RS Aggarwal Solutions Class 6 Ex 9B

Question 1:

If 9 is added to a certain number, the result is 36. Find the number.

Solution:

Let the required number be x.

According to the question:

9 + x = 36

or, x + 9 – 9 = 36 – 9 [Subtracting 9 from both the sides]

or, x = 27

Thus, the required number is 27.

Question 2:

If 11 is subtracted from 4 times a number, the result is 89. Find the number.

Solution:

Let the required number be x.

According to the question:

4x – 11 = 89

or, 4x – 11 + 11 = 89 + 11 [Adding 11 on both the sides]

or, 4x = 100

or, \(\frac{4x}{4}=\frac{100}{4}\) [Dividing both the sides by 4]

or, x = 25

Thus, the required number is 25.

Question 3:

Find a number which when multiplied by 5 is increased by 80.

Solution:

Let the required number be x.

According to the question:

or, 5x = x + 80

or, 5x – x = 80 [Transposing x to the L.H.S.]

or, 4x = 80

or, \(\frac{4x}{4}=\frac{80}{4}\) [Dividing both the sides by 4]

or, x = 20

Thus, the required number is 20.

Question 4:

The sum of three consecutive natural numbers is 114. Find the numbers.

Solution:

Let the three consecutive natural numbers be x, (x + 1), (x + 2).

According to the question:

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

or, x + x + 1 + x + 2 = 114

or, 3x + 3 = 114

or, 3x + 3 – 3 = 114 – 3 [Subtracting 3 from both the sides]

or, 3x = 111

or, \(\frac{3x}{3}=\frac{111}{3}\) [Dividing both the sides by 3]

or, x = 37

Required numbers are:

x = 37

or, x + 1 = 37 + 1 = 38

or, x + 2 = 37 + 2 = 39

Thus, the required numbers are 37, 38 and 39.

Question 5:

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find the number.

Solution:

Let the required number be x.

When Raju multiplies it with 17, the number becomes 17x.

According to the question:

17x + 4 = 225

or, 17x + 4 – 4 = 225 – 4 [Subtracting 4 from both the sides]

or 17x = 221

or, \(\frac{17x}{17}=\frac{221}{17}\) [Dividing both the sides by 17]

or, x = 13

Thus, the required number is 13.

Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

Solution:

Let the required number be x.

According to the question, the number is tripled and 5 is added to it.

Therefore, 3x + 5

or, 3x + 5 = 50

or, 3x + 5 – 5 = 50 – 5 [Subtracting 5 from both the sides]

or, 3x = 45

or, \(\frac{3x}{3}=\frac{45}{3}\) [Dividing both the sides by 3]

or, x = 15

Thus, the required number is 15.

Question 7:

Find two umbers such that one of them exceeds the other by 18 and their sum is 92.

Solution:

Let one of the number be x.

Therefore, The other number = (x + 18)

According to the question:

x + (x + 18) = 92

or, 2x + 18 – 18 = 92 – 18 [Subtracting 18 from both the sides]

or, 2x = 74

or, \(\frac{2x}{2}=\frac{74}{2}\) [Dividing both the sides by 2]

or, x = 37

Required numbers are:

x = 37

or, x + 18 = 37 + 18 = 55

Question 8:

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

Solution:

Let one of the number be ‘x’

Therefore, Second number = 3x

According to the question:

x + 3x = 124

or, 4x = 124

or, \(\frac{4x}{4}=\frac{124}{4}\) [Dividing both the sides by 4]

or, x = 31

Thus, the required number is x = 31 and 3x = 3 x 31 = 93.

Question 9:

Find two numbers such that one of them is five times the other and their difference is 132.

Solution:

Let one of the number be x.

Therefore, Second number = 5x

According to the question:

5x – x = 132

or, 4x = 132

or, \(\frac{4x}{4}=\frac{132}{4}\) [Dividing both the sides by 4]

or, x = 33

Thus, the required numbers are x = 33 and 5x = 5 x 33 = 165.

Question 10:

The sum of two consecutive even numbers is 74. Find the numbers.

Solution:

Let one of the even number be x.

Then, the other consecutive even number is (x + 2).

