## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 2 Chapter Name Linear Equations in One Variable Exercise Ex 2.6 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Question 1.
Solve the following equations:

Solution.
1. $$\frac { 8x-3 }{ 3x } =2$$

2. $$\frac { 9x }{ 7-6x } =15$$

3. $$\frac { z }{ z+15 } =\frac { 4 }{ 9 }$$

4. $$\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }$$

5. $$\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }$$

Question 2.
The ages of Hari and Harry are in the ratio 5: 7. Four years from now the ratio of their ages will be 3 :4. Find their present ages.
Solution.
Let the present ages of Hari and Harry be 5x years and 7x years respectively.

Question 3.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is $$\frac { 3 }{ 2 }$$. Find the rational number.
Solution.

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## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 2 Chapter Name Linear Equations in One Variable Exercise Ex 2.5 Number of Questions Solved 2 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

Question 1.
Solve the following linear equations:
1. $$\frac { x }{ 2 } -\frac { 1 }{ 5 } =\frac { x }{ 3 } +\frac { 1 }{ 4 }$$
2. $$\frac { n }{ 2 } -\frac { 3n }{ 4 } +\frac { 5n }{ 6 } =21$$
3. $$x+7-\frac { 8x }{ 3 } =\frac { 17 }{ 6 } -\frac { 5x }{ 2 }$$
4. $$\frac { x-5 }{ 3 } =\frac { x-3 }{ 5 }$$
5. $$\frac { 3t-2 }{ 4 } -\frac { 2t+3 }{ 3 } =\frac { 2 }{ 3 } -t$$
6. $$m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 }$$
Solution.
1. $$\frac { x }{ 2 } -\frac { 1 }{ 5 } =\frac { x }{ 3 } +\frac { 1 }{ 4 }$$

2. $$\frac { n }{ 2 } -\frac { 3n }{ 4 } +\frac { 5n }{ 6 } =21$$

3. $$x+7-\frac { 8x }{ 3 } =\frac { 17 }{ 6 } -\frac { 5x }{ 2 }$$

4. $$\frac { x-5 }{ 3 } =\frac { x-3 }{ 5 }$$

5. $$\frac { 3t-2 }{ 4 } -\frac { 2t+3 }{ 3 } =\frac { 2 }{ 3 } -t$$

6. $$m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 }$$
We have $$m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 }$$

Question 2.
Simplify and solve the following linear equations:
7. 3(t-3)=5(2t+1)
8. 15(y-4)-2(y-9)+5 (y+6) = 0
9. 3(5z-7)-2 (9z-11)=4 (8z-13)-17
10. 0.25(4f-3) = 0.05 (10f-9).
Solution.
7. 3(t-3)=5(2t+1)

8. 15(y-4)-2(y-9)+5 (y+6) = 0

9. 3(5z-7)-2 (9z-11)=4 (8z-13)-17

10. 0.25(4f-3) = 0.05 (10f-9)

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## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 2 Chapter Name Linear Equations in One Variable Exercise Ex 2.4 Number of Questions Solved 10 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.
Amina thinks of a number and subtracts $$\frac { 5 }{ 2 }$$ from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution.
Let the number be x.
Then, according to the question,

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution.
Let the numbers be x and 5x.
If 21 is added to both the numbers, then first new number = x + 21
and, second new number = 5x + 21

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution.
Let the unit’s digit of the two-digit number be x.
Then, the ten’s digit of the two-digit number = 9 – x

Question 4.
One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution.
Let in the original number

Question 5.
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution.
Let the present age of Shobo be x  year. Then, the present age of Shobo’s mother = 6x years.
Five years from now
Shobo’s age = (x + 5) years
According to the question,

Hence, their present ages are 5 years and 30 years.

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate of ₹ 100 per meter, it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution.
Let the length and breadth of the plot be 11x m and 4x m respectively.
Then, the perimeter of the plot

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per meter and trouser material that costs him ₹ 90 per meter. For every 2 meters of the trouser material, he buys 3 meters of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?
Solution.
Suppose that he bought x meters of trouser material.

Question 8.
Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution.
Let the number of deer in the herd be x.

