहान एवं उनका प्राकृतिक परिवेश सहज आकर्षित कर लेता है। इसी बात को वह कविता को द्वारा दर्शा रहे हैं|

पाठ 15 - मेघ आए

कविता में कवि ने मेघों के आने की तुलना सजकर आए प्रवासी अतिथि (दामाद) से की है। ग्रामीण संस्कृति में दामाद के आने पर उल्लास का जो वातावरण बनता है, मेघों के आने का सजीव वर्णन करते हुए कवि ने उसी उल्लास को दिखाया है।

पाठ 16 - यमराज की दिशा

कविता में कवि सभ्यता के विकास की खतरनाक दिशा की ओर इशारा करते हुए कहना चाहता है कि जीवन-विरोधी ताकतें चारों तरफ़ फैलती जा रही हैं।

पाठ 17 बच्चे काम पर जा रहे हैं

प्रस्तुत कविता में बच्चों से बचपन छीन लिए जाने की पीड़ा व्यक्त हुई है। कवि ने उस सामाजिक-आर्थिक विडंबना की ओर इशारा किया है जिसमें कुछ बच्चे खेल, शिक्षा और | जीवन की उमंग से वंचित हैं।

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पाठ 4 - धर्म, जाति और लिंग लोकतांत्रिक राजनीति के नोट्स| Class 10th

December 26, 2018, 5:31 am

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पठन सामग्री और नोट्स (Notes)| पाठ 2 - धर्म, जाति और लिंग (Dharm, Jati aur Ling) Loktantrik Rajniti Class 10th

लैंगिक विभाजन

• श्रम के लैंगिक विभाजन- एक प्रणाली जिसमें घर के अंदर के सभी काम परिवार की महिलाओं द्वारा किया जाता है, जबकि पुरुषों से उम्मीद की जाती है की वो पैसा कमाने के लिए घर से बाहर काम करें।

• यह अवधारणा जैविक विज्ञान पर आधारित नहीं है बल्कि सामाजिक अपेक्षाओं और रूढ़िवादों पर आधारित है।

नस्लवादी आंदोलन

• सामाजिक आंदोलनों का उद्देश्य पुरुषों और महिलाओं के बीच समानता स्थापित करना है, जिन्हें नारीवादी आंदोलन कहा जाता है।

विभिन्न तरीकों से महिलाओं का दमन

• साक्षरता दर: जहाँ पुरुष 82.14% साक्षर है वहीं महिलायें मात्रा 65.46% साक्षर है।

• नौकरियां: उच्च भुगतान और उच्च पदों की नौकरियों में महिलाओं का प्रतिशत बहुत कम है क्योंकि कुछ ही लड़कियों को उच्च शिक्षा ग्रहण के लिए प्रोत्साहित किया जाता है।

• मजदूरी: समान मजदूरी अधिनियम लागू होने के बावजूद भी खेल, सिनेमा, कृषि और निर्माण आदि कार्य क्षेत्रों में महिलाओं को पुरुषों के तुलना में कम वेतन दिया जाता है।

• लिंग अनुपात: अधिकांश माँ-बाप में बेटियों के अपेक्षाकृत बेटों की इच्छा ज्यादा होती हैं, इसी मानसिकता का कारण हैं की हमारे देश में जन्म से पहले और जन्म के बाद बेटियों की हत्या आम बात हैं। जिसके परिणामस्वरूप लिंग अनुपात पर प्रतिकूल असर हुआ है।

• सामाजिक बुराई: विशेष रूप से शहरी क्षेत्र महिलाओं के लिए सुरक्षित नहीं है। छेड़खानी, दहेज उत्पीड़न और यौन उत्पीड़न आम बात हैं।

• भारत में महिलाओं का राजनीतिक प्रतिनिधित्व बहुत कम है। अभी भी महिलाओं का प्रतिनिधित्व विधानसभा में 5% और लोकसभा में 12% से कम है।

धर्म, सांप्रदायिकता और राजनीति

राजनीति में धार्मिक मतभेद

• मानवाधिकार कार्यकर्ताओं का मानना है कि सांप्रदायिक हिंसा में मरने वाले ज्यादातर लोग अल्पसंख्यक समुदाय के होते हैं।

सांप्रदायिकता क्या है?

• जब एक धर्म के विचारों को दूसरे से श्रेष्ठ माना जाने लगता है।

सांप्रदायिक राजनीति क्या है?

• समाज में समस्या तब शुरू होती है जब धर्म का इस्तेमाल राजनीतिक फायदे के लिये किया जाता है।

• समस्या तब ज्यादा गंभीर हो जाती है जब अपनी मांगों के लिये विभिन्न धार्मिक समूह आमने-सामने आ जाते है।

• ठीक इसी मौके पर जब सरकार केवल एक धार्मिक समूह की मांगों को पूरा करने के लिए अपनी शक्ति का उपयोग करती है। तब समस्या और भी विकराल रूप ले लेती है ?

• इस तरह धर्म का इस्तेमाल राजनीति फायदे के लिये करना सांप्रदायिक राजनीति कहा जाता है।

सांप्रदायिक राजनीति का सिद्धांत

• धर्म सामाजिक संरचना का मुख्य आधार है।

• एक धर्म के अनुयायियों को एक समुदाय बनाना चाहिए।

• उनके मौलिक हित समान होते हैं।

सांप्रदायिक राजनीति का सिद्धांत गलत क्यों है?

• सामान धर्म के सभी लोग हर बिषय-वस्तु पर समान रुचि और विचार नहीं रखते हैं।

• प्रत्येक व्यक्ति की पहचान विभिन्न संदर्भों में भिन्न-भिन्न होती हैं।

सांप्रदायिकता से निजात पाने लिए कदम उठाए गए कदम

• भारत एक धर्मनिरपेक्ष राज्य है। भारत का कोई अपना आधिकारिक धर्म नहीं है।

• सभी नागरिको एवं सुमदाय को अपने धर्म को अपनाने और प्रचार करने की आजादी है।

• संविधान धर्म के आधार पर भेदभाव को असंबैधानिक घोषित करता है।

जाति और राजनीति

जातिगत असमानताएँ

• समान जाति या व्यवसाय के लोग या मिलते -जुलते व्यवसाय के लोग आपस में सामाजिक समुदाय का गठन कर अपने समुदाय के भीतर ही बेटी -रोटी का संबंध रखते हैं। अन्य जाति समूहों के सदस्यों के साथ उनका नजदीकी पारिवारिक संबंध नहीं होता।

जाति व्यवस्था क्यों बनी रहती है?

• ज्यादातर लोग अपनी जाति के लोगों के साथ ही शादी करना पसंद करते हैं।

• अभी तक छुआछूत पूरी तरह समाप्त नहीं हुई है।

• जिस जाति समूहों के पास शिक्षा की पहुंच सुलभ थी, उसी जाति समूह का उच्च शिक्षा में भी बोलबाला जारी है।

राजनीति जाति से कैसे प्रभावित होती है

• जब पार्टियां चुनाव में उम्मीदवारों का चयन करती हैं, तो जातिबाहुल उम्मीदवार का चयन प्राथमिकता में होती है।

• राजनीतिक दल और उम्मीदवार समर्थन हासिल करने के लिये जातिगत भावनाओं को उकसाते हैं।

सिर्फ जातिगत राजनीति ही भारतीय चुनाव को प्रभावित नहीं करती है

• देश के किसी भी संसदीय क्षेत्र में किसी एक जाति का इतना जनसंख्या नहीं है की वो स्पष्ट बहुमत दिला सके ।

• कोई भी पार्टी अपने समुदाय के सभी मतदाताओं के वोट नहीं पाते हैं।

• अगर किसी जाति समूह के पास अपने ही जाति के उम्मीदवार चयन करने के एक से अधिक विकल्प हो और अन्य जाति समूहों का कोई उम्मीदवार नहीं हो अगर वे जातिगत आधार पर मतदान करना पसंद करते हैं।

• मतदाताओं का लगाव पार्टी और पार्टी के विचारधारा से ज्यादा अपनी जाति के उम्मीदवार से होता है।

जातिगत राजनीति के परिणाम

• पार्टियाँ सत्ता वंचित समूहों को सत्ता में अपनी भागीदारी की मांग करने के लिए प्रेरित करते हैं।

• पार्टियाँ उन्हें सामाजिक न्याय के लिए लड़ने में मदद करते हैं।

• जाति आधारित राजनीति निश्चित रूप से लोकतंत्र के लिए फायदेमंद नहीं है।

• यह गरीबी , विकास और भ्रष्टाचार जैसे महत्वपूर्ण मुद्दों से ध्यान हटाता है।

• इससे तनाव, संघर्ष और हिंसा उत्पन्न हो सकती है।

NCERT Solutions of पाठ 4 - धर्म, जाति और लिंग

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RD Sharma Class 10 Solution - Math

December 26, 2018, 8:49 am

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Solutions of RD Sharma for Class 10 All Chapters - Download PDF

Get Detailed Solution of RD Sharma Class 10. All the solutions of the chapters of RD Sharma is based on the latest textbook based on CBSE pattern. Our faculty has tried to make the solution in easiest and accurate way. Below are the name of the chapters from RD Sharma Mathematics Textbooks. Click on the Exercise name to get the solution.

RD Sharma Class 10 Chapter 1: Real Numbers Solutions
There are total 6 exercises in this chapter and one exercise for MCQ based questions.
RD Sharma Class 10 Real Numbers System Exercise 1.1 Solutions
RD Sharma Class 10 Real Numbers System Exercise 1.2 Solutions

RD Sharma Class 10 Real Numbers System Exercise 1.3 Solutions

RD Sharma Class 10 Real Numbers System Exercise 1.4 Solutions

RD Sharma Class 10 Real Numbers System Exercise 1.5 Solutions

RD Sharma Class 10 Real Numbers System Exercise 1.6 Solutions

RD Sharma Class 10 Real Numbers MCQ's Solutions

RD Sharma Class 10 Chapter 2: Polynomials Solutions
In this chapter, there are total 3 exercise and one is MCQ's type exercise.
RD Sharma Class 10 Polynomials Exercise 2.1 Solutions
RD Sharma Class 10 Polynomials Exercise 2.2 Solutions

RD Sharma Class 10 Polynomials Exercise 2.3 Solutions

RD Sharma Class 10 Polynomials MCQ's Solutions

RD Sharma Class 10 Chapter 3: Pair of Linear Equations in Two Variables Exercise 3.1 Solutions
RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.2 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.3 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.4 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.5 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.6 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.7 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.8 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.9 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.10 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables Exercise 3.11 Solutions

RD Sharma Class 10 Pair of Linear Equations in Two Variables MCQ's Solutions

RD Sharma Class 10 Ch 4: Triangles Solutions
RD Sharma Class 10 Triangles Exercise 4.1 Solutions
RD Sharma Class 10 Triangles Exercise 4.2 Solutions

RD Sharma Class 10 Triangles Exercise 4.3 Solutions

RD Sharma Class 10 Triangles Exercise 4.4 Solutions

RD Sharma Class 10 Triangles Exercise 4.5 Solutions

RD Sharma Class 10 Triangles Exercise 4.6 Solutions

RD Sharma Class 10 Triangles Exercise 4.7 Solutions

RD Sharma Class 10 Triangles MCQ's Solutions

RD Sharma Class 10 Ch 5: Trigonometric Ratios Solutions
RD Sharma Class 10 Trigonometric Ratios Exercise 5.1 Solutions

RD Sharma Class 10 Trigonometric Ratios Exercise 5.2 Solutions

RD Sharma Class 10 Triangles Exercise 5.3 Solutions

RD Sharma Class 10 Triangles MCQ's Solutions

RD Sharma Class 10 Ch 6: Trigonometric Identities Solutions
RD Sharma Class 10 Trigonometric Identities Exercise 6.1 Solutions

RD Sharma Class 10 Trigonometric Identities Exercise 6.2 Solutions

RD Sharma Class 10 Trigonometric Identities MCQ's Solutions

RD Sharma Class 10 Ch 7: Statistics Solutions
RD Sharma Class 10 Statistics Exercise 7.1 Solutions
RD Sharma Class 10 Statistics Exercise 7.2 Solutions

RD Sharma Class 10 Statistics Exercise 7.3 Solutions

RD Sharma Class 10 Statistics Exercise 7.4 Solutions

RD Sharma Class 10 Statistics Exercise 7.5 Solutions

RD Sharma Class 10 Statistics Exercise 7.6 Solutions

RD Sharma Class 10 Statistics MCQ's Solutions

RD Sharma Class 10 Ch 8: Quadratic Equations Solutions
RD Sharma Class 10 Quadratic Equations Exercise 8.1 Solutions
RD Sharma Class 10 Quadratic Equations Exercise 8.2 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.3 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.4 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.5 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.6 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.7 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.8 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.9 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.10 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.11 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.12 Solutions

