Quantcast
Channel: Study Rankers
Viewing all 6266 articles
Browse latest View live
↧

R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.5

February 17, 2019, 2:16 am
$
0
0

Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.5

1. Find the mode of the following data:
(i) 3,5,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15    

Solution

(i)

(ii) 
 

(iii) 


2. The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:
Shirt size: 37  38  39  40  41  42  43  44
Number of persons: 15  25  39  41  36  17  15  12
Find the model shirt size worn by the group. 

Solution  


3. Find the mode of the following distribution.
(i) Class-interval: 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70-80
Frequency: 5   8  7  12  28  20  10  10
(ii) Class-interval: 10-15  15-20  20-25  25-30  30-35  35-40
Frequency: 30  45  75  35  25  15
(iii) Class-interval: 25-30  30-35  35-40  40-45  45-50  50-60
Frequency: 25  34  50  42  38  14

Solution

(i)

(ii) 

(iii) 

4. Compare the modal ages of two groups of students appearing for an entrance test:
Age (in years): 16-18  18-20  20-22  22-24  24-26
Group A: 50  78  46  28  23
Group B: 54  89  40  25  17

Solution 


5. The marks in science of 80 students of class X are given below: Find the mode of the
marks obtained by the students in science.
Marks: 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70-80  80-90  90-100
Frequency: 3  5  16  12  13  20  5  4  1  1



Solution


6. The following is the distribution of height of students of a certain class in a certain city:
Height (in cm): 160-162  163-165  166-168  169-171  172-174
No. of students: 15  118  142  127  18
Find the average height of maximum number of students.  



Solution


7. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years): 5-15  15-25  25-35  35-45  45-55  55-65
No. of students: 6  11  21  23  14  5
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.




Solution


8. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in hours): 0-20  20-40  40-60  60-80  80-100  100-120
No. of components: 10  35  52  61  38  29
Determine the modal lifetimes of the components.

Solution


9. The following table gives the daily income of 50 workers of a factory:
Daily income (in Rs) : 100-120  120-140  140-160  160-180  180–200
Number of workers:  12   14   8   6  10
Find the mean, mode and median of the above data

Solution


10. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

Solution


11. Find the mean, median and mode of the following data:
Classes: 0-50  50-100  100-150  150-200  200-250  250-300  300-350
Frequency:  2   3   5   6  5   3  1

Solution  


12.  A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data :
Number of cars : 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70-80 
Frequency : 7  14  13  12   20   11   15   8  

Solution


13. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption : 65-85  85-105  105-125  125-145  145-165  165-185  185-205
(in units)
No. of consumers:  4  5  13  20  14  8  4

Solution 


14. 100 surnames were randomly picked up from a local telephone directly and the frequency
distribution of the number of letters in the English alphabets in the surnames was obtained
as follows:
Number of letters: l-4  4-7  7-10  10-13  13-16  16-19
Number surnames: 6  30  40  16  4  4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution 


15. Find the mean, median and mode of the following data:
Classes:  0-20  20-40  40-60  60-80  80-100  100-120  120-140
Frequency: 6  8  10  12   6   5   3

Solution


16. The following data gives the distribution of total monthly houshold expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution



17. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. 



Solution


18. The frequency distribution table of agriculture holdings in a village is given below :
Area of land (in hectares) : 1-3  3-5 5-7 7-9 9-11 11-13
Number of families : 20  45  80  55  40  12
Find the modal agriculture holdings of the village .


Solution

The maximum class frequency is 80. The class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l(the lower limit of modal class) = 55
f1 (frequency of the modal class) = 80
fo (frequency of the class preceding the modal class) = 45
f2 (frequency of the class succeeding the modal class) = 55
h (class size) = 2
Mode = l+(f1-fo/2f1-fo-f2) × h
= 5 + (80-45/2×80-45-55) × 2
= 5 + 35/60 × 2
= 5 + 35/60
= 6.2

19. The monthly income of 100 families are given as below :
Income inNumber of families
0-50008
5000 - 1000026
10000 - 1500041
15000 - 2000016
20000 - 250003
25000 - 300003
30000 - 350002
35000 - 400001
Calculate the modal income .

Solution

The maximum class frequency is 41. The class corresponding to this frequency is 5-7.
So, the modal class is 10000-15000.
l (the lower limit of modal class) = 10000
f1 (frequency of the modal class) = 41
fo (frequency of the class preceding the modal class) = 26
f2 (frequency of the class succeeding the modal class) = 16
h (class size) = 5000
Mode = l+(f1-fo/2f1-fo-f2) × h
= 10000 + (41-26/2 × 41-26 - 16) × 5000
= 10000 + 15/40 × 5000
= 10000 +1875
= 11875
↧
Search

R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.6

February 17, 2019, 6:17 am
$
0
0

Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.6

1. 1. Draw an given by less than method for the following data:
No.of rooms: 1  2  3  4  5  6  7  8  9  10
No. of houses: 4  9  22  28  24  12  8  6  5  2 

Solution


2. The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Solution


3. Draw an ogive to represent the following frequency distribution:
Class-interval: 0-4  5-9  10-14  15-19   20-24
No. of students:  2   6   10   5   3

Solution


4. The monthly profits (in Rs.) of 100 shops are distributed as follows:
Profits per shop: 0-50  50-100  100-150  150-200   200-250   250–300
No. of shops:  12  18   27   20  17  6
Draw the frequency polygon for it.

Solution



5. The  following distribution the daily income of 50 workers  of  a factory :
Daily income in : 100-120  120-140  140-160  160-180  180-200
Number of workers : 12  14   8   6  10
Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive .

Solution

6. The following table gives production yield per hectare of wheat of 100 farms of a village:
Production yield in kg per hectare: 50-55  55-60  60-65  65-70  70-75  75-80 
Number of farms: 2   8  12   24   38   16
Draw ‘less than’ ogive and ‘more than’ ogive.

Solution


7. During the medical check-up of 35 students of a class, their weights were recorded as follows:
Draw a less than type ogive for the given data. Hence, obtain the median weight from the
graph and verify the result by using the formula .  

Solution


8. The annual rainfall record of a city for 66 days is given in the following table : 
Calculate the median rainfall using ogives of more than type and less than type .

Solution

Prepare a table for less than type .
Now, plot the less than ogive using suitable points .

Here, N = 66 
N/2 = 33
In order to find the median rainfall , we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting to curve at Q. From this point Q, draw a line parallel to the y - axis and meeting the x-axis at the point R. The x - coordinate of R is 21.25 .
Thus, median rainfall is is 21.25 cm . 
Here , N = 66 
N/2 =  33
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis.Let the point be P.From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinates of R is 21.25.  

9. The following table gives the height of tress . 
Draw ' less than 'ogive and more than 'ogive'

Solution

Consider the following table .
Now draw less than ogive using suitable points . 
Now, prepare the cumulative frequency table for men than series .  
Now draw more than ogive using suitable points .
10. The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution . 
Draw both ogives for the above data and hence obtain the median. 

Solution

Firstly , we prepare the cumulative frequency table for less than type .
Again, prepare the cumulative frequency table for more than type . 
Now, ''more than ogive'' and ''less than ogive'' can be drawn as follows : 
The x - coordinate of the point of intersection of the '' more than ogive'' and less- than ogive'' gives the median of the given distribution..
So, the corresponding median is 17.5 lakh .
↧
↧

Summary of Dust of Snow by Robert Frost NCERT Class 10th

February 18, 2019, 2:50 pm
$
0
0

Summary of Dust of Snow with Quick Revision Notes  First Flight Class 10

Summary of the Poem

Dust of Snow by Robert Frost

The poem consists of 2 stanzas. Each of these stanzas is made up of 4 lines. Thus, the entire poem consists of 8 lines in total.

The way a crow
Shook down on me
The dust of snow
From a hemlock tree

In this poem, the poet is talking about a winter’s day when all the treetops were covered with snowflakes. The poet was walking under one such snow-covered tree. This was a Hemlock tree. The hemlock tree is usually associated with poison and toxicity as a poison named Hemlock is also made. Thus, this tree cannot be associated with anything positive. However, a crow came and sat down on this tree suddenly, and because of this sudden motion, the mass of snowflakes from the top fell down on the poet like rain. The snowflakes were so light and small that the poet thought they looked like white and immaculate dust particles (if indeed dust particles could ever be white). This shower of snowflakes from the hemlock tree is evidently the subject matter of this poem, since it touched the poet in some way.

Has given my heart
A change of mood
And saved some part
Of a day I had rued.

Now in second stanza, the poet explains exactly in what way the shower of snowflakes had an effect on him. He says that it changed his mood. Earlierr, he had been in a despairing mood but the shower of the snowflakes falling on him all of a sudden gladdened his heart. In addition to this, there was another effect that the shower of snowflakes had on the poet. He says that that particular day was one that he had not been looking forward to, and in fact, it was not going too well either.
He had decided, in his mind, that that particular day would go to waste. However, the shower of dust like snow on his shoulder changed his mind. It made him feel that the day had not been a complete waste. At least some part of that day had been pleasant, since it had given him a new experience that he could treasure as well as the material for another one of his wonderful poetic compositions.


Quick Revision Notes

• The poet was sitting under the hemlock tree. 

• He was upset as his days did not go too well. 

• Suddenly, a crow sitting on the tree shakes the tree and the fine particles of snow from tree falls on the poet.  

