## RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C

**Question 1.**

**Solution:**

Let the required number be x.

Then, x + 9 = 36

=> x = 36 – 9

(Transposing 9 to R.H.S.)

=> x = 27

The required number = 27.

**Question 2.**

**Solution:**

Let the required number be x. Then,

4x – 11 = 89

=> 4x = 89 + 11

**Question 3.**

**Solution:**

Let the required number be x. Then,

x x 5 = x + 80

=> 5x = x + 80

=> 5x – x = 80

**Question 4.**

**Solution:**

Let the three consecutive numbers be x,

x + 1 and x + 2. Then,

x + (x + 1) + (x + 2) = 114

=> x – x + 1 + x + 2 = 114

**Question 5.**

**Solution:**

Let the required number be x. Then

x x 17 + 4 = 225

=> 17x + 4 = 225

=> 17x = 225 – 4

**Question 6.**

**Solution:**

Let the required number be x. Then

3 x + 5 = 50

=> 3 x = 50 – 5

**Question 7.**

**Solution:**

Let the smaller number be x.

Then the other number = x + 18

According to question,

x + (x + 18) = 92

=> 2 x + 18 = 92

=> 2 x = 92 – 18

(Transposing 18 to R.H.S.)

=> 2 x = 74

=> x = 37

(Dividing both sides by 2)

One number = 37

Another number = 37 + 18

= 55.

**Question 8.**

**Solution:**

Let the smaller number be x.

Then, the other number = 3 x

According to question, x + 3 x = 124

=> 4 x = 124

**Question 9.**

**Solution:**

Let the smallest number be x.

Then, the other number = 5 x

According to question,

5 x – x = 132

=> 4 x = 132

**Question 10.**

**Solution:**

Let the two consecutive even number be x and x + 2.

Then x + x + 2 = 74

=> 2 x + 2 = 74

=> 2 x = 74 – 2

(Transposing 2 to RHS.)

=> 2x = 72

=> x = 36

(Dividing both sides by 2)

The required numbers are 36 and (36 + 2) i.e. 36 and 38.

**Question 11.**

**Solution:**

Let the three consecutive odd numbers be x, (x + 2) and (x + 4).

According to the question,

x + (x + 2) + (x + 4) = 21

3x + 6 = 21

=> 3 x = 21 – 6

(Transposing 6 to R.H.S.)

=> 3 x = 15

=> x = 5

(Dividing both sides by 3)

The required numbers are 5, 5 + 2 and 5 + 4 i.e. 5, 7 and 9.

**Question 12.**

**Solution:**

Let the present age of Ajay be x years. Then, the present age of Reena = (x + 6) years

According to the question,

x + (x + 6) = 28

2x + 6 = 28

=> 2x = 28 – 6

(Transposing 6 to R.H.S.)

=> 2 x = 22

=> x = 11

(Dividing both sides by 2)

Present age of Ajay = 11 years

and present age of Reena = 11 + 6

= 17 years.

**Question 13.**

**Solution:**

Let the present age of Vikas be x years.

Then, the present age of Deepak = 2x years

According to the question,

2x – x = 11

=> x = 11

Present age of Vikas = 11 years

and present age of Deepak = 2 x 11

= 22 years.

**Question 14.**

**Solution:**

Let the present age of Rekha be x years

Then, the present age of Mrs. Goel = (x + 27) years

Rekha’s age after 8 years = (x + 8) years

Mrs. Goel’s age after 8 years = (x + 27 + 8) years

= (x + 35) years

According to the question,

x + 35 = 2 (x + 8)

=> x + 35 = 2x + 16

=> x – 2x = 16 – 35

(Transposing 2x to L.H.S. and 35 to R.H.S.)

=> – x = – 19

=> x = 19

Present age of Rekha =19 years and present age of Mr. Goel = 19 + 27 = 46 years.

**Question 15.**

**Solution:**

Let the present age of the son be A years.

Then, the present age of the man

= 4 x years

Son’s age after 16 years = (x + 16) years Man’s age after 16 years

= (4 x + 16) years According to the question,

4x + 16 = 2 (x + 16)

=> 4x + 16 = 2x + 32

=> 4x – 2x = 32 – 16

(Transposing 2 x to L.H.S. and 16 to R.H.S.)

=> 2x = 16

=> x = 8 (Dividing both sides by 2)

Present age of the son = 8 years and present age of the man = 8 x 4 = 32 years.

**Question 16.**

**Solution:**

Let the present age of the son be x years.

