## RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
- RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
- RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
- RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

**Question 1.**

**Solution:**

Let the required number = x

Then 2x – 7 = 45

2x = 45 + 7 = 52

x = 26

Required number = 26

**Question 2.**

**Solution:**

Let the required number = x Then

3x + 5 = 44

⇒ 3x = 44 – 5 = 39

x = 13

Required number = 13

**Question 3.**

**Solution:**

Let the required fraction = x

then 2x + 4 = \(\frac { 26 }{ 5 }\)

**Question 4.**

**Solution:**

Let the required number = x

and half of .the number = \(\frac { x }{ 2 }\)

**Question 5.**

**Solution:**

Let the required number = x

Two third of the number = \(\frac { 2 }{ 3 }\) x

**Question 6.**

**Solution:**

Let the required number = x

Then, 4x = x + 45

⇒ 4x – x = 45

⇒ 3x = 45

⇒ x = 15

Required number = 15

**Question 7.**

**Solution:**

Let the required number = x

Then x – 21 = 71 – x

⇒ x + x = 71 + 21

⇒ 2x = 92

⇒ x = 46

**Question 8.**

**Solution:**

Let the original number = x

Then \(\frac { 2 }{ 3 }\) of the number = \(\frac { 2 }{ 3 }\) x

**Question 9.**

**Solution:**

Let the second number = x

then first number = \(\frac { 2 }{ 5 }\) x

their sum = 70

**Question 10.**

**Solution:**

Let the required number = x

**Question 11.**

**Solution:**

Let the required number = x

Fifth part of the number = \(\frac { x }{ 5 }\)

Fourth part of the number = \(\frac { x }{ 4 }\)

**Question 12.**

**Solution:**

Let first natural number = x then

next number = x + 1

x + x + 1 = 63

⇒ 2x = 63 – 1 = 62

x = 31

first number = 31

and second number = 31 + 1 = 32

Numbers are 31, 32

**Question 13.**

**Solution:**

Let first odd number = 2x + 1

second odd number = 2x + 3

2x + 1 + 2x + 3 = 76

⇒ 4x + 4 = 76

⇒ 4x = 76 – 4 = 72

x = 18

First number = 2x + 1 = 2 x 18 + 1 = 36 + 1 = 37

Second number = 2x + 3 = 2 x 18 + 3 = 36 + 3 = 39

Numbers are 37, 39

**Question 14.**

**Solution:**

Let first positive even number = 2x

Second number = 2x + 2

Third number = 2x + 4

2x + 2x +2 + 2x + 4 = 90

⇒ 6x + 6 = 90

⇒ 6x = 90 – 6 = 84

x = 14

First even number = 2x = 2 x 14 = 28

Second number = 2x + 2 = 2 x 14 + 2 = 28 + 2 = 30

Third number = 30 + 2 = 32

Required numbers are 28, 30, 32

**Question 15.**

**Solution:**

Sum of two numbers = 184

Let first number = x

Then second number = 184 – x

First part = 72

Second part = 184 – 72 = 112

Hence parts are 72, 112

**Question 16.**

**Solution:**

Total number of notes = 90

Let number of notes of Rs. 5 = x

Then number of notes of Rs.10 = 90 – x

Then x x 5 + (90 – x) x 10 = 500

⇒ 5x + 900 – 10x = 500

⇒ -5x = 500 – 900 = -400

x = 8

Number of 5 rupees notes = 80

and ten rupees notes = 90 – 80 = 10

**Question 17.**

**Solution:**

Amount of coins = Rs. 34

Let 50 paisa coins = x

then 25 paisa coins = 2x

Number of 50 paisa coins = 34

and number of 25 paisa coins = 2x = 2 x 34 = 68

**Question 18.**

**Solution:**

Let present age of Raju’s cousin = x years

then age of Raju = (x – 19) years

After 5 years,

Raju’s age = x – 19 + 5 = (x – 14) years

and his cousin age = x + 5

(x – 14) : (x + 5) = 2 : 3

⇒ \(\frac { x – 14 }{ x + 5 }\) = \(\frac { 2 }{ 3 }\)

⇒ 3(x – 14) = 2 (x + 5) (By cross multiplication)

⇒ 3x – 42 = 2x + 10

⇒ 3x – 2x = 10 + 42

⇒ x = 52

Raju’s age = x – 19 = 52 – 19 = 33 years

and his cousin age = 52 years.