• The soft and cold touch of snow changes the poet’s mood from sad to happy. 

• He starts feeling soothed and refresh. 

• In this way a simple moment proves to be very significant and saves rest of the day of poet from being wasted and held in regret. 

• The poem describes a very simple happening in very simple words.

• It tells us that sometimes even a small incident may prove to be of a larger significance.

• The poet is upset in the beginning of the poem but a small incident of falling of snow changes his mood totally.

• In the end of the poem the poet feels relaxed and thankful to the nature for saving his day from being wasted.

↧

Summary of Fire and Ice by Robert Frost NCERT Class 10th

February 19, 2019, 9:15 am
$
0
0

Summary of Fire and Ice with Quick Revision Notes  First Flight Class 10

Summary of the Poem Fire and Ice

Dust of Snow by Robert Frost

Some say the world will end in fire,
Some say in ice.

In these lines, the poet says that there are two theories held by the general people about how the world will be destroyed. The first group of people thinks that fire will cause the apocalypse to happen. It means that earth’s core will heat up to an extremely high temperature, then the heat will eventually reach the surface of the planet, and then everything on the surface will be destroyed. It will be destroyed beyond repair. The second theory says just the opposite. Instead of destruction by heat, some people believes in the theory that the world will freeze till all the species of flora and fauna i.e. plants and wildlife inhabiting the earth become extinct once and for all. Hence, this theory is based on the power of ice which will change the living conditions on earth. The poet is aware of both these theories and will now proceed to compare them against each other.

From what I’ve tasted of desire
I hold with those who favor fire.

In these lines, the poet tries to outline the merits of the first theory. He praised the theory about the destruction of the earth by fire. He compares the fire with human passion and desire. He also says that he is quite familiar with this concept of desire. He also knows about the capability of desire which can be produced in human beings. With this knowledge in mind, the poet frankly confesses that he agrees with those people who believe that the world will be burnt to ash.

But if it had to perish twice,
I think I know enough of hate
To say that for destruction ice
Is also great
And would suffice.

In these lines, the poet stops discussing the first theory and goes on to talk about the theory about the destruction of earth by ice. Though, he does not discredit the first theory about fire, or withdraw his support for its greater possibility in causing the apocalypse. He simply considers what might happen if the earth were to be destroyed a second time. He is sure that the second time ice and the freezing temperatures will be enough to destroy the earth as nearly and effectively as fire previously had. The poet then compares ice with hatred. He says that hatred is also an emotion that he is familiar with, and that he knows what kind of danger can arise from hatred. With this knowledge, he comes to the conclusion that both an ice age and the coldness in man’s heart can cause the apocalypse for a second time.

Quick Revision Notes

• The poet talks about the two different beliefs regarding the end of this world.

• He says that he is in the favour of those who say this world will end in fire as he has seen the effect and result of uncontrolled and unending desires of humans.

• He finds the human desires same as fire in its nature.

• On the other hand, the second belief tells that ice is sufficient for destroying this world and the poet compares the nature of ice with hatred of humans.

• As ice can make a part of body numb with its prolonged contact like hatred can also give numbness to our mind and thoughts and make us insensitive and cruel.

• The speaker brings us into the middle of an argument between people who think the world will come to a fiery end and people who think the world will freeze.

• Author is talking about the literal end of the world, but he's also talking about the power that human beings have to harm or "destroy" one another.

↧

Summary of A Tiger in the Zoo by Leslie Norris NCERT Class 10th

February 19, 2019, 10:41 am
$
0
0

Summary of A Tiger in the Zoo with Quick Revision Notes  First Flight Class 10

Summary of the Poem A Tiger in the Zoo

A Tiger in the Zoo by Leslie Norris

He stalks in his vivid stripes
The few steps of his cage,
On pads of velvet quiet,
In his quiet rage.


In this stanza, the poet describes the appearance of the tiger in the zoo and its daily movements. He says that the tiger moves up and down of his cage. The stripes on its body can be distinguished even seen from far as they are darker in colour than the rest of its coat. The tiger walks quietly without making a sound because of his smooth velvet pads. The tiger is able to walk only few steps because of the length of its cage. It covers that distance in a few mere steps only. The tiger is full of anger but it is suppressed because he knows that he is helpless here.

He should be lurking in shadow,
Sliding through long grass
Near the water hole
Where plump deer pass.

In this stanza, the poet is filled with pity  while looking at the imprisoned tiger.  The poet imagines an another life for the tiger other than the zoo. He says that the poor tiger should have been in his natural habitat that is in the wild forest. He says that under normal conditions, this tiger would have been moving in the darkness not in the clear daylights. It would have been lying in the shadows of trees  or hiding itself behind long grass so that its prey will not be able to detect its movement. In this way, the tiger would have moved near the water hole where all the animals of the forest come to drink on sunny afternoons. There, near the water hole, he would be waiting for some fat deer to pass that way. Thus, he would be lying there in expectation of a heavy feast.

He should be snarling around houses
At the jungle’s edge,
Baring his white fangs, his claws,
Terrorising the village!

In this stanza, the poet says that even though he would prefer the tiger to live in the wild, it is not as if it would lose all touch with human civilization. He imagines what the tiger will do in case he failed to find any prey in his natural habitat. He says that the  tiger would be growling at the edge of jungle near some village. He would be showing his white fangs  and claws while  moving here and  there  with the intention of scaring all the inhabitants of that village.. He would thus become a cause of terror  for the villagers. The poetess here gives  a  hint  that  if  we  destroy the  natural habitat  of  tigers, they  will  be  forced to turn to our towns and  villages to find their  food.

But he’s locked in a concrete cell,
His strength behind bars,
Stalking the length of his cage,
Ignoring visitors.

In this stanza, the poet stops imaging the tiger’s alternate life and returns to its real one. The poet sees the tiger caged in a small concrete cell in the zoo like a prisoner in the jail.  In spite of all his strength, it now lies imprisoned behind the bars. The tiger pays no attention to the visitors who come to see him every day at the zoo. Instead of this, the tiger Very slowly and silently, moves up and down along the length of the cage without any disturbance in the form of human intervention.

He hears the last voice at night,
The patrolling cars,
And stares with his brilliant eyes
At the brilliant stars.

In this stanza, the poet describes about the daily routine of the tiger. The tiger does not rest during the day because of the visitors. Even at night, it stays awake till the very last voice of the zookeeper locking up and going home can be heard. After that too, it does not go to sleep. All night, it remains disturbed due to the noise of the patrolling cars. He keeps staring at the brilliant stars with his brilliant eyes. Perhaps he is asking heaven why he has been imprisoned there.

Quick Revision Notes

• This poem contrasts a tiger when it is in its natural habitat and when it is imprisoned in a zoo.

• On a starry night, the poet sees a tiger in a zoo.

• The tiger moves slowly up and down in his cage.

• He is full of rage but is quiet in his helplessness.

• The poet is moved to pity for the tiger.

• He says that the tiger should have been in the jungle.

• He should have been moving quietly in the shadows near some water hole, and waiting for a plump deer to pass that way.

• But sadly he is locked behind bars in a concrete cell.

• He does not take any notice of the visitors and is looking at the brilliant stars with his brilliant eyes.

• Now, he has to be content with merely looking at the stars .

• He is no longer free to move in his natural habitat and look at the brilliant stars from there .

↧
↧

Summary of Chapter 6 The Browning Version NCERT Class 11th English

February 21, 2019, 12:26 pm
$
0
0

Summary of Chapter 6 The Browning Version Role (Summary, Character Sketch and Word Meanings) Class 11 Hornbill

Summary of the Chapter

 This one-act-play is a story of two school teachers and a student named Taplow. Mr. Crocker-Harris is a middle aged teacher while Mr. Frank is a young colleague of Mr. Harris. Taplow is a student of lower fifth class. He is a boy of 16. Both the teachers provide a striking contrast. They have only one thing in common. They belong to the same school.

Young Frank is quite open to his students. He doesn't keep any distance while dealing with them. Taplow is not a student of his class. But still he takes a lot of interest in him. His long conversation with Taplow reveals his open nature. On the other hand, Mr. Crocker-Harris is quite reserved. He doesn't mix up with his students. He maintains a respectable distance with them.

Mr. Frank doesn't believe in observing formalities. He cares little regarding rules and regulations. Mr. Crocker-Harris follows them.

Mr. Crocker-Harris is feared and even respected. He has a wonderful hold over his students. They are scared of him. Mr. Frank admits that he is envious of Mr. Crocker-Harris. Perhaps he lacks that 'effect' which Harris has left over his students. Mr. Frank encourages Taplow to criticise Crocker-Harris. Actually, he urges him to imitate Harris. This clearly reveals the working of his mind. He even asks Taplow to 'cut' Crocker-Harris. He lacks Harris' devotion. He teaches science but shows no interest in his subject. This shows his lack of dedication and commitment towards his profession.

Taplow says that Crocker-Harris is shriveled inside like a nut. It is quite true. He is not open. He doesn't like flattery. Nor does he like anyone who likes him. He is a hard task-master. He gives extra work to Taplow even on the last day of school. This he does to punish Taplow for being absent for a day last week. Sometimes he cracks jokes. His 'classical' jokes lack humour. No one understands them except him. They are as dry and humourless as he himself is. But in spite of everything he likes Crocker-Harris. Taplow, however, praises Frank as a teacher. He says that he is young.