RD Sharma Class 10 Quadratic Equations Exercise 8.13 Solutions

RD Sharma Class 10 Quadratic Equations MCQ's Solutions

RD Sharma Class 10 Ch 9: Arithmetic Progressions Solutions
RD Sharma Class 10 Arithmetic Progressions Exercise 9.1 Solutions

RD Sharma Class 10 Arithmetic Progressions Exercise 9.2 Solutions

RD Sharma Class 10 Arithmetic Progressions Exercise 9.3 Solutions

RD Sharma Class 10 Arithmetic Progressions Exercise 9.4 Solutions

RD Sharma Class 10 Arithmetic Progressions Exercise 9.5 Solutions

RD Sharma Class 10 Arithmetic Progressions Exercise 9.6 Solutions

RD Sharma Class 10 Arithmetic Progressions MCQ's Solutions

RD Sharma Class 10 Ch 10: Circles Solutions
RD Sharma Class 10 Circles Exercise 10.1 Solutions
RD Sharma Class 10 Circles Exercise 10.2 Solutions
RD Sharma Class 10 Circles MCQ's Solutions

RD Sharma Class 10 Ch 11: Constructions Solutions
RD Sharma Class 10 Constructions Exercise 11.1 Solutions
RD Sharma Class 10 Constructions Exercise 11.2 Solutions

RD Sharma Class 10 Constructions Exercise 11.3 Solutions

RD Sharma Class 10 Ch 12: Some Applications of Trigonometry Solutions

RD Sharma Class 10 Some Applications of Trigonometry Exercise 12.1 Solutions

RD Sharma Class 10 Some Applications of Trigonometry MCQ's Solutions

RD Sharma Class 10 Ch 13: Probability Solutions
RD Sharma Class 10 Probability Exercise 13.1 Solutions
RD Sharma Class 10 Probability Exercise 13.2 Solutions
RD Sharma Class 10 Probability MCQ's Solutions

RD Sharma Class 10 Ch 14: Co-ordinate Geometry Solutions
RD Sharma Class 10 Co-ordinate Geometry Exercise 14.1 Solutions
RD Sharma Class 10 Co-ordinate Geometry Exercise 14.2 Solutions

RD Sharma Class 10 Co-ordinate Geometry Exercise 14.3 Solutions

RD Sharma Class 10 Co-ordinate Geometry Exercise 14.4 Solutions

RD Sharma Class 10 Co-ordinate Geometry Exercise 14.5 Solutions

RD Sharma Class 10 Co-ordinate Geometry MCQ's Solutions

RD Sharma Class 10 Ch 15: Areas Related to Circles Solutions
RD Sharma Class 10 Areas Related to Circles Exercise 15.1 Solutions
RD Sharma Class 10 Areas Related to Circles Exercise 15.2 Solutions

RD Sharma Class 10 Areas Related to Circles Exercise 15.3 Solutions

RD Sharma Class 10 Areas Related to Circles Exercise 15.4 Solutions

RD Sharma Class 10 Areas Related to Circles MCQ's Solutions

RD Sharma Class 10 Ch 16: Surface Areas and Volumes Solutions
RD Sharma Class 10 Surface Areas and Volumes Exercise 16.1 Solutions
RD Sharma Class 10 Surface Areas and Volumes Exercise 16.2 Solutions

RD Sharma Class 10 Surface Areas and Volumes Exercise 16.3 Solutions

RD Sharma Class 10 Surface Areas and Volumes MCQ's Solutions

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R.D. Sharma Solutions Class 10th: Ch 2 Polynomials Exercise 2.1

December 29, 2018, 2:18 am

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Chapter 2 Polynomials R.D. Sharma Solutions for Class 10th Math Exercise 2.1

Exercise 2.1

1. Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their co efficient:
(i) f(x) = x² - 2x - 8
(ii) g(s) = 4s² - 4x + 1
(iii) h(t) = t² - 15
(iv) p(x) = x² +2√2x + 6
(v) q(x) = √3x² + 10x + 7√3
(vi) f(x) = x² - (√3+1)x + √3
(vii) g(x) = a(x² +1) - x(a² +1)
(viii) 6x2 - 3 - 7x

Solution

2. For each of the following , find a quadratic polynomial whose sum and product respectively of the zeroes are as give . Also, find the zeroes of these polynomials by factorization.
(i) -8/3, 4/3
(ii) 21/8, 5/16
(iii) -2√3, -9
(iv) -3/2√5, -1/2

Solution

3. If α and β are the zeros of the quadratic polynomial f(x) = x² – 5x + 4, find the value of 1/α + 1/ β - 2 αβ.

Solution

4. If α and β are the zeros of the quadratic polynomial p(y) = 5y² - 7y + 1, find the value of 1/α + 1/β - 2αβ .

Solution

5. If α and β are the zeros of the quadratic polynomial f(x) = x² - x - 4, find the value of 1/α + 1/ β - αβ .

Solution

6. If α and β are the zeros of the quadratic polynomial f(x) = x² + x - 2, find the value of 1/α - 1/ β .

Solution

7. If one zeros of the quadratic polynomial f(x) = 4x² - 8kx - 9 is negative of the other , find the value of k .

Solution

8. If the sum of the zeros of the quadratic polynomial f(t) = kt² + 2t + 3k is equal to their product, find the value of k. product, find the value of k.

Solution

9. If α and β are the zeros of the quadratic polynomial p(x) = 4x² - 5x - 1, find the value of α²β + αβ².

Solution

10. If α and β are the zeros of the quadratic polynomial f(t) = t² - 4t + 3 , find the value of α⁴β³ + α³β⁴.

Solution

11. If α and β are the zeros of the quadratic polynomial f(x) = 6x² + x - 2 , find the value of α/β + β/α.

Solution

12. If α and β are the zeros of the quadratic polynomial p(s) = 3s² – 6s + 4, find the value α/β + β/α + 2 (1/α + 1/β) + 3αβ .

Solution

13. If the squared difference of the zeros of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p.

Solution

14. If α and β are the zeros of the quadratic polynomial f(x) = x² – px + q , prove that α²/β² + β²/ α² = p⁴/q² – 4p²/q + 2

Solution

15. If α and β are the zeros of the quadratic polynomial f(x) = x² – p(x+1) – c, show that (α + 1) (β+1) = 1 – c .

Solution

16. If α and β are the zeros of a quadratic polynomial such that α + β = 24 and α − β = 8, find a quadratic polynomial have α and β as its zeros.

Solution

17. If α and β are the zeros of the quadratic polynomial f(x) = x² − 1, find a quadratic polynomial whose zeroes are 2 α/β and 2 β/α

Solution

18. If α and β are the zeros of the quadratic polynomial f(x) = x² − 3x − 2, find a quadratic polynomial whose zeroes are 1/2α + β + 1/2β + α .

Solution

19. If α and β are the zeros of the polynomial f(x) = x² + px + q , form a polynomial whose zeroes are (α+ β)² and (α – β)²

Solution

20. If α and β are the zeros of the quadratic polynomial f(x) = x² – 2x + 3, find a polynomial whose roots are (i) α+2, β+2 (ii) α–1/α+1, β–1/ β+1

Solution

21. If α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c, the evaluate :-

Solution

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R.D. Sharma Solutions Class 10th: Ch 2 Polynomials Exercise 2.2

December 29, 2018, 3:12 pm

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Chapter 2 Polynomials R.D. Sharma Solutions for Class 10th Math Exercise 2.2

Exercise 2.2

1. Verify that the numbers given alongside of the cubic polynomials below are their zeros.
Also, verify the relationship between the zeros and coefficients in each case:
(i) f(x) = 2x³ + x² - 5x + 2 ; 1/2 , 1, -2
(ii) g(x) = x³ - 4x² + 5x - 2; 2,1,1

Solution

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, −1 and −3 respectively.

Solution

3. If the zeros of the polynomial f(x) = 2x³ − 15x² + 37x − 30 are in A.P., find them.

Solution

4. Find the condition that the zeros of the polynomial f(x) = x³ + 3px² + 3qx + r may be in A.P.

Solution

5. If the zeroes of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ - 3abc + a²d = 0

Solution

6. If the zeroes of the polynomial f(x) = x³ - 12x² + 39x + k are in A.P., find the value of k.

Solution

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R.D. Sharma Solutions Class 10th: Ch 2 Polynomials Exercise 2.3

December 30, 2018, 10:20 am

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Chapter 2 Polynomials R.D. Sharma Solutions for Class 10th Math Exercise 2.3

Exercise 2.3

1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following :
(i) f(x) = x³ - 6x² + 11x - 6, g(x) = x² + x + 1
(ii) f(x) 10x⁴ + 17x³ - 62x² + 30x - 105(x) = 2x² + 7x + 1
(iii) f(x) = 4x³ + 8x² + 8x + 7:9(x) = 2x² - x + 1
(iv) f(x) = 15x³ - 20x² + 13x - 12; g(x) = x² - 2x + 2

Solution

2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(i) g(t) = t² - 3; f(t) = 2t⁴ + 3t³ - 2t² - 9t
(ii) g(x) = x² - 3x + 1, f(x) = x⁵ - 4x³ + x² + 3x + 1
(iii) g(x) = 2x² - x + 3, f(x) = 6x⁵ - x⁴ + 4x³ - 5x² - x - 15

Solution

3. Obtain all zeros of the polynomial f(x) = 2x⁴ + x³ - 14x² - 19x - 6, if two of its zeors are -2 and -1.

Solution

4. Obtain all zeros of f(x) = x³ + 13x² + 32x + 20, if one of its zeros is -2.

Solution

5. Obtain all zeros of the polynomial f(x) = x⁴ - 3x² = x² + 9x - 6 if two of its zeros are -√3, and √3 .

Solution

Solution

7. Find all the zeroes of the polynomial x⁴ + x³ - 34x² - 4x + 120, if two of its zeroes are 2 and -2 .

Solution

8. Find all zeros of the polynomial 2x⁴ + 7x³ - 19x² - 14x + 30, if two of its zeros are √2 and -√2 .

Solution

9. Find all the zeros of the polynomial 2x³ + x² - 6x - 3, if two of its zeros are -√3 and √3.

Solution

10. Find all the zeros of the polynomial x³ + 3x² - 2x - 6, if two of its zeros are -√2 and √2.

Solution

11. What must be added to the polynomial f(x) = x⁴ + 2x³ - 2x² + x -1 so that the resulting polynomial is exactly by x² + 2x - 3 ?

Solution

12. What must be subtracted from the polynomial f(x) = x⁴ + 2x³ - 13x² - 12x + 21 so that the resulting polynomial is exactly divisible by x² - 4x + 3 ?

Solution

13. Given that √2 is a zero of the cumbic polynomial 6x³ + √2x² – 10x - 4√2 , find its other two zeroes.
Solution

We know that if x = a is a zero of a polynomial, and then x -1 is a factor of f (x).
It is given that √2 is a zero of the cubic polynomial f (x) = 6x³ + √2x² – 10x - 4√2 .
Therefore, x - √2 is a factor of f(x) . Now, we divide 6x³ + √2x² – 10x - 4√2 by x - √2 to find the other zeroes of f(x).

∴ Quotient = 6x² + 7√2x + 4 and remainder = 0 .
By using division algorithm , we have f(x) = g(x) × q(x) + r(x) .
f(x) = (x - √2) (6x² + 7√2x + 4) + 0
= (x - √2) (√2x + 1) (3√2x + 4)
Hence, the other two zeroes of the given polynomial are – 1/√2 and – 4/3√2 .

14 . Given that x - √5 is a factor of the cubic polynomial x³ – 3√5x² + 13x - 3√5 , find all the zeroes of the polynomial .
Solution

We know that if x = a is a zero of a polynomial , and then x –a is a factor of f(x) .
It is given that x - √5 is a factor of f(x) = x³ – 3√5x² + 13x - 3√5.
Now, we divide x³ – 3√5x² + 13x - 3√5 by x -√5 to find the other zeroes of f(x) .

∴ Quotient = x² – 2√5x + 3 and remainder = 0 .
By using division algorithm , we have f(x) = g(x) × q(x) + r(x) .
f(x) = (x- √5 ) (x² - 2√5x + 3) + 0
= (x- √5) [x-(√5 + √2)] [x-(√5 - √2)]
Hence, the zeroes of the given polynomial are √5 , √5 + √2 and √5 - √2 .