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution.
Let the present age of granddaughter be x years
Then, the present age of grandfather is 10x years
According to the question,

Question 10.
Am.an’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution.
Let the present age of Aman’s son be x years
Then, the present age of Aman

We hope the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 2 Chapter Name Linear Equations in One Variable Exercise Ex 2.3 Number of Questions Solved 1 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

Question 1.
Solve the following equations and check your results:
1. 3x=2x+18
2. 5t-3=3t-5
3. 5x+9=5+3x
4. 4z+3=6+2z
5. 2x-1=14-x
6. 8x+4=3(x-1)+7
7. $$x=\frac { 4 }{ 5 } \left( x+10 \right)$$
8. $$\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } +3$$
9. $$2y+\frac { 5 }{ 3 } =\frac { 26 }{ 3 } -y$$
10. $$3m=5m-\frac { 8 }{ 5 }$$
Solution.
1. 3x=2x+18

2. 5t-3=3t-5

3. 5x+9=5+3x

4. 4z+3=6+2z

5. 2x-1=14-x

6. 8x+4=3(x-1)+7

7. $$x=\frac { 4 }{ 5 } \left( x+10 \right)$$

8. $$\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } +3$$

9. $$2y+\frac { 5 }{ 3 } =\frac { 26 }{ 3 } -y$$

10. $$3m=5m-\frac { 8 }{ 5 }$$

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## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 2 Chapter Name Linear Equations in One Variable Exercise Ex 2.2 Number of Questions Solved 16 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

Question 1.
If you subtract $$\frac { 1 }{ 2 }$$ from a number and multiply the result $$\frac { 1 }{ 2 }$$ by you get $$\frac { 1 }{ 8 }$$ What is the number ?
Solution.

Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution.

Question 3.
The base of an isosceles triangle is $$\frac { 4 }{ 3 }$$cm. The perimeter of the triangle is $$4\frac { 2 }{ 15 }$$cm. What is the length of either of the remaining equal sides ?
Solution.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution.

Question 5.
Two numbers are in the ratio 5 :3. If they differ by 18, what are the numbers?
Solution.

Question 6.
Three consecutive integers add up to 51. What are these integers ?
Solution.

Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution.

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later, the sum of their ages will be 56 years. What are their present ages ?
Solution.

Question 10.
The numbers of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength ?
Solution..

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them ?
Solution.

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age ?
Solution.

Question 13.
A rational number is such that when you multiply it by $$\frac { 5 }{ 2 }$$ and add $$\frac { 2 }{ 3 }$$ to the product, you get$$-\frac { 7 }{ 12 }$$. What is the number ?
Solution.

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have ?
Solution.

Question 15.
I have a total of oft 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me ?
Solution.

Hence, I have 80, 60, and 20 coins of denomination ₹ 1, ₹ 2 and ₹ 5 respectively.

Question 16.
The organizers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Solution.

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## NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 1 Chapter Name Rational Numbers Exercise Ex 1.2 Number of Questions Solved 7 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2

Question 1.
Represent these numbers on the number line.
(i) $$\frac { 7 }{ 4 }$$
(ii) $$\frac { -5 }{ 6 }$$
Solution.
(i) $$\frac { 7 }{ 4 }$$
We make 7 markings of distance $$\frac { 1 }{ 4 }$$ each on the right of 0 and starting from 0. The seventh marking represents $$\frac { 7 }{ 4 }$$

(ii) $$\frac { -5 }{ 6 }$$
We make 5 markings of distance $$\frac { 1 }{ 6 }$$ each on the right of 0 and starting from 0. The seventh marking represents $$\frac { -5 }{ 6 }$$

Question 2.
Represent $$\frac { -2 }{ 11 } ,\frac { -5 }{ 11 } ,\frac { -9 }{ 11 }$$ on the number line.
Solution.
We make 9 markings of distance $$\frac { 1 }{ 11 }$$ each on the left of 0 and starting from 0.
The second marking represents $$-\frac { 2 }{ 11 }$$ the fifth marking represents $$-\frac { 5 }{ 11 }$$ and the ninth marking represents $$-\frac { 9 }{ 11 }$$

Question 3.
Write five rational numbers which are smaller than 2.
Solution.
Five rational numbers which are smaller than 2 are 1, $$\frac { 1 }{ 2 }$$,0, -1, $$-\frac { 1 }{ 2 }$$

Question 4.
Find ten rational numbers between $$-\frac { 2 }{ 5 }$$ and $$\frac { 1 }{ 2 }$$
Solution.

Question 5.
Find five rational numbers between:

Solution.

Question 6.
Write five rational numbers greater than -2.
Solution.
Five rational numbers greater than -2 are:
$$\frac { -3 }{ 2 }$$, -1, $$\frac { -1 }{ 2 }$$, 0, $$\frac { 1 }{ 2 }$$
There can be many more such rational numbers

Question 7.
Find ten rational numbers between $$\frac { 3 }{ 5 }$$ and $$\frac { 3 }{ 4 }$$
Solution.