Millie Crocker Millie Crocker is the wife of Crocker-Harris. She is a thin woman in her late thirties. She is rather more smartly dressed than the `general run of schoolmasters' wives'. Both Frank and Taplow feel her presence. Taplow feels uneasy. He fears that perhaps she heard him talking ill of Mr. Harris. She sends Taplow to a chemist. She agrees to take the blame on her in case Mr. Crocker-Harris comes in Taplow's absence.

NCERT Solutions of The Browning Version
↧

Notes of Ch 11 Marketing| Class 12th Business Studies

February 24, 2019, 8:36 am
$
0
0

Summary and Notes of Ch 11 Marketing| Class 12th Business Studies 

Meaning of Market

Market- Market refers to the place where buyers and sellers gather to enter into transactions involving the exchange of goods and services.

Marketing- Marketing has been referred to as performance of business activities that direct the flow of goods and services from producers to consumers.

Following important features of marketing

• Needs and Wants: The process of marketing helps individuals and groups in obtaining what they need and want.

• Creating a Market Offering: creating offers like size, quality, taste available at a given outlet or location.

• Customer Value: The process of marketing facilitates exchange of product and service between buyers and sellers.

• Exchange Mechanism: The process of marketing involves exchange of products and services for money or something.

Following conditions are satisfied for exchange:

• Involvement of two parties.

• Each party should be capable of offering something of value to the other.

• Each party is free to either accept or reject the offer of other party.

• The parties should be willing to enter into transaction with each other.

• Each party should have the ability to communicate and deliver the product or service.

What can be marketed?

• Physical products, services, ideas, persons place, Experience, Properties, Events, information, organisations.

• Marketing Management

It refers to planning, organising, directing and control of the activities which facilitate exchange of goods and services between producers and consumers or users of products and services.

Difference between selling and marketing

Basis 
Selling
Marketing
Starting pointThe selling starts after production has taken place.It starts before goods are produced.
Means It is achieved through advertisement personal selling and sales promotion.Various marketing activities.
Means It is achieved through advertisement personal selling and sales promotion.Various marketing activities.
Main focus Attracting consumers.Consumer satisfaction.
Ends/profit maximisation  Profit maximisation is achieved through volume of sales.Profit maximisation is achieved through customer satisfaction.

Function of marketing

The following activities are undertaken in respect of the exchange of the goods and services.

• Gathering and Analysing Market Information- It is an important function of marketing. Through it easily identifies the needs of customers and also takes various decisions related to outside marketing, like latest trend, strengths and weaknesses of the organisation.

• Marketing planning- To develop an appropriate marketing plans, so that the marketing objectives of the organisation can be achieved.

• Product Designing and Development- Product designing plays an important role in product selling. The company whose product is better and attractively designed sells more than the product of those companies whose design happens to be weak and unattractive.

• Standardisation and Grading- Standardisation refers to determine of standard regarding size, quality, design, colour, weight and raw material to be used. Grading is the process of classification of products into different groups on the basis of quality, size, design etc.

• Packaging and labelling- Packaging refers to designing and developing the package for the products. Labelling refers to designing and developing the label to be put on the package.

• Branding - Branding is very important decision of marketing because brand gives a product differentiation.

• Customer Support Services-It refers to service which is given to the customers such as sales services, handling customer complaints and adjustments. All these services aim at providing maximum satisfaction to the customers.

• Transportation- It means physical movement of goods and services from one place to another place.

• Storage or warehousing- In order to maintain smooth flow of products in the market, there is a need for proper storage of the products. Further, there is a need for storage of adequate stock of goods to protect against unavoidable delays in delivery or to meet ‘out contingencies in the demand. Wholesalers and retailers are playing an important role.

Marketing mix

Marketing mix refers to the sum total of policies formulated with a view of successfully completing the different marketing activities. The main objective of this policy may be to attract all types of customers.

Elements of Marketing Mix

• Product – It refers to all decisions relating to the product(branding, packaging, labelling, colour)

• Price-Price mix refers to all those decisions which are related to price fixation of product.

• Place –It refers to activities that make firm’s products available to the targets customers.

• Promotion –It refers to combination of promotional tools used by the business to inform and get settled customers about the products.
Product – Anything that can be value to the buyer can be termed as a product.

Shopping Efforts Involved

The consumer products are classified into three categories:

•Convenience Products-Those products which are purchased frequently, immediately and with least time and efforts are referred to as convenience goods example- cigarettes, ice-creams, medicines, newspaper, stationary items.

•Shopping products-It is those products which buyers devote considerable time, to compare the quality, price, style, suitability, etc.., at several stores before making final purchase.

•Speciality Products-It is a high level product which has brand loyalty of the highest order, with a significant number of buyers.
B. Durability of products

On the basis of durability consumer products have been classified into three categories:

• Non-durable Products: The consumer products which are normally consumed in one or few uses are called non- durable products example-toothpaste, detergents, bathing soap, kitchen products, stationary products etc.

• Durable Products: The consumer products which are generally used for a longer period, command a higher per unit margin, required greater personal-selling efforts, guarantees etc. Example-radio, bicycle, sewing machine.

• Services: It is intangible in form and gives a customer benefits or satisfaction, which are offered for sale. Examples: dry cleaning, watch repairs, hair cutting, postal- services.

• Industrial Products: Industrial products are used as a raw material for producing other products in industry. The market for industrial products consists of manufacturers, transport agencies, banks and insurance companies, mining companies and public utilities.

Classification of Industrial Products

•Materials and Parts: It includes all those goods which are completely entered in manufacturer’s products. Such goods are of two types:

•Raw material, manufactured material and parts.

•Capital Items: These are such goods that are used in the production of finished goods. These include:
(a) Installations like elevators, mainframe computers
(b) equipment like Hand Tools, Personal Computer, Fax Machine, etc.

• Branding: The process of giving a name or a sign or a symbol etc., to a product is called branding.
The various terms relating to branding are as follows:

•Brand: Brand is used to identify the products and differentiate them from those of the competitors. A brand is a name, term, sign, symbol, design.

•Brand Name: The name of product which can be spoken is called a brand name. Example- Asian Paints, Safola, Maggie.

• Brand Mark: That part of a brand which can be recognised but which is not utter able is called brand mark.

•Trade Mark: A Trade Mark is a part of a brand which gives legal protection of brand name; brand mark etc. so, no other firm could use it. Trade mark provides several advantages to the sellers as well as the consumers.

Advantages to the Marketers

•Branding helps in product differentiation from other competitors.

•Branding helps in advertisement of product and help in display of programmes.

•Branding enables firms to charge different price for its products than that charged by its competitors.

•Branding helps in introduction of new products in the same name.

Advantages to Customers

•Branding helps the customers in identifying the products easily and they satisfied with a particular brand of a product.

•The quality of branded product undoubtedly is better.

•Branding gives a status symbol because of their quality.

Characteristics of a Good Brand Name

•The good brand name should be simple and short, easily pronounceable, suggestive, distinctive.

•The good brand name should be capable of being registered and protected legally.
Packaging: It refers to the groups of those activities which are related to the designing and production of containers.

Levels of Packaging

There are generally three levels of Packaging:

•Primary Package: Primary package is very close to the product. It keeps till the consumer is ready to use and also keeps throughout the entire life of the product.

•Secondary Packaging: Secondary Packaging refers to additional layers of protection that are kept till the product is ready to use.

•Transportation Packaging: Transportation Packaging refers to that packaging which is necessary to use for transportation, identification and storage.

Importance of Packaging

•When customers purchased packed products. They feel safe and hygienic because the chances of adulteration in such packed goods are minimised.

•Self service outlets are very popular, particularly in major cities and towns.

•With the help of Packaging many products are kept for several days.

•Packaging creates product differentiation.

Functions of packaging

•Packaging helps in identification of product.

•Packaging protects the product from spoilage, breakage, leakage.

•The size and shape of the package should be such that it should be convenient to open, handle and use for the consumers.

•Packaging is also used for promotion purpose.

Labelling: It is a process in which companies prepare a label.

Function of labelling

•It provides complete information related to the product like weight, price and ingredients of product.

• It makes easier to identify a particular product among many with the help of labelling.

•It helps in grading of the products.

•It helps in promotion of products.

•It provides information related to the products required by law.

Pricing

The value which is paid for some product or services is known as pricing.

Factors Affecting Price Determination

•Product Cost: Cost of production is the main component of price. This includes the cost of producing, distributing and selling the product. There are broadly three types of costs:

→ Fixed costs- Fixed costs are those costs which do not vary with the level of activity of a firm. Example, Rent of building, salary of a sales manager.

→ Variable costs: Variable Costs are those cost which vary in direct proportion with the level of activity.

→ Semi-Variables costs: Those costs which vary with the level of activity but not in direct proportion.

•The Utility and Demand:  If demand of a product is relatively more than supply than the higher price should be fixed.

•Extent of Competition in the Market: The price will tend to reach the upper limit in case there is lesser degree of competition while under conditions of free competition, the price will tend to be set at the lowest level.

•Government and Legal Regulations: Some product’s price will be fixed by government because of unfair practices in the field of price fixing.

•Marketing Method Used: Price Fixation process is also affected by other elements of marketing such as distribution system, quality of salesman employed Salesmen promotion.

