NCERT Exemplar Solutions: Simple Equations

# NCERT Exemplar Solutions: Simple Equations | Mathematics (Maths) Class 7 PDF Download

## Exercise Page: 104

In the Questions 1 to 18, there are four options out of which, one is correct. Choose the correct one.
Q.1. The solution of the equation ax + b = 0 is
(a) a/b

(b) –b
(c) –b/a
(d) b/a
Ans: c
Solution:
Consider the given equation ax + b = 0
Transpose ‘b’ to right hand side then it becomes –b.
ax = -b
x = -b/a

Q.2. If a and b are positive integers, then the solution of the equation ax = b will always be a
(a) positive number
(b) negative number
(c) 1
(d) 0

Ans: a
Solution: The solution of the given equation ax = b
x = b/a

Q.3. Which of the following is not allowed in a given equation?
(a) Adding the same number to both sides of the equation.
(b) Subtracting the same number from both sides of the equation.
(c) Multiplying both sides of the equation by the same non-zero number.
(d) Dividing both sides of the equation by the same number.

Ans: d

Q.4. The solution of which of the following equations is neither a fraction nor an integer?
(a) 2x + 6 = 0
(b) 3x – 5 = 0
(c) 5x – 8 = x + 4
(d) 4x + 7 = x + 2

Ans: d

Solution: Consider the equation 4x + 7 = x + 2
Transpose 7 to right hand side then it becomes –7 and x to Left hand side then it becomes – x.
4x – x = 2 – 7
3x = -5
x = -5/3

Q.5. The equation which cannot be solved in integers is
(a) 5y – 3 = – 18
(b) 3x – 9 = 0
(c) 3z + 8 = 3 + z
(d) 9y + 8 = 4y – 7
Ans:
c
Solution: Consider the equation 3z + 8 = 3 + z
Transpose 8 to right hand side then it becomes –8 and z to Left hand side then it becomes – z.
3z – z = 3 – 8
2z = -5
z = -5/2

Q.6. If 7x + 4 = 25, then x is equal to
(a) 29/7
(b) 100/7
(c) 2
(d) 3

Ans: d
Solution: Consider the given equation 7x + 4 = 25
Transpose 4 to right hand side then it becomes –4
7x = 25 – 4
7x = 21
x = 21/7
x = 3

Q.7. The solution of the equation 3x + 7 = – 20 is
(a) 17/7
(b) -9
(c) 9
(d) 13/3
Ans:
b
Solution: Consider the given equation 3x + 7 = – 20
Transpose 7 to right hand side then it becomes –7
3x = -20 – 7
3x = -27
x = -27/3
x = -9

Q.8. The value of y for which the expressions (y – 15) and (2y + 1) become equal is
(a) 0
(b) 16
(c) 8
(d) – 16

Ans: d
Solution: Consider the given equation (y – 15) = (2y + 1)
Transpose -15 to right hand side then it becomes 15 and 2y to Left hand side then it becomes – 2y.
y – 2y = 15 + 1
-y = 16
y = -16

Q.9. If k + 7 = 16, then the value of 8k – 72 is
(a) 0
(b) 1
(c) 112
(d) 56

Ans: a
Solution: Consider the given equation k + 7 = 16
To find out the value of K.
Transpose 7 to right hand side then it becomes -7
k = 16 – 7
k = 9
Then, the value of 8k – 72 will be,
= (8 × 9) – 72
= 72 – 72
= 0

Q.10. If 43m = 0.086, then the value of m is
(a) 0.002
(b) 0.02
(c) 0.2
(d) 2

Ans: a
Solution: Consider the given equation 43m = 0.086
To find out the value of m.
m = 0.086/43
m = 0.002

Q.11. x exceeds 3 by 7, can be represented as
(a) x + 3 = 2
(b) x + 7 = 3
(c) x – 3 = 7
(d) x – 7 = 3

Ans: c

Q.12. The equation having 5 as a solution is:
(a) 4x + 1 = 2
(b) 3 – x = 8
(c) x – 5 = 3
(d) 3 + x = 8

Ans: c
Solution: Consider the equation 3 + x = 8.
Transpose 3 to right hand side then it becomes -3
x = 8 – 3
x = 5

Q.13. The equation having – 3 as a solution is:
(a) x + 3 =1
(b) 8 + 2x = 3
(c) 10 + 3x = 1
(d) 2x + 1 = 3

Ans: c
Solution: Consider the equation 10 + 3x = 1.
Transpose 10 to right hand side then it becomes -10
3x = 1 – 10
x = -9/3
x = -3

