RS Aggarwal Class 8 Math Fifth Chapter Playing with Numbers Exercise 5A Solution

RS Aggarwal Class 8 Math Fifth Chapter Playing with Numbers Exercise 5A Solution

EXERCISE 5A

(1) The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.

Solution: Let the digit be x. Then, the units place digit = 3

∴ Number = (10x + 3)

∴ 7 (x+3) = (10x + 3)

⇒ 7x + 21 = 10x + 3

⇒10x + 3 = 7x + 21

⇒ 10x – 7x = 21 – 3

⇒ 3x = 18

⇒ x = 6

Therefore the number is {(10 × 6) + 3}=63.

(2) In a two-digit number, the digit at the units place is double the digit in the tens place. The number exceeds the sum of its digits by 18. Find the number.

Solution: Let the tens digit be x. The unit place digit = 2x.

∴ Number = 10x + 2x

∴ (x + 2x) + 18 = (10x + 2x)

⇒ 3x + 18 = 12x

⇒ 12x = 3x + 18

⇒ 12x – 3x = 18

⇒ 9x = 18

⇒ x = 2

Therefore, the number {(10 × 2)+(2×2)}=24.

(3) A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Solution: Let the tens digit be a. The unit place digit = b.

∴ Number = (10a + b)

∴ 4(a + b) + 3 = (10a + b)

⇒ 4a + 4b + 3 = 10a + b

⇒ 10a + b – 4a – 4b = 3

⇒ 6a – 3b = 3

⇒ 3(2a – b) =3

⇒ 2a – b = 1……………….(i)

Given, 18 is added to the number, its digit are reversed.

∴ Number = (10b + a)

(10a + b) + 18 = (10b + a)

⇒ 10a + b – 10b – a = -18

⇒ 9a – 9b = – 18

⇒ 9 (a – b) = – 18

⇒ a – b = – 2…………………(ii)

Subtract (i) – (ii)

2a – b = 1

a – b = – 2
–     +       +
 a        = 3

Using a = 3 in equation (i)

∴ (2 × 3) – b = 1

⇒ 6 – b = 1

⇒ b = 6 – 1

⇒ b = 5

Therefore, the number = {(10 × 3) +5} = 35.

(4) The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits is 63. What is the difference between the digits of the number?

Solution: Let the tens and unit digits of the required number be a and b respectively. Then,

a + b = 15………….(i)

Original number = 10a + b

Number obtained by interchanging its digits = 10b + a.

∴ (10a + b) + 9 = (10b + a)

⇒ 10a + b + 9 = 10b + a

⇒ 10a + b – 10b – a = – 9

⇒ 9a – 9b = – 9

⇒ 9 (a – b) = – 9

⇒ a – b = – 1……………(ii)

Adding the equation (i) and (ii)

a + b = 15

a – b = – 1
2a       = 14

⇒ a = 7

Using a = 7 in equation (i)

a + b = 15

⇒ 7 + b = 15

⇒ b = 15 – 7

⇒ b = 8

Therefore, the number = {(10 × 7) + 8] = 78.

(5) The difference between a 2-digit number and the number obtained by interchanging its digit is 63. What is the difference between the digits of the number?

Solution: Let the tens and unit digits of the number be a and b respectively. Then,

∴ Number = 10a + b

The number obtained by interchanging its digits = (10b + a)

∴ (10a + b) – (10b + a) = 63

⇒ 10a + b – 10b – a = 63

⇒ 9a – 9b = 63

⇒ 9 (a – b) = 63

⇒ a – b = 7

Therefore, the different between the digits is 7.

(6) In a 3-digit number, the tens digit is thrice the units digit and the hundreds digit is four times the units digit. Also, the sum of its digits is 16. Find the number.

Solution: Let the unit place digit be x. Then, the tens place digit = 3x and the hundred place digit = 4x.

∴ 4x + 3x + x = 16

⇒ 8x = 16

⇒ x = 2

Therefore, Unit place digit = 2. The tens place digit = (3 × 2) = 6. The hundreds place digit = (4 × 2) = 8.

Hence, the number is 862.


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