## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1

Question 1.

Write the following numbers in generalized form:

(i) 89

(ii) 207

(iii) 369

Solution:

(i) 89 = 8 × 10 + 9

(ii) 207 = 2 × 100 + 0 × 10 + 7 × 1

(iii) 369 = 3 × 100 + 6 × 10 + 9 × 1

Question 2.

Write the quotient, when the sum of a 2-digit number 34 and number obtained by reversing the digits is divided by

(i) 11

(ii) sum of digits

Solution:

Sum of two-digit number 34 and the number

obtained by reversing the digit 43 = 34 + 43 = 77

(i) 77 ÷ 11 = 7

(ii) 77 ÷ (Sum of digit) = 77 ÷ (4 + 3)

= 77 ÷ 7 = 11

Question 3.

Write the quotient when the difference of a 2-digit number 73 and number obtained by reversing the digits is divided by

(i) 9

(ii) a difference of digits.

Solution:

Difference of two digit number 73 and the number

obtained by reversing the digits

= 73 – 37 = 36

(i) Now, 36 ÷ 9 = 4

(ii) 36 ÷ (7 – 3) = 36 + 4 = 9

Question 4.

Without actual calculation, write the quotient when the sum of a 3-digit number abc and the number obtained by changing the order of digits cyclically i.e. bca and cab is divided by

(i) 111

(ii) (a + b + c)

(iii) 37

(iv) 3

Solution:

Sum of 3-digit number abc and the number

obtained by changing the order of digits i.e. bca and cab.

∴ abc + bca + cab

= 100a + 10b + c + 1006 + 10c + a + 100c + 10a + b

= 111a + 111b + 111c = 111 (a + b + c)

(i) When divided by 111 = 111 (a + b + c) ÷ 111 = a + b + c

(ii) When divided by (a + b + c) = 111 (a + b + c) ÷ (a + b + c) = 111

(iii) When divided by 37 = 111 (a + b + c) ÷ 37 = 3(a + b + c)

(iv) When divided by 3 = 111 (a + b + c) ÷ 3 = 37(a + b + c)

Question 5.

Write the quotient when the difference of a 3-digit number 843 and number obtained by reversing the digits is divided by

(i) 99

(ii) 5

Solution:

Difference of 3-digit number 843 and the number

obtained by reversing the digit 348

= 843 – 348 = 495

(i) Divided by 99, then \(\frac{495}{99}\) = 5

(ii) Divided by 5, then = \(\frac{495}{5}\) = 99

Question 6.

The sum of digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.

Solution:

Sum of two digit number = 11

Let unit’s digit = x

and tens digit = y,

then x + y = 11 …(i)

and number = x + 10y

By reversing the digits,

Unit digit = y

and tens digit = x

and number = y + 10x

Now, y + 10x + 9 = x + 10y

10x + y – 10y – x = -9

9x – 9y = -9

x – y = -1 …(ii)

Adding (i) and (ii),

2x= 10 ⇒ x= \(\frac{10}{2}\) = 5

∴ y = 1 + 5 = 6

and number = x + 10y = 5 + 10 × 6

= 5 + 60 = 65

Question 7.

If the difference of two-digit number and the number obtained by reversing the digits is 36, find the difference between the digits of the 2-digit number.

Solution:

Let unit digit = x

and tens digit =y

∴ Number = x + 10y

and by reversing the digits

Unit digit =y

and tens digit = x

Then number = y + 10x

∴ x + 10y – y – 10x = 36

⇒ -9x + 9y = 36

(y – x) = \(\frac{36}{9}\)

⇒ y – x = 4

Question 8.

If the sum of two-digit number and number obtained by reversing the digits is 55, find the siun of the digits of the 2-digit number.

Solution:

Let unit digit = x

and tens digit = y

∴ Number = x + 10y

and by reversing the digits

Unit digit =y

and tens digit = x

Then number = y + 10x

∴ x + 10y + y + 10x = 55

⇒ 11x + 11y = 55

⇒ x + y = \(\frac{55}{11}\)

⇒ x + y = 5

∴ Difference of the digits of the number = 4

Question 9.

In a 3-digit number, unit’s digit, ten’s digit and hundred’s digit are in the ratio 1 : 2 : 3. If the difference of original number and the number obtained by reversing the digits is 594, find the number.

Solution:

Ratio in the digits of a three digit number

= 1 : 2 : 3

Let unit digit = x

Then tens digit = 2x

and hundreds digit = 3x

and number = x + 10 × 2x + 100 × 3x

= x + 20x + 300x = 321x

and reversing the digits,

Unit digit = 3x

Ten’s digit = 2x

Hundreds digit = x

∴ Number = 3x + 10 × 2x + 100 × x

= 3x + 20x + 100 = 123x

According to the condition,

∴ 321x – 123x = 594

⇒ 198x = 594 ⇒ x = \(\frac{594}{198}\)= 3

∴ Number = 321x = 321 × 3 = 963

Question 10.

In a 3-digit number, unit’s digit is one more than the hundred’s digit and ten’s digit is one less than the hundred’s digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the number.

Solution:

Let hundreds digit = x

Then unit digit = x + 1

and ten’s digit = x – 1

Number = (x + 1) + 10(x – 1) + 100 × x

= x + 1 + 10x – 10 + 100x

= 111x – 9

and by reversing the digits,

Unit digit = x – 1

Tens digit = x

Hundred digit = x + 1

∴ Number = x – 1 + 10x + 100x + 100

= 111x + 99

and number = x + 10(x + 1) + 100(x – 1)

= x + 10x + 10 + 100x – 100

= 111x – 90

Now according to the condition,

111x – 9 + 111x + 99 + 111x – 90 = 2664

⇒ 333x + 99 – 99 = 2664

333x = 2664

x = \(\frac{2664}{333}\) = 8

∴ Original number = 111x – 9

= 888 – 9 = 879