Author name: Prasanna

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.3

Question 1.
List the outcomes you can see in these experiments
(a) Spinning a wheel
(b) Tossing two coins together
Solution:
(a) Outcomes in spinning the given wheel are A, B, C, or D.
(b) Outcomes in tossing two coins together are HH, HT, TH, TT
[HT means Head on first coin + Tail on the second coin]

Question 2.
When a die is thrown, list the outcomes of an event of getting.
(i) (a) prime number
(b) not a prime number
(ii) (a) a number greater than 5
(b) a number not greater than 5
Solution:
Possible outcomes are 1, 2, 3, 4, 5 and 6
(i) (a) Outcomes are 2, 3 and 5
(b) Outcomes are 1, 4 and 6
(ii) (a) Outcomes greater than 5 is 6
(b) Outcomes of getting a number not greater than 5 are 1, 2, 3, 4 and 5

Question 3.
Find the
(a) Probability of the pointer stopping on D in (Question 1-(a))?
(b) Probability of getting an ace from a well-shuffled deck of 52 playing cards?
(c) Probability of getting a red apple. (See figure below)

Solution:
(a) On the spinning wheel there are 5 sectors containing A, B, C, and D.
Since, there is only one sector containing D,
Number of possible outcome = 1
Number of equally likely outcomes = 5
∴ Probability = \(\frac{1}{5}\)

(b) Number of possible outcomes = 52
Since there are 4 aces in a pack of 52 cards, and out of the one ace can be obtained in 4 ways
∴ Probability of getting an ace = \(\frac{4}{52}=\frac{1}{13}\)

(c) There are 7 apples in all
Possible number of ways = 7
Since there are 4 red apples
No. of equally likely outcomes = 4
∴ Probability of getting a red apple = \(\frac{4}{7}\)

Question 4.
Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box, and mixed well. One slip is chosen from the box without looking into it. What is the probability of?
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
Solution:
Since there are 10 slips,
Total number of outcomes = 10
(i) We can get a slip containing the number ‘6’ only once
Number of favourable outcome = 1
∴ Probability of getting the number = \(\frac{6}{10}\)

(ii) Numbers less than 6 are 1, 2, 3, 4 and 5
No. of Favourable outcomes are 5
∴ Probability of getting a no. less than 6 = \(\frac{5}{10}=\frac{1}{2}\)

(iii) Numbers greaterthan 6 are 7, 8, 9 and 10
No. of favourable outcomes = 4
∴ Probability of getting a number greater than 6 = \(\frac{4}{10}=\frac{2}{5}\)

(iv) One-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9
No. of favourable outcomes = 9
∴ Probability of getting a one-digit number = \(\frac{9}{10}\)

Question 5.
If you have a spinning wheel with 3 green sectors, 1 blue sector, and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:
There are 5 sectors is all (3 green + 1 blue + 1 red)
Total possible outcomes = 5
Since there are 3 green sectors
Number of favourable outcomes = 3
∴ Probability of getting a green sector = \(\frac{3}{5}\)
Again there are 4 non-blue sectors.
Number of favourable outcomes = 4
∴ Probability of getting a non blue sector = \(\frac{4}{5}\)

Question 6.
Find the probabilities of the events given in question 2.
Solution:
When a die is thrown there are 6 outcomes in all (1, 2, 3, 4, 5 and 6)
(i) Since there are 3 prime numbers 2, 3 and 5
Number of favourable outcomes = 3
∴ Probability of getting a prime number = \(\frac{3}{6}=\frac{1}{2}\)

(ii) Since there are 3 non-prime numbers (1, 4 and 6)
Number of favourable outcomes = 3
∴ Probability of getting a non prime number = \(\frac{3}{6}=\frac{1}{2}\)

(iii) Since there is 1 number greater than 5 (i.e., 6)
Number of favourable outcome = 1
∴ Probability of number greater than 5 = \(\frac{1}{6}\)