Q.14. Which of the following equations can be formed starting with x = 0?
(a) 2x + 1 = – 1
(b) x/ 2 + 5 = 7
(c) 3x – 1 = – 1
(d) 3x – 1 = 1
Ans: c
Solution:
Consider the equation 3x – 1 = – 1.
Transpose -1 to right hand side then it becomes 1
3x = -1 + 1
x = 0/3
x = 0

15. Which of the following equations cannot be formed using the equation x = 7?
(a) 2x + 1 = 15
(b) 7x – 1 = 50
(c) x – 3 = 4
(d) (x/7) – 1 = 0

Ans: b
Solution: Consider the equation 7x – 1 = 50.
Transpose -1 to right hand side then it becomes 1.
7x = 50 + 1
7x = 51
x = 51/7

Q.16. If (x/2) = 3, then the value of 3x + 2 is
(a) 20
(b) 11
(c) 13/2
(d) 8

Ans: a
Solution: Consider the given equation (x/2) = 3
To find out the value of x.
x = 3 × 2
x = 6
Then, the value of 3x + 2 will be,
= (3 × 6) + 2
= 18 + 2
= 20

Q.17. Which of the following numbers satisfy the equation –6 + x = –12 ?
(a) 2
(b) 6
(c) – 6
(d) – 2

Ans: c
Solution: Consider the given equation –6 + x = –12.
x = -12 + 6
x = -6
Left hand side is equal to Right hand side.

Q.18. Shifting one term from one side of an equation to another side with a change of sign is known as
(a) commutativity
(b) transposition
(c) distributivity
(d) associativity

Ans: b

Transposition

In Questions 19 to 48, fill in the blanks to make the statements true.
Q.19. The sum of two numbers is 60 and their difference is 30.
(a)
If smaller number is x, the other number is .(use sum)
Ans: From the question it is given that smaller number is x,
Then, the other number is 60 – x

(b) The difference of numbers in term of x is.
Ans:
From the question, smaller number be x, and other number be (60 – x)
Then,
Difference between the numbers = (60 – x) – x
= 60 – x – x
= 60 – 2x

(c) The equation formed is.
Ans:
As per the condition given in the question, difference of the two numbers is 30.
So, 60 – 2x = 30
Transpose, -2x to RHS then it becomes 2x and 30 to LHS it becomes -30,
60 – 30 = 2x
30 = 2x
2x – 30 = 0

(d) The solution of the equation is.
Ans: 30 = 2x
x = 30/2
x = 15

(e) The numbers are and.
Ans: x = 15
60 – x = 60 – 15
= 45
The numbers are 15 and 45.

Q.20. Sum of two numbers is 81. One is twice the other.

(a) If smaller number is x, the other number is.
Ans:
From the question it is given that smaller number is x and one is twice the other.
Then, the other number is 2x

(b) The equation formed is.
Ans: From the question, smaller number be x, and other number be 2x
Then,
Sum of two numbers is 81,
2x + x = 81

(c) The solution of the equation is.
Ans:
2x + x = 81
3x = 81
x = 81/3
x = 27

(d) The numbers are and.
Ans:
x = 27
2x = 2 × 27
= 54
The numbers are 27 and 54.

Q.21. In a test Abha gets twice the marks as that of Palak. Two times Abha’s marks and three times Palak’s marks make 280.

(a) If Palak gets x marks, Abha gets marks.
Ans:
From the question it is given that Palak gets x marks and Abha gets twice the marks as that of Palak.
Then, Abha gets 2x marks.

(b) The equation formed is.
Ans:
As per the condition given in the question, two times Abha’s marks and three times Palak’s marks make 280.
So, (2 × 2x) + (3 × x) = 280
Therefore, 4x + 3x = 280

(c) The solution of the equation is.
Ans:
Consider the equation,
4x + 3x = 280
7x = 280
x = 280/7
x = 40

(d) Marks obtained by Abha are.
Ans:
Marks obtained by Abha are 80.
2x = 2 × 40
= 80 marks

Q.22. The length of a rectangle is two times its breadth. Its perimeter is 60 cm.
(a) If the breadth of rectangle is x cm, the length of the rectangle is.
Ans:
From the question it is given that the breadth of rectangle is x cm and length of a rectangle is two times its breadth.
Then, length of the rectangle is 2x.

(b) Perimeter in terms of x is.
Ans:
We know that, perimeter of rectangle = 2(Length + breadth)
= 2(2x + x)
= 4x + 2x

(c) The equation formed is.
Ans:
As per the condition given in the question, perimeter is 60 cm
The equation formed is, 4x + 2x = 60
6x = 60

(d) The solution of the equation is.
Ans:
4x + 2x = 60
6x = 60
x = 60/6
x = 10

Q.23. In a bag there are 5 and 2 rupee coins. If they are equal in number and their worth is Rs. 70, then

(a) The worth of x coins of Rs. 5 each.
Ans:
The worth of x coins of Rs. 5 each 5x.