According to the question:

x + (x + 2) = 74

or, 2x + 2 = 74

or, 2x + 2 – 2 = 74 – 2 [Subtracting 2 from both the sides]

or, 2x = 72

or, \(\frac{2x}{2}=\frac{72}{2}\) [Dividing both the sides by 2]

or, x = 36

Thus, the required numbers are x = 36 and x + 2 = 38.

Question 11:

The sum of three consecutive odd numbers is 21. Find the numbers.

Solution:

Let the first odd number be x.

Then, the next consecutive odd numbers will be (x + 2) and (x + 4).

According to the question:

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

or, 3x + 6 = 21

or, 3x + 6 – 6 = 21 – 6 [Subtracting 6 from both the sides]

or, 3x = 15

or,\(\frac{3x}{3}=\frac{15}{3}\) [Dividing both the sides by 3]

or, x = 5

Therefore, Required numbers are:

x = 5

x + 2 = 5 + 2 = 7

x + 4 = 5 + 4 = 9

Question 12:

Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

Solution:

Let the present age of Ajay be x years.

Since Reena is 6 years older than Ajay, the present age of Reena will be (x + 6) years.

According to the question:

x + (x + 6) = 28

or, 2x + 6 = 28

or, 2x + 6 – 6 = 28 – 6 [Subtracting 6 from both the sides]

or 2x = 22

or, \(\frac{2x}{2}=\frac{22}{2}\) [Dividing both the sides by 2]

or, x = 11

Therefore, Present age of Ajay = 11 years

Present age of Reena = x + 6 = 11 + 6 = 17 years.

Question 13:

Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.

Solution:

Let the present age of Vikas be x years.

Since Deepak is twice as old as Vikas, the present age of Deepak will be 2x years.

According to the question:

2x – x = 11

x = 11

Therefore, Present age of Vikas = 11 years

Present age of Deepak = 2x = 2 x 11 = 22 years.

Question 14:

Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

Solution:

Let the present age of Rekha be x years.

As Mrs Goel is 27 years older than Rekha, the present age of Mrs Goel will be (x + 27) years.

After 8 years:

Rekha’s age = (x + 8) years

Mrs Goel’s age = (x + 27 + 8) = (x + 35) years

According to the question:

(x + 35) = 2(x + 8)

or, x + 35 = 2x + 16 [On expanding the brackets]

or, 35 – 16 = 2x – x [Transposing 16 to the L.H.S. and x to the R.H.S.]

or, x = 19

Therefore, Present age of Rekha = 19 years

Present age of Mrs Goel = x + 27 = x19 + 27 = 46 years.

Question 15:

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present age.

Solution:

Let the present age of the son be x years.

As the man is 4 times as old as his son, the present age of the man will be 4x years.

After 16 years:

Son’s age = (x + 16) years

Man’s age = (4x + 16) years

According to the question:

(4x + 16) = 2(x + 16)

or, 4x + 16 = 2x + 32 [On expanding the brackets]

or, 4x – 2x = 32 – 16 [Transposing 16 to the R.H.S. and 2x to the L.H.S.]

or, 2x = 16

or, \(\frac{2x}{2}=\frac{16}{2}\) [Dividing both the sides by 2]

or, x = 8

Therefore, Present age of the son = 8 years

Present age of the man = 4x = 4 x 8 = 32 years

Question 16:

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

Solution:

Let the present age of the son be x years.

As the man is 3 times as old as his son, the present age of the man will be (3x) years.

5 years ago:

Son’s age = (x – 5) years

Man’s age = (3x – 5) years

According to the question:

(3x – 5) = 4(x – 5)

or, 3x – 5 = 4x – 20 [On expanding to the brackets]

or, 20 – 5 = 4x – 3x [Transposing 3x to the R.H.S. and 20 to the L.H.S.]

or x = 15

Therefore, Present age of the son = 15 years

Present age of the man = 3x = 3 x 15 = 45 years

Question 17:

After 16 years, Fatima will be three times as old as she is now. Find her present age.

Solution:

Let the present age of Fatima be x years.

After 16 years:

Fatima’s age = (x + 16) years

According to the question:

x + 16 = 3(x)

or, 16 = 3x – x [Transposing x to the R.H.S.]

or, 16 = 2x

or, \(\frac{2x}{2}=\frac{16}{2}\) [Dividing both the sides by 2]

or, x = 8

Therefore, Present age of Fatima = 8 years

Question 18:

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

Solution:

Let the present age of Rahim be x years.