(iv) Since there are 5 numbers which are not greater than 5 (i.e., 1, 2, 3, 4 and 5)
Number of favourable outcomes = 5
∴ The probability of a number which is not greater than 5 = \(\frac{5}{6}\)

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions

NCERT In-text Question Page No. 78
Question 1.
The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.
Answer:

Thus the condition 10y – 20 is true for y = 7

NCERT In-text Question Page No. 80
Question 1.
Write atleast one other form for each equation (ii), (iii) and (iv)

Statement of other form of the equation:
(a) Taking away 4 from five times p gives 16.
(b) Add 9 to four times n to get 45.
(c) Add 8 to one-third of m to get 17.

NCERT In-text Question Page No. 88
Question 1.
Start with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5.
Answer:
(i) x = 5
Multiplying both sides by 2, we have
2x = 10
Adding 6 to both sides, we have
or 2x + 6 = 10 + 6
or 2x + 6 = 16 is an equation.

(ii) x = 5
Dividing both sides by 3, we have
\(\frac{x}{3}=\frac{5}{3}\)
Subtracting 2 from both sides, we have
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5}{3}\) – 2
or
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5-6}{3}=\frac{-1}{3}\)
Thus,
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\) is an equation.

Solution to I
2x + 6 = 16
Subtracting 6 from both sides, we have
2x + 6- 6 = 16 – 6
or 2x = 10
Dividing both sides by 2, we have
\(\frac{2 \mathrm{x}}{2}=\frac{10}{2}\) or x = 5

Solution to II
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\)
Adding 2 to both sides, we have
\(\frac{\mathrm{x}}{3}\) – 2 + 2 = \(\frac{-1}{3}\) + 2 or \(\frac{x}{3}=\frac{5}{3}\)
Multiplying both sides by 3, we have
\(\frac{\mathrm{x}}{3}\) × 3 = \(\frac{5}{3}\) × 3 or x = 5

NCERT In-text Question Page No. 88
Question 1.
Try to make two number puzzles, one with the solution 11 and another with 100.
Answer:
I. A puzzle having the solution as 11:
Think of a number. Multiply bit by 3 and add 7. Tell me the sum.
If the sum is 40, then the number is 11.

II. A puzzle having the solution as 100:
Think of a number. Divide it by 4 and add 5. Tell me what you get.
If you get 30, then the number is 100.

Note:
Instead of making the same operation on both sides, we can move a number from one side to another by changing its sign from (+) to (-) and (-) to (+). This is called ‘transposing a number’. Thus, transposing a number is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed.

NCERT In-text Question Page No. 90
Question 1.
(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?

(ii) What is that number one-third of which added to 5 gives 8?
Answer:
(i) Let the number be x.
∴ Multiply the number by 6, we have 6x.
Now, according to the condition, we have
6x = 7 + 5 = 12
Dividing both sides by 6, we have
\(\frac{6 x}{6}=\frac{12}{6}\)
or x = 2
∴ The required number = 2

(ii) Let the required number be x.
∵ One-third of the number = \(\frac{1}{3}\)x
∴ According to the condition, we have
5 + \(\frac{1}{3}\)x = 8
Transposing 5 from L.H.S. to R.H.S., we have
\(\frac{1}{3}\)x = 8 – 5 = 3
Multiplying both sides by 3, we have
3 × \(\frac{1}{3}\)x = 3 × 3
or x = 9
Thus, the required number = 9.

NCERT In-text Question Page No. 90
Question 1.
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of he smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?
Answer:
Let the number of mangoes contained in the smaller box be x.
∴ Number of mangoes in 8 smaller boxes = 8x
Now, according to the condition, we have
8x + 4 = 100
Transposing 4 to R.H.S., we have
8x = 100 – 4 or 8x = 96
Dividing both sides by 8, we have
\(\frac{8 x}{8}=\frac{96}{8}\)
or x = 12
Thus, the number of mangoes in the smaller box = 12.