Physical Distribution

It is concerned with making the goods and services available at the right place and right time so people easily purchase.

Channels of Distribution

Channel of Distribution refers to that path through which the products reach the consumers with the help of intermediaries. The channels bring economy of effort. It includes mainly Agents, wholesalers and retailers.

Functions of Distribution Channels

•Sorting: Middlemen supplies of goods from variety of sources, which is often not of the same nature, quality, size etc.

•Accumulation: This function involves accumulation of goods into larger homogeneous stocks, which help in maintaining continuous flow of supply.

•Allocation: Allocation involves breaking homogenous stock into smaller, marketable lots.

•Assorting: Middleman build assortment of products for resale.

•Product Promotion: Sales promotion and advertising are main idea for product promotion.

•Negotiation: Both parties can negotiate with each other related to price, quality, guarantee and other related matters.

Types of Channels

Direct Channel or zero level channels

It is the shortest mode of distribution in which goods are directly available by the manufactures to customers, without involving any intermediaries.

Indirect Channels

When a manufacturer gets the help of one or more middlemen to move goods from the production place to the place of consumption, this distribution network is called indirect channels.

This may take any of the following forms:

•Manufacturer-Retailer Consumer( one Level channel)

•Manufacturer-Wholesaler-Retailer Consumer( Two level Channel)

•Manufacturer- Agent- wholesaler-Retailer-Consumer(Three Level Channel)

Factors Determining Choice of Channels

The choice of channels depends on various factors:

•Product Related Factors: The choice also depends on whether the product is perishable in nature or non- perishable.

•Company Characteristics: The choice of channel also depends on the financial conditions of the company.

•Competitive Factors: The choice of channel also depends on the channels which are elected by competitors in the same industry.

•Market Factors: The choice of channel also depends on the size of market, geographical concentration of potential buyers and quantity purchased like when a number of buyers are small then a company adopt short channel of distribution.

•Environmental Factors: Other important factors affecting the choice of channels of distribution include environmental factor such as economic conditions and legal constraints. Depressed economy marketers use shorter channels to distribute their goods.

Physical Distribution

The physical handling and movement of goods from place of production to the place of distribution is known as physical distribution. Physical distribution covers all those activities which are related to movement of goods from manufacturers to the customers.

Components of Physical Distribution

•Order Processing: It is a first step of physical Distribution. Order processing means the process which is followed to fulfil the material order of the customer.

•Transportation: Transportation is the means of carrying goods and raw materials from production place to sale point.

•Warehousing: warehousing is the place where company stores goods.

•Inventory Control: It means the stock of raw material, semi-finished goods and finished goods held in anticipation of sales or use.

Advertising

It is that art by means of which people are informed about a particular idea/service/product so as to induce them to buy that product or services or ideas.

The features of advertising:

•Paid form: It is a paid form of communication means sponsor has to bear the cost of communicating with the prospectus.

•Impersonality: It means there is no a personal communication and relation between prospect and the advertiser. Advertising creates a monologue and not a dialogue.

•Identified Sponsor: It means advertising is taken by some identified individual or company.

Merits of advertising

•Advertising is a medium through which a large number of people can be reached over a vast geographical area.

•Advertising enhance customer satisfaction and confidence about the quality and quantity.

•It a forceful medium of communication with a special effect of even simple product.

•It is a very economical mode of communication in which when it spread over the numerous number of people, as a result more people know about the product.

Limitations of advertising

•It is less forceful than the personal selling as there is no compulsion on the prospectus to pay attention to the message.

•The evaluation of the effectiveness of advertising message is very difficult as there is no immediate and accurate feedback mechanism of the message that is delivered.

•Advertising is less flexible as comparison to the personal sales because customers take message according to the own use.

•When the advertising message is expanded more and more it is difficult to make advertising message heard by target which is set by organisation.

Objections to advertising

•An organisation has to spend large amount on advertising, it increases the cost of the products.

• Advertising undermines social values and promotes materialism, It breed discontentment among people as they come to know about new products and feel dissatisfied with their present state of affairs. This criticism is not entirely time. Advertisement in fact helps buyer by informing them about the new products which may be improvement over the existing products

•Another criticism against advertisement is that so many products are being advertised which makes similar claims that the buyer gets confused as to which one is true and which one should be relied upon.

•Advertisement encourages a sale of inferior product even consumer pays a handsome amount of inferior products.

•Some advertisements are bad in taste they show something which is not acceptable like women dancing, man without shirt.

Personal Selling

It involves personal presentation of message in the form of conversation with one or more prospective customers for the purpose of making sales.

Features of personal selling

•Personal form: Under personal selling, seller directs face-to-face contact between the buyers for making a sale.

•Development of Relationship: Personal selling results in the development of personal relationship between the sales person and possible buyer.

Merits of Personal Selling

•Flexibility: There is lot of flexibility in personal selling because the sales presentation can be adjusted to fit the specific needs of the individual customers.

•Direct Feedback: When seller direct face to face presentation in front of consumer it is possible to take a direct feedback from customers.

•Minimum Wastage: The wastage of efforts in personal selling can be minimised as company can decide the target customers before making any contact with them.

Role of Personal Selling

Personal selling plays a very important role in the marketing of goods and services. It is important for businessmen, customers and society.

Important to Businessmen

•Effective Promotional Tool: Personal selling is very effective promotional tool, which helps in influencing the prospectus about the merits of a product.

•Flexible Tool: Personal Selling helps business persons in adopting their offer in varying purchase situations.

• Minimises Wastage of Efforts: When company compared with other promotional tools, the possibility of wastage of efforts in personal selling is minimum. It helps the business persons in bringing economy in their efforts.

•Consumer Attention: With the help of personal selling a seller easily detect the loss of consumer attention and interest in a personal selling situation. This may help the business person in successfully completing the sale.

•Lasting Relationship: Personal selling helps to develop lasting relationship between the sales persons and the customers, which is very important for achieving the objectives of business.

•Personal Rapport: Development of personal rapport with customers increases the competitive strength of a business organisation.

•Role in Introduction Stage: It plays very important role in the introduction stage of a new product.
Importance to Customers

•Help in identifying Needs: Personal selling helps the customers in identifying their needs and wants.

•Latest Market Information: Customers get latest market information about product like price, their use, product availability and storage, new product introduction; with help them in taking the purchase decisions in a better way.

•Expert Advice: With the help of personal selling customers get expert advice and guidance in purchasing various goods and services, which help them in making better purchase.

•Motivate Customers: Personal selling motivate customers to purchase new products.
Importance to society

•Converts Latest Demand: With the help of personal selling economic activity in the society is fostered, leading to more jobs, more incomes and more products and services.

•Employment Opportunities: Personal selling helps to create more opportunities to the unemployed youth.

•Career opportunities: Personal selling provides greater carrier opportunity, job satisfaction as well as security, respect, variety, interest and independence to young men and women.

•Product Standardisation: Personal selling creates product standardisation and uniformity in consumption pattern in a diverse society.

Sales Promotion

It is short-term incentives which are designed to encourage the buyers to make immediate purchase of a product or service. Sales promotion activities include offering cash discounts, sales contest and free gift offers etc.

Merits of Sales Promotion

•Sales promotion activities attract attention of the people because of the use of incentives.

•At the time of the launching a new product, sales promotion tools can be very effective.

•It is designed to supplement the personal selling and advertisement efforts used by a firm.

Limitations of Sales Promotion

•Sales promotion tools may affect the image of a product than buyers start feeling dissatisfied with product’s quality and price.

•If a firm have full faith on sales promotion, it may give sometimes bad impression that it is unable to manage its sales target by firms.

Commonly used sales Promotion Activities

Some common used sales Promotion Activities are Rebate, Discount, Refunds, Product Combinations, Quantity Gift, Instant Draws and Assigned Gift, Lucky Draw, Usable Benefit, Full finance@ 0%, Sampling, Contests.

Publicity

•It is a non-personal form of communication.

•It is a non- paid form of communication. It does not involve any direct expenditure by the marketing firm.

•It generally takes place when favourable news is presented in the mass media about a product or service.

Public Relations

•The main affairs of Public Relations are managing public opinion of an organisation.

•Their task is to disseminate information and build goodwill about the business.

•They also advise top management to adopt certain programmes which will add to their public image and ensure that negative publicity does not take place at all.
↧

Summary of How to Tell Wild Animals by Carolyn Wells NCERT Class 10th

February 24, 2019, 2:46 pm
$
0
0

Summary of How to Tell Wild Animals by Carolyn Wells with Quick Revision Notes  First Flight Class 10

Summary of the Poem The Tiger

The Tiger by Carolyn Wells

If ever you should go by chance
To jungles in the east;
And if there should to you advance
A large and tawny beast,
If he roars at you as you’re dyin’
You’ll know it is the Asian Lion....

In this stanza, the poet says that this is an extreme situation to identify a lion. The poet says that if the roar of the animal is so fearsome that you feel like you are dying then you can be sure that you are near a lion.

Or if some time when roaming round,
A noble wild beast greets you,
With black stripes on a yellow ground,
Just notice if he eats you.
This simple rule may help you learn
The Bengal Tiger to discern.

In this stanza, the poet says that what to do if you see a wild animal with black stripes on yellow body. If he starts eating you then you can be sure that it is a tiger. But it is better not to wait to be a tiger’s lunch to identify it.