↧

R.D. Sharma Solutions Class 10th: Ch 2 Polynomials MCQ's

January 1, 2019, 10:45 pm

≫ Next: R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.1

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Chapter 2 Polynomials R.D. Sharma Solutions for Class 10th Math MCQ's

Multiple Choice Questions

1. If α, β are the zeros of the polynomial f(x) = x² + x + 1 , then 1/α + 1/ β =

(a) 1

(b) -1

(d) None of these

Solution

Since α and β are the zeros of the quadratic polynomial f(x) = x² + x + 1

α + β = - coefficient of x/coefficient of x²

= -1/1 = 1

α × β = constant term / coefficient of x²

= 1/1 = 1

We have

1/α + 1/β =

= β+α/ αβ

= -1/1

= -1

The value of 1/α + 1/β is -1

Hence , the correct choice is b .

2. If α, β are the zeros of the polynomial p(x) = 4x² + 3x + 7, then 1/α + 1/β is equal to

(a) 7/3

(b) – 7/3

(d) -3/7

Solution

Since α and β are the zeros of the quadratic polynomial p(x) = 4x² + 3x + 7
α + β = -coefficient of x / coefficient of x²
= -3/4
= αβ = Constant term / Coefficient of x²
= 7/4
We have

3. If one zero of the polynomial f(x) = (k² + 4)x² + 13x + 4k is reciprocal of the other, then k =

(a) 2

(b) -2

(d) -1

Solution

We are given f(x) = (k² + 4)x² 13x + 4k then

α + β = -Coefficient of x/ Coefficient of x²

= -13/k² + 4

α × β = Constant term / Coefficient of x²

= 4k/k² + 4

One root of the polynomial is reciprocal of the other. Then , we have

α × β = 1

= 4k/k²+4

= 1

= k² – 4k + 4 = 0

= (k-2)² = 0

= k = 2

Hence the correct choice is a .

4. If the sum of the zeros of the polynomial f(x) = 2x³ − 3kx² + 4x − 5 is 6, then the value of k is

(a) 2
(b) 4
(c) −2
(d) −4

Solution

Let α, β be the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x -5 and we are that α + β + γ = 6 .

Then,

α + β + γ = -Coefficient of x/Coefficient of x2

= - (-3k)/2 = 3k/2

It is given that

α + β + γ = 6

Substituting α + β + γ = 3k/2 , we get

+3k/2 = 6

+ 3k = 6 × 2

+3k = 12

k = 12/+3

k = +4

The value of k is 4 .

Hence, the correct alternative is b .

5. If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having 1/α and 1/ β is its zero is

(a) x₂ + qx + p

(b) x₂ – px + q

(d) px₂ + qx + 1

Solution

Let α and β are the zeros of the polynomial f(x) = x² + px + q . Then ,

α + β = -Coefficient of x/ Coefficient of x²

= -p/1

= -p

And

αβ = Constant term/Coefficient of x²

= q/1

= q

Let S and R denote respectively the sum and the product of the zeros of a polynomial

Whose zeros are 1/α and 1/β . then

S = 1/α + 1/β

= α+ β/ αβ

= -p/q

R = 1/α × 1/β

= 1/ αβ

= 1/q

Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by

x² - Sx + R = 0

x² + P/q x + 1/q = 0

qx² + Px + 1/q = 0

qx² + Px + 1

So g(x) = qx² + Px + 1

Hence , the correct choice is c .

6. If α,β are the zeros of polynomial f(x) = x² – p(x+1) – c , then (α + 1) (β + 1) =

(a) c – 1

(b) 1 – c

(d) 1 + c

Solution

Since, α and β are the zeros of quadratic polynomial

f(x) = x² – p(x+1)-c

= x²

α + β = -Coefficient of x/Coefficient of x²

= -(-p/1)

= p

α×β = Constant term/Coefficient of x²

= -p-c/1

= -p-c

We have

(α +1)( β+1)

= αβ + β + α + 1

= αβ + (α+β) + 1

= - p - c + (p) + 1

= -c + 1

= 1 –c

The value of (α +1)( β+1) is 1 – c .

Hence, the correct choice is b .

7. If α, β are the zeros of the polynomial f(x) = x² − p(x + 1) − c such that (α +1) (β + 1) = 0, then c =

(a) 1
(b) 0
(c) −1
(d) 2

Solution

Since, α and β are the zeros of quadratic polynomial

f(x) = x² – p(x+1) – c

f(x) = x² – px – p – c

α+β = -Coefficient of x/Coefficient of x²

= -(-p/1)

= p

αβ = Constant term/Coefficient of x²

= -p - c/1

= -p-c

We have

0 = (α+1)(β+1)

0 = αβ + (α+β) + 1

0 = -p-c+p+1

0 = -c + 1

c = 1

The value of c is 1 .

Hence, the correct alternative is a.

8. If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these

Solution

If f(x) = ax² + bx + c has no real zeros and a + b + c < 0 then c < 0

Hence, the correct choice is c .

9. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax² + bx + c, then

(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0

Solution

Clearly, f(x) = ax² + bx + c represent a parabola opening upwar .

Therefore, f(x) = ax² + bx + c cuts Y axis at P which lies on OY. Putting x = 0 in y = ax² + bx + c, we get y = c. So the coordinates of P is (0,c). Clearly , P lies on OY. Therefore c > 0
Hence , the correct choice is a .

10. Figure 2.23 show the graph of the polynomial f(x) = ax² + bx + c for which

(a) a < 0, b > 0 and c > 0
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0

Solution

Clearly, f(x) = ax² + bx + c represent a parabola opening downwards. Therefore, a<0
y = ax² + bx + c cuts y - axis at P which lies on OY. Putting x = 0 in y = ax² + bx + c, we get y = c.So the coordinates P are (0,c). Clearly , P lies on OY. Therefore c > 0
The vertex (-b/2a, -D/4a) of the parabola is in the second quadrant. Therefore -b/2a<0, b<0

Therefore a<0,b<0, and c>0
Hence , the correct choice is b

11. If the product of zeros of the polynomial f(x) ax³ − 6x² + 11x − 6 is 4, then a =
(a) 3/2
(b) -3/2
(c) 2/3
(d) -2/3

Solution

Since α and β are the zeros of quadratic polynomial f(x) = ax³ - 6x² + 11x - 6
αβ = -Constant term/Coefficient of x²
So we have
4 = -(-6/a)
4 = 6/a
4a = 6
a = 6/4
a = 3×2/2×2
a = 3/2
The value of a is 3/2
Hence, the correct alternative is a .

12. If zeros of the polynomial f(x) = x³ − 3px² + qx − r are in A.P., then

(a) 2p³ = pq − r

(b) 2p³ = pq + r

(d) None of these

Solution

Let a-d,a,a+d be the zeros of the polynomial f(x) = x³ - 3px² + qx - r then

Sum of zeros = -Coefficient ofx²/Coefficient of x³

(a-d)+a+(a+d) = -(-3p)/1

a-d + a + a + d = 3p

3a = 3p

a = 3/3p

a = p

Since a is a zero of the polynomial f(x)

Therefore,

f(a) = 0

a³ - 3pa³ + qa - r = 0

Substituting a = p . we get

p³ - 3p(p)² + q×p-r=0

p³ - 3p³ + qp - r = 0

-2p^{3 + qp - r = 0
qp - r = 2p³}Hence, the correct choice is a13. If the product of two zeros of the polynomial f(x) = 2x³ + 6x² − 4x + 9 is 3, then its third zero is (a) 3/2
(b) -3/2
(c) 9/2
(d) -9/2
Solution
^{Let α,β,γ be the zeros of polynomial f(x) = 2x³ + 6x² - 4x + 9 such that αβ = 3
We have,
αβγ = Constant term/Coefficient of x³
= -9/2
Putting αβ = 3 in αβγ = -9/2, we get
αβγ = -9/2
3γ = -9/2
γ = -9/2 × 1/3
γ = -3/2
Therefore, the value of third zero is -3/2
Hence, the correct alternative is b.}

14. If the polynomial f(x) = ax³ + bx − c is divisible by the polynomial g(x) = x² + bx + c, then ab =

(a) 1

(b) 1/c

(d) -1/c

Solution

We have to find the value of ab

Given f(x) = ax³ + bx - c is divisible by the polynomial g(x) = x² + bx + c

We must have

bx - acx + ab²x + abc - c = 0, for all x

So put x = 0 in this equation

x(b-ac+ab²)+c(ab-1) = 0

c(ab-1) = 0

Since c≠0 , so

ab - 1 = 0

⇒ ab = 1

Hence, the correct alternative is a.

15. If the polynomial f(x) = ax³ + bx − c is divisible by the polynomial g(x) = x² + bx + c, then c =

(a) b

(b) 2b

(d) −2b

Solution

We have to find the value of c

Given f(x) = ax³ + bx -c is divisible by the polynomial g(x) = x² + bx + c

We must have,

bx - acx + ab²x + abc - c = 0 for all x

x(b-ac+b²) + c(ab - 1) = 0 ...(1)

c(ab-1) = 0

Since c≠0, so

ab - 1 = 0

ab = 1

Now in the equation (1) the condition is true for all x. So put x =1 and also we have ab = 1

Therefore we have

b-ac+ab²=0

b+ab² -ac=0

b(1+ab)-ac=0

Substituting a = 1/b and ab = 1 we get,

b(1+1) - 1/b×c=0

2b - 1/b×c = 0

-1/b×c = -2b

c=2b×b/1

c=2b²

Hence, the correct alternative is c .

16. If one root of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other, then the value of k is

(a) 0

(b) 5

(d) 6

Solution

If one zero of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other . So β = 1/α ⇒ αβ = 1

Now we have

α×β = Constant term/Coefficient of x²

= k/5

Since αβ = 1

Therefore we have

αβ = k/5

1= k/5

⇒ k = 5

Hence the correct choice is b .

17. If α,β,γ are the zeros of the polynomial f(x) ax³ + bx² + cx + d, then 1/α + 1/β + 1/γ =

(a) -b/d

(b) c/d

(d) -c/a

Solution

We have to find the value of 1/α + 1/β + 1/γ =

Given α,β,γ be the zeros of the polynomial f(x) = ax³ + bx² + cx + d

we know that

αβ +βγ + γα = coefficient of x/coefficient of x³

= c/a

αβγ = -Constant term/Coefficient of x³

= -d/a

1/α + 1/β + 1/γ = βγ + αγ + αβ/αβγ

1/α + 1/β + 1/γ = c/a/-d/a

1/α + 1/β + 1/γ = c/a × (-a/d)

1/α + 1/β + 1/γ = -c/d

Hence, the correct choice is c.

18. If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d, then α² + β² + γ² =

(a) b²-ac/a²

(b) b²-2ac/a

(d) b² - 2ac/a²

Solution

We have to find the value of α² + β² + γ²

Given α,β,γ be the zeros of the polynomial f(x) - ax³ + bx² + cx + d

α+β+γ = -Coefficient of x²/Coefficient of x³

= -b/a

αβ + βγ + γα = Coefficient of x/Coefficient of x³

= c/a

Now

α² + β² + γ² = (α+β+γ)-2(αβ + βγ + γα)

α² + β² + γ² = (-b/a)² -2(c/a)

α² + β² + γ² = b²/α² - 2c/a

α² + β² + γ² = b²/α² - 2ca/α²

α² + β²

+ γ² = b²-2ac/a²

The value of α² + β² + γ² = b²-2ac/a²

Hence, the correct choice is d.

19. If α, β, γ are the zeros of the polynomial f(x) = x³ - px² + qx - r , then 1/αβ + 1/βγ + 1/γα =

(a) r/p

(b) p/r

(d) -r/p

Solution

If α, β, γ are the zeros of the polynomial f(x) = x³ - px² + qx - r , then 1/αβ + 1/βγ + 1/γα =

We have to find the value of 1/αβ + 1/βγ + 1/γα

Given α,β,γ be the zeros of the polynomial f(x) = x³ - px² + qx - r

α+β+γ = -Coefficient of x²/Coefficient of x³

= -(p)/1

= p

αβγ = -Constant term/Coefficient of x³

= -(r)/1

= r

Now we calculate the expression

1/αβ + 1/βγ + 1/γα = γ/αβγ + α/αβγ + β/αβγ

1/αβ + 1/βγ + 1/γα = α+γ+β/αβγ

1/αβ + 1/βγ + 1/γα = p/r

Hence, the correct choice is b.

20. If α,β are the zeros of the polynomial f(x) = ax² + bx + c, then 1/α + 1/β =

(a) b²-2ac/a²

(b) b²-2ac/c²

(c)b²+2ac/a²

(d) b²+2ac/c²

Solution

We have to find the value of 1/α²+ 1/β²

Given α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c

α+β = -Coefficient of x/Coefficient of x³

= -b/a

αβ = Constant term/Coefficient of x³

= c/a

We have,

21. If two of the zeros of the cubic polynomial ax³ + bx² + cx + d are each equal to zero, then the third zero is

(a) -d/a

(b) c/a

(d) b/a

Solution

Let α=0, β=0 and γ be the zeros of the polynomial

f(x) = ax³ + bx² + cx + d

Therefore

α+β+γ = -Coefficient of x²/Coefficient of x³

= -(b/a)

α+β+γ = -b/a

0+0+γ = -b/a

γ = - b/a

The value of γ = -b/a

Hence, the correct choice is c.