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## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 16 Chapter Name Playing with Numbers Exercise Ex 16.1 Number of Questions Solved 1 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

Question 1.
Find the values of the letters in each of the following and give reasons for the steps involved.

Solution.
1.
Here, there are two letters A and B whose values are to be found out.
Let us see the sum in unit’s column. It is A + 5 and we get 2 from this. So,

2.
Here, there are three letters A, B and C whose values are to be found out.
Let us see the sum in unit’s column. It is A + 8 and we get 3 from this. So A has to be 5

That is, A = 5, B = 4 and C = 1.

4.
Here, there are two letters A and B whose values are to be found out.

5.
Here, there are three letters A, B and C whose values are to be found out.

∴ A = 5, B = 0 and C = 1

6.
Here, there are three letters A, B and C, whose values are to be found out.

7.
Here, there are two letters A and B whose values are to be found out. We have

8.
Here, there are two letters A and B whose values are to be found out.

10.
We are to find out values of A and B

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## NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 15 Chapter Name Introduction to Graphs Exercise Ex 15.1 Number of Questions Solved 7 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1

Question 1.
The following graph shows the temperature of a patient in a hospital, recorded every hour.
(a) What was the patient’s temperature at 1 p.m.?
(b) When was the patient’s temperature 38.5°C?
(c) The patient’s temperature was the same two times during the period given. What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patient’s temperature showed an upward trend?

Solution.
(а) The patient’s temperature at 1 p.m. was 36.5°C.
(b) The patient’s temperature was 38.5°C at 10.50 a.m. and 12 noon.
(c) The two times when the patient’s temperature was the same were 1 p.m. and 2 p.m.
(d) The temperature at 1.30 p.m. was 36.5°C.
From the graph, we see that the temperature was constant from 1 p.m. to 2 p.m. Since 1.30 p.m. comes in between 1 p.m. and 2 p.m., therefore we arrived at our answer.
(e) The patient’s temperature showed an upward trend during the periods 9 a.m. to 10 a.m., 10 a.m. to 11 a.m. and 2 p.m. to 3 p.m.

Question 2.
The following line graph shows the yearly sales figures for a manufacturing company.
(a) What were the sales in
(i) 2002
(ii) 2006?

(b) What were the sales in
(i) 2003
(ii) 2005?

(c) Compute the difference between sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?

Solution.
(а) The sales in
(i) 2002 were ₹ 4 crores and in
(ii) 2006 were ₹ 8 crores.

(b) The sales in
(i) 2003 were ₹ 7 crores and in
(ii) 2005 were ₹ 10 crores.

(c) The difference between the sales in 2002 and 2006
= ₹ 8 crores – ₹ 4 crores = ₹ 4 crores

(d) The difference between sales in 2002 and 2003
= ₹ 7 crores – ₹ 4 crores = ₹ 3 crores
The difference between sales in 2003 and 2004
= ₹ 7 crores – ₹ 6 crores = ₹ 1 crore
The difference between the sales in 2004 and 2005
= ₹ 10 crores – ₹ 6 crores = ₹ 4 crores
The difference between sales in 2005 and 2006
= ₹ 10 crores – ₹ 8 crores = ₹ 2 crores
Therefore, in year 2005 the difference between the sales as compared to its previous year was the greatest.

Question 3.
For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.

(a) How high was Plant A after
(i) 2 weeks
(ii) 3 weeks?

(b) How high was Plant B after
(i) 2 weeks
(ii) 3 weeks?

(c) How much did Plant A grow during
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did Plant A grow most?
(f) During which week did Plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.
Solution.
(а) The Plant A
after (i) 2 weeks was 7 cm high and
after (ii) 3 weeks was 9 cm high.

(b) The Plant B
after (i) 2 weeks was 7 cm high and
after (ii) 3 weeks was 10 cm high.

(c) The Plant A grew 9 cm – 7 cm = 2 cm during the 3rd week.

(d) From the end of the 2nd week to the end of the 3rd week, Plant B grew
= 10 cm – 7 cm = 3 cm.

(e) The Plant A grew in 1st week
= 2 cm – 0 cm = 2 cm
The Plant A grew in 2nd week
= 7 cm – 2 cm = 5 cm
The Plant A grew in 3rd week
= 9 cm – 7 cm = 2 cm
Therefore, Plant A grew mostly in the second week.