If strolling forth, a beast you view,
Whose hide with spots is peppered,
As soon as he has lept on you,
You’ll know it is the Leopard.
’Twill do no good to roar with pain,
He’ll only lep and lep again.

In this stanza, the poet asks what do you think if you happen to see an animal with spots which gives it a salt and pepper appearance, wait until he leaps on you. Because no matter how hard you cry, it will continue pouncing on you. So, be careful and don’t allow it to leap on you.

If when you’re walking round your yard
You meet a creature there,
Who hugs you very, very hard,
Be sure it is a Bear.
If you have any doubts, I guess
He’ll give you just one more caress.

In this stanza, the poet says that years are thought to be good wrestlers and can give a really tight hug. Although a friendly hug is referred to as bear hug. But if a real bear will hug you, then it may not feel friendly at all.

Though to distinguish beasts of prey
A novice might nonplus,
The Crocodile you always may
Tell from the Hyena thus:
Hyenas come with merry smiles;
But if they weep they’re Crocodiles.

In this stanza, the poet describes that some animals are famous for their weird behaviour. For example, a laughing hyena’s voice resembles human’s laughing sound. A crocodile is said shed tears while eating its pray. So don’t wait for a hyena to laugh or for a crocodile to weep.

The true Chameleon is small,
A lizard sort of thing;
He hasn’t any ears at all,
And not a single wing.
If there is nothing on the tree,
’Tis the chameleon you see.

A chameleon or garden lizard is an expert at camouflage. If you are unable to see a thing on tree then chances are chameleon is sitting there. As it changes colour as per its surrounding it is difficult to see. This capacity of camouflage helps the lizard in saving it from hunters.

Quick Revision Notes

• If by chance you go to any forest in the East and if a huge terrible animal comes to you and if it roars loudly and if you are dying then you will come to know that it is an Asian Lion.

• If a wild animal with black stripes on yellow hide welcomes you and eats you then this simple rule will teach you that it is a Bengal tiger.

• When strolling if you see whose hide is covered with spots and it jumps at you and starts eating you then you understand that it is a leopard.  There will be no use of shouting with pain because it will continuously eat you.

• If you are just walking around in your yard and an animal meets you and it hugs you tightly then believe that it is a bear.

• If you have any doubt then I think that it will kiss you, a novice may be confused to recognize animals.

• You can differentiate between crocodile and hyena. Hyena always smiles but if tears come from its eyes then it is a crocodile.

• An original chameleon is like a lizard.  It doesn’t have ears or wings.  If there is nothing on the tree then you will find there a chameleon.
↧
Search

Summary of The Ball Poem by John Berryman NCERT Class 10th

February 27, 2019, 11:37 am
$
0
0

Summary of The Ball Poem by John Berryman with Quick Revision Notes  First Flight Class 10

Summary of the Poem The Tiger

The Tiger by Carolyn Wells

What is the boy now, who has lost his ball,
What, what is he to do? I saw it go
Merrily bouncing, down the street, and then
Merrily over — there it is in the water!
No use to say ‘O there are other balls’:
An ultimate shaking grief fixes the boy
As he stands rigid, trembling, staring down
All his young days into the harbour where
His ball went. I would not intrude on him;
A dime, another ball, is worthless. Now
He senses first responsibility
In a world of possessions. People will take
Balls, balls will be lost always, little boy.
And no one buys a ball back. Money is external.
He is learning, well behind his desperate eyes,
The epistemology of loss, how to stand up
Knowing what every man must one day know
And most know many days, how to stand up.

The poem is about losing something that you love, and learning to grow up without it. It is about a little boy who for the first time in his young life, is learning what it is like to experience grief at the loss of a much beloved possession-his ball. To us, the loss of a ball is of minor consequence, and our reaction to it is to say 'O there are other balls'. But to a little boy, this is not so. A dime, another ball, is worthless. Money is external; it cannot buy back our love, nor replace the things that we love: the things that really matter.
In this poem, the boy's ball personifies his young days and happy innocence. In this world, people will take balls just as they will take away our innocence and force us to grow up. And once we lose this innocence, we can never get it back. Balls will be lost always, little boy, and no one buys a ball back. This poem goes to show how, all throughout your life, you will be forced to do things that you don't want to do; and you will lose or have to give up the things that you love. But, despite this, you have to learn to stand up - to be strong and get on with your life - no matter how much it hurts inside. Because that is the only way you will survive; you have to learn to accept and let go - and not cling onto something that you can never have.
The poet uses imagery when describing how the ball personifies the spirit of the boy's childish innocence. In the last five lines, we visualize how the spirit of this little boy, like the ball, is sinking into the dark waters of the harbour. As it drifts further away, the boy learns to grow up, and that part of him that is linked to that ball grows up as well, until it is no longer a little boy. 

Quick Revision Notes

• The little boy has lost his ball.

• He can buy many new balls but he has a very deep emotional attachment and memories associated with that ball.

• So, he regrets over the loss of the ball.

• The poet is indirectly trying to tell us that we should learn how to cope up with the loss.

• Money is external which means that we can’t buy memories with money.

• We all should learn our responsibility and learn to cope up with the loss.
↧
↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.1

March 2, 2019, 8:59 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.1

1. Which of the following are quadratic equations ?

Solution


2. In each of the following , determine whether the given values are solutions of the given equation or not :

Solution

(i)

(ii) 

(iii)

(iv)

(v)

(vi)

(vii)

3. In each of the following , find the value of k for which the given value is a solution of the given equation :

Solution

(i)

(ii) 

(iii) 

(iv) 

4. Determine if, 3 is a root of the equation given below:

Solution


5. If x = 2/3 and x = -3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b .   

Solution

We know that x = -2/3 and x = -3 are the roots of the equation ax2 + 7x + b. When we the value of x in this equation, we will get the result as 0.
Putting x = -2/3
a(2/3)2 + 7(2/3) + b = 0
⇒ 4/9 a +14/3 + b =0
⇒ 4/9 a  + b = -14/3 --- (i)

Putting x = -3
a(-3)2 + 7(-3) + b = 0
⇒ 9a - 21 + b =0
⇒ 9a  + b = 21 --- (ii)

Subtracting (i)  from (ii), we get
9a  + b - (4/9 a  + b) = 21 + 14/3
⇒ 9a  + b - 4/9 a  -b = 77/3
⇒ 77/9 a = 49/3
⇒  a = 77/3 × 9/77
⇒  a = 3

Now, putting a =3 in equation (ii),
9×3 + b = 21
⇒ 27 + b = 21
⇒  b = 21 - 27
⇒  b = -6

Thus, a = 3 and b = -6

↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.2

March 3, 2019, 12:07 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.2

1. The product of two consecutive positive integers is 306. From the quadratic equation to find the integers , if x denotes the smaller integer.

Solution


2. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each , and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if john had x marbles .

Solution


3. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.

Solution


4. The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.

Solution


5. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 1 1 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.

Solution


6. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train

Solution

↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.3

March 3, 2019, 11:43 pm
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.3

Solve the following quadratic equations by factorization: 

1. (x+4)(x+2) = 0 

Solution

We have,
(x-4) (x+2) = 0
⇒ either (x-4) = 0 or (x+2) = 0
⇒ x = 4 or x = -2
Thus, x = 4 and x = -2 are two roots of the equation (x-4) (x+2) = 0

2. (2x+3)(3x-7) = 0

Solution


3. 3x2 – 14x – 5 = 0 

Solution


4. 9x2-3x-2 = 0

Solution


5. 1/(x-1) - 1/(x+5) = 6/7, x≠1, -5

Solution


6. 6x2 + 11x + 3 = 0


Solution


7. 5x2 – 3x – 2 = 0

Solution


8. 48x2 - 13x – 1= 0

Solution


9. 3x2 = -11x – 10



Solution


10. 25x(x+1) = -4


Solution


11. 16x - 10/x = 27

Solution

12. 1/x - 1/(x-2) = 3, x ≠ 0, 2

Solution


13. x - 1/x = 3, x ≠ 0

Solution


14. 1/(x+4) - 1/(x-7) = 11/30, x ≠ 4, 7

Solution


15. 1/(x-3) + 2/(x-2) = 8/x, x ≠ 0, 2, 3

Solution 

.
16. a2x2-3abx + 2b2 = 0

Solution


17. 9x2 – 6b2x-(a4 - b4) = 0


Solution


18. 4x2 +4bx – (a2 – b2)