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Answer:
(a) Let the required number be x
∴ Eight times of the number = 8x
According to the given question, we get
8x + 4 =60
Transposing 4 to R.H.S
8x = 60 – 4
8x = 56
Dividing both sides by 8, we get
\(\frac{8 x}{8}=\frac{56}{8}\)
x = 7
∴ The required number = 7

(b) Let the number be x
one-fifth of the number = \(\frac{1}{5}\) x
According to the given question, we get
\(\frac{1}{5}\) x – 4 = 3
Transposing -4 to R.H.S
\(\frac{1}{5}\)x = 3 + 4
\(\frac{1}{5}\)x = 7
Multiplying both sides by 5, we get 1
\(\frac{1}{5}\)x × 5 = 7 × 5
x = 35
∴ The required number = 35

(c) Let the number be x
Three-fourths of the number = \(\frac{3}{4}\) x
According to the given question, we get
\(\frac{3}{4}\) x + 3 = 21
Transposing 3 to R.H.S
\(\frac{3}{4}\) x = 21 – 3
\(\frac{3}{4}\) x = 18
Multiplying both sides by 4, we get
\(\frac{3}{4}\) x × 4 = 18 × 4
3x = 72
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{72}{3}\)
x = 24
∴ Thus, the required number = 24

(d) Let the number be x
Twice a number = 2x
According to the given question
2x – 11 = 15
Transposing -11 to R.H.S
2x = 15 + 11
2x = 26
Dividing both sides by 2, we get
\(\frac{2 \mathrm{x}}{2}=\frac{26}{2}\)
x = 13
∴ The required number 13.

(e) Let the number of notebooks with Munna be ‘x’
Thrice the number of notebooks = 3x
According to the given question, we get
50 – 3x = 8
Transposing 50 to R.H.S
-3x = 8 – 50
-3x = -42
Dividing both sides by (-3), we get
\(\frac{-3 x}{-3}=\frac{-42}{-3}\)
x = 14
∴ The required number of notebooks is 14.

(f) Let the number be x
According to the given question, we get
\(\frac{x+19}{5}\) = 8
Multiplying both sides by 5, we get
\(\frac{x+19}{5}\) × 5 = 8 × 5
x + 19 = 40
Transposing 19 to R.H.S
x = 40 – 19
x = 21
∴ The required number = 21

(g) Let the number be x
\(\frac{5}{2}\) of the number = \(\frac{5}{2}\) x
According to the given question, we get
\(\frac{5}{2}\) x – 7 = 23
Transposing -7 to R.H.S
\(\frac{5}{2}\)x = 23 + 7
\(\frac{5}{2}\)x = 30
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x × 2 = 30 × 2
5x = 60
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{60}{5}\)
x = 12
The required number = 12.

Question 2.
Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Answer:
(a) Let the lowest marks scored be x
Twice the lowest marks = 2x
According to the given question, we get
2x + 7 = 87
Transposing 7 to R.H.S
2x = 87 – 7
2x = 80
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{80}{2}\)
x = 40
∴ The lowest marks = 40

(b) Let the base angle be x°
∴ The other base angle = x°
The vertex angle = 40°
∴ Sum of the angles = x + x + 40
= 2x + 40
According to the given question, we get
2x + 40 = 180°
(Sum of the three angles of a triangle)
2x = 180° – 40°
2x = 140°
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{140}{2}\)
x = 70°
∴ The base angle of the triangle
= 70°

(c) Let the runs scored by Rahul be x
∴ Sachins runs = 2x
According to the given question, we get
x + 2x =2(100) -2
3x = 200 – 2
3x = 198
Dividing both sides by 3, we get
\(\frac{3 x}{x}=\frac{198}{3}\)
x = 66
∴ Rahul score = 66 runs and Sachins score = 2 × 66 = 132 runs