(b) The worth of x coins of Rs. 2 each.
Ans:
The worth of x coins of Rs. 2 each 2x.

(c) The equation formed is.
Ans:
As per the condition given in the question, If 5 and 2 rupee coins are equal in number and their worth is Rs. 70.
So, the equation is 5x + 2x = Rs. 70

(d) There are 5 rupee coins and 2 rupee coins.
Ans:
Consider the equation,
5x + 2x = 70
7x = 70
x = 70/7
x = 10
There are 10 Rs. 5 coins and 10 Rs. 2 coins.

Q.24. In a Mathematics quiz, 30 prizes consisting of 1st and 2nd prizes only are to be given. 1st and 2nd prizes are worth Rs. 2000 and Rs. 1000 respectively. If the total prize money is Rs. 52,000 then show that:

(a) If 1st prizes are x in number the number of 2nd prizes are .
Ans:
From the question it is given that, 30 prizes consisting of 1st.
So, If 1st prizes are x in number the number of 2nd prizes are 30 – x.

(b) The total value of prizes in terms of x are.
Ans:
Given, 1st and 2nd prizes are worth Rs. 2000 and Rs. 1000 respectively
= (Rs. 2000 × x) + (Rs. 1000 × (30 – x))

(c) The equation formed is.
Ans:
As per the condition given in the question, 30 prizes consisting of 1st and 2nd prizes only are to be given, the total prize money is Rs. 52,000.
Then, the equation is = Rs.1000x + Rs.30,000 = Rs. 52,000

(d) The solution of the equation is.
Ans:
Consider the equation, Rs.1000x + Rs.30,000 = Rs. 52,000
Transpose, 30000 to RHS then it becomes -30,000
1000x = 52,000 – 30,000
1000x = 22,000
x = 22,000/1000
x = 22

(e) The number of 1st prizes are and the number of 2nd prizes are.
Ans:
The number of 1st prize = x = 22
The number of 2nd prize = 30 – x = 30 – 22 = 8
The number of 1st prizes are 22 and the number of 2nd prizes are 8.

Q.25. If z + 3 = 5, then z =.
Ans:
Consider the equation, z + 3 = 5
Transpose 3 to RHS it becomes -3
z = 5 – 3
z = 2

Q.26. is the solution of the equation 3x – 2 = 7.
Ans:
Consider the given equation, 3x – 2 = 7
Transpose -2 to RHS then it becomes 2.
3x = 7 + 2
3x = 9
x =9/3
x =3
x = 3 is the solution of the equation 3x – 2 =7.

Q.27. is the solution of 3x + 10 = 7.
Ans:
Consider the given equation, 3x + 10 = 7
Transpose 10 to RHS then it becomes -10.
3x = 7 – 10
3x = -3
x = -3/3
x = -1
x = -1 is the solution of the equation 3x + 10 = 7.

Q.28. If 2x + 3 = 5, then value of 3x + 2 is .
Ans:
Consider the given equation, 2x + 3 = 5
Transpose 3 to RHS then it becomes -3.
2x = 5 – 3
2x = 2
x = 2/2
x = 1
then value of 3x + 2 is = (3 × 1) + 2
= 3 + 2
= 5

Q.29. In integers, 4x – 1 = 8 has solution.
Ans:
Consider the given integers, 4x – 1 = 8
Transpose -1 to RHS then it becomes 1.
4x = 8 + 1
4x = 9
x = 9/4
In integers, 4x – 1 = 8 has no solution.

Q.30. In natural numbers, 4x + 5 = – 7 has solution.
Ans:
Consider the given integers, 4x + 5 = – 7.
Transpose 5 to RHS then it becomes -5.
4x = – 7 – 5
4x = – 12
x = -12/4
x = – 3
In natural numbers, 4x + 5 = – 7 has no solution.

Q.31. In natural numbers, x – 5 = – 5 has solution.

Ans: Consider the given integers, x – 5 = – 5.
Transpose – 5 to RHS then it becomes 5.
x = – 5 + 5
x = 0
In natural numbers, x – 5 = – 5 has no solution.

Q.32. In whole numbers, x + 8 = 12 – 4 has solution.
Ans:
Consider the given integers, x + 8 = 12 – 4.
x + 8 = 8
Transpose 8 to RHS then it becomes – 8.
x = 8 – 8
x = 0
In natural numbers, x + 8 = 12 – 4 has one solution.

Q.33. If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as.
Ans:
Let us assume the number be x,
Given, If 5 is added to three times a number = 3x + 5
7 is subtracted from four times the same number = 4x – 7
If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as 3x + 5= 4x – 7.