**Question 19.**

**Solution:**

Let present age of son = x years

Age of father = (x + 30) years

12 years after,

Father’s age = x + 30 + 12 = (x + 42) years

and son’s age = (x + 12) years

(x + 42) = 3(x + 12)

⇒ x + 42 = 3x + 36

⇒ 3x + 36 = x + 42

⇒ 3x – x = 42 – 36

⇒ 2x = 6

⇒ x = 3

Son’s age = 3 years

Father’s age = 3 + 30 = 33 years

**Question 20.**

**Solution:**

Ratio in present ages of Sonal and Manoj = 7 : 5

Let Sonal’s age = 7x

then Manoj’s age = 5x

10 years hence,

Sonal’s age will be = 7x + 10

and Manoj’s age = 5x + 10

Sonal’s present age = 7x = 7 x 5 = 35 years

and Manoj’s age = 5x = 5 x 5 = 25 years

**Question 21.**

**Solution:**

Five years ago,

Let Son’s age = x years

and father’s age = 7x years

Present age of son = (x + 5) years

and age of father = (7x + 5) years

5 years hence,

father’s age = 7x + 5 + 5 = 7x + 10

and Son’s age = x + 5 + 5 = x + 10

(7x + 10) = 3(x + 10)

⇒ 7x + 10 = 3x + 30

⇒ 7x – 3x = 30 – 10

⇒ 4x = 20

⇒ x = 5

Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years

and son’s age = x + 5 = 5 + 5 = 10 years

**Question 22.**

**Solution:**

Let age of Manoj 4 years ago = x

then his present age = x + 4

After 12 years his age will be = x + 4 + 12 = x + 16

x + 16 = 3(x)

x + 16 = 3x

⇒ 16 = 3x – x

⇒ 2x = 16

x = 8

His present age = 8 + 4 = 12 years

**Question 23.**

**Solution:**

Let total marks = x

Pass marks = 40% of x = \(\frac { 40x }{ 100 }\) = \(\frac { 2 }{ 5 }\) x

No. of marks got by Rupa = 185

No. of marks by which she failed = 15

Pass marks = 185 + 15 = 200

\(\frac { 2 }{ 5 }\) x = 200

⇒ x = \(\frac { 200 x 5 }{ 2 }\) x

⇒ x = 500

Hence total marks = 500

**Question 24.**

**Solution:**

Sum of digits = 8

Let units digit = x

Then tens digit = 8 – x

and number will be x + 10 (8 – x) ….(i)

By adding 18, the digits are reversed then

units digit = 8 – x

and tens digit = x

Number = (8 – x) = 10x

According to the condition,

(8 – x) + 10x = 18 + x + 10 (8 – x)

⇒ 8 – x + 10x = 18 + x + 80 – 10x

⇒ 10x – x – x + 10x = 18 + 80 – 8

⇒ 18x = 90

⇒ x = 5

Number is

x + 10(8 – x) = 5 + 10(8 – 5) = 5 + 10 x 3 = 35

**Question 25.**

**Solution:**

Cost of 3 tables and 2 chairs = 1850

Cost of table = Rs. 75 + cost of a chair

Let cost of chair = Rs. x,

then Cost of table = Rs. 75 + x

According to the condition,

3 (75 + x) + 2x = 1850

⇒ 225 + 3x + 2x = 1850

⇒ 225 + 5x = 1850

⇒ 5x = 1850 – 225 = 1625

x = 325

Cost of chair = Rs. 325

and cost of table = Rs. 325 + 75 = Rs. 400

**Question 26.**

**Solution:**

S.P of article = Rs. 495

gain = 10%

Let cost price = Rs. x

**Question 27.**

**Solution:**

Perimeter of field = 150 m

Length + Breadth = \(\frac { 150 }{ 2 }\) = 75 m

[Perimeter = 2(l + b)]