Question 3.
Solve the following:
(i) Ifran says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Answer:
(i) Let the number of marbles with Parmit be x
Number of marbles with Irfan = 37
According to the given question, we get
5x + 7 = 37
Transposing 7 to R.H.S
5x = 37 – 7
5x = 30
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{30}{5}\)
\(\frac{5 x}{5}=\frac{30}{5}\) ⇒ x = 6
∴ Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years
Three times Laxmi s age = 3x
According to the given question, we get
3x + 4 = 49
Transposing 4 to R.H.S
3x = 49 – 4
3x = 45
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{45}{3}\)
x = 15
∴ Laxmi’s age= 15 years

(iii) Let the number of fruit trees be x
Three times fruit trees = 3x
Number of non-fruit trees = 2 + 3x
According to the given question, we get
2 + 3 x = 77
Transposing 2 to R.H.S
3x = 77 – 2
3x = 75
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{75}{3}\)
⇒ x = 25
∴ The number of fruit trees = 25

Question 4.
Solve the following riddle :
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Answer:
Let the required number be x
7 times the number = 7x
each triple century = 3 × 100 = 300
According to the given riddle, we get
7x + 50 = 300 – 40
7x + 50 = 260
Transposing 50 to R.H.S
7x = 260 – 50
7x = 210
Dividing both sides by 7, we get
\(\frac{7 \mathrm{x}}{7}=\frac{210}{7}\)
x = 30
∴ The required number is 30.

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3

Question 1.
Solve the following equations:
(a) 2y + \(\frac{5}{2}=\frac{37}{2}\)
(b) 5t + 28 = 10
(c) \(\frac{\mathrm{a}}{5}\) + 3 = 2
(d) \(\frac{\mathrm{q}}{4}\) + 7 = 5
(e) \(\frac{5}{2}\) x = -10
(f) \(\frac{5}{2}\) x = \(\frac{25}{4}\)
(g) 7m + \(\frac{19}{2}\) = 13
(h) 6z + 10 = -2
(i) \(\frac{31}{2}=\frac{2}{3}\)
(j) \(\frac{2 \mathrm{~b}}{3}\) – 5 = 3
Answer:
(a) we have
2y + \(\frac{5}{2}=\frac{37}{2}\)
Transposing \(\frac{5}{2}\) from L.H.S to R.H.S
2y = \(\frac{37}{2}-\frac{5}{2}\)
2y = \(\frac{37-5}{2}\)
2y = \(\frac{32}{2}\)
2y = 16.

(b) we have
5t + 28 = 10
Transposing 28 from L.H.S to R.H.S
5t = 10 – 28
5t = – 18
Dividing both sides by 5, we get
\(\frac{5 \mathrm{t}}{5}=\frac{-18}{5}\)
t = \(\frac{-18}{5}\)
The solution is t = \(\frac{-18}{5}\) or -3 \(\frac{3}{5}\)
Dividing both sides by 2, we get
\(\frac{2 y}{2}=\frac{16}{2}\)
y = 8
y = 8 is the required solution.

(c) we have a
\(\frac{\mathrm{a}}{5}\) + 3 = 2
Transposing 3 from L.H.S to R.H.S
\(\frac{\mathrm{a}}{5}\) = 2 – 3
\(\frac{\mathrm{a}}{5}\) = -1
Multiplying both sides by 5, we get
\(\frac{\mathrm{a} \times 5}{5}\) = -1 x 5
a = – 5
The solution is a = – 5.

(d) We have
\(\frac{\mathrm{q}}{4}\) + 7 = 5
Transposing 7 to R.H.S
\(\frac{\mathrm{q}}{4}\) = 5 – 7
\(\frac{\mathrm{q}}{4}\) = -2
Multiplying both sides by 4, we get
\(\frac{\mathrm{q}}{4}\) × 4 = – 2 × (4)
q = -8
The solution is q = – 8.

(e) We have
\(\frac{5}{2}\) x = – 10
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x x 2 = -10 × 2
5x = – 20
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{-20}{5}\)
The solution is x = – 4.