Q.34. x + 7 = 10 has the solution.
Ans:
Consider the given equation, x + 7 = 10.
Transpose 7 to RHS then it becomes -7.
x = 10 – 7
x = 3

Q.35. x – 0 = ; when 3x = 12.
Ans:
Consider the given equation, 3x = 12
x = 12/3
x = 4
Then,
= x – 0
= 4 – 0
= 4

Q.36. x – 1= ; when 2x = 2.

Ans: Consider the given equation, 2x = 2
x = 2/2
x = 1
Then,
= x – 1
= 1 – 1
= 0

Q.37. x – = 15; when x/2 = 6.
Ans:
Consider the given eqaution, x/2 = 6
x = 6 × 2
x = 12
Then,
= x –
= 12 – (- 3)
= 12 + 3
= 15
x – (-3) = 15; when x/2 = 6

Q.38. The solution of the equation x + 15 = 19 is .

Ans: Consider the given equation, x + 15 = 19
Transpose 15 to RHS then it becomes -15.
x = 19 – 15
x = 4

Q.39. Finding the value of a variable in a linear equation that the equation is called a of the equation.
Ans: Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.

Q.40. Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the of the term.
Ans:
Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the sign of the term.

Q.41. If (9/5)x = (18/5), then x =
Ans:
Consider the given equation, (9/5)x = (18/5)
By cross multiplication x = (18/5) × (5/9)
= (18/9)
= 2
Then, the value of x = 2

Q.42. If 3 – x = -4, then x =
Ans: Consider the given equation, 3 – x = -4
Transpose –x to RHS then it becomes x and – 4 to LHS then it becomes 4,
3 + 4 = x
x = 7
Then, the value of x = 7

Q.43. If x – ½ = – ½, then x = .
Ans: Consider the given equation, x – ½ = – ½
Transpose – ½ to RHS then it becomes ½
x = – ½ + ½
x = 0
Then, the value of x = 0

Q.44. If (1/6) – x = 1/6, then x =.
Ans:
Consider the given equation, (1/6) – x = 1/6
Transpose – x to RHS then it becomes x and 1/6 to LHS then it becomes – 1/6
x = 1/6 – 1/6
x = 0
Then, the value of x = 0

Q.45. If 10 less than a number is 65, then the number is .
Ans: Given 10 less than a number is 65
Then, the equation will be x – 10 = 65
Transpose, – 10 to RHS then it becomes 10
x = 65 + 10
x = 75

Q.46. If a number is increased by 20, it becomes 45. Then the number is .
Ans: Let us assume the number be x.
If a number is increased by 20 = x + 20
If a number is increased by 20, it becomes 45, x + 20 = 45
Transpose, 20 to RHS then it becomes -20.
x = 45 – 20
x = 25

Q.47. If 84 exceeds another number by 12, then the other number is .
Ans: Let us assume the number be x.
If a number is increased by 12 = x + 12
If a number is increased by 12, it becomes 84, x + 12 = 84
Transpose, 12 to RHS then it becomes -12.
x = 84 – 12
x = 72

Q.48. If x – (7/8) = 7/8, then x = .
Ans:
Consider the equation, x – (7/8) = 7/8
Transpose, -7/8 to RHS then it becomes 7/8.
Then, x = (7/8) + (7/8)
x = (14/8) [divide by 2]
x = 7/4

In Questions 49 to 55, state whether the statements are True or False.
Q.49. 5 is the solution of the equation 3x + 2 = 17.

Ans: True.
Consider the equation, 3x + 2 = 17
Transpose 2 to RHS then it becomes – 2.
3x = 17 – 2
3x = 15
x = 15/3
x = 5

Q.50. 9/5 is the solution of the equation 4x – 1 = 8.
Ans:
False.
Consider the equation, 4x – 1 = 8
Transpose -1 to RHS then it becomes 1.
4x = 8 + 1
4x = 9
x = 9/4

Q.51. 4x – 5 = 7 does not have an integer as its solution.
Ans: False.
Consider the equation, 4x – 5 = 7
Transpose -5 to RHS then it becomes 5.
4x = 7 + 5
4x = 12
x = 12/4
x = 3

Q.52. One third of a number added to itself gives 10, can be represented as (x/3) + 10 = x
Ans:
False.
One third of a number added to itself = (x/3) + x
One third of a number added to itself gives 10, can be represented as (x/3) + x = 10

Q.53. 3/2 is the solution of the equation 8x – 5 = 7.
Ans:
True.
Consider the equation, 8x – 5 = 7
Transpose -5 to RHS then it becomes 5.
8x = 7 + 5
8x = 12
x = 12/8
x = 3/2