After 32 years:

Rahim’s age = (x + 32) years

8 years ago:

Rahim’s age = (x – 8) years

According to the question:

x + 32 = 5(x – 8)

or, x + 32 = 5x – 40

or, 40 + 32 = 5x – x [Transposing x to the R.H.S and 40 to the L.H.S.]

or, 72 = 4x

or, \(\frac{4x}{4}=\frac{72}{4}\) [Dividing both the sides by 4]

or, x = 18

Thus, the present age of Rahim is 18 years.

Question 19:

A bag contains 25-paisa and 50-paisa coins whose total values is Rs. 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

Solution:

Let the number of 50 paisa coins be x.

Then, the number of 25 paisa coins will be 4x.

According to the question:

0.50(x) + 0.25(4x) = 30

or, 0.5x + x = 30

or, 1.5x = 30

or, \(\frac{1.5x}{1.5}=\frac{30}{1.5}\) [Dividing both the sides by 1.5]

or, x = 20

Thus, the number of 50 paisa coins is 20.

Number of 25 paisa coins = 4x = 4 x 20 = 80

Question 20:

Five times the price of a pen is Rs. 17 more than three times its price. Find the price of the pen.

Solution:

Let the price of pen be Rs. x.

According to the question:

5x = 3x + 17

or, 5x – 3x = 17 [Transposing 3x to the L.H.S.]

or, 2x = 17

or, \(\frac{2x}{2}=\frac{17}{2}\) [Dividing both the sides by 2]

or, x = 8.50

Therefore, Price of one pen = Rs. 8.50

Question 21:

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

Solution:

Let the number of girls in the school be x.

Then, the number of boys in the school will be (x + 334).

Total strength of the schools = 572

Therefore, x + (x + 334) = 572

or, 2x + 334 = 572

or, 2x + 334 – 334 = 572 – 334 [Subtracting 334 from both the sides]

or, 2x = 238

or, \(\frac{2x}{2}=\frac{238}{2}\) [Dividing both the sides by 2]

or, x = 119

Therefore, Number of girls in the school = 119

Question 22:

The length of a rectangular park is thrice its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

Solution:

Let the breadth of the park be x metres.

Then, the length of the park will be 3x metres.

Perimeter of the park = 2 (Length + Breadth) = 2 (3x + x) m

Given perimeter = 168 m

Therefore, 2 (3x + x) = 168

or, 2(4x) = 168

or, 8x = 168 [On expanding the brackets]

or, \(\frac{8x}{8}=\frac{168}{8}\) [Dividing both the sides by 8]

or, x = 21 m

Therefore, Breadth of the park = x = 21 m

Length of the park = 3x = 3 x 21 = 63 m

Question 23:

The length of a rectangle hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

Solution:

Let the breadth of the hall be x metres.

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

Perimeter of the hall = 2(Length + Breadth) = 2(x + 5 + x) metres

Given perimeter of the rectangular hall = 74 metres

Therefore, 2(x + 5 + x) = 74

or, 2(2x + 5) = 74

or, 4x + 10 = 74 [On expanding the brackets]

or, 4x + 10 – 10 = 74 – 10 [Subtracting 10 from both the sides]

or, 4x = 64

or, \(\frac{4x}{4}=\frac{64}{4}\) [Dividing both the sides by 4]

or, x = 16 metres

Therefore, Breadth of the park = x = 16 metres

Length of the park = x + 5 = 16 + 5 = 21 metres.

Question 24:

A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

Solution:

Let the breadth of the rectangle be x cm.

Then, the length of the rectangle will be (x + 7) cm.

Perimeter of the rectangle = 2(Length + Breadth) = 2(x + 7 + x) cm

Given perimeter of the rectangle = Length of the wire = 86 cm

Therefore, 2(x + 7 + x) = 86

or, 2(2x + 7) = 86

or, 4x + 14 = 86 [On expanding the brackets]

or, 4x + 14 – 14 = 86 – 14 [Subtracting 14 from both the sides]

or, 4x = 72

or, \(\frac{4x}{4}=\frac{72}{4}\) [Dividing both the sides by 4]

or, x = 18

Breadth of the hall = x = 18 cm

Length of the hall = x + 17 = 18 + 7 = 25 cm


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