## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2

Question 1.

If 7 is added to five times a number, the result is 57. Find the number.

Solution:

Let the number be x

Five times a number is = 5x

If 7 is added, it becomes 7 + 5x

According to given condition,

7 + 5x = 57

⇒ 5x = 57 – 7

⇒ 5x = 50

⇒ x = 10

Question 2.

Find a number, such that one-fourth of the number is 3 more than 7.

Solution:

Let number = x

According to the condition,

\(\frac { 1 }{ 4 }\) x – 3 = 7

\(\frac { 1 }{ 4 }\) x = 7 + 3 = 10

x = 10 × 4 = 40

Number = 40

Question 3.

A number is as much greater than 15 as it is less than 51. Find the number.

Solution:

Let the number be x

If it is greater than 15, it becomes x – 15.

If it is less than 51, it becomes 51 – x

According to statement,

x – 15 = 51 – x

⇒ x + x = 51 + 15

⇒ 2x = 66

⇒ x = 33

Question 4.

If \(\frac { 1 }{ 2 }\) is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?

Solution:

Let the number = x

According to the condition,

(x – \(\frac { 1 }{ 2 }\)) × 4 = 5

⇒ 4x – 2 = 5

⇒ 4x = 5 + 2

⇒ 4x = 7

⇒ x = \(\frac { 7 }{ 4 }\)

Number = \(\frac { 7 }{ 4 }\)

Question 5.

The sum of two numbers is 80 and the greater number exceeds twice the smaller by 11. Find the numbers.

Solution:

Let the numbers be x and y

Smaller number = x

Greater number = y

If greater number exceeds twice the smaller by 11,

It becomes y = 2x + 11

According to statement,

x + 2x + 11 = 80

⇒ 3x + 11 = 80

⇒ 3x = 80 – 11

⇒ 3x = 69

⇒ x = 23

Smaller number = 23

Greater number = 2x + 11 = 2 × 23 + 11 = 46 + 11 = 57

Question 6.

Find three consecutive odd natural numbers whose sum is 87.

Solution:

Let the three consecutive odd natural numbers be

x, x + 2, x + 4.

According to statement,

x + x + 2 + x + 4 = 87

⇒ 3x + 6 = 87

⇒ 3x = 81

⇒ x = 27

x + 2 = 27 + 2 = 29 and x + 4 = 27 + 4 = 31

Three consecutive odd natural numbers are 27, 29 and 31.

Question 7.

In a class of 35 students, the number of girls is two-fifths of the number of boys. Find the number of girls in the class.

Solution:

Let the number of boys = x

The number of girls = \(\frac { 2x }{ 5 }\)

According to statement,

Question 8.

A chair costs ₹ 250 and the table costs ₹ 400. If a housewife purchased a certain number of chairs and two tables for ₹ 2800, find the number of chairs she purchased.

Solution:

Cost of a chair = ₹ 250

and cost of a table = ₹ 400

Let number of chairs = x

and number of tables = 2

Total cost = ₹ 2800

x × 250 + 2 × 400 = 2800

⇒ 250x + 800 = 2800

⇒ 250 x = ₹ 2800 – ₹ 800 = ₹ 2000

⇒ x = 8

Number of chairs = 8

Question 9.

Aparna got ₹ 27840 as her monthly salary and over-time. Her salary exceeds the overtime by ₹ 16560. What is her monthly salary?

Solution:

Let Apama’s monthly salary = ₹ x

Then over-time payment = ₹ (27840 – x)

According to the condition,

x – (27840 – x) = 16560

⇒ x – 27840 + x = 16560

⇒ 2x = 16560 + 27840 = ₹ 44400

⇒ x = 22200

Monthly salary = ₹ 22200

Question 10.

Heena has only ₹ 2 and ₹ 5 coins in her purse. If in all she has 80 coins in her purse amounting to ₹ 232, find the number of ₹ 5 coins.

Solution:

Total number of coins = 80

Let the number of ₹ 2 coins = x

The number of ₹ 5 coins = 80 – x

According to given statement,

2x + 5 (80 – x) = 232

⇒ 2x + 400 – 5x = 232

⇒ -3x = 232 – 400

⇒ -3x = -168

⇒ x = 56

Number of ₹ 5 coins = 80 – x = 80 – 56 = 24

Number of ₹ 5 coins = 24

Question 11.

A purse contains ₹ 550 in notes of denominations of ₹ 10 and ₹ 50. If the number of ₹ 50 notes is one less than that of ₹ 10 notes, then find the number of ₹ 50 notes.

Solution:

Total amount in a purse = ₹ 550

Let number of notes of ₹ 10 = x

The number of notes of ₹ 50 = x – 1

According to the condition,

x × 10 + (x – 1) × 50 = 550

⇒ 10x + 50x – 50 = 550

⇒ 10x = 550 + 50 = 600

⇒ x = 10

50 rupees notes = 10 – 1 = 9

Question 12.

After 12 years, 1 shall be 3 times as old as I was 4 years ago. Find my present age.

Solution:

Let my present age = x years

After 12 years, I will be = (x + 12) years

and 4 years ago, I was = (x – 4) years

According to the condition,

(x – 4) × 3 = x + 12

⇒ 3x – 12 = x + 12

⇒ 3x – x = 12 + 12

⇒ 2x = 24

⇒ x = 12

My present age = 12 years

Question 13.

Two equal sides of an isosceles triangle are 3x – 1 and 2x + 2. The third side is 2x units. Find x and the perimeter of the triangle.

Solution:

Two equal sides of an isosceles Δ are 3x – 1 and 2x + 2

3x – 1 = 2x + 2

3x – 2x = 2 + 1

x = 3

We know that

Perimeter of a Δ = (3x – 1) + (2x + 2) + (2x)

= (3 × 3 – 1) + (2 × 3 + 2) + (2 × 3)

= (9 – 1) + (6 + 2) + (6)

= 8 + 8 + 6

= 22 units

Question 14.

The length of a rectangle plot is 6 m less than thrice its breadth. Find the dimensions of the plot if its perimeter is 148 m.

Solution:

Let the breadth of a rectangle = x m.

Thrice its breadth = 3x m

Length of a rectangle = 3x – 6 m

Perimeter of a rectangle = 2 (l + b)

= 2 (3x – 6 + x)

= 2 (4x – 6)

= 8x – 12

But we are given, perimeter = 148 m

8x – 12= 148

8x = 148 + 12

8x = 160

x = 20 metres

Breadth = x = 20 metres

and Length = 3x – 6 = 3 × 20 – 6 = 60 – 6 = 54 metres.

Question 15.

Two complementary angles differ by 20°. Find the measure of each angle.

Solution:

We know that

Sum of measures of two complementary angles = 90°

⇒ x + y = 90° ……. (i)

But we are given x – y = 20° …… (ii)

2x = 110° [On comparing (i) and (ii)]

⇒ x = 55°

Now, x + y = 90°

⇒ y = 90° – x

⇒ y = 90° – 55° = 35°