(f) Plant B grew in 1st week
= 1 cm – 0 cm = 1 cm
Plant B grew in 2nd week
= 7 cm – 1 cm = 6 cm
Plant B grew in 3rd week
= 10 cm – 7 cm = 3 cm
Therefore, Plant B grew least in the first week.

(g) At the end of 2nd week, the two plants shown here were of the same height.

Question 4.
The following graph shows the temperature forecast and the actual temperature for each day of a week:

(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?
Solution.
(a) The forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday.
(b) The maximum forecast temperature during the week was 35°C.
(c) The minimum actual temperature during the week was 15°C.
(d)

Therefore, the actual temperature differed the most from the forecast temperature on Thursday.

Question 5.
Use the tables below to draw linear graphs.
(a) The number of days a hill side city received snow in different years.

(b) Population (in thousands) of men and women in a village in different years.

Solution.
(a)

(b)

Question 6.
Courier-person cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?

Solution.
(a) The scale taken for the time axis is 4 units = 1 hour.
(b) The time taken by the person for the travel 8 a.m. to 11.30 a.m. = $$3\frac { 1 }{ 2 }$$ hours.
(c) The place of the merchant from the town in 22 km.
(d) Yes. This is indicated by the hori¬zontal part of the graph (10 a.m. – 10.30 a.m.)
(e) He rides fastest between 8 a.m. and 9 a.m. (As line is more steep in this period).

Question 7.
Can there be a time-temperature graph as follows ? Justify your answer.

Solution.
(i) Yes; it can be
It shows a time-temperature graph. It shows an increase in temperature with an increase in time.
(ii) Yes; it can be
It shows a time-temperature graph.
It shows a decrease in temperature with increase in time.
(iii) It cannot be a time-temperature graph because it shows infinitely many different temperatures at one particular time which is not possible.
(iv) Yes; it can be
It shows a time-temperature graph,
It shows a fixed temperature at different times.

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## NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 14 Chapter Name Factorisation Exercise Ex 14.1, Ex 14.2, Ex 14.3, Ex 14.4 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1

Question 1.
Find the common factors of the given terms:

Solution.

Question 2.
Factorise the following expressions:

Solution.

Question 3.
Factorise:

Solution.

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## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 13 Chapter Name Direct and Indirect Proportions Exercise Ex 13.1 Number of Questions Solved 10 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1

Question 1.
Following are the car parking charges near a railway station up to
4 hours ₹ 60
8 hours ₹ 100
12 hours ₹ 140
24 hours ₹ 180
Check if the parking charges are in direct proportion to the parking time.
Solution.
We have
$$\frac { 60 }{ 4 } =\frac { 15 }{ 1 }$$
$$\frac { 100 }{ 8 } =\frac { 25 }{ 2 }$$
$$\frac { 140 }{ 12 } =\frac { 35 }{ 3 }$$
$$\frac { 180 }{ 24 } =\frac { 15 }{ 2 }$$
Since all the values are not the same, therefore, the parking charges are not in direct proportion to the parking time.

Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. In the following table, find the parts othe f base that need to be added.

Solution.
Sol. Let the number of parts of red pigment is x and the number of parts of the base is y.
As the number of parts of red pigment increases, a number of parts of the base also increases in the same ratio. So it is a case of direct proportion.
We make use of the relation of the type

Question 3.
In Question 2 above, if 1 part of a red pigment requires 75 mL of the base, how much red pigment should we mix with 1800 mL of base?
Solution.
Let the number of parts of red pigment is x and the amount of base be y mL.
As the number of parts of red pigment increases, the amount of base also increases in the same ratio. So it is a case of direct proportion. We make use of the relation of the type.

Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution.
Let the machine fill x bottles in five hours. We put the given information in the form of a table as shown below :
Number of bottles filled 840 x2
Number of hours 6 5
More the number of hours, more the number of bottles would be filled. So, the number of bottles filled and the number of hours are directly proportional to each other.
So, $$\frac { { x }_{ 1 } }{ { x }_{ 2 } } =\frac { { y }_{ 1 } }{ { y }_{ 2 } }$$
⇒ $$\frac { 840 }{ { x }_{ 2 } } =\frac { 6 }{ 5 }$$
⇒ $$6{ x }_{ 2 }=840\times 5$$
⇒ $${ x }_{ 2 }=\frac { 840\times 5 }{ 6 }$$
⇒ $${ x }_{ 2 }=700$$
Hence, 700 bottles will be filled.

Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria ? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Solution.
Actual length of the bacteria
= $$\frac { 5 }{ 50000 }$$cm
= $$\frac { 1 }{ 10000 }$$ = $${ 10 }^{ -4 }$$cm
10000
Let the enlarged length be y2 cm. We put the given information in the form of a table as shown below:
Number of times Length attained
photograph enlarged (in cm)
50.000 5
20.000 y2
More the number of times a photograph of a bacteria is enlarged, more the length attained. So, the number of times a photograph of a bacteria is enlarged and the length attained are directly proportional to each other.
So,$$\frac { { x }_{ 2 } }{ { y }_{ 2 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } }$$
⇒ $$\frac { 50000 }{ 5 } =\frac { 20000 }{ { y }_{ 2 } }$$
⇒ $$50000{ y }_{ 2 }=5\times 20000$$
⇒ $${ y }_{ 2 }=\frac { 5\times 20000 }{ 50000 }$$
⇒ $${ y }_{ 2 }=2$$
Hence, its enlarged length would be
2 cm.

Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship ?

Solution.

Question 7.
Suppose 2 kg of sugar contains 9 x 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution.
Suppose the amount of sugar is x kg and the number of crystals is y.
We put the given information in the form of a table as shown below:

As the amount of sugar increases, the number of crystals also increases in the same ratio. So it is a case of direct proportion. We make use of the relation of the type $$\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } }$$

Question 8.
Rashmi has a roadmap with a scale of 1 cm representing 18 km. She drives on a T’oad for 72 km. What would be her distance covered in the map?
Solution.
Let the distance covered in the map be x cm. Then,
1 : 18 = x : 72
⇒ $$\frac { 1 }{ 18 } =\frac { x }{ 72 }$$
⇒ $$x=\frac { 72 }{ 18 }$$
⇒ x = 4
Hence, the distance covered in the map would be 4 cm.

Question 9.
A5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long.
Solution.
Let the height of the vertical pole be x m and the length of the shadow be y m.
We put the given information in the form of a table as shown below:

Question 10.
A loaded truck travels 14 km. in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution.
Two quantities x and y which vary in direct proportion have the relation

We hope the NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 11 Chapter Name Mensuration Exercise Ex 11.1, Ex 11.2, Ex 11.3, Ex 11.4 Number of Questions Solved 5 Category NCERT Solutions

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

Question 1.
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Solution.
Area of the square field = a x a
= 60 m x 60 m = 3,600 $${ m }^{ 2 }$$
Perimeter of the square field = 4a
= 4 x 60 m = 240 m
∴ Perimeter of rectangular field = 240 m
⇒ 2(l + b) = 240
⇒ 2(80 + b) = 240
where b m is the breadth of the rectangular field
⇒ 80 + b = $$\frac { 240 }{ 2 }$$ ⇒ 80 + bx = 120
⇒ b = 120 – 80 = 40
∴ Area of rectangular field
= l x b = 80 m x 40 m = 3,200 $${ m }^{ 2 }$$
So, the square field (a) has a larger area

Question 2.
Mrs. Kaushik has a square plot ‘ with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a: garden around the house at the rate of ₹ 55 per $${ m }^{ 2 }$$.
Solution.
Area of the square plot = a x a
= 25 x 25 $${ m }^{ 2 }$$ = 625 $${ m }^{ 2 }$$
Area of the house = a x b
= 20 x 15 $${ m }^{ 2 }$$ = 300 $${ m }^{ 2 }$$
∴ Area of the garden
= Area of the square plot – Area of the house
= 625 $${ m }^{ 2 }$$ – 300 $${ m }^{ 2 }$$
= 325 $${ m }^{ 2 }$$
∵ The cost of developing the garden per square metre = ₹ 55.
∴ Total cost of developing the garden
= ₹ 325 x 55
= ₹ 17,875.

Question 3.
The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden.

Solution.

Question 4.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the cor-responding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Solution.
Area of a flooring tile = bh
= 24 x 10 $${ cm }^{ 2 }$$
= 240 $${ cm }^{ 2 }$$
Area of the floor
= 1080 $${ m }^{ 2 }$$
= 1080 x 100 x 100 $${ cm }^{ 2 }$$
∵ m2 = 100 x 100 $${ cm }^{ 2 }$$

∴ Number of tiles required to cover the floor
=$$\frac { Area\quad of\quad the\quad floor\quad }{ Area\quad of\quad a\quad flooring\quad tile }$$
= $$\frac { 1080\times 100\times 100\quad }{ 240 }$$
= 45000.

Question 5.
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1, drop a comment below and we will get back to you at the earliest.