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

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

ML Aggarwal Class 8 Solutions for ICSE Maths

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