22. If two zeros x³ + x² - 5x - 5 are √5 and -√5, then its third zero is

(a) 1

(b) -1

(d) -2

Solution

Let α = √5 and β = -√5 be the given zeros and γ be the third zero of x³ + x² - 5x - 5 = 0 then

By using α+β+γ = -Coefficient of x²/Coefficient of x³

α+β+γ = +(+1)/1

α+β+γ = -1

By substituting α = √5 and β = -√5 in α+β+γ = -1

√5 - √5 + γ = -1

γ = -1

Hence, the correct choice is b .

23. The product of the zeros of x³ + 4x² + x − 6 is

(a) −4

(b) 4

(d) −6

Solution

Given α,β,γ be the zeros of the polynomial f(x) = x³ + 4x² + x -6

Product of the zeros = Constant term/Coefficient of x³ = -(-6)/1 =6

The value of product of the zeros is 6.

24. What should be added to the polynomial x² − 5x + 4, so that 3 is the zero of the resulting polynomial?

(a) 1

(b) 2

(d) 5

Solution

If x = a, is a zero of a polynomial then x - a is a factor of f(x)

Since 3 is the zero of the polynomial f(x) = x² - 5x + 4,

Therefore x -3 is a factor of f(x)

Now we divide f(x) = x² - 5x + 4 by (x-3) we get

Therefore we should add 2 to the given polynomial

Hence, the correct is b.

25. What should be subtracted to the polynomial x² − 16x + 30, so that 15 is the zero of the resulting polynomial ?

(a) 30

(b) 14

(d) 16

Solution

We know that, if x = a, is zero of a polynomial then x -a is a factor of f(x)

Since 15 is zero of the polynomial f(x) = x² - 16x + 30, therefore (x-15) is a factor of f(x)

Now, we divide f(x) = x² - 16x + 30 by (x-15) we get

Thus we should subtract the remainder 15 from x² - 16x + 30 .

Hence, the correct choice is c.

26. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is

(a) x² − 9

(b) x² + 9

(d) x² − 3

Solution

Since α and β are the zeros of the quadratic polynomials such that

0 = α+β

If one of zero is 3 then

α+β = 0

3+β = 0

β = 0-3

β = -3

Substituting β = -3 in α + β = 0 we get

α - 3 = 0

α = 3

Let S and P denote the sum and the product of the zeros of the polynomial respectively then

S = α+β

S = 0

P = αβ

P = 3×-3

P = -9

Hence, the required polynomial is

= (x² -Sx + P)

= (x² - 0x - 9)

= x² - 9

Hence, the correct choice is a .

27. If two zeroes of the polynomial x³ + x² − 9x − 9 are 3 and −3, then its third zero is

(a) −1

(b) 1

(d) 9

Solution

Let α = 3 and β = -3 be the given zeros and γ be the third zero of the polynomial x³ + x² - 9x -9 then,

By using α+β+γ = -Coefficient of x²/Coefficient of x³

α+β+γ = -1/1

α+β+γ = -1

Substituting α = 3 and β = -3 in α+β+γ = -1, we get

3-3+γ = -1

γ = -1

Hence the correct choice is a .

28. If √5 and -√5 are two zeroes of the polynomial x³ + 3x² - 5x - 15, then its third zero is

(a) 3

(b) -3

(d) -5

Solution

Let α = √5 and β = -√5 be the given zeros and γ be the third zero of the polynomial x³ + 3x² - 5x - 15. Then,

By using α+β+γ = -Coefficient of x²/Coefficient of x³

α+β+γ = -3/1

α+β+γ = -3

Substituting α = √5 and β = -√5 in α+β+γ = -3

We get

√5 - √5 + γ = -3

γ = -3

Hence, the correct choice is b.

29. If x + 2 is a factor of x² + ax + 2b and a + b = 4, then

(a) a = 1, b = 3

(b) a = 3, b = 1

(d) a = 5, b = −1

Solution

Given that x+2 is a factor of x² + ax + 2b and a + b = 4

f(x) = x² + ax + 2b

f(-2) = (-2)² + a(-2) + 2b

0 = 4 - 2a + 2b

-4 = -2a + 2b

By solving -4 and -2a+2b and a+b = 4 by elimination method we get

Multiply a+b = 4 by 2 we get,

2a + 2b = 8. So

-4 = -2a + 2b

+8 = +2a + 2b

4 = 4b

4/4 = b

1 = b

By substituting b =1 in a+b = 4 we get

a+1 = 4

a = 4-1

a = 3

Then a = 3, b = 1

Hence, the correct choice is b.

30. The polynomial which when divided by - x² + x - 1 gives a quotient x -2 and remainder 3, is

(a) x³ - 3x² + 3x - 5

(b) -x³ - 3x² -3x - 5

(d) x³ -3x² -3x + 5

Solution

We know that,

f(x) = g(x)q(x)+r(x)

= (-x²+x-1)(x-2)+3

= -x³ + x² - x + 2x² - 2x + 2 + 3

= -x³ + x² + 2x² - x - 2x + 2 + 3

= -x³ + 3x² - 3x + 5

Therefore,

The polynomial which when divide by -x² + x - 1 gives a quotient x-2 and remainder 3, is x³+3x²-3x+5

Hence, the correct choice is c .

31. The number of polynomials having zeros -2 and 5 is

(a) 1

(b) 2

(d) More than 3.

Solution

Polynomials having zeros -2 and 5 will be of the form

p(x) = a(x+2)ⁿ(x-5)^m

Here, n and m can take any value from 1,2,3,...

Thus, the number of polynomials will be more than 3.

Hence, the correct answer is option D.

32. If one of the zeroes of the quadratic polynomial (k-1)x² + kx + 1 is -3, then the value of k is

(a) 4/3

(b) -4/3

(d) -2/3

Solution

The given polynomial is f(x) = (k-1)x² + kx + 1.

Since -3 is one of the zeros of the given polynomial, so f(-3) = 0

(k-1)(-3)² + k(-3) + 1 = 0

⇒ 9(k-1)-3k + 1 = 0

⇒ 9k - 9 - 3k + 1 = 0

⇒ 6k - 8 = 0

⇒ k = 4/3

Hence, the correct answer is option A.

33. The zeroes of the quadratic polynomial x² + 99x + 127 are

(a) Both positive

(b) Both negative

(d) One positive and one negative

Solution

Let f(x) = x² + 99x + 127.

Product of the zeroes of f(x) = 127 × 1 = 127 [Product of zeroes = c/a when f(x) = ax² + bx + c]

Since the product of zeroes is positive , we can say that it is only possible when both zeroes are positive or both zeroes are negative.

Also, sum of the zeroes = -99 [sum of zeroes = b/a when f(x) = ax² + bx + c]

The sum being negative implies that both zeroes are positive is not correct.

So, we conclude that both zeroses are negative.

Hence, the correct answer is option B.

34. If the zeroes of the quadratic polynomial x² +(a+1) x+b are 2 and -3, then

(a) a = -7,b = -1

(b) a = 5, b = -1

(d) a = 0, b = -6

Solution

The given quadratic equation is x² + (a+1)x+b = 0.

Since the zeroes of the given equation are 2 and -3.

So,

α = 2 and β = -3

Now,

Sum of zeroes = -Coefficient of x/Coefficient of x²

⇒ 2+(-3)= -(a+1)/1

⇒ -1 = -a-1

⇒ a = 0

↧

R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.1

January 2, 2019, 9:56 pm

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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.1

Exercise 3.1

1. Akhila went to a fair in her village. She wanted to enjoy rides in the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.

Solution

The pair of equations is formed :

Let us represent these equations graphically. For this, we need at least two solutions for
each equation. We give these solutions in Table

Recall from Class IX that there are infinitely many solutions of each linear equation. So

each of you choose any two values, which may not be the ones we have chosen. Can you
guess why we have chosen x = 0 in the first equation and in the second equation? When
one of the variables is zero, the equation reduces to a linear equation is one variable, which
can be solved easily. For instance, putting x = 0 in Equation (2), we get 4y = 20 i.e,

y = 5.

Similarly , putting y = 0 in Equation(2), we get 3x = 20 i.e., x = 20/3 . But as 20/3 is

not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2 which gives x = 4, an integral value.

Plot the points A(O,O) B(2,1) and P(O,5), Q(412), corresponding to the draw the lines AB and PQ, representing the equation x-2y = 0 and 3x+4y = 20, as shown in figure
In fig., observe that the two lines representing the two equations are intersecting at the point(4,2),

2. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be". Is not this interesting? Represent this situation algebraically and graphically.

Solution

Let the present age of Aftab and his daughter be x and y respectively. Seven years ago.
Age of Ahab = x - 7
Age of his daughter = y - 7
According to the given condition.

(x-7) = 7 (y-7)

⇒ x-7 = 7y - 49

⇒ x-7 y = -42

Three years hence

Age of Aftab = x+3

Age of his daughter = y+3

According to the given condition,

(x+3) = 3 (x+3)

⇒ x+3 = 3y + 9

⇒ x - 3y = 6

Thus, the given condition can be algebraically represented as

x-7y = -42

x - 3y = 6

x-7y = -42 ⇒ x = -42 + 7y

Three solution of this equation can be written in a table as follows:

x -3y = 6 ⇒ x = 6 + 3y

Three solution of this equation can be written in a table as follows:

The graphical representation is as follows:

Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here. Aftab and his daughters present age needs to be found so, so the ages will be represented by variables z and y. The problem talks about their ages seven years ago and three years from now. Here, the words ’seven years ago’ means we have to subtract 7 from their present ages. and ‘three years from now’ or three years hence means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

3. The path of a train A is given by the equation 3x+4y-12=0 and the path of another train B is given by the equation 6x+8y-48=0. Represent this situation graphically .

Solution

The paths of two trains are giver by the following pair of linear equations.

3x+4y-12=0 ...(1)
6x+8y-48=0 ...(2)

In order to represent the above pair of linear equations graphically. We need two points on
the line representing each equation. That is, we find two solutions of each equation as given
below:

We have,

Thus, two solution of equation 3x+4y-12=0 are (0,3) and (4,0)
We have ,
6x+8y-48=0
Putting x = 0, we get
6×0+8y-48=0
⇒ 8y= 48

Thus, two solution of equation 6x+8y-48=0 are (0,6) and (8,0)

Clearly, two lines intersect at (-1,2)
Hence, x = -1, y=2 is the solution of the given system of equations .

4. Gloria is walking along the path joining (− 2, 3) and (2, − 2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.

Solution

It is given that Gloria is walking along the path joining (-2,3) and (2,-2), while Suresh is walking along the path joining (0,5) and (4,0).

We observe that the lines are parallel and they do not intersect anywhere.

5. On comparing the ratios a₁/a₂ , and c₁/c₂ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide:
(i) 5x-4y+8 = 0, 7x+6y-9 = 0
(ii) 9x+3y+12 = 0, 18x+6y+24 = 0,
(iii) 6x-3y+10 = 0, 2x-y+9 = 0

Solution

(i) We have,
5x-4y+8=0
7x+6y-9=0
Here,
a₁=5,b₁=-4,c₁=8
a₂=7,b₂=6,c₂=-9
We have,

∴ Two lines are intersecting with each other at a point.

(ii) 9x+3y+12=0
18+6y+24=0
Here
a₁=9,b₁=3,c₁=12

a₂=18,b₂=6,c₂=24

Now,

∴ Both the lines coincide .

(iii) 6x-3y+10=0

2x-y+9=0
Here,
a₁=6, b₁=-3, c₁=10
a₂=2, b₂=-1, c₂=9
Now,

∴ The lines are parallel

6. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such
that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines.

Solution

We have ,
2x+3y-8=0
Let another equation of line is :
4x+9y-4=0
Here,
a₁=2,b₁=3,c₁=-8
a₂=4,b₂=9,c₂=-4
Now,

∴ 2x+3y-8=0 and 4x+9y-4=0 intersect each other at one point.
Hence, required equation of line is 4x+9y-4=0

(ii) We have,

2x+3y-8=0
Let another equation of line is:
4x+6y-4=0
Here,
a₁=2,b₁=3,c₁=-8
a₂=4,b₂=6,c₂=-4
Now,

∴ Lines are parallel to each other.

Hence, required equation of line is 4x+6y-4=0.