Q.54. If 4x – 7 = 11, then x = 4.
Ans: False.
Consider the equation, 4x – 7 = 11
Transpose -7 to RHS then it becomes 7.
4x = 11 + 7
4x = 18
x = 18/4
x = 9/2

Q.55. If 9 is the solution of variable x in the equation ((5x – 7)/2) = y, then the value of y is 28.
Ans: False.
Consider the equation, ((5x – 7)/2) = y
Then, from the question it is given that the value of x = 9
So, (((5 × 9) – 7)/2) = y
((45 – 7)/2) = y
38/2 = y
y = 19

Q.56. Match each of the entries in Column 1 with the appropriate entries in Column II.

Solution:
(i) → (C);
(ii) → (E);
(iii) → (F)
(iv) → (D);
(v) → (B);
(vi) → (A)
(i) x + 5 = 9 ⇒ x = 9 – 5 = 4
(ii) x – 7 = 4 ⇒ x = 4+ 7 = 11
(iii)
(iv)
(v)
(vi)Given that p = 2

Q.57. 13 subtracted from twice of a number gives 3.
Solution:
Let the number be x. So, twice of number – 2x On subtracting 13 from it, we get 2x – 13 Therefore, 2x – 13 = 3 is the required equation

Q.58. One-fifth of a number is 5 less than that number.
Solution:
Let the number be x.
So, one-fifth of number
Therefore, is the required equation.

Q.59. A number is 7 more than one-third of itself.
Solution:
Let the number be x.
So, one-third of number
7 more than
Therefore, is the required equation

Q.60. Six times a number is 10 more than the number.
Solution:
Let the number be x.
So, six times of the number = 6x
Therefore, 6x – x + 10 is the required equation.

Q.61. If 10 is subtracted from half of a number, the result is 4.
Solution:
Let the number be x.
So, half of the number
On subtracting 10 from it, we get
Thereforeis the required equation.

Q.62. Subtracting 5 from p, the result is 2.
Solution:
The number is p.
On subtracting 5 from it, we get p – 5
∴ p – 5 = 2 is the required equation.

Q.63. Five times a number increased by 7 is 27.
Solution:
Let the number be x.
So, five times of the number = 5x
When increased by 7, it gives the expression 5x +7
∴ 5x + 7 = 27 is the required equation.

Q.64. Mohan is 3 years older than Sohan. The sum of their ages is 43 years.
Solution:
Let Sohan is x years old.
So, Mohan is x + 3 years old.
∴ Sum of their ages be x + (x + 3).
∴ x + (x + 3) = 43 is the required equation.

Q.65. If 1 is subtracted from a number and the difference is multiplied by 1/2 the result is 7.
Solution:
Let the number be x.
On subtracting 1 from it, we get x – 1
Multiplying it by 1/2 we get
is the required equation.

Q.66. A number divided by 2 and then increased by 5 is 9.
Solution:
Let the number be x.
Dividing the number by 2, we get x/2
When, increased by 5, it gives the expression

is the required equation.

Q.67. The sum of twice a number and 4 is 18.
Solution:
Let the number be x.
So, twice of the number = 2x
On adding 4 to it, we get 2x + 4
∴ 2x + 4 = 18 is the required equation.

Q.68. The age of Sohan Lal is four times that of his son Amit. If the difference of their ages is 27 years, find the age of Amit.
Solution:
Let the age of Amit be x years.
So, age of Sohan Lal = 4x years
According to question, 4x – x = 27

Thus, Amit is 9 years old.

Q.69. A number exceeds the other number by 12. If their sum is 72, find the numbers.
Solution:
Let the number be x.
∴ Other number = x +12
According to question,
x + x + 12 = 72
⇒ 2x + 12 = 72 ⇒ 2x = 72 – 12 = 60

Hence, the numbers are 30 and 42

Q.70. Seven times a number is 12 less than thirteen times the same number. Find the number.
Solution:
Let the number be x.
So, seven times of the number = 7x
Thirteen times of the number = 13x
According to question,
7x = 13x – 12
⇒ 12 = 13x – 7x
⇒ 6x = 12

Thus, r -2 is the required number.

Q.71. The interest received by Karim is Rs. 30 more than that of Ramesh. If the total interest received by them is Rs. 70, find the interest received by Ramesh.
Solution:
Let the interest received by Ramesh be x.
So, interest received by Karim = Rs. (30+ x)
According to question,
x + x + 30 = 70
⇒ 2x = 70 – 30

Thus, Rs. 20 is the interest received by Ramesh

Q.72. Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is Rs. 125 more than that of Subramaniam. If the total money paid by them is Rs. 975, find the amount of money donated by Subramaniam.
Solution:
Let the amount of money donated by Subramaniam be x.
So, the amount paid by Naidu is= (x + 125)
According to question,
x + x + 125 = 975
⇒ 2x = 975 – 125 = 850

Thus, Rs.425 is donated by Subramaniam.