(f) We have
\(\frac{5}{2}\) x = \(\frac{25}{4}\)
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x × 2 = \(\frac{25}{4}\) × 2
5x = \(\frac{25}{2}\)
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{25}{2 \times 5}\)
x = \(\frac{5}{2}\)
The solution is x= \(\frac{5}{2}\) or x = 2 \(\frac{1}{2}\).

(g) We have
7m + \(\frac{19}{2}\) = 13
Transposing \(\frac{19}{2}\) to R.H.S
7m = 13 – \(\frac{19}{2}\)
7m = \(\frac{26-19}{2}\)
7m = \(\frac{7}{2}\)

(h) We have
6z + 10 = – 2
Transposing 10 to R.H.S
6z = – 2 – 10
6z = – 12
Dividing both sides by 6, we get
\(\frac{6 z}{6}=\frac{-12}{6}\)
z = – 2
The solution is z = – 2.
Dividing both sides by 7, we get
\(\frac{7 \mathrm{~m}}{7}=\frac{7}{2 \times 7}\)
m = \(\frac{1}{2}\)
The solution is m = \(\frac{1}{2}\) .

(i) We have
\(\frac{31}{2}=\frac{2}{3}\)
Multiplying both sides by 2, we get
\(\frac{31}{2}\) x 2 = \(\frac{2}{3}\) x 2
31 = \(\frac{4}{3}\)
Dividing both sides by 3, we get
\(\frac{31}{3}=\frac{4}{3 \times 3}\)
1 = \(\frac{4}{9}\)
The solution is 1 = \(\frac{4}{9}\)

(j) We have
\(\frac{2 \mathrm{~b}}{3}\) – 5 = 3
Transposing – 5 to R.H.S
\(\frac{2 \mathrm{~b}}{3}\) = 3 + 5
\(\frac{2 \mathrm{~b}}{3}\) = 8
Multiplying both sides by 3, we get
\(\frac{2 \mathrm{~b}}{3}\) × 3 = 8 × 3
2b = 24
Dividing both sides by 2, we get
\(\frac{2 b}{2}=\frac{24}{2}\)
b = 12
The solution is b = 12.

Question 2.
Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
Answer:
(a) 2(x + 4) = 12
Dividing both sides by 2, we get
\(\frac{2(x+4)}{2}=\frac{12}{2}\) ⇒ x + 4 = 6
Transposing 4 to R.H.S
x = 6 – 4 = 2

(b) 3(n – 5) = 21
Dividing both sides by 3, we get
\(\frac{3(n-5)}{3}=\frac{21}{3}\)
⇒ n – 5 = 7
Transposing – 5 to R.H.S
n = 7 + 5 = 12

(c) 3(n – 5) = -21
Dividing both sides by 3, we get
\(\frac{3(n-5)}{3}=\frac{-21}{3}\)
⇒ n – 5 = -7
Transposing – 5 to R.H.S
n = -7 + 5 = -2

(d) – 4(2 + x) =8
Dividing both sides by -4, we get
\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)
⇒ 2 + x = – 2
Transposing 2 to R.H.S
x = -2 -2 = -4

(e) 4(2 – x) = 8
Dividing both sides by 4, we get 4(2-x) _ 8
\(\frac{4(2-\mathrm{x})}{4}=\frac{8}{4}\)
⇒ 2 – x = 2
Transposing 2 to R.H.S
-x = 2 – 2 = 0
Multiplying both sides by (- 1), we get
-x × (-1) = 0 × (- 1)
x = 0

Question 3.
Solve the following equations:
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
Answer:
(a) 4 = 5 (p – 2)
Interchanging the sides, we get 5 ( p – 2) =4
Dividing both sides by 5, we get
\(\frac{5(p-2)}{5}=\frac{4}{5}\)
⇒ p – 2 = \(\frac{4}{5}\)
Transposing -2 to R.H.S
p = \(\frac{4}{5}\) + 2
= \(\frac{4+10}{5}=\frac{14}{5}\) = 2\(\frac{4}{5}\)