7. The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After amonth, the cost of 4kg of apples and 2kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Solution

Let the cost of 1 kg of apples and 1 kg grapes be Rs x and Rs y.
The given conditions can be algebraically represented as
2x+y=160
4x+2y=300
2x+y=160⇒y=160-2x
Three solutions of this equation can be written in a table as follows:

4x+2y=300⇒y=300-4x/2
Three solutions of this equation can be written in a table as follows:

The graphical representation is as follows:

Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1kg grapes will be taken as the variables from the given condition of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then In order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are Large so take the suitable scale.

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NCERT Solutions for Class 12th: Ch 7 Directing (MCQ and Short Questions)

January 3, 2019, 8:33 am

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NCERT Solutions for Class 12th: Ch 7 Directing (MCQ and Short Questions)

Exercises

Page No. 211

Multiple Choice

1. Which one of the following is not an element of direction?

(a) Motivation
(b) Communication
(c) Delegation
(d) Supervision
► (c) Delegation

2. The motivation theory which classifies needs in hierarchical order is developed by
(a) Fred Luthans
(b) Scott
(c) Abraham Maslow
(d) Peter F. Drucker
► (b) Scott

3. Which of the following is a financial incentive?
(a) Promotion
(b) Stock Incentive
(c) Job Security
(d) Employee Participation
► (c) Job Security

4. Which of the following is not an element of communication process?
(a) Decoding
(b) Communication
(c) Channel
(d) Receiver
► (b) Communication

5. Grapevine is
(a) Formal communication
(b) Barrier to communication
(c) Lateral communication
(d) Informal communication
► (c) Lateral communication

6. Status comes under the following type of barriers
(a) Semantic barrier
(b) Organisational barrier
(c) Non Semantic barrier
(d) Psychological barrier
► (c) Non Semantic barrier

7. The software company promoted by Narayana Murthy is
(a) Wipro
(b) Infosys
(c) Satyam
(d) HCL
► (b) Infosys

Page No. 212

8. The highest level need in the need Hierarchy of Abraham Maslow:
(a) Safety need
(b) Belongingness need
(c) Self actualisation need
(d) Prestige need
► (a) Safety need

9. The process of converting the message into communication symbols is known as-
(a) Media
(b) Encoding
(c) Feedback
(d) Decoding
► (d) Decoding

10. The communication network in which all subordinates under a supervisor communicate through supervisor only is:
(a) Single chain
(b) Inverted V
(c) Wheel
(d) Free flow
► (d) Free flow

Page No: 175

Short Answer Type

1. Distinguish between leaders and managers.

Answer

Basis	Manager	Leader
Existence	Exist in formal organisation structure.	Exist in both formal and informal organisation structure.
Focus	The focus of a manager is on achieving the organisational goals.	The focus of a leader is on fulfilling aspirations and expectations of his followers.
Authority	A manager possesses formal authority.	A leader enjoys acceptance authority.
Functions	Managers perform all functions of organisations like planning, organising, directing, controlling.	Leader performs only one function like directing.

2. Define Motivation.

Answer

Motivation is the process of stimulating people to action to accomplish desired goals. Motivation depends upon satisfying needs of people.

3. What is informal communication?

Answer

Communication that takes place without following the formal lines of communication is said to be informal communication. Informal system of communication is generally referred to as the ‘grapevine’ because it spreads throughout the organisation with its branches going out in all directions in utter disregard to the levels of authority. The informal communication arises out of needs of employees to exchange their views, which cannot be done through formal channels. Workers chit chating in a canteen about the behaviour of the superior, discussing about rumours that some
employees are likely to be transferred are some examples of informal communications.

4. What are semantic barriers of communications?

Answer

Semantics barriers are concerned with problems and obstructions in the process of encoding and decoding of message into words or impressions.
(i) Badly expressed message
(ii) Symbols with different meanings
(iii) Faulty translations
(iv) Unclarified assumptions
(v) Technical jargon
(vi) Body language and gesture decoding

5. Who is a supervisor?

Answer

Supervisor is the link between management and subordinates. He maintains day-to-day contact and maintains friendly relations with workers. A good supervisor acts as a guide, friend and philosopher to the workers. He conveys management ideas to the workers on one hand and workers problems to the management on the other.

6. What are the elements of directing?

Answer

(i) Supervision: The process of guiding and instructing employees.

(ii) Motivation: It means the process of making subordinates to act in a desired manner to achieve
certain organisational goals.

(iii) Leadership: It is the process of influencing the behaviour of people by making them strive voluntarily towards achievement of organisational goals.

(iv) Communication- It is understood as a process of exchange of ideas, views, facts, feelings etc.,

7. Explain the process of motivation.

Answer

Motivation means incitement or inducement to act or move. In the context of an organisation, it means the process of making subordinates to act in a desired manner to achieve certain organisational goals.

8. Explain different networks of grapevine communications.

Answer

Grapevine communications may follow different types of network they are
(i) Single strand network: Each person communicates to the other in sequence.
(ii) Gossip network: Each person communicates with all on non- selective basis.
(iii) Probability network: The individual communicates randomly with other individual.
(iv) Cluster network: The individual communicates with only those people whom we trust.

Go to Index of NCERT Solutions of Business Studies Class 12th Part I

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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.3

January 4, 2019, 12:09 am

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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.3

Exercise 3.3

1. Solve the following systems of equations:

11x + 15y + 23 = 0

7x – 2y – 20 = 0

Solution

The given system of equation is

Hence, the solution of the given system of equations is x = 2,y = - 3.

2. 3x – 7y + 10 = 0

y – 2x – 3 = 0

Solution

The given system of equation is

Hence, the solution of the given system of equations is x = -1, y = 1.

3. 0.4x + 0.3y = 1.7

0.7x + 0.2y = 0.8

Solution

The given system of equation is

Hence, the solution of the given system of equation is x = 2, y = 3.

4. x/2 + y = 0.8

Solution

5. 7(y + 3) – 2 (x + 3) = 14

4(y – 2) + 3 (x – 3) = 2

Solution

The given system of equations is

Hence, the solution of the given system of equations is x = 5,y = 1.

6. x/7 + y/3 = 5
x/2 - y/9 = 6

Solution

The given system of equation is

Hence, the solution of thee given system of equations is x=14, y=9.

7. x/3 + y/4 = 11
5x/6 - y/3 = 7

Solution

The given system of equations is

Let us eliminate y from the given equations. The coefficients of y in the equations(iii) and

(iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y

equal to 6 in the two equations.

Multiplying (iii) by 2 and (iv) by 3, we get

Hence, the solution of the given system of equations is x = 6, y = 36.

8. 4u + 3y = 8

6u - 4y = -5

Solution

So, the solution of the given system of equation is x=2, y=2.

9. x + y/2 = 4

x/3 + 2y = 5

Solution

The given system of equation is

Hence, solution of the given system of equation is x=3, y=2.

10. x + 2y = 3/2

2x + y = 3/2

Solution

The given system of equation is

Let us eliminate y from the given equations. The Coefficients of y in the given equations

are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal

to 2 in the two equations.

Hence, solution of the given system of equation is x = 1/2, y = 1/2.

11. √2 x + √3 y = 0

√3 x - √8 y = 0

Solution

12. 3x - (y+7)/11 + 2 = 10

2y + (x+11)/7 = 10

Solution

The given systems of equation is

Hence, solution of the given system of equation is x=3, y=4.

13. 2x - 3/y = 9
3x + 7/y = 2, y ≠ 0

Solution

The given systems of equation is

Hence, solution of the given system of equation is x=3,y=1.

14. 0.5x + 0.7y = 0.74

0.3x + 0.5y = 0.5

Solution

The given systems of equations is

Hence, solution of the given system of equation is x = 0 5, y = 0.7.

15 . 1/7x + 1/6y = 3
1/2x - 1/3y = 5

Solution

16. 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Solution

The given equations are:

Multiply equation (i) by 1/2 and (ii) by 1/3 and subtract equation (ii) from (i) we get

Hence the value of x= 1/2 and y = 1/3 .

17. 15/u + 2v = 17
1/u + 1/v = 36/5

Solution

The given equations are:

Multiply equation (ii) by 2 and subtract (ii) from (i), we get

Put the value of u in equation (i), we get

Hence the value of u=5 and v=1/7

18. 3/x - 1/y = -9
2/x + 3/y = 5

Solution

The given equations are:

Multiply equation (i) by 3 and add both equations, we get

Put the value of x in equation(i), we get

Hence the value of x=1/2 and y=1/3

19. 2/x + 5/y = 1
60/x + 40/y = 19

Solution

The given equations are:

Multiply equation (i) by 8 and subtract (ii) from equation (i), we get

Put the value of x in equation(i), we get

Hence the value x=4 and y=10.

20. 1/5x + 1/6y = 12
1/3x - 3/7y = 8

Solution

The given equations are:

Multiply equation (i) by 3/7 and equation (ii) by 1/6, add both equation, we get

Put the value of x in equation (i), we get

Hence the value of = x - 89/4080 and y=89/1512

21. 4/x + 3/y = 14
3/x - 4y = 23

Solution

The given equations are:

Multiply equation (i) by 4 and equation (ii) by 3, add both equations, we get

Put the value of x in equation (i), we get

Hence the value of x = 1/5 and y = -2

22. 4/x + 5y = 7
3/x + 4y = 5

Solution

The given equations are:

Multiply equation (i) by 4 and equation (ii) by 5 and subtract (ii) from (i) we get

Put the value of x in equation (i), we get

Hence the value of x=1/3 and y=-1

23. 2/x + 3/y = 13
5/x - 4y = -2

Solution

Multiply equation (i) by 4 and equation (ii) by 3 and add both equations we get

Put the value of x in equation (i), we get

Hence the value of x=1/2 and y=1/3.

24. 2/√x + 3/√y = 2
4/√x - 9/√y = -1

Solution

The given equations are:

Multiply equation (i) by 3 and add both equations we get

Put the value x in equation (i), we get

Hence the value of x = 4 and y = 9

25. (x+y)/xy = 2
(x-y)/xy = 6

Solution

The given equations are:

Adding both equations, we get

Put the value y in equation (i), we get

Hence the value of x=-1/2 and y=1/4

26. 2/x + 3/y = 9/xy
4/x + 9/y = 21/xy

Solution

The given equations are:

Multiply equation (i) by 3 and subtract (ii) from (i), we get

Put the value of x in equation (i), we get

Hence the value of x=1 and y=3 .

27. 6/(x+y) = 7/(x-y) + 3
1/2(x+y) = 1/3(x-y)

Solution

The given equations are:

28. xy/(x+y) = 6/5

xy/(y-x) = 6

Solution

The given equations are:

29. 22/(x+y) + 15/(x-y) = 5

55/(x+y) + 45/(x-y) = 14

Solution

The given equations are:

30. 5/(x+y) - 2/(x-y) = -1

15/(x+y) + 7/(x-y) = 10

Solution

The given equations are:

31. 3/(x+y) + 2/(x-y) = 2
9/(x+y) - 4/(x-y) = 1

Solution

The given equations are:

32. 1/2(x+2y) + 5/3(3x-2y) = -3/2
4(x+2y) - 3/5(3x-2y) = 61/60

Solution

The given equations are:

33. 5/(x+1) - 2/(y-1) = 1/2
10/(x+1) + 2/(y-1) = 5/2

Solution

The given equations are:

34. x+y=5xy

3x+2y=13xy

Solution

The given equations are:

35. x+y = 2xy

x-y/xy = 6

Solution

The given equations are:

36. 2(3u-v) = 5uv

2(u+3v) = 5uv

Solution

The given equations are:

37. 2/(3x+2y) + 3/(3x-2y) = 17/5

5/(3x+2y) + 1/(3x-2y) = 2

Solution

The given equations are:

38. 44/(x+y) + 30/(x-y) = 10

55/(x+y) + 40/(x-y) = 13

Solution

The given equations are:

39. 5/(x-1) + 1/(y-2) = 2

6/(x-1) - 3(y-2) = 1

Solution

The given equations are:

40. 10/(x+y) + 2/(x-y) = 4

15/(x+y) - 9/(x-y) = -2

Solution

The given equations are:

41. 1/(3x+y) + 1/(3x-y) = 3/4
1/2(3x+y) - 1/2(3x-y) = -1/8

Solution

The given equations are:

42. (7x-2y)/xy = 5
(8x+7y)/xy = 15

Solution

The given equations are:

43. 152x-378y=-74
-378x + 152y = -604

Solution

The given equations are:

44. 99x + 101y = 499

101x + 99y = 501

Solution

45. 23x − 29y = 98

29x − 23y = 110

Solution

The given equations are:

46. x -y + z = 4

x - 2y - 2z = 9

2x + y + 3z = 1

Solution

The given equations are:

47. x − y + z = 4

x + y + z = 2

2x + y − 3z = 0

Solution

The given equations are:

48. 21x + 47y = 110

47x + 21y = 162

Solution

21x + 47y = 110 ...(i)

47x + 21y = 162 ...(ii)

Adding (i) and (ii), we get

68x + 68y = 272

⇒ x + y = 4 ...(iii)

Subtracting (i) from (ii), we get

26x - 26y = 52

⇒ x - y = 2 ...(iv)

Adding (iii) and (iv), we get

2x = 6

⇒ x = 3

Putting x = 3 in (iv), we get

3 - y = 2

⇒ y = 1

49. If x+1 is a factor of 2x³ + ax² + 2bx + 1, then find the values of a and b given that 2a - 3b = 4

Solution

Since (x+1) is a factor of 2x³ + ax² + 2bx + 1, so

2(-1)³ + a(-1)² + 2b(-1) + 1 = 0

⇒ -2 + a - 2b + 1 = 0

⇒ a - 2b - 1 = 0

⇒ a - 2b = 1 ...(i)

Also, we are given

2a - 3b = 4 ...(ii)

From (i) and (ii) we get

a = 1 + 2b ...(iii)

Substituting the value of a in (ii), we get

2(1 + 2b) - 3b = 4

⇒ 2 + 4b - 3b = 4

⇒ b = 2

Putting b = 2 in (iii), we get

a = 1 + 2 × 2 = 5

Thus, the value of a = 5 and b = 2.