Solution



19. ax2 + (4a2 – 3b) x – 12ab = 0

Solution



20 . 2x2 + ax-a2 = 0

Solution


21. 16/x - 1 = 15/(x+1), x≠ 0, -1

Solution


22. (x+3)/(x+2) = (3x-7)/(2x-3), x ≠ 2, 3/2

Solution


23. 2x/(x-4) + (2x-5)/(x-3) = 25/3, x ≠ 3, 4

Solution


24. (x+3)/(x-2) - (1-x)/x = 17/4, x ≠ 0, 2

Solution


25. (x-3)/(x+3) - (x+3)/(x-3) = 48/7, x ≠ 3, x ≠ -3

Solution


26. 1/(x-2) + 2/(x-1) = 6/x, x ≠ 0

Solution


27. (x+1)/(x-1) - (x-1)/(x+1) = 5/6, x ≠ 1, -1

Solution


28. (x-1)/(2x+1) + (2x+1)/(x-1) = 5/2, x ≠ -1/2, 1 

Solution


29. 4/3 - 3 = 5/(2x+3), x ≠ 0, -3/2

Solution


30. (x-4)/(x-5) + (x-6)/(x-7) = 10/3; x≠5, 7

Solution


31. (x-2)/(x+3) + (x-4)/(x-5) = 10/3; x≠3,5

Solution

32. (5+x)/(5-x) - (5-x)/(5+x) = 3¾; x≠ 5, -5

Solution


33. 3/(x+1) - 1/2 = 2/(3x-1), x≠-1, 1/3

Solution


34. 3/(x+1) + 4/(x-1) = 29/(4x-1), x≠ 1, -1, 1/4

 Solution


35. 2/(x+1) + 3/2(x-2) = 23/5x, x≠ 0, -1, 2

Solution


36. x2 - (√3 +1) x + √3 = 0

Solution


37. 3√5x2 + 25x - 10 √5 = 0

Solution


38. √(3x2) - 2√2x - 2 √3 = 0

Solution

 
39. 4√3x2 + 5x - 2 √3 = 0

Solution


40. √2x2 - 3x - 2√2 = 0

Solution 


41. x2 - (√2 + 1)x + √2 = 0

Solution


42. 3x2 - 2√6 x + 2 = 0

Solution


43. x2 - 4√2 x + 6 = 0 

Solution


44. m/n x2 + m/n = 1- 2x

Solution


45. (x-a)/(x-b) + (x-b)/(x-a) = a/b + b/a

Solution


46. 1/(x-1)(x-2) + 1(x-2)(x-3) + 1/(x-3)(x-4) = 1/6

Solution


47. a/(X-b) - b/(x-a) = 2, x ≠ a,b 

Solution


48. (x+1)/(x-1) + (x-2)/(x+2) = 4 - (2x+3)/(x-2); x≠ 1,-2, 2

Solution


49. a(x-a) + b/(x-b) = 2c/(x-c)

Solution


50. x2 + 2ab = (2a+b)x

Solution


51. (a+b)2x2 - 4abx - (a-b)2 = 0

Solution


52. a(x2+1) - x(a2+1) = 0 

Solution


53. x2 - x - a(a+1) = 0 

Solution 


54. x2+ (a+1/a)x + 1 = 0 

Solution


55. abx2+ (b2-ac)x - bc = 0

Solution


56. a2b2x2 + b2x - a2x -1 = 0

Solution


57. (x-1)/(x-2) + (x-3)/(x-4) = 3⅓; x≠ 2, 4

Solution


58. 1/(2a+b+2x) = 1/2a + 1/b + 1/2x

Solution


59. 3{(3x-1)/(2x+3) -2{(2x+3)/3x-1)} = 5; x≠ 1/3, -3/2

Solution


60. 3{(7x-1)/(5x-3)} - 4{(5x-3)/(7x+1)} = 11; x≠ 3/5, -1/7

Solution


61. (x-5) (x-6) = 25/(24)2

Solution


63. 7x + 3/x = 35 3/5

Solution

↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.4

March 4, 2019, 9:10 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.4

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

1. x2 - 4√2x + 6 = 0

Solution



2. 2x2 - 7x + 3 = 0

Solution



3. 3x2 + 11x + 10 = 0

Solution



4. 2x2 + x - 4 = 0

Solution



5. 2x2 + x + 4 = 0

Solution



6. 4x2 + 4√3 x + 3 = 0

Solution



7. √2 x2 -3 x - 2√2 = 0

Solution



8. √3 x2 + 10 x + 7√3 = 0

Solution


9. x2 - (√2 + 1) x + √2 = 0

Solution



10. x2 - 4ax + 4a2 = 0

Solution

↧
↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.5

March 4, 2019, 11:23 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.5

1. Write the discriminant of the following quadratic equation:

(i) 2x2 - 5x + 3 = 0

Solution


(ii) x2 + 2x + 4 = 0

Solution


(iii) (x-2) (2x-1) = 0

Solution


(iv)  x2 - 2x + k = 0, k∈R

Solution


(v) √3 x2 + 2√2 x + 2√3 = 0

Solution


(vi) x2 - x + 1 = 0

Solution


(vii) 3x2 + 2x + k = 0

Solution


(viii) 4x2 - 3kx + 1 = 0

Solution


2. In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

(i) 16x2 = 24x + 1

Solution


(ii) x2 + x + 2 = 0

Solution


(iii) √3 x2 + 10x - 8√3 = 0

Solution


(iv)  3x2 - 2x + 2 = 0

Solution


(v) 2 x2 - 2√6 x + 3 = 0

Solution


(vi) 3a2x2 + 8abx + 4b2 = 0, a≠0

Solution


(vii) 3x2 + 2√b x - b = 0

Solution


(viii) x2 - 2x + 1 = 0

Solution


(vii) 2x2 + 5√3 - 6 = 0

Solution


(x) √2 x2 + 7x + 5√2 = 0

Solution

   

(xi) 2x2 + 2√x + 1 = 0

Solution 


(xii) 3x2 + bx + 2 = 0

Solution


3. Solve for x : 

(i) (x-1)/(x-2) + (x-3)/(x-4) = 3⅓; x ≠ 2, 4

Solution


(ii) 1/x + 2/(2x-3) = 1/(x-2); x ≠ 0, 3/2, 2

Solution

(iii) x + 1/x = 3; x ≠ 0

Solution

(iv) 16/x -1 = 15/(x+1), x ≠ 0, -1

Solution


(v) 1/(x-3) - 1/(x+5) = 1/6, x ≠ 3, -5

Solution

 
↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.6

March 6, 2019, 5:11 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.6

1. Determine the nature of the roots of the following quadratic equations:

(i) 2x2 - 3x + 5 = 0

Solution


(ii) 2x2 - 6x + 3 = 0

Solution

The given quadric equation is 2x2
Here , a = 2, b = -6 and c = 3
As we know that D = b2 – 4ac
Putting the value of a = 2, b = -6 and c = 3
= (-6)2 - 4 × 2 × 3
= 36 – 24
= 12
Since, D > 0
Therefore, root of the given are real and distinct .


(iii) 3/5 x2 - 2/3 x + 1 = 0

Solution



(iv) 3x2 - 4√3 x + 4 = 0

Solution




(v) 3x2 - 2√6 x + 2 = 0

Solution



2. Find the values of k for which the roots are real and equal in each of the following equation: 

(i) kx2 + 4x + 1 = 0

Solution


(ii) kx2 - 2√5 x + 4 = 0

 Solution


(iii) 3x2 - 5x + 2k = 0

Solution

(iv) 4x2 + kx + 9 = 0

Solution


(v) 2kx2 - 40 x + 25 = 0

Solution


(vi) 9x2 - 24x + k = 0

Solution


(vii) 4x2 - 3kx + 1 = 0

Solution

(viii) x2 - 2(5+2k)x + 3(7+10k) = 0

Solution


(ix) (3k+1)x2 + 2(k+1) x + k = 0

Solution


(x) kx2 + kx + 1 = -4x2 - x

Solution


(x) (k+1)x2 + 20 = (k+3)x + (k+8) = 0 

Solution


(xii) x2 - 2kx + 7k - 12 = 0

Solution




(xiii) (k+1)x2 - 2(3k+1)x + 8k + 1 = 0

Solution




(xiv) 5x2 - 4x + 2 + k(4k2 - 2x + 1) = 0

Solution




(xv) (4-k)x2 + (2k + 4)x + (8k + 1) = 0

Solution




(xvi) (2k+1)x2 + 2(k+3)x + (k+5) = 0

Solution




(xvii) 4x2 - 2(k+1)x + (k+4) = 0

Solution




3.In the following determine the set of values of k foe which the green quadratic equation has
real roots:
(i) 2x2 + 3x + k = 0

Solution



(xvii) 2x2 + x + k = 0

Solution


(iii) 2x2 - 5x - k = 0

Solution


(iv) kx2 + 6x + 1 = 0

Solution


(v) 3x2 + 2x + k = 0

Solution


4. Find the least positive value of k for which the equation have real and equal roots : 

Solution

(i) x2 - 2(k+1)x + k2 = 0

Solution

(i) The given quadric equation is x2 -2 (k+1) x+k2 = 0, and roots are real and equal . Then find the value of k.
Here ,
a = 1, b = -2(k+1) and, c = k2
As we know that D = b2 – 4ac
Putting the value of a = 1, b = -2 (k+1) and, c =
= {-2(k+1)}2 -4 × 1 × k2
= {4(k2+2k+1)} – 4k2
= 4k2 + 8k + 4 – 4k2
= 8k + 4
The given equation will have real and equal roots, if D = 0
8k + 4 = 0
8k = -4
k = -4/8
= -1/2
Therefore, the value of k = -1/2

(ii) k2x2 - 2(2k-1)x + 4 = 0 

Solution

The given quadric equation k2x2-2(2k-1)x+4 = 0 , and roots are real and equal 
Then find the value of k.
Here,
a = k2, b = -2(2k-1) and , c = 4
As we know that D = b2 – 4ac
Putting the value of a = k2 , b = -2(2k-1) and, c = 4
= {-2(2k-1)}2 -4 × k2 × 4
= {4(4k2 – 4k + 1)} – 16k2
= 16k2 – 16k + 4 – 16k2
= -16k + 4
The given equation will have real and equal roots , if D = 0
-16k + 4 = 0
16k = 4
k =4/16
= 1/4
Therefore , the value of k = 1/4