Let length = x Then breadth = 75 – x

Then x = 2(75 – x)

⇒ x = 150 – 2x

⇒ x + 2x = 150

⇒ 3x = 150

⇒ x = \(\frac { 150 }{ 3 }\) = 50

Length = 50 m

and breadth = 75 – 50 = 25 m

**Question 28.**

**Solution:**

Perimeter of an isosceles triangle = 55 m

Let the third side of an isosceles triangle = x

Then each equal side = (2x – 5) m

According to the condition,

x + 2 (2x – 5) = 55

⇒ x + 4x – 10 = 55

⇒ 5x = 55 + 10

⇒ 5x = 65

⇒ x = 13

and 2x – 5 = 2 x 13 – 5 = 21 m

Sides will be 13m, 21m, 21m

**Question 29.**

**Solution:**

Sum of two complementary angles = 90°

Let first angle = x

then second = 90° – x

x – (90 – x) = 8

⇒ x – 90 + x = 8

⇒ 2x = 8 + 90

⇒ 2x = 98

⇒ x = 49

first angle = 49°

and second angle = 90° – 49° = 41°

Hence angles are 41°, 49°

**Question 30.**

**Solution:**

Sum of two supplementary angles = 180°

Let first angle = x

Then second angle = 180° – x

x – (180° – x) = 44°

⇒ x – 180° + x = 44°

⇒ 2x = 44° + 180° = 224°

⇒ 2x = 224°

⇒ x = 112°

First angle = 112°

and second angle = 180° – 112° = 68°

Hence angles are 68°, 112°

**Question 31.**

**Solution:**

In an isosceles triangle

Let each equal base angles = x

Then vertex angle = 2x

According to the condition,

x + x + 2x = 180° (sum of angles of a triangle)

⇒ 4x = 180°

⇒ x = 45°

Then vertex angle = 2x = 2 x 45° = 90°

Angles of the triangle are 45°, 45° and 90°

**Question 32.**

**Solution:**

Let length of total journey = x km

According to the condition,

⇒ 39x + 80 = 40x

⇒ 40x – 39x = 80

⇒ x = 80

Total journey = 80km

**Question 33.**

**Solution:**

No. of days = 20 Let no. of days he worked = x

Then he will receive amount = x x Rs. 120 = Rs. 120x

No. of days he did not work = 20 – x

Fine paid = (20 – x) x Rs. 10 = Rs. 10(20 -x)

120x – 10 (20 – x) = 1880

⇒ 120x – 200 + 10x = 1880

⇒ 130x = 1880 + 200 = 2080

x = 16

No. of days he remained absent = 20 – x = 20 – 16 = 4 days

**Question 34.**

**Solution:**

Let value of property = x

**Question 35.**

**Solution:**

Solution = 400 mL

Quantity of alcohol = 15% of 400 mL

= \(\frac { 400 x 15 }{ 100 }\) = 60 mL

Let pure alcohol added = x mL

Total solution = 400 + x

and total alcohol = (x + 60)

Now (400 + x) x 32% = x + 60

⇒ (400 + x) x \(\frac { 32 }{ 100 }\) = x + 60

⇒ 32 (400 + x) = 100 (x + 60)

⇒ 12800 + 32x = 100x + 6000

⇒ 12800 – 6000 = 100x – 32x

⇒ 6800 = 68x

⇒ x = 6800

Pure alcohol added = 100 mL

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B are helpful to complete your math homework.

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