Frank ICSE Class 8 Solutions Chapter 5

Frank ICSE Mathematics Class 8 Solutions Chapter 5 Playing with Numbers

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 5, Playing with Numbers. Here students can easily find step by step solutions of all the problems for Playing with Numbers, Exercise 5.1, 5.2 and 5.3 Also our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 5 solutions. Here in this post all the solutions are based on latest Syllabus.

Playing with Numbers Exercise 5.1 Solution :

Question no – (1)

Solution :

Correct option – (d)

100 × 8 + 10 × 9 is the generalized form of 890

Given, 100 × 8 + 10 × 9

= 800 + 90 (after multiplying)

= 890

Question no – (3)

Solution :

The quotient obtained when the sum of 87 and 78 is divided by 11 is 15

= 87 + 78

= 11 (7 + 8)

= 11 (15)

Therefore, 15 will be quotient.

Question no – (3)

Solution :

The quotient obtained when the sum of 87 and 78 is divided by 11 is 15

= 87 + 78

= 11 (7 + 8)

= 11 (15)

Therefore, 15 will be quotient.

Question number – (4)

Solution :

If the difference of 85 and 58 is divided by 3 the quotient is 9

= 85 – 58 …(according to the question)

= 9 (8 – 5)

= 9 (3)

Therefore, 9 will be the quotient.

Question no – (5)

Solution :

The sum of 24 and 42 is divided by (i) 11 (ii) 6, Now we have to the find the quotient.

The sum of 24 and 42 are divisible by both 11 and 6

So, we can write,

= 24 + 42

= 11 (2 + 4)

= 11 (6)

Therefore,

(i) Dividing by 11, the quotient will be 6.

(ii) Dividing by 6, the quotient will be 11.

Question no – (6)

Solution :

We can solve the above problem without performing actual computation.

The difference of 92 and 29 are divisible by both 9 and 7

So, we can write 92 – 29 = 9 (9 – 2) = 9 × 7

Therefore,

(i) Dividing by 7, the quotient will be 9

(ii) Dividing by 9, the quotient will be 7

Question no – (7)

Solution :

As per the question here we have to find quotient in each given case.

According to the question, first we find the sum of

637, 376 and 763

Now, 637 + 763 + 376

= 100 (6 + 7 + 3) + 10 (6 + 7 + 3) + 1 (6 + 7 + 3)

= 100 × (16) + 10 (16) + 1 (16)

= 16 (100 + 10 + 1)

= (16 × 111)

Divided by 111, the quotient will be = 16

Divided by 16, the quotient will be = 111

Divided by 37, the quotient will be

= 16 × 3

= 48

Question no – (8)

Solution :

According to the given question, here we have to find the quotient in each case,

987 + 879 + 798

= 100 (9 + 8 + 7) + 10 (9 + 8 + 7) + 1 (9 + 8 + 7)

= 100 (24) + 10 (24) + 1 (24)

= 24 (100 + 10 + 1)

= 24 (111)

Dividing by 111, the quotient will be = 24

Dividing by 24, the quotient will be = 111

Dividing by 37, the quotient will be

= 24 × 3

= 72

Question no – (9)

Solution :

According to the given question here we have to find the quotient.

= 753 – 357

= 99 (7 – 3)

= 99 × 4

Dividing by 4, the quotient will be = 99

Dividing by 9, the quotient will be

= 11 × 4

= 44

Dividing by 33, the quotient will be

= 3 × 4

= 12

Question no – (10)

Solution :

According to the question here we have to find quotient in each case

= 892 – 298

= 99 (8 – 2)

= 99 × 6

Now,

Dividing by 6, the quotient will be = 99

Dividing by 9, the quotient will be

= 11 × 6

= 66

∴ Dividing by 33, the quotient will be

= 3 × 6

= 18

Playing with Numbers Exercise 5.2 Solution :

Question number – (3)

Solution :

Correct option is – (a)

If the number 45m8 is divisible by 9, the value of m is 1

= 4 + 5 + 8

= 17 + 1

= 18 (divisible)

Question number – (4)

Solution :

Correct option – (c)

In the given numbers only 1517 is not divisible by 7

As we know that,

A number is divisible by 7 if the twice of unit digit and its subtraction from the remaining number (except unit digit) is divisible by 7.

(a) 308 → 8 × 2 = 16 → 30 – 16 = 14

14 is divided by 7

So, 308 is divisible by 7

(b) 525 → 5 × 2 = 10 → 52 – 10 = 42

42 is divisible by 7

So, 525 is divisible by 7

(c) 1517 → 7 × 2 = 14 → 151 – 14 = 137

Now, 137 → 7 × 2 = 14 → 13 – 14 = -1

= -1 is not divisible by 7

So, 1517 is not divisible by 7

(d) 2996 → 6 × 2 = 12 → 299 – 12 = 287

Now, 287 is divisible by 7

Henec, 2996 is divisible by 7

Question no – (5)

Solution :

A number is divisible by 3 if the sum of digits is divisible by 3

A number is divisible by 9 if the sum of all digits is divisible by 9

(a) 298 → 2 + 9 + 8 = 19,

So, 298 is not divisible by 3, 9

(b) 426 → 4 + 2 + 6 = 12,

Thus, 426 is divisible by 3, but not by 9

(c) 405 → 4 + 5 = 9,

Hence, 405 is divisible by both 3 and 9

(d) 927 → 9 + 2 + 7 = 18,

Therefore, 927 is divisible by both 3 and 9

Question no – (6)

Solution :

According to the question, the value of n will be 1, 4, 7

If 5n6 be divisible by 3, then (5 + 6 + n) must be divisible by 3

Now, 5 + 6 + n = (11 + n) will be divisible by 3 if n = 1, 4, 7