Q.73. In a school, the number of girls is 50 more than the number of boys. The total number of students is 1070. Find the number of girls.
Solution:
Let the number of girls be x.
So, the number of boys = x – 50

According to question,
x + x – 50 = 1070
⇒ 2x = 1070 + 50 = 1120

Thus, 560 are girls.

Q.74. Two times a number increased by 5 equals 9. Find the number.
Solution:
Let the number be x.
So, two times of the number = 2x
When, increased by 5, it gives the expression 2x + 5
∴ 2x + 5 = 9
⇒ 2x = 9 – 5 = 4

Thus, x = 2 is the required number.

Q.75. 9 added to twice a number gives 13. Find the number.
Solution:
Let the number be x.
So, twice of the number = 2x
On adding 9 to it, we get 2x + 9
∴ 2x + 9 = 13
⇒ 2x = 13 – 9 = 4

Thus, x = 2 is the required number.

Q.76. 1 subtracted from one-third of a number gives 1. Find the number.
Solution:
Let the number be x.
So, one third of the number
On subtracting 1 from it, we get

Thus, x = 6 is the required number.

Q.77. After 25 years, Rama will be 5 times as old as he is now. Find his present age.
Solution:
Let present age of Rama be x years.
So, five times of his age = 5x.
According to question,
5x = x + 25
⇒ 5x – x = 25
⇒ 4x = 25

Thus, at present Rama is years old.

Q.78. After 20 years, Manoj will be 5 times as old as he is now. Find his present age.
Solution:
Let present age of Manoj be x years.
So, five times of his age = 5x
According to question,
5 = x + 20
⇒ 5x – x = 20
⇒ 4x = 20

Thus, at present Manoj is 5 years old.

Q.79. My younger sister’s age today is 3 times, what it will be 3 years from now minus 3 times what her age was 3 years ago. Find her present age.
Solution:
Let present age of my younger sister be x years.
After 3 years, her age will be = (x + 3) years
Before 3 years, her age was = (x – 3) years
According to question,
x = 3 (x + 3) – 3 (x – 3)
⇒ x = 3 (x+3 – x + 3)
⇒ x = 3(6) = 18
Thus, at present my younger sister is 18 years old.

Q.80. If 45 is added to half a number, the result is triple the number. Find the number.
Solution:
Let the number be x.
So, half of the number
On adding 45 to it, we get

Q.81. In a family, the consumption of wheat is 4 times that of rice. The total consumption of the two cereals is 80 kg. Find the quantities of rice and wheat consumed in the family.
Solution:
Let the quantity of rice consumed in the family be x kg.
So, quantity of wheat consumed = 4x kg
According to question, x + 4x = 80
⇒ 5x = 80

Thus, 16 kg rice consumed in the family and 4 × 16 = 64 kg wheat consumed in the family

Q.82. In a bag, the number of one rupee coins is three times the number of two rupees coins. If the worth of the coins is 120, find the number of 1 rupee coins.
Solution:
Let two rupees coins are x in numbers.
So, one rupee coins are 3x in numbers.
According to question, 1(3x) + 2(x) = 120
⇒ 3x + 2x = 120
⇒ 5x = 120

Thus, 3 × 24 – 72 coins are of one rupee coins.

Q.83. Anamika thought of a number. She multiplied it by 2, added 5 to the product and obtained 17 as the result. What is the number she had thought of?
Solution:
Let the number Anamika thought be x.
Multiplying it by 2, we get 2x
On adding 5 to it, we get 2x + 5
∴ 2x + 5 = 17
⇒ 2x = 17 – 5 = 12

Thus, she had thought of number 6.

Q.84. One of the two numbers is twice the other. The sum of the numbers is 12. Find the numbers.
Solution:
Let the one number be x.
So, other number = 2x
According to question,
x + 2x = 12
⇒ 3x = 12

∴ Other number is 2x = 2 × 4 = 8

Q.85. The sum of three consecutive integers is 5 more than the smallest of the integers. Find the integers.
Solution:
Let the smallest integer be x
So, next two consecutive integer would be x + 1 and x + 2 respectively.
According to question,
x + x + 1 + x + 2 = x + 5
⇒ 3x + 3 = x + 5 ⇒ 3x – x = 5 – 3
⇒ 2x – 2 ⇒ x = 1
Thus, next two consecutive integers are 2 and 3 respectively.