50. Find the solution of the pair of equations x/10 + y/5 - 1 = 0 and x/8 + y/6 = 15 . Hence, find λ, if y = λ x + 5.

Solution

The given equations are

Thus, the value of λ = -1/2 .

51. Find the values of x and y in the following rectangle .

Solution

ABCD is the given rectangle. So, AB = CD and AD = BC .
Thus,
x + 3y = 13 ....(i)
3x + y = 7 ....(ii)
Adding (i) and (ii), we get
4x + 4y = 20
⇒ x + y = 5 ....(iii)
Subtracting (i) from (ii), we get
2x - 2y = -6
⇒ x - y = -3 ....(iv)
Adding (iii) and (iv), we get
2x = 2
⇒ x = 1
Putting x = 1 in (iii), we get
1+y = 5
⇒ y = 4
Thus, x = 1 and y = 4.

52. Write an equation of a line passing through the point representing solution of the pair of linear equation x + y = 2 and 2x - y = 1. How many such lines can we find?

Solution

The given equations are

x + y = 2 ....(i)

2x - y = 1 ....(ii)

Adding (i) and (ii), we get

3x = 3

⇒ x = 1

Putting x = 1 in (i), we get

1 + y = 2

⇒ y = 1

Thus, the solution of the given equations is (1, 1).

We know that, infinitely many straight lines pass through a single point.

So, the equation of one such line can be 3x + 2y = 5 or 2x + 3y = 5.

53. Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?

Solution

The unique solution is given as x = −1 and y = 3.

The one pair of linear equations having x = −1 and y = 3 as unique solution can be

12x + 5y = 3

2x + y = 1

Similarly, infinitely many pairs of linear equations are possible.

↧

R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.4

January 4, 2019, 10:42 pm

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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.4

Exercise 3.4

1. Solve each of the following systems of equations by the method of cross-multiplication:-

1. x + 2y + 1 = 0
2x - 3y - 12 = 0

Solution

The given system of equation is

Hence, the solution of the given system of equations is x = 3, y = -2.

2. 3x + 2y + 25 = 0
2x + y + 10 = 0

Solution

The given system of equation is

Hence, x = 5, y = -20 is the solution of the given system of equations.

3. 2x + y - 35 = 0
3x + 4y - 65 = 0

Solution

The given system of equations may be written as

Hence, x = 15, y = 5 is the solution of the given system of equations.

4. 2x - y - 6 = 0
x - y - 2 = 0

Solution

The given system of equations may be written as

Hence , x = 4, y = 2 is the solution of the given system of the equations.

5. (x+y)/xy = 2
(x-y)/xy = 6

Solution

The given system of equations is

6. ax + by = a - b
bx - ay = a + b

Solution

The given system of equations is

Hence, x = 1, y = 1 is the solution of the given system of the equations.

7. x + ay - b = 0
ax - by - c = 0

Solution

The given system of equations may be written as

8. ax + by = a²
bx + ay = b²

Solution

The system of the given equations may be written as

9. 5/(x+y) - 2/(x-y) = -1

15/(x+y) + 7/(x-y) = 10

where x ≠ 0 and y ≠ 0

Solution

10. 2/x + 3/y = 13

5/x - 4/y = -2

where x ≠ 0 and y ≠ 0

Solution

The given system of equation is

11. 57/(x+y) + 6/(x-y) = 5
38/(x+y) + 21/(x-y) = 9

Solution

Now adding eq.(3) and (4) we get x = 11

And after substituting the value of x in eq. (4) we get y = 8
Hence we get the value of x = 11 and y = 8 .

12. x/a + y/b = 2

ax-by = a² - b²

Solution

The system of the given equations may be written as .

13. x/a + y/b = a+b

x/a² + y/b² = 2

Solution

Hence we get the value = a² and y = b²

14. x/a = y/b

ax + by = a² + b²

Solution

15. 2ax + 3by = a + 2b
3ax + 2by = 2a + b

Solution

The given system of equations is

16. 5ax + 6by = 28
3ax + 4by - 18 = 0

Solution

The given system of equation is

17. (a+2b)x + (2a-b)y = 2
(a-2b)x + (2a+b)y = 3

Solution

The given system of equations may be written as

18. x[a-b + ab/(a-b)] = y[a+b - ab/(a+b)]

x+y = 2a²

Solution

The given system of equation is

19. bx + cy = a + b
ax[1/(a-b) - 1/(a+b)] + cy[1/(b-a) - 1/(b+a) = 2a/(a+b)

Solution

The given system of equation is

20. (a-b)x + (a+b)y = 2a² - 2a²

(a+b) (x+y) = 4ab

Solution

The given system of equation is

21. a²x + a²y = c²

b²x + a²y = d²

Solution

The given system of equations may be written as

22. 57/(x+y) + 6/(x-y) = 5

38/(x+y) + 21/(x-y) = 9

Solution

23. 2(ax-by) + a + 4b = 0
2(bx+ay) + b - 4a = 0

Solution

The given system of equation may be written as

24. 6(ax + by) = 3a + 2b

6(bx - ay) = 3b - 2a

Solution

25. a²/x - b²/y = 0

a²b/x + c²a/y = a + b, x,y ≠ 0

Solution

26. mx - my = m² + n²
x + y = 2m

Solution

The given system of equations may be written as

27. ax/b - by/a = a+b

ax - by = 2ab

Solution

The given system of equation may be written as

28. bx/a + ay/b - (a² + b²) = 0
x + y - 2ab = 0

Solution

The given system of equation may be written as

↧

पाठ 1- यूरोप में राष्ट्रवाद का उदय इतिहास के नोट्स| Class 10th

January 5, 2019, 3:46 am

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पठन सामग्री और नोट्स (Notes)| पाठ 1- यूरोप में राष्ट्रवाद का उदय (Europe me Rashtravaad ka Uday) Itihas Class 10th

• उन्नीसवीं शताब्दी के दौरान, राष्ट्रवाद के विचारों ने यूरोप के राजनीतिक और मानसिक दुनिया में अनेक परिवर्त्तन किया।

फ्रांसीसी क्रांति और राष्ट्र का विचार

• राष्ट्रवाद की पहली स्पष्ट अभिव्यक्ति 1789 में फ्रांसीसी क्रांति के साथ हुई।

• फ्रांसीसी क्रांतिकारियों द्वारा फ्रांसीसी लोगों के बीच सामूहिक पहचान की भावना पैदा करने के लिए उठाए गए कदम।

→ La patrie और Le citoyen के विचार

→ नया फ्रांसीसी झंडा

→ एस्टेट्स जनरल को चुना गया और उसका नाम बदलकर नेशनल असेंबली रखा गया

→ नई स्तुतियाँ रची गई और शपथ ली गई

→ केंद्रीकृत प्रशासनिक व्यवस्था प्रणाली

→ आंतरिक आयात -निर्यात शुल्क समाप्त कर दिए गए

→ भार और माप की एकसमान प्रणाली लागू की गई

→ फ्रांसीसी को आम लोगों की भाषा बनायी गई

नेपोलियन

• फ्रांस में 1799 से 1815 तक शासन किया।

• 1799 में प्रथम राजदूत बनकर पूर्ण अधिकार प्राप्त किया।

1804 की नागरिक संहिता अथवा नेपोलियन की संहिता

• कानून के समक्ष बराबरी और सम्पति के अधिकार को सुरक्षित बनाया।

• प्रशासनिक विभाजन को सरल बनाया।

• सामंती व्यवस्था को समाप्त किया।

• किसानों को भू -दासत्व और जागीदारी शुल्कों से मुक्ति दिलाई।

• यातायात और संचार प्रणालियों में सुधार किया गया।

• नेपोलियन ने राजनीतिक स्वतंत्रता छीन ली, करों में वृद्धि की, सेंसरशिप लगाई और लोगों को फ्रांसीसी सेना में शामिल होने के लिए मजबूर किया।

यूरोप में राष्ट्रवाद का निर्माण

सामूहिक पहचान या समान संस्कृति नहीं होने के कारण कोई भी राष्ट्र राज्य यूरोप में नहीं थे।

• अलग -अलग क्षेत्रों में रहने वाले लोग अक्सर अलग -अलग भाषाएँ बोलते थे।

• उदाहरण: हंगरी की आधी आबादी मैग्यार भाषा बोलते थे, जबकी बाकी आधी आबादी दूसरी विभिन्न भाषाऐं की भाषा बोलते थे। गालीसिया के कुलीन वर्ग पोलिश भाषा बोलते थे।

• कुलीन वर्ग और नया मध्यवर्ग

→ कुलीन वर्ग :- सामाजिक और राजनातिक रूप से भूमि का मालिक

→ आपस में विवाहिक संबंधों से जुड़ा हुआ था और बोलचाल में फ्रेंच भाषा प्रयोग करते थे।

→ जनसंख्या की दृष्टि से यह समूह छोटा था।

• किसान

→ अधिकांश जनसंख्या

• मध्यवर्ग

→ कस्बों के विकास और वाणिज्यिक वर्गों के उदय के साथ ही नया सामाजिक वर्ग उभरा।

→ शिक्षित और उदारवादी मध्यवर्गों के बीच ही राष्ट्रीयता के विचारों को लोकप्रियता मिली।

उदारवादी राष्ट्रवाद के क्या मायने थे ?

• उदारवाद व्यक्ति की आजादी और कानून के समक्ष बराबरी लिए खड़ा था।

→ निरंकुश शासक और पादिरीवर्ग विशेषाधिकारों की समाप्ति।

→ संसद के माध्यम से एक संविधान और प्रतिनिधि सरकार।

→ आर्थिक क्षेत्र में उदारवाद, बाजारों की मुक्ति और चीजों और पूंजी के आगमन पर राज्य द्वारा लगाए गए प्रतिबंधों को समाप्त करने के पक्ष में था।

→ ज़ोल्वरिन(एक सीमा शुल्क संघ) ने टैरिफ बाधाओं को समाप्त कर दिया, मुद्राओं की संख्या को दो तक कम कर दिया और गतिशीलता को प्रोत्साहित करने के लिए रेल के नेटवर्क को बढ़ावा दिया।

1815 के बाद एक नया रूढ़िवाद

• नेपोलियन द्वारा शुरू किए गए परिवर्तनों से यह मान लिया गया की राज्य और समाज के स्थापित संस्थानों को मजबूत बनाने में सक्षम है।

1815 की विएना की संधि
• फ्रांस में बूर्बों राजवंश को बहाल किया गया था।

• फ्रांस की सीमाओं पर कई राज्य कायम कर दिये गए ताकि भविष्य में फ्रांस विस्तार न कर सके।

• जर्मन परिसंघ को बरकरार रखा गया।

• इसका मुख्य उद्देश्य नेपोलियन द्वारा बर्खास्त किये गए राजतंत्रों को बहाल करना था।

क्रांतिकारी

• विएना कांग्रेस के बाद स्थापित किए गए राजतंत्रीय रूपों का विरोध करना और स्वतंत्रता की लड़ाई के लिए प्रतिबद्धता।

ज्युसेपी मेत्सिनी

• 1807 में जेनोआ में पैदा हुए।

• कार्बोनारी के गुप्त संगठन का सदस्य

• मार्सेई में युवा इटली एवं बर्न में यंग यूरोप की स्थापना।

• एक गणतंत्र राष्ट्र के रूप में इटली के एकीकरण पर विश्वास किया।

क्रांतियों की आयु: 1830-1848

• फ्रांस में जुलाई 1830 में बूर्बों राजाओं को उखाड़ फेंका गया और एक संवैधानिक राजतंत्र स्थापित किया गया।

