RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C - Learn Hool

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

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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