Q.86. A number when divided by 6 gives the quotient 6. What is the number?
Solution:
Let the number be x.
Dividing it by 6, we get x/6

Thus, x = 36 is the required number.

Q.87. The perimeter of a rectangle is 40 m. The length of the rectangle is 4 m less than 5 times its breadth. Find the length of the rectangle.
Solution:
Let the breadth of rectangle be x m.
So, five times of breadth = 5x
Therefore, length of rectangle = (5x – 4) m
Perimeter of rectangle = 40 m
⇒ 2(x + 5x – 4) = 40
⇒ 2(6x – 4) = 40 ⇒ 6x – 4 = 20
⇒ 6x = 20 + 4 = 24

Thus, length of rectangle = (5 × 4 – 4) = 16 m

Q.88. Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.
Solution:
Let length of equal sides of an isosceles triangle be 2x cm.
So, the third side of triangle x cm
Perimeter of triangle = 30 cm
⇒ x + 2x + 2x = 30
⇒ 5x = 30

Thus, 6 cm, 12 cm and 12 cm are the required sides of the triangle.

Q.89. The sum of two consecutive multiples of 2 is 18. Find the numbers.
Solution:
Let first multiple of 2 be x.
So, next multiple of 2 = x + 2
According to question,
x + x + 2 = 18 ⇒ 2x = 18 – 2 = 16
⇒ x = 8
Thus, 8 and 10 are required numbers.

Q.90. Two complementary angles differ by 20<sup>o</sup>. Find the angles.
Solution:
Let one angle be x.
So, complement of x = 90°- x
According to question,
x – (90°- x) = 20°
⇒ x – 90° + x = 20°
⇒ 2x = 20° + 90° = 110°

∴ Complement of x = 90°- x = 90°- 55° = 35°
Thus, 55° and 35° are required complementary angles.

Q.91. 150 has been divided into two parts such that twice the first part is equal to the second part. Find the parts.
Solution:
Let first part be x
So, other part = 150 – x
According to question, 2x = 150 – x
⇒ 2x + x = 150 ⇒ 3x = 150

Thus, other part = 150 – x = 150 – 50 = 100
Hence, 150 has been divided into 50 and 100.

Q.92. In a class of 60 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.
Solution:
Let the number of boys in class = x
So, the number of girls in class = 60 – x
According to question,

⇒ 180 – 3x = x ⇒ 180 = 3x + x
⇒ 4x = 180

Thus, number of boys in the class = 45
And number of girls in the class = 60 – 45 = 15

Q.93. Two-third of a number is greater than one third of the number by 3. Find the number.
Solution:
Let the number be x.
So, two-third of numbers
According to question,

Thus, x = 9 is the required number.

Q.94. A number is as much greater than 27 as it is less than 73. Find the number.
Solution:
Let the number be x.
According to question,
x – 27 = 73 – x
⇒ x + x = 73 + 27 ⇒ 2x = 100

Thus, x = 50 is the required number.

Q.95. A man travelled two fifth of his journey by train, one-third by bus, one-fourth by car and the remaming 3 km on foot. What is the length of his total journey?
Solution:
Let the length of total journey be x km.
∴ Journey by train = x km
∴ Joumey by bus=
∴ Journey by car
And journey on foot

Thus, x = 180 km is the length of total journey

Q.96. Twice a number added to half of itself equals 24. Find the number.
Solution:
Let the number be x.
According to question,

Thus, x = 9.6 is the required number.

Q.97. Thrice a number decreased by 5 exceeds twice the number by 1. Find the number.
Solution:
Let the number be x.
So, thrice of the number = 3x
When it decreased by 5, we get 3x – 5
According to question,
3x – 5 = 2x + 1
⇒ 3x – 2x = 1 + 5 ⇒ x = 6
Thus, x = 6 is the required number.

Q.98. A girl is 28 years younger than her father. The sum of their ages is 50 years. Find the ages of the girl and her father.
Solution: Let the age of father be x years.
So, age of his girl = (x – 28) years
According to question,
x + x – 28 = 50
⇒ 2x = 50 + 28 = 78

Thus, age of father = 39 years
And age of his girl = 11 years

Q.99. The length of a rectangle is two times its width. The perimeter of the rectangle is 180 cm. Find the dimensions of the rectangle.
Solution:
Let the width of rectangle be x cm.
So, the length of rectangle = 2x
Perimeter of rectangle = 180 cm
⇒ 2(2x + x) = 180

Thus, width of rectangle = 30 cm
And length of rectangle = 2 × 30 = 60 cm

Q.100. Look at this riddle?
If she answers the riddle correctly however will she pay for the pencils?