• बेल्जियम नीदरलैंड के संयुक्त राज्यों से अलग हो गया।

• पंद्रहवीं शताब्दी के बाद से ग्रीस, जो ओटोमन साम्राज्य का हिस्सा था, आजादी के लिए संघर्ष आरम्भ किया।

• कुस्तुनतुनिया की संधि ने 1832 में ग्रीस को एक स्वतंत्र राष्ट्र के रूप में मान्यता दे दी।

रूमानी कल्पना और राष्ट्रीय भावना

• एक सांस्कृतिक आंदोलन जिसने राष्ट्रवादी भावना के एक विशेष रूप को विकसित की, जिसमें तर्क- वितर्क और विज्ञान के महिमामंडन की आलोचना की और उसकी जगह भावनाओं, अंतर्दृष्टि और रहस्यवादी भावनाओं पर जोर दिया।

• जर्मन दार्शनिक योहान गॉटफ्रीड ने लोक गीत, लोक कविता और लोक नृत्यों के माध्यम से आम लोगों के बीच जर्मन संस्कृति को जगाने की कोशिश की।

भूख, कठिनाइयाँ और जन विद्रोह

• अधिकांश देशों में उपलब्ध रोजगार की तुलना में नौकरियों ढूँढने वाले अधिक थे।

• ग्रामीण इलाकों की अतिरिक्त आबादी शहर जाकर भीड़ -भाड़ वाले गरीब वस्तियों में रहने लगी।

• खाद्य पदार्थ की कीमत बढ़ने या फसल के खराब होने के कारण शहरों और गाँवों में व्यापक गरीबी फैल जाती थी।

• वर्ष 1848 में, पेरिस के लोग सड़क पर उतर आए और लुई फिलिप को भागने पर मजबूर किया गया अन्तः नेशनल असेंबली ने पेरिस को एक गणराज्य घोषित किया।

• वर्ष 1845 में, सिलेसिया में बुनकरों ने ठेकेदारों के खिलाफ विद्रोह कर दिया।

1848: उदारवादियों की क्रांति

• क्रांति का नेतृत्व शिक्षित मध्यवर्गों ने किया इन्होंने संविधानवाद की माँग को राष्ट्रीय एकता के साथ जोड़ दिया।

फ्रैंकफर्ट संसद

• 18 मई 1848 को, राजनीतिक संगठन के 831 निर्वाचित सदस्यों ने फ्रैंकफर्ट संसद की बैठक सेंट पॉल के चर्च में आयोजित किया और जर्मन राष्ट्र के लिए एक संविधान का मसौदा तैयार किया।

• यह प्रशिया के राजा द्वारा अस्वीकार किया गया था और वही संसद का सामाजिक आधार कमजोर हो गया था क्योंकि श्रमिकों ,मजदूरों और महिलाओं को कोई अधिकार नहीं दिया गया था।

• इन्होंने निरंकुश राजतंत्रों में कुछ बदलाव लाने के लिए मजबूर किया - भूदासत्व और बंधुआ श्रम को समाप्त कर दिया गया।

• हंगरी को अधिक स्वायत्तता दी गई।

जर्मनी और इटली का निर्माण

जर्मनी

• प्रशा की सेना और नौकरशाही की मदद से ओटो वैन बिस्मार्क ने राष्ट्रीय एकीकरण के लिए आंदोलन का नेतृत्व किया था।

• सात वर्षों के दौरान तीन युद्धों में प्रशिया की जीत हुई और एकीकरण की प्रक्रिया पूरी हुई।

• प्रशा के कैसर विलियम प्रथम ने नए जर्मन साम्राज्य का नेतृत्व किया।

इटली

• इटली सात राज्यों में बॉंटा हुआ था जिनमें से केवल एक सार्डिनिया पीडमॉन्ट पर एक इतालवी राजघराने का शासन था।

• ज्यूसेपे मेत्सिनी ने द्वारा एक एकीकरण कार्यक्रम का शुरुआत किया गया था लेकिन यह असफल रहा।

• मंत्री प्रमुख कैवोर ने ग्यूसेप गैरीबाल्डी की मदद से आंदोलन का नेतृत्व किया था।

• 1861 में विक्टर इमैनुएल द्वतीय को एकीकृत इटली का राजा घोषित किया गया था।

ब्रिटेन का अजीब दास्तान

• 1688 में संसद के माध्यम से एक राष्ट्र-राज्य का निर्माण हुआ जिसके केंद्र में इंग्लैंड था।

• अंग्रेजी संसद ने राजशाही से ताकत छीन ली थी।

• यूनियन अधिनियम 1707 के परिणामस्वरूप 'यूनाइटेड किंगडम ऑफ ग्रेट ब्रिटेन’ का गठन हुआ।

• असफल क्रांति के बाद 1801 में आयरलैंड को बलपूर्वक यूनाइटेड किंगडम में शामिल कर लिया गया।

• नया 'ब्रितानी राष्ट्र' का निर्माण किया गया और एक प्रमुख एंग्लो संस्कृति का प्रचार-प्रसार किया गया।

राष्ट्र की दृश्य-कल्पना

• राष्ट्र को महिला के आकृति (रूपक) के रूप में चित्रित किया गया था।

• राष्ट्र को स्त्री के रूप में परिभाषित करने के लिए चुना गया था जो वास्तविक जीवन में किसी विशेष महिला के लिए नहीं थी , बल्कि उसने राष्ट्र के अमूर्त विचार को ठोस रूप देने की प्रयास था।

• फ्रांस में मारीआन और जर्मनी में जर्मनिया को राष्ट्र का रूपक माना गया।

राष्ट्रवाद और साम्राज्यवाद

• बाल्कन क्षेत्र में आधुनिक रोमानिया, बुल्गारिया, अल्बानिया, ग्रीस, मैसेडोनिया, क्रोएशिया, बोस्निया-हर्जेगोविना, स्लोवेनिया, सर्बिया और मोंटेनेग्रो शामिल थे।

• बाल्कन भौगोलिक और जातीय भिन्नता का एक क्षेत्र था जो ओटोमन साम्राज्य के नियंत्रण में था।

• रूमानी राष्ट्रवाद के विचारों और ऑटोमन साम्राज्य के विघटन से स्थिति काफी विस्फोटक हो गई थी।

• बाल्कन राज्य एक-दूसरे से भारी ईर्ष्या करते थे और प्रेत्यक राज्य ज्यादा से ज्यादा इलाका हथियाना चाहता था।

• यूरोपीय शक्तियाँ भी इस क्षेत्र पर अपना नियंत्रण बढ़ाने की कोशिश कर रही थीं।

• इसके कारण इस क्षेत्र में कई युद्ध हुए और अंततः प्रथम विश्व युद्ध हुआ।

NCERT Solutions of पाठ 1- यूरोप में राष्ट्रवाद का उदय

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NCERT Solutions for Class 12th: Ch 7 Directing (Long Answer Questions)

January 6, 2019, 2:16 pm

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NCERT Solutions for Class 12th: Ch 7 Directing (Long Answer Questions) Business Studies I

Exercises

Long Answer Type Questions:

1. Explain the principles of Directing.

Answer

(i) Maximum Individual Contribution: This principle emphasises that directing techniques must help every individual in the organisation to contribute to his maximum potential for achievement of
organisational objectives. It should bring out untappted energies of employees for the efficiency of
organisation.

(ii) Harmony of Objectives: Very often, we find that individual objectives of employees and the organisational objectives as understood are conflicting to each other. Good directing should provide harmony by convincing that employee rewards and work efficiency are complimentary to each other.

(iii) Unity of Command: This principle insists that a person in the organisation should receive
instructions from one superior only. If instructions are received from more than one, it creates confusion, conflict and disorder in the organisation. Adherence to this principle ensures effective
direction.

(iv) Appropriateness of Direction Technique: According to this principle, appropriate motivational and leadership technique should be used while directing the people based on subordinate needs, capabilities, attitudes and other situational variables. For example, for some people money can act as
powerful motivator while for others promotion may act as effective motivator.

(v) Managerial Communication: Directing should convey clear instructions to create total understanding to subordinates. Through proper feedback the manager should ensure that subordinate understands his instructions clearly.

(vi) Use of informal organisation: A manager should realise that informal groups or organisations
exist within every formal organisation. He should spot and make use of such organisations for
effective directing.

(vii) Leadership: While directing the subordinates, managers should exercise good leadership as it
can influence the subordinates positively without causing dissatisfaction among them.

(viii) Follow through: Mere giving of an order is not sufficient. Managers should follow it up by
reviewing continuously whether orders are being implemented accordingly or any problems are
being encountered. If necessary, suitable modifications should be made in the directions.

2. Explain the qualities of a good leader. Do the qualities alone ensure leadership success?

Answer

(i) Physical features: Physical features like height, weight, health, appearance determine the physical personality of an individual. It is believed that good physical features attract people.

(ii) Knowledge: A good leader should have required knowledge and competence. Only such person can instruct subordinates correctly and influence them.

(iii) Integrity: A leader should posses high level of integrity and honesty. He should be a role model to others regarding the ethics and values.

(iv) Initiative: A leader should not wait for opportunities come to his way, rather he should grab the opportunity and use it to the advantage of organisation.

(v) Communication skills: A leader should be a good communicator. He should have the capacity to
clearly explain his ideas and make the people to understand his ideas.

(vi) Motivation skills: A leader should be an effective motivator. He should understand the needs
of people and motivate them through satisfying their needs.

(vii) Self Confidence: A leader should have high level of self confidence. He should not loose his confidence even in most difficult times.

(vii) Decisiveness: Leader should be decisive in managing the work. Once he is convinced about
a fact, he should be firm and should not change opinions frequently.

(ix) Social skills: A leader should be sociable and friendly with his colleagues and followers. He
should understand people and maintain good human relations with them.

No, the qualities cannont alone ensure leadership success. The followers who have different skills, knowledge, commitment, willingness to co-operate team spirit etc make a person an effective leader. Therefore both followers and leaders are playing a vital role in leadership process.

3. Discuss Maslow’s need Hierarchy theory of motivation.

Answer

Maslow’s Need Hierarchy Theory is considered fundamental to understanding of motivation. His theory was based on human needs. He felt that within every human being, there exists a hierarchy of five needs.

(i) Basic Physiological Needs: These needs are most basic in the hierarchy and corresponds to primary needs. Hunger, thirst, shelter, sleep and sex are some examples of these needs.

(ii) Safety/Security Needs: These needs provide security and protection from physical and
emotional harm. Examples: job security, stability of income, Pension plans etc.,

(iii) Affiliation/Belonging Needs: These needs refer to affection, sense of belongingness, acceptance and friendship.

(iv) Esteem Needs: These include factors such as self-respect, autonomy status, recognition and attention.

(v) Self Actualisation Needs: It refers to the drive to become what one is capable of becoming. These needs include growth, self-fulfillment and achievement of goals.

4. What are the common barriers to effective communication suggest measures to overcome them?

Answer

The managers face several problems due to communication breakdowns or barriers. The barriers to communication in the organisations can be broadly grouped as: semantic barriers, psychological
barriers, organisational barriers, and personal barriers.

(i) Semantic Barriers: Semantic barriers are concerned with problems and obstructions in the process of encoding and decoding of message into words or impressions. Normally, such barriers result on account of use of wrong words, faulty translations, different interpretations etc. These are discussed below

(a) Badly Expressed Message: Sometimes the message is not communicated correctly by the manager because of inadequate vocabulary, usage of wrong words, omission of needed words etc.

(b) Symbols with Different Meaning: A word may have several meanings. Receiver has to perceive one such meaning for the word used by communicator.

(c) Faulty Translations: Sometimes while translating if incorrect translation is done due to poor command over both the languages then meaning of the message changes. This leads to cause different meanings to the communications.

(d) Unclarified Assumptions: Sometimes communication may have certain assumptions which are subject to different interpretations The one should always clear the meaning of what he is instructing the worker to do, so that the worker has no doubts in his mind.

(e) Technical Jargon: Sometimes specialists may use technical words in their communication by which the receiver is not aware. Therefore, they may not understand the complete conversation.

(f) Body Language and Gesture Decoding The body movement and body gestures plays an important role in conveying the message. If there is no match between what is said and what is expressed in body movements, communications may be wrongly perceived.

(ii) Psychological Barriers: Emotional or psychological factors acts as barrier to communications e.g., a person who is worried cannot understand what is being told. Some of the psychological barriers are

(a) Premature Evaluation: Sometimes people evaluate the meaning of message before the sender
completes his message. Such premature evaluation may be due to pre-conceived notions.

(b) Lack of Attention: If the mind is pre-occupied then the result is non-listening of message by receiver act as a major psychological barrier.