(b) – 4 = 5(p – 2)
Interchanging the sides, we get }
5 ( p – 2) = -4
Dividing both sides by 5, we get
\(\frac{5(p-2)}{5}=\frac{-4}{5}\)
p – 2 = \(\frac{-4}{5}\)
Transposing – 2 to R.H.S
p = \(\frac{-4}{5}\) + 2 = \(\frac{-4+10}{5}\)
= \(\frac{6}{5}\) = 1\(\frac{1}{5}\)

(c) 16 = 4 + 3(t + 2)
Interchanging the sides, we get 4 + 3 (t + 2) = 16
Transposing 4 to R.H.S
3(t + 2) – 16 – 4 = 12
Dividing both sides by 3, we get
\(\frac{3(t+2)}{3}=\frac{12}{3}\)
⇒ t + 2 = 4
Transposing 2 to R.H.S
t = 4 – 2 = 2

(d) 4 + 5 (p – 1) = 34
Transposing 4 to R.H.S
5 (p – 1) = 34 – 4 = 3
Dividing both sides by 5, we get
\(\frac{5(\mathrm{p}-1)}{5}=\frac{30}{5}\)
p – 1 = 6
Transposing -1 to R.H.S
p = 6 + 1 = 7

(e) 0 = 16 + 4(m – 6)
Interchanging the sides we get
16 + 4(m – 6) = 0
Transposing 16 to R.H.S
4(m – 6) = 0 – 16 = – 16
Dividing both sides by 4 we get
\(\frac{4(\mathrm{~m}-6)}{4}=\frac{-16}{4}\)
⇒ m – 6 = – 4
Transposing – 6 to R.H.S
m = -4 + 6 = 2

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2
Answer:
(a) Starting with x = 2
(1) Multiplying both sides by 3, we get
3x = 6
Subtracting 7 from both sides, we get
3x – 7 = 6 – 7
3x – 7 = – 1

(2) x = 2
Adding 5 on both sides, we get
x + 5 = 2 + 5
x + 5 = 7
Multiplying both sides by 4, we get
4(x + 5) = 7 × 4
4 (x + 5) = 28

(b) Starting with x = – 2
(1) x = – 2
Adding 11 to both sides, we get
x + 11 = – 2 + 11
x + 11 = 9

(2) x = – 2
Multiplying both sides by 5, we get
5x = – 10
Adding 3 on both sides, we get
5x + 3 = – 10 + 3
5x + 3 = – 7

(3) x = 2
Dividing both sides by 5, we get
\(\frac{x}{5}=\frac{2}{5}\)
Subtracting 2 from both sides, we get
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{2}{5}\) -2
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{2-10}{5}\)
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{-8}{5}\)

(3) x = – 2
Dividing both sides by 7, we get
\(\frac{x}{7}=\frac{-2}{7}\)
Add 3 on both sides, we get
\(\frac{\mathrm{X}}{7}\) + 3 = \(\frac{-2}{7}\) + 3
= \(\frac{-2+21}{7}\)
\(\frac{\mathrm{X}}{7}\) + 3 = \(\frac{19}{7}\)

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.2

Question 1.
A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD’s how many of each type would they make?

Solution:
Let the required number of young people be ‘x’
∴ 10% of x = 20
\(\frac{10}{100}\) × x = 20
x = \(\frac{20 \times 100}{10}\) = 200
(ii) Light music is liked by the maximum number of people
(iii) Total number of CD’s = 1000
Number of CD’s of Semi classical music = 20% of 1000
= \(\frac{20}{100}\) × 1000
= 200
Number of CD’s for classical = 10% of 1000
= \(\frac{10}{100}\) × 1000
= 100
Number of CD’s for folks = 30% of 1000
= \(\frac{30}{100}\) × 1000
= 300
Number of CD’s for light music = 40% of 1000
= \(\frac{40}{100}\) × 1000
= 400

Question 2.
A group of 360 people were asked to vote for their favourite season from the three season rainy, winter and summer.