(iii) (k+1)x2 - 2(k-1)x + 1 = 0 

Solution

The given quadratic equation is (k+1)x2 – 2 (k-1)x+1 = 0, and roots are real and equal
Then find the value of k .
Here,
a = k+1, b = -2(k-1) and , c = 1
As we know that D = b2 – 4ac
Putting the value of a = k+1, b = -2(k-1) and , c = 1
= {-2(k-1)}2 – 4 × (k+1) × 1
= {4(k2 – 2k + 1)} – 4k – 4
= 4k2 – 8k + 4 – 4k – 4
= 4k2 – 12k = 0
The given equation will have real and equal roots, if  D = 0
4k2 – 12k + 0 = 0
4k2 – 12k = 0
Now factorizing of the above equation
4k(k-3) = 0
k(k-3) = 0
So, either
(k-3) = 0
k = 0 or
k = 3
Therefore, the value of k = 0, 3

5. Find the value of k for which the following equation have real roots

(i) 2x2 + kx + 3 = 0

Solution




(ii) kx(x-2) +6 = 0

Solution




(iii) x2 - 4kx + k = 0

Solution




(iv) kx(x-2√5)+ 10 = 0

Solution



(v) kx(x-3)+ 9 = 0

Solution




(vi) 4x2 + kx + 3 = 0

Solution



6. Find the values of k for which the given quadratic equation has real and distinct roots :

(i) kx2 + 2x + 1 = 0

Solution


(ii) kx2 + 6x + 1 = 0

Solution




7. For what value of k, (4-k)x2 + (2k+4)x+(8k+1) = 0, is a perfect square . 

Solution


8. Find the least positive value of k for which the equation x2+kx+4 = 0 has real roots .

Solution


9. Find the values of k for (3k + 1)x2 +2(k+1) x+1 = 0 has equal roots . Also, find the roots .

Solution



10. Find the values of p for which the quadratic equation (2p+1)x2 – (7p+2)x+(7p-3)=0 has equal roots . Also find these roots .

Solution



11. If -5 is a root of the quadratic equation 2x2 + px -15 = 0 and the quadratic equation p(x2+x)+k = 0 has equal roots , find the value of k.

Solution



12. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2-2px + k = 0 has equal roots , find the value of k .

Solution


13. If 1 is a root of the quadratic equation is 3x2 + ax – 2 = 0 and the quadratic equation a(x2 + 6x) – b = 0 has equal roots, find the value of b .

Solution

The given quadratic equation is3x2 + ax – 2 = 0 and one root is 1.
Then , it satisfies the given equation .
3(1)2 + a(1)-2=0
⇒ 3 + a – 2 = 0
⇒ 1+ a = 0
⇒ a = -1
The quadratic equation a(x2 + 6x) – b = 0 , has equal roots .
Putting the value of a , we get
-1(x2 + 6x) – b = 0
⇒ x2 + 6x + b = 0
Here . A = 1, B = 6 and C = b
D = (6)2 – 4(1) (b)
= 36 – 4b
The given equation will have real and equal roots , if   D = 0
Thus, 36 – 4b = 0
⇒ 4b = 36
⇒ b = 9
Therefore , the value of b is 9 .

14. Find the value of p for which the quadratic equation  (p+1)x2- 6(p+1) x + 3(p+9) = 0 , p ≠ -1 has equal roots . Hence, find the roots of the equation .

Solution 

The given quadratic equation (p+1)x2- 6(p+1) x + 3(p+9) = 0 , has equal roots .
Here, a = p + 1, b = -6p -6 and c = 3p + 27.
As we know that D = b2 – 4ac
Putting the values of a = p + 1 , b = -6p – 6 and c = 3p +27.
D = [-6(p+1)]2 – 4(p+1)[3(p+9)]
= 36(p2 + 2p + 1) – 12(p2 + 10p + 9)
= 36p2 – 12p2 + 72p – 120p + 36 - 108
= 24p2 – 48 – 72
The given equation will have real and equal roots, if D = 0
Thus , 24p2 – 48p – 72 = 0
⇒ p2 – 2p – 3 = 0
⇒ p2 – 3p + p – 3 = 0
⇒ p(p -3) + 1 (p-3) = 0
⇒ (p+1)(p -3) = 0
⇒ p+1 = 0 or p – 3 = 0
⇒ p = -1 or p = 3
Therefore , the value of p is -1,3.
It is given that p ≠ 1, thus p = 3 only .
Now the equation becomes
4x2 -24x + 36 = 0
⇒ x2 – 6x + 9 = 0
⇒ x2 – 3x – 3x + 9 = 0
⇒ x(x-3)-3(x-3) = 0
⇒ (x-3)2 = 0
⇒ x = 3 , 3
Hence , the root of the equation is 3.

15. Determine the nature of the roots of the following quadratic equations :
(i) (x-2a) (x – 2b) = 4ab
(ii) 9a2b2x2-24abcdx + 16c2d2 = 0 , a ≠ 0, b ≠ 0
(iii) 2(a2+b2)x2 + 2(a+b)x+1 = 0
(iv) (b+c)x2 – (a+b+c) x + a = 0

Solution

(i) The given quadratic equation is (x-2a)(x-2b) = 4ab
⇒ x2 – 2(a+b)x + 4ab – 4ab = 0
⇒ x2 – 2(a+b) x = 0
Here, a = 1, b = -2(a+b) and, c = 0
As we know that D = b2 – 4ac
Putting the value of a = 1, b = -2(a+b) and , c = 0
= (-2(a+b))2-4×1×0
= 4(a2+2ab + b2) – 0
Since , D > 0
Therefore , root of the given equation are real and distinct .

(ii) The given quadratic equation is 9a2b2x2-24abcdx + 16c2d2 = 0
Here, a = 9a2b, b = -24abcd and, c = 16c2d2
As we know that D = b2 – 4ac
Putting the value of a = 9a2b2, b = -24abcd and, c = 16c2d2
= (24abcd)2 – 4 × 9a2b2×16c2d2
= (576a2b2c2d2) – 576a2b2c2d2
= 0
Since, D = 0 
Therefore, root of the given equation are real and equal .

(iii) The given quadratic equation is 2(a2+b2)x2 + 2(a+b)x+1=0
Here , a = 2(a2+b2), b = 2(a+b) and, c = 1
As we know that D = b2 – 4ac
Putting the value of a = 2(a2+b2), b = 2(a+b) and, c = 1
= (2(a+b))2 – 4 × 2(a2 + b2) × 1
= (4a2 + 4b2 + 8ab) – 8a2 – 8b2
= 8ab – 4a2 – 4b2
Since, D < 0
Therefore , root of the given equation are not real. 


(iv) The given quadratic equation is (b+c)x2 – (a+b+c)x+a = 0
Here, a = (b+c), b = -(a+b+c) and, c = a
As we know that D = b2 – 4ac
Putting the value of a = (b+c), b = -(a+b+c) and, c = a
= (-a(a+b+c))2 – 4 × (b+c) × a
= (a2 + b2 + c2 + 2ab + 2bc + 2ca) – 4ab – 4ca
= a2 + b2 + c2 – 2ab + 2bc – 2ca
Since, D > 0
Therefore, root of the given equation are real and unequal .

16.  Determine the set of values of k for which the following quadratic equations have real roots: 
(i) x2-kx+9=0
(ii) 2x2 + kx + 2 = 0
(iii) 4x2 – 3kx + 1 = 0
(iv) 2x2 + kx – 4 = 0

Solution

(i) The given quadratic equation x2 – kx + 9 = 0, and roots are real
Then find the value of k .
Here, a= 1 , b = -k and, c = 9
As we know that D = b2 – 4ac
Putting the value of a = 1, b = -k and c = 9
= (-k)2 – 4 × 1 × 9
= k2 – 36
The given equation will have real roots , if D ≥ 0
k2 – 36 ≥ 0
k2 ≥ 36
k ≥ √36 or k ≤ - √36
k ≤ - 6 or k ≥ 6
Therefore, the value of k ≤ - 6 or k ≥ 6

(ii) The given quadratic equation is 2x2 + kx + 2 = 0, and roots are real.
Then find the value of k.
Here , a = 2 , b = k and , c = 2
As we know that D = b2 – 4ac
Putting the value of a = 2 , b = k and , c = 2
= (k)2 - 4×2×2
= k2 - 16
The given equation will have real roots , if D ≥ 0
k2 – 16 ≥ 0
k2 ≥ 16
k ≥ √16 or k ≤ - √16
k ≤ -4 or k ≥ 4
Therefore , the value of k ≤ -4 or k ≥ 4


(iv) The given quadratic equation is 2x2 + kx – 4 = 0, and roots are real
Then find the value of k .
Here, a = 2, b = k and c , = -4
As we know that D = b2 – 4ac
Putting the value of a = 2 , b = k and c = -4
= (k)2 – 4 × 2 × (-4)
= k2 + 32
The given equation will have real roots , if D ≥ 0
k2 + 32 ≥ 0
since left hand side is always positive . So k ∈ R
Therefore , the value of k ∈ R

17. If the roots of the equation (b-c) x2 + (c-a)x+(a-b) = 0 are equal , then prove that 2b = a + c.

Solution


18. If the roots of the equation (a2 + b2) x2 - 2(ac + bd) x + (c2 + d2) = 0 are equal ,  prove that a/b = c/d . 

Solution


19. If the roots of the equations ax2 + 2bx + c = 0 and bx2 – 2√acx + b = 0 are simultaneously real , then prove that b2 = ac .