Solution:
Let the cost of one pencil be Rs. x
So, the cost of 5 pencils = Rs. 5x
Cost of 7 pencils = Rs. 7x
According to question,
7x = 5x + 6
⇒ 7x – 5x = 6
⇒ 2x = 6

Cost of 10 pencils = Rs. (3 × 10) = Rs. 30

Q.101.In a certain examination, a total of 3768 students secured first division in the years 2006 and 2007. The number of first division in 2007 exceeds those in 2006 by 34. How many students got first division in 2006?
Solution:
Let number of students got first division in 2006 be x.
So, the number of students got first division in 2007 = 3768 – x.
According to question,
3768 – x = x + 34
⇒ 3768 – 34 = x + x ⇒ 2x = 1867

Thus, 1867 students got first division in 2006

Q.102. Radha got Rs. 17,480 as her monthly salary and over time. Her salary exceeds the over-time by Rs. 10,000. What is her monthly salary?
Solution:
Let Radha’s monthly salary = Rs. x
So, money got by her in over time = Rs. (17480 – x)
According to question,
x = 17480 – x + 10000
⇒ x + x = 17480 + 10000
⇒ 2x = 27480

Thus, 13740 is her monthly salary.

Q.103. If one side of a square is represented by 18 – 20 and the adjacent side is represented by 42 – 13x, find the length of the side of the square.
Solution:
Since, square has all sides equal
∴ 18x – 20 = 42 – 13x
⇒ 18x + 13x = 42 + 20
⇒ 31x= 62

∴ Side of square = 18 × 2 – 20 = 36 – 20 = 16
Thus, length of the side of the square is 16 units.

Q.104. Follow the directions and correct the given incorrect equation, written in Roman numerals:
(a) Remove two of these matchsticks to make a valid equation:

(b) Remove one matchstick to make the equation valid. Find two different solutions.

Solution:
(a) Incorrect equation is

After removing two matchsticks the correct equation is

(b) Incorrect equation is

After removing one matchstick the correct equation is

Q.105. What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below:
If i + 69 = 70, then i = ?
If 8u = 6 +8, then u = ?
If 4a = -5a + 45, then a = ?
If 4q + 5 = 17, then q = ?
If – 5t – 60 = -70, then t = ?
If then s = ?
Ifthen p=_____?
If 3c = c + 12, then c = ______?
If 3 (k+ 1) = 24, then k= _____?
For riddle answer : substitute the number for the letter it equals

Solution:
i + 69 = 70 ⇒ i = 70 – 69 = 1
8u = 6u +8 ⇒ 8u – 6u = 8
⇒ 2u = 8 ⇒ u = 4
4a = -5a + 45 ⇒ 4a + 5a = 45
⇒ 9a – 45 ⇒ a = 5
4q + 5 = 17 ⇒ 4q = 17 – 5
⇒ 4q = 12 ⇒ q = 3
-5t – 60 = – 70 ⇒ -5t = -70 + 60
⇒ -5t = -10 ⇒ t = 2

Thus, 1 = i, 2 = t, 3 = q, 4 = u, 5 = a, 6 = c, 7 = k, 8 = s, 9 = P
Substituting the letter for corresponding number in equation, we get ‘it quacks up’.

Q.106. The three scales below are perfectly balanced if . = 3. What are the values of Δ and *?

Solution:

Q.107. The given figure represents a weighing balance. The weights of some objects in the balance are given. Find the weight of each square and the circle.

Solution:
From the weighing balance, we have

The document NCERT Exemplar Solutions: Simple Equations | Mathematics (Maths) Class 7 is a part of the Class 7 Course Mathematics (Maths) Class 7.
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## Mathematics (Maths) Class 7

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## FAQs on NCERT Exemplar Solutions: Simple Equations - Mathematics (Maths) Class 7

 1. How can simple equations be solved?
Ans. Simple equations can be solved by isolating the variable on one side of the equation using inverse operations such as addition, subtraction, multiplication, and division.
 2. What is the difference between an equation and an expression?
Ans. An equation contains an equal sign, showing that two expressions are equal, while an expression does not have an equal sign and represents a mathematical phrase.
 3. How can we verify if a solution to a simple equation is correct?
Ans. To verify if a solution to a simple equation is correct, substitute the value back into the original equation and check if both sides are equal.
 4. Can a simple equation have more than one solution?
Ans. Yes, a simple equation can have more than one solution, especially when dealing with variables that can take on multiple values.
 5. What are some common mistakes to avoid when solving simple equations?
Ans. Common mistakes when solving simple equations include forgetting to perform the same operation on both sides of the equation, incorrectly applying inverse operations, and not simplifying expressions before solving.

## Mathematics (Maths) Class 7

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