(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
Solution:
(i) Winter season got the most votes.
(ii) Total votes = 90 + 120 + 150 = 360
Central angle of the sector corresponding to summer season
No. of people who vote for summer
= \(\frac{\text { No. of people who vote for summer }}{\text { No. of people }} \times 360\)
= \(\frac{90}{360}\) × 360°
= 90°
rainy sector = \(\frac{120}{360}\) × 360° = 120°
winter sector = \(\frac{150}{360}\) × 360° = 150°

Question 3.
Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Find the proportion of each sector. For example, Blue is \(\frac{18}{36}=\frac{1}{2}\); Green is \(\frac{9}{36}=\frac{1}{4}\) and so on. Use this to find the corresponding angles.
Solution:

The required pie chart is given below:

Question 4.
The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

(i) In which subject did the student score 105 marks? (Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than in Science and Hindi. (Hint: Just study the central angles).
Solution:
(i) Total marks = 540
∴ For 540 marks, the central angle = 360°
∴ For 105 marks, the central angle = \(\frac {360}{540}\) × 105 = 70°
Since the sector having central angle 70° is corresponding to Hindi. Thus, the student scored 105 marks in Hindi.
(ii) Marks obtained in Mathematics = \(\frac{90^{\circ}}{360^{\circ}} \times 540\) = 135
Students in Mathematics scored = 135 – 105 = 30
(iii) Sum of the central angles for social science and mathematics = 65° + 90° = 155°
Sum of the central angles for science and Hindi = 80 + 70 = 150°
∴ Marks obtained are proportional to the central angles corresponding to various items and 155° > 150°
∴ Marks obtained in Science and Mathematics are more than the marks obtained in Science and Hindi.

Question 5.
The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Solution:
Central angle of the sector representing
(a) Hindi language = \(\frac{40}{72}\) × 360° = 200°
(b) English language = \(\frac{12}{72}\) × 360° = 60°
(c) Marathi Language = \(\frac{9}{72}\) × 360° = 45°
(d) Tamil language = \(\frac{7}{72}\) × 360° = 35°
(e) Bengali language = \(\frac{4}{72}\) × 360 = 20°
Thus, the required pie chart is given below.

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.1

Question 1.
For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produces by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station. Give reason for each.
Solution:
We represent those data by a histogram which can be grouped into class intervals so, for (b) and (d), the data can be represented by histograms.

Question 2.
The shoppers who come to a departmental store are marked as man (M) woman (W), boy (B) or girl (G). The following list gives the shoppers who come during the first hour in the morning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution:
The frequency distribution table for the above data can be:

We can represent the above data by a bar graph as given below:

Question 3.
The weekly wages (in ₹) of 30 workers in factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks, make a frequency table with intervals as 800 – 810, 810 – 820, and so on.
Solution:

Question 4.
Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions.
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
Solution:
The histogram for the above frequency table is given below. Here we have represented the class intervals on the horizontal axis and frequencies of the class intervals along the y-axis (vertical axis).

Now, we can answer the question.
(i) The group 830 – 840 has the maximum number of workers.
(ii) Number of workers earning ₹ 850 or more = 1 + 3 + 1 + 1 + 4 = 10
(iii) Number of workers earning less than ₹ 850 = 3 + 2 + 1 + 9 + 5 = 20

Question 5.
The number of hours for which students of a particular class watched television during holidays is shown through the given graph. Answer the following.

(i) For how many hours did the maximum number of students watched TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours in watching TV?
Solution:
(i) The maximum number of students watch TV for 4 to 5 hours.
(ii) 4 + 8 + 22 = 34 students watch TV for less than 4 hours.
(iii) 8 + 6 = 14 students spend more than 5 hours in watching TV.