(c) Lon by Transmission and Poor Retention: When message passes through various levels, successive transmission of message results in loss of information. It happens mostly with oral communication. Also, people cannot retain the information for a long time if they are inattentive or not interested.

(d) Distrust: If the communicator and communicatee do not believe on each other, they can not understand each others message in its original sense as they are not giving importance to the information exchanged.

(iii) Organisational Barriers: The factors related to organisation structure, authority relationships, rules and regulations may sometimes act as barriers to effective communication some of these barriers are

(a) Organisational Policy: If the organisational policy is not supportive to free flow of communication, it may hamper effectiveness of communications.

(b) Rules and Regulations Rigid rules and cumbersome procedures may be a hurdle to communication similarly, communication through prescribed channel may result in delays.

(c) Status Status of superior may create psychological distance between him and his subordinates. The people working at higher level may not allow his subordinates to express their feelings freely.

(d) Complex Organisational Structure In an organisation where there are number of managerial levels, communications gets delayed and distorted as number of filtering points are more.

(e) Organisational Facilities For smooth clear and timely communication proper facilities are required like frequent meetings suggestion box, internet connection, inter-com facility. Lack or ineffectiveness of these facilities may create communication problems.

(iv) Personal Barriers: The personal factors of both sender and receiver may exert influence on effective communication. Some of the personal barriers are:

(a) Fear of Challenge to Authority: If a superior feels that a particular communication may affect his authority negatively then he/she may not speak it out clearly and openly.

(b) Lack of Confidence of Superior on his Subordinate: If superiors do not have confidence on their subordinates, they may not seek their advice or opinions.

(c) Unwillingness to Communicate: Sometime subordinator may not be prepared to communicate with their superiors, if they think that it may adversely affect their interests.

(d) Lack of Proper Incentives: If there is no reward for communication then employees may not be motivated to communication, e.g., if there is no reward or appreciation for a good suggestion, the subordinate may not be willing to offer useful suggestions again.

Some measures which can be adopted by organisations to improve communications are

(i) Clarify the ideas before communication: The problem to be communicated to subordinates
should be clear in all its perspective to the executive himself.

(ii) Communicate according to the needs of receiver: The level of understanding of receiver should be crystal clear to the communicator.

(iii) Consult Others Before Communicating: Before communicating anything, others who are linked with it in some way or the other should be taken into confidence for developing a better plan.

(iv) Beaware of Languages, Tone and Content of Message: The language used for communication should be understandable to the listener. The tone of the appropriate and the matter should not be offending to anyone.

(v) Convey Things of Help and Value to Listener: It is always better to know the interests of the people with whom you are communicating. If the message relates directly or indirectly to such interests and needs it certainly evokes response from communicatee.

(vi) Ensure Proper Feedback: The receiver of communication may be encouraged to respond to communication. The communication process may be improved by the feedback received to make it more responsive.

(vii) Follow up Communication: There should be a regular follow up and review on the instructions given to subordinates. Such follow up measures help in removing hurdles if any in implementing the instructions.

(viii) Be a Good Listener: Manager should be a good listener. Patient and attentive listening solves half of the problems. Managers should also give indications of their interest in listening to their subordinates

5. Explain different financial and non-financial incentives used to motivate employees of a company.

Answer

Financial incentives refer to incentives which are in direct monetary form or measurable in monetary term and serve to motivate people for better performance.The financial incentives are:

(i) Pay and Allowances: For every employee, salary is the basic monetary incentive. It includes
basic pay, dearness allowance and other allowances. Salary system consists of regular increments in the pay every year and enhancement of allowances from time-to-time.

(ii) Productivity Linked Wage Incentives: Several wage incentives aim at linking payment of wages to increase in productivity at individual or group level.

(iii) Profit Sharing: Profit sharing is meant to provide a share to employees in the profits of the organisation. This serves to motivate the employees to improves their performance and contribute to increase in profits.

(v) Co-Partnership/Stock Option: Under these incentives schemes, employees are offered company shares at a set price which is lower than market price. The allotment of shares creates a peeling of ownership to the employees and makes them to contribute more for the growth of the organisation.

(vi) Retirement Benefits: Several retirement benefits such as provident fund, pension and gratuity provide financial security to employees after their retirement. This act as an incentive when they are in service in the organisation.

(vii) Perquisites: In many companies perquisites and fringe benefits are offered such as car allowance, housing, medical aid, and education etc over and above the salary. These measures help to provide motivation to the employees/managers.

Non-financial Incentives help in fulfilling our psychological, emotional and social needs are known as non-financial incentives. The non-financial incentives are

(i) Status: Status means ranking or high positions in the organisation. Whatever power position prestige an employee enjoys in the organisation are indicated by his status. Psychological, social and esteem needs of an individual are satisfied by status given to their job.

(ii) Organisational Climate: This indicates the characteristics which describe an organisation and distinguish one from the other. Individual autonomy, reward orientation, consideration to employees, etc are some of the positive features of an organisation. If managers try and include more of these in an organisation helps to develop better organisational climate.

(iii) Career Advancement Opportunity: Managers should provide opportunity to employees to improve their skills and be promoted to the higher level jobs appropriate skill development programmes and sound promotion policy will help employees to achieve promotions. Promotions have always worked as tonic and encourages employees to exhibit improved performance.

(iv) Job Enrichment: It is concerned with designing jobs that include a greater variety of work contentment, requires higher level of knowledge and skill, gives workers more autonomy and responsibility and provide opportunity for personal growth and a meaningful work experience.

(v) Employee Recognition: Programmes Recognition means acknowledgement with a show of appreciation. When such appreciation is given to the work performed by employees, they feel motivated to perform/work at higher level. These are:

(a) Congratulate the employee

(b) Displaying names of star performers

(d) Distributing mementos

(vi) Job Security

(vi) Job Security: Employees want their job to be secure. They want certain stability about future income and work so that they do not feel worried on these aspects and work with greater zeal. There is only one problem with this incentive i.e. when people feel that they are not likely to lose their jobs, they may become relaxed.

(vii) Employee Participation: It means involving employees in decision making of the issues related to them. In many companies, these programmes are in practice in the form of joint management committees, work committees canteen committees etc.

(viii) Employee Empowerment: Empowerment means giving more autonomy and powers to subordinates. Empowerment makes people feel that their jobs are important. This feeling contributes positively to the use of skills and talents in the job performance.

NCERT Solutions of Ch 7 Directing (MCQs and Short Answer Questions)

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R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.5

January 7, 2019, 10:29 am

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Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.5

Exercise 3.5

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. x - 3y - 3 = 0

3x - 9y - 2 = 0

Solution

The given system of equations may be written as

2. 2x + y - 5 = 0

4x + 2y - 10 = 0

Solution

The given system of equation may be written as

3. 3x - 5y = 20

6x - 10y = 40

Solution

4. x - 2y - 8 = 0

5x - 10y - 10 = 0

Solution

The given system of equation may be written as

Find the value of k for which the following system of equations has a unique solution

5. kx + 2y - 5 = 0

3x + y - 1 = 0

Solution

The given system of equation is

6. 4x + ky + 8 = 0

2x + 2y + 2 = 0

Solution

7. 4x - 5y = k

2x - 3y = 12

Solution

The given system of equation is

So, the given system of equations will have a unique solution for all real values of k.

8. x + 2y = 3

5x + ky + 7 = 0

Solution

The given system of equation is

So, the given system of equations will have a unique solution for all real values of k other than 10.

Find the value of k for which each of the following systems of equations have infinitely many solution: (9-19)

9. 2x + 3y - 5 = 0

6x - ky - 15 = 0

Solution

The given system of equation is

Hence, the given system of equations will have infinitely many solutions, if 9.

10. 4x + 5y = 3

kx + 15y = 9

Solution

The given system of equation is

Hence, the given system of equations will have infinitely many solutions, if k = 12.

11. kx - 2y + 6 = 0

4x + 3y + 9 = 0

Solution

The given system of equation is

Hence, the given system of equations will have infinitely many solutions, if k = 8/3.

12. 8x + 5y = 9

kx + 10y = 18

Solution

The given system of equation is

Hence, the given system of equations will have infinitely many solutions, if k = 16

13. 2x - 3y = 7

(k+2) x - (2k + 1) y - 3 (2k-1)

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 4.

14. 2x + 3y = 2

(k+2) x + (2k+1) y - (k-1)

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 4 .

15. x = (k+1) y = 4

(k+1) x + 9y - (5k+2)

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 2.

16. kx + 3y - 2k + 1

2(k+1) x + 9y - (7k+1)

Solution

The given system of equation may be written as

17. 2x + (k-2) y = k

6x + (2k-1) y- (2k+5)

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 5.

18 . 2x + 3y = 7

(k+1) x + (2k-1) y - (4k+1)

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 5.

19. 2x + 3y = k

(k-1) x + (k+1) y - 3k

Solution

The given system of equation may be written as

Hence, the given system of equations will have infinitely many solutions, if k = 7.

Find the value of k for which the following system of equations has no solution: (20 – 25)

20. kx - 5y = 2

6x + 2y = 7

Solution

Given

21. x + 2y = 0

2x + ky - 5 = 0

Solution

The given system of equation may be written as

Hence, the given system of equations has no solutions, when k = 4.

22. 3x - 4y + 7 = 0

kx + 3y - 5 = 0

Solution

The given system of equation may be written as

23. 2x - ky + 3 = 0

3x + 2y - 1 = 0

Solution

The given system of equation may be written as

24. 2x + ky = 11

5x - 7y = 5

Solution

The given system of equation is

25. kx + 3y = 3

12x + ky = 6

Solution

26. For what value of a, the following system of equations will be inconsistent?

4x + 6y - 11 = 0

2x + ky -7 = 0

Solution

The given system of equation may be written as

Hence, the given system of equation is inconsistent, when k = 3

27. For what value of a, the system of equations

ax + 3y = a - 3

12x + ay = a

will have no solution ?

Solution

The given system of equation may be written as

ax + 3y - (a - 3) = 0

12x + ay - a = 0

Hence, the given system of equation will have no solution, if a = -6 .

28. Find the value of k for which the system

kx + 2y = 5

3x + y = 1

has (i) a unique solution, and (ii) no solution.

Solution

The given system of equation may be written as

29. Prove that there is a value of c (≠ 0) for which the system

6x + 3y = c - 3

12x + cy = c

has infinitely many solutions. Find this value.

Solution

The given system of equation may be written as

30. Find the values of k for which the system

2x + ky = 1

3x – 5y = 7

will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions ?

Solution

The given system of equation may be written as

31. For what value of k, the following system of equations will represent the coincident lines ?

x + 2y + 7 = 0

2x + ky + 14 = 0

Solution

The given system of equations may be written as

Hence, the given system of equations will represent coincident lines, if k = 4

32. Obtain the condition for the following system of linear equations to have a unique solution

ax + by = c

lx + my = n

Solution

The given system of equations may be written as

Hence, am ≠ bl is the required conidition.

33. Determine the values of a and b so that the following system of linear equations have infinitely many solutions:

(2a - 1) x + 3y - 5 = 0

3x + (b - 1)y - 2 = 0

Solution

The given system of equations may be written as

34. Find the values of a and b for which the following system of linear equations has infinite number of solutions:

2x - 3y = 7

(a+b) x - (a+b-3) y = 4a + b

Solution

The given system of equations may be written as

Hence, the given system of equations will have infinitely many solutions, If a = -5 and b = -1 .

35. Find the values of p and q for which the following system of linear equations has infinite number of solutions:

2x - 3y = 9

(p+q) x + (2p - q) y = 3 (p+q+1)

Solution

The given system of equations may be written as

36. Find the values of a and b for which the following system of equations has infinitely many solutions:

(i) 2x = 3y = 7

(a-b)x + (a+b)y = 3a+b-2

Solution

Hence, the given system of equation will have infinitely many solutions,if a = 3 and b = 1/5.

(ii) 2x-(2a+5)y = 5

(2b + 1) x - 9y = 15

Solution

Hence, the given system of equations will have infinitely many solutions, If a = -1 and b = 5/2.

(iii) (a-1) x + 3y = 2

6x + (1+2b) y = 6

Solution

The given system of equations is

Hence, the given system of equations will have infinitely many solutions, If a = 3 and b = -4.

(iv) 3x + 4y = 12

(a+b) x + 2 (a-b) y = 5a - 1

Solution

The given system of equations is

Putting b = 1 in a = 5b, we get

a = 5 × 1 = 5

Hence, the given system of equations will have infinitely many solutions, If a = 5 and b = 1.

(v) 2x + 3y = 7

(a-1) x + (a+1)y = (3a-1)

Solution

The given system of equations is

Hence, the given system of equations will have infinitely many solutions, If a = 5.

(vi) 2x + 3y = 7

(a-1) x + (a+2) y = 3a

Solution

The given system of equations is

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