Then, the present age of the man = 3x years

five years ago, the age of the son = (x – 5) years

five years ago, the age of the man = (3x – 5) years

According to the question, 3x – 5 = 4(x – 5)

=>3x – 5 = 4x – 20

=>3x – 4x = – 20 + 5

(Transposing 4 x to L.H.S. and – 5 to R.H.S.)

=> – x = – 15 => x = 15

Present age of the son = 15 years and present age of the man = 3 x 15 = 45 years.

**Question 17.**

**Solution:**

Let the present age of Fatima be A years. According to the question,

x + 16 = 3 x

=> x – 3x = – 16

(Transposing 3 x to L.H.S. and 16 to R.H.S.)

=> – 2 x = – 16

=> x = 8

(Dividing both sides by – 2)

Present age of Fatima = 8 years.

**Question 18.**

**Solution:**

Let the present age of Rahim be x years

Rahim’s age after 32 years = (x + 32) years

Rahim’s age 8 years ago = (x – 8) years

According to the question, x + 32 = 5 (x – 8)

=> x + 32 = 5 x – 40

=> x – 5x = – 40 – 32

(Transposing 5 x to L.H.S. and 32 to R.H.S.)

– 4 x = – 72

x= 18

(Dividing both sides by – 4)

Present age of Rahim =18 years

**Question 19.**

**Solution:**

Let the number of 50 paisa coins be x.

Then, the number of 25 paisa coins = 4x

Total value of 50 paisa coins = 50 x paisa

and total value of 25 paisa coins

= 25 x 4 = 100 x paisa

But total value of both the coins

= Rs. 30 (Given)

= 30 x 100 paisa

= 3000 paisa

According to the question,

50 x + 100 x = 3000

=> 150 x = 3000

\(\\ \frac { 150x }{ 150 } \) = \(\\ \frac { 3000 }{ 150 } \)

(Dividing both sides by 150)

x = 20

Number of 50 paisa coins = 20

and number of 25 paisa coins = 4 x 20

= 80

**Question 20.**

**Solution:**

Let the price of the pen be x rupees. According to the question,

5 x = 3 x + 17

5 x – 3 x = 17

(Transposing 3 x to L.H.S.)

=> 2 x = 17

=> x = \(\\ \frac { 17 }{ 2 } \)

(Dividing both sides by 2)

Price of the pen = \(\\ \frac { 17 }{ 2 } \) rupees

= Rs. 8.50.

**Question 21.**

**Solution:**

Let the number of girls in the school be x.

Then, the number of boys in the school = (x + 334)

According to the question, x + (x + 334) = 572

=> 2 x + 334 = 572

=> 2 x = 572 – 334

(Transposing 334 to L.H.S.)

2 x = 238

=> x = \(\\ \frac { 238 }{ 2 } \)

(Dividing both sides by 2) => x = 119

Number of girls in the school = 119.

**Question 22.**

**Solution:**

Let the breadth of the park be x metres.

Then, the length of die park=3x metres.

According to the question,

Perimeter of the park = 168 metres

=> 2 (x + 3 x) = 168

=> 2 x 4x = 168

=> 8 x = 168

=> x = 21

(Dividing both sides by 8)

Breadth of the park = 21 metres and length of the park = 3 x 21 = 63 metres.

**Question 23.**

**Solution:**

Let the breadth of the hall be x metres.

Then, the length of the hall = (x + 5) metres .

According to question,

Perimeter of the hall = 74 metres

=> 2 (x + x + 5) = 74

=> 2 (2 x + 5) = 74

=> 4 x + 10 = 74

=> 4 x = 74 – 10

(Transposing 10 to R.H.S.)

4x = 64

=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 64 }{ 4 } \)

(Dividing both sides by 4)

=> x = 16

Breadth of the hall = 16 metres

and length of the hall = 16 + 5 = 21 metres.

**Question 24.**

**Solution:**

Since a wire of length 86 cm is bent to form die rectangle, so the perimeter of the rectangle = 86 cm.

Let the breadth of the rectangle = x cm

Then, die length of the rectangle = (x + 7) cm

2 (x + x + 7) = 86

=> 2 (2 x + 7) = 86 .

=> 4 x + 14 = 86

=> 4x = 86 – 14

(Transposing 14 to R.H.S.)

=> 4 x = 72

\(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 72 }{ 4 } \) = 18

(Dividing both sides by 4)

Breadth of the rectangle = 18 cm

Length of the rectangle = (18 + 7) = 25 cm.

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C are helpful to complete your math homework.

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