ML Aggarwal CBSE Solutions Class 8 Math Fifth Chapter Playing with Numbers Exercise 5.1

ML Aggarwal CBSE Solutions Class 8 Math 5th Chapter Playing with Numbers Exercise 5.1

(1) (i) 89 = 10 x 8 + 9

(ii) 408 = 100 x 4 + 10 x 0 + 8

(iii) 369 = 100 x 3 + 10 x 6 + 9

(2) (i) Given number 23

Number obtained by reversing the digit = 32

Therefore, sum = 23 + 32 = 55 = 11 x 5

(i) When sum is divided by 11,

Quotient is 5,

(ii) Sum of digits = 2 + 3 = 5

Therefore, when sum 55 is divided by 5,

Quotient is 11.

(3) Given number = 83

Number obtained by reversing the digit is 38.

Therefore, difference = 83 – 38 = 45 = 5 x 9

(i) When sum is divided by 9,

Then the quotient will be ‘5’

(ii) Difference digit = 8 – 3 = 5

When difference ‘45’ is divided by ‘5’ then the quotient will be ‘9’

(4) Given number = a b c

Changing the order of digit = b c a

Changing the order of digit = c a b

Therefore, a b c = 100a + 10b + c

bca = 100a + 10c + a

Therefore, c a b = 100c + 10a + b

On adding number, we get —

abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

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

= 111a + 111b + 111c

= 111 x (a + b + c) = 3 x 37 x (a + b + c)

(i) When divided by 111, the quotient will be (a + b + c)

(ii) When divided by (a + b + c), the quotient will be 111

(iii) When divided by ‘37’, the quotient will be 3 x (a + b + c)

(iii) When divided by ‘3’, the quotient 37 x (a + b + c)

(5) Given number = 943

Reversing number = 349

Therefore difference = 943 – 349

= 594

= 99 x 6

(i) When dividing by 99

The equitant will be 6.

(ii) When dividing by ‘6’

the equitant will be ‘99’

(6) let,

One digit = a

Another digit = b

Original number = 10a + b

Reversing the digits number obtained = 10b + a

According to que——

(1) a + b = 11 ——(i)

(2) (10a + b) – (10b + a) = 9

We know that –

(10a + b) – (10b + a) = 9

Or, 10a + b – 10b – a = 9

Or, 9a – 9b = 9

Or, 9(a – b) = 9

Or, a – b = 1 —— (ii)

Now, (i) & (ii) are equating.

We get,

a = b = 11

a – b = 1

_________

2a = 12

Or, a = 12/2

= 6

Therefore, putting the value of a in eq. (ii) we get,

a – b = 1

Or, a – 1 = b

Or, b = 6 – 1

Therefore, a = 6

a = 5

Therefore, the original number = 65.

= 5

(7) Let, one digit = a

Another digit = b

Therefore, original number = 10a + b

Number obtained by reversing number = 10b + a

According to question——–

(10a + b) – (10b + a) = 27

Or, 10a + b – 10b – a = 27

Or, 9a – 9b = 27

Or, 9(a – b) = 27

Or, a – b = 27/9

= 3

Therefore, the difference between the digits of ‘2’ digit number = a – b = 3

(8) Let,

One digit = a

Another digit = b

Therefore, original number = 10a + b

Number obtained by reversing number = 10b + a

According to question—

(10a + b) + (10b + a) = 44

Or, 10a + b + 10b + a = 44

Or, 11a + 11b = 44

Or, 11(a + b) = 44

Or, a + b = 44/11

= 4

(9) Let,

1^st digit = a

Last digit = C

Middle digit = b = 0

Therefore, original number = 100a + b + c

Given, a + c = 11 —- (i)

Reversing digit number obtain = 100c + b + a

According to question —

(100c + b + a) – (100a + b + c) = 495

Or, (100c + 0 + a) – (100a + 0 + c) = 495

Or, 100c + a – 100a – c = 495

Or, 99c – 99a = 495

Or, 99(c – a) = 495

Or, c – a = 495/99

= 5 ——(ii)

Equating (i) & (ii) we get –

C + a = 11

C – a = 5

_______

2c = 16

Therefore, a + c = 11

Or, a = 11 – c

Or, a = 11 – 8

Or, a = 3

Therefore, a = 3

C = 8

Therefore, original number = 308.

(10) Ratio = 1:2:3

Let, common factor = x

Therefore, units digits = x

Ten’s digit = 2x

Hundreds digit = 3x

Therefore, Original number [(100 X 3x) + (100 x 2x)+ x]

= (300x + 20x + x) = 321x

Therefore, reversing digit number obtained = [(100 X x) + (10 X 2x) + 3x]

= 123x

According to question —-

321x – 123x = 594

Or, 198x = 594

Or, x = 594/198 = 3

Therefore, units digit = 3

Ten’s digit = 2 x 3 = 6

Hundred’s digit = 3 x 3= 9

Therefore, original number = 963.

(11) Let,

Hundred’s unit (a) = x

Ten’s unit (b) = x – 1

Unit’s unit (c) = x + 1

Original number => a + b + c

= (100 X x) + 10(x – 1) + x +1

= 100x + 10x – 10 + x + 1

= 111x – 9

2^nd reversing digit obtained number => c + a + b

= [100 (x +1) + 10 X x + (x – 1)]

= 100x + 100 + 10x + x – 1

= 111x + 99

3^rd, reversing digit obtained number => b + c + a

= [100 (x – 1) + 10 ( x + 1) + x]

= 100x – 100 + 10x + 10 + x

= 111x – 90

According to question—–

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

Or, (111x + 111x + 111x) – (9 – 99 +90) = 2664

Or, 333x – (-90 + 90) = 2662

Or, 333x = 2664

Or, x = 2664/333

Or, x = 8

Ten’s units = 8 – 1 = 7 Therefore, original number = 879.

Unit’s unit = 8 + 1 = 9