Solution


20. If p ,q are real and p ≠ q, then show that the roots of the equation (p-q)x2 +5(p+q) x – 2(p-q) = 0 are real unequal .
             
Solution


 21. If the roots of the equation (c2 – ab) x2 – 2(a2 – bc) x + b2 – ac = 0 are equal , prove that either a = 0 or a3 + b3 + c3 = 3abc.  

Solution


22. Show that the equation 2(a2 + b2) x2 + 2(a+b) x + 1 = 0 has no real roots, when a ≠ b .  

Solution


23. Prove that both the roots of the equation (x-a) (x-b) + (x-b) (x-c ) + (x-c) (x-a) = 0 are real but they are equal only when a = b = c . 

Sol. 


24. If a,b,c are real numbers such that ac ≠ 0, then show that at least one of the equations ax2 + bx + c = 0 and - ax2 + bx + c = 0 has real roots .

Solution


25. If the equation (1+m2)x2 + 2mcx + (c2 – a2) = 0 has equal roots , prove that c2 = a2(1+m2).

Solution

The given equation (1+m2)x2 + 2mcx + (c2 – a2) = 0, has equal roots
Then prove that c2 = (1+m2).
Here,
a = (1+m2) , b = 2mc and , c = (c2 – a2)
As we know that D = b2 – 4ac
Putting the value of a = (1+m2) , b = 2mc and, c = (c2 – a2)
D = b2 – 4ac
= {2mc}2 – 4 × (1+m2) × (c2 – a2)
= 4 (m2c2) – 4(c2 a2 + m2c2 – m2a2)
= 4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2
= 4a2 + 4m2a2 – 4c2
The given equation will have real roots , if D = 0
4a2 + 4m2a2 – 4c2 = 0
4a2 + 4m2a2 = 4c2
4a2 (1+m2) = 4c2
a2(1+m2) = c2
Hence , c2 = a2 (1+m2)

↧
Search

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.7

March 6, 2019, 10:06 pm
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.7

1. Find the consecutive numbers whose squares have the sum 85.

Solution



2. Divide 29 into two parts so that the sum of the squares of the parts is 425.

Solution



3. Two squares have sides x cm and (x + 4)cm. The sum of this areas is 656 cm 2 . Find the sides of the squares.

Solution



4. The sum of two numbers is 48 and their product is 432. Find the numbers . 

Solution



5. If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.

Solution



6. Find the whole numbers which when decreased by 20 is equal to 69 times the reciprocal of the members.

Solution



7. Find the two consecutive natural numbers whose product is 20.

Solution



8. The sum of the squares of the two consecutive odd positive integers as 394. Find them.

Solution



9. The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.

Solution



10. The sum of a numbers and its positive square root is 6/25 . Find the numbers.

Solution



11. The sum of a number and its square is 63/4. Find the numbers.

Solution



12. There are three consecutive integers such that the square of the first increased by the product of the first increased by the product of the others the two gives 154. What are the integers ?

Solution



13. The product of two successive integral multiples of 5 is 300. Determine the multiples.

Solution



14. The sum of the squares of two numbers as 233 and one of the numbers as 3 less than twice the other number find the numbers.

Solution



15. Find the consecutive even integers whose squares have the sum 340.

Solution



16. The difference of two numbers is 4. If the difference of their reciprocals is 4/21. Find the numbers.

Solution



17. Let us find two natural numbers which differ by 3 and whose squares have the sum 117.

Solution



18. The sum of the squares of three consecutive natural numbers as 149. Find the numbers.

Solution



19. The sum of two numbers is 16. The sum of their reciprocals is 1/3 . Find the numbers.

Solution



20. Determine two consecutive multiples of 3, whose product is 270.

Solution



21. The sum of a number and its reciprocal is 17/4. Find the number.

Solution



22. A two-digit number is such that the products of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number ?

Solution



23. A two digits number is such that the product of the digits is 12. When 36 is added to the number, the digits inter change their places determine the number

Solution



24. A two digit number is such that the product of the digits is 16. When 54 is subtracted from the number the digits are interchanged. Find the number .

Solution



25. Two numbers differ by 3 and their product is 504. Find the number .

Solution



26. Two number differ by 4 and their product is 192. Find the numbers ?

Solution



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

Solution



28. The differences of the square of two positive integers is 180 . The square of the smaller number is 8 times the larger , find the numbers .

Solution



29. The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers .

Solution



30. The sum of two numbers and . a and b is 15. and the sum of their reciprocals 1/a and 1/b is 3/10. Find the numbers a and b.

Solution



31. The sum of two numbers is 9. The sum of their reciprocals is 1/2 . Find the numbers.

Solution



32. Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46. Find the integers

Solution



33. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers .

Solution



34. The difference of square of two numbers is 180 . the square of the smaller number is 8 times the large numbers find two numbers

Solution



35. Find two consecutive odd positive integers, sum of whose squares is 970 .

Solution



36. The differences of two natural numbers is 3 and the difference of their reciprocals is 3/28 .

Solution



37. The sum of the squares of two consecutive odd numbers is 394. Find the numbers .

Solution



38. The sum of the square of two consecutive multiples of 7 is 637. Find the multiples Sol.

Solution



39. The sum of the squares of two consecutive even numbers is 340. Find the numbers .

Solution



40. The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is 29/20. Find the original fraction.

Solution



41. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number .

Solution



42. A natural number when increased by 84 equals to 160 times its reciprocal . Find the number.

Solution

↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.8

March 7, 2019, 3:11 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.8

1. The speed of a boat in still water is 8 km/hr It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Solution 



2. A train , travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km / hr more. Find the original speed of the train.

Solution

Given,
A train is travelling at an unifrom speed for 360 km would have taken 48 min less to travel the same distance.
Let x be the original speed of the train.
time = distance / speed
A/q,

360/x = 360/(x + 5) + 48 / 60
⇒ 360/x - 360/(x + 5) = 4/5
⇒ 360[ 1/x - 1/( x +5) ] = 4/5
⇒ 450 [x + 5 - x  ] / x ( x +5) = 1
⇒ 2250 = x ( x +5)
⇒ x2 + 5x - 2250 = 0
⇒ x2 + 50x - 45x - 2250 = 0
⇒ x( x + 50) - 45( x + 50) = 0
⇒ (x - 45)( x + 50) = 0
x = 45 km /h

∴ The original speed of the train = 45 km /h.

3. A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

Solution



4. A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

Solution



5. The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction ?.

Solution



6. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

Solution



7. An aeroplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.

Solution



8. A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed . If it takes 3 hours to complete total journey, what is its original as average speed ?

Solution



9. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Solution



10. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution



11. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kin/hr more than that of the passenger train, find the average speeds of the two trains.

Solution



12. An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Solution



13. While boarding an aeroplane , a passenger got hurt . The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination , 1500 km away in time , the pilot increased the speed by 100 km/hr . Find the original speed/hour of the plane .

Solution



14. A motor boat whose speed in still water is 18/km/hr takes 1 hour more to go 24 km up stream that to return down stream to the same spot. Find the speed of the stream .

Solution

↧
↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.9

March 7, 2019, 9:30 pm
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.9

1. Ashu is x years old while his mother Mrs Veena is x years old. Five years hence Mrs Veena will be three times old as Ashu. Find their present ages.

Solution



2. The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at the time. Find their present ages.

Solution 



3. The product of Shikha’s age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.

Solution



4. The product of Ramu’s age (in years) five years ago and his age (in years) nine years later is 15. Determine Ramu’s present age.

Solution



5. Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution



6. A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.

Solution



7. he sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age .

Solution



8. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age . What is her age now ?

Solution



9. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha . Find the present ages of both Asha and Nisha .

Solution

↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.10

March 8, 2019, 1:22 am
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.10

1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.

Solution



2. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solution



3. The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle?

Solution



4. A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected ?

Solution


↧

R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.11

March 8, 2019, 11:51 pm
$
0
0

Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.11

1. The perimeter of a rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle.

Solution



2. The length of a hail is 5 m more than its breadth. If the area of the floor of the hail is 84 m2 , what are the length and breadth of the hail ?

Solution



3. Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2 . Find the sides of the squares.

Solution



4. The area of a right angled triangle is 165 m2. Determine its base and altitude if the latter exceeds the former by 7m.

Solution




5. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2? If so, find its length and breadth.

Solution



6. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Solution



7. Sum of the areas of two squares is 640 2. If the difference of their perimeters is 64 m, find the sides of the two squares.

Solution



8. sum of the areas of two squares is 400 c2. If the differences of their perimeters is 16 cm, find the sides of two squares.

Solution



9. The area of a rectangle plot is 528 cm2. The length of the plot (in metres) is one metre more then twice its breadth . Find the length and the breadth of the plot .

Solution



10. In the centre of a rectangular lawn of dimensions 50 m × 40 m , a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 2. Find the length and breadth of the pond .

Solution

↧
Viewing all 6266 articles
Browse latest View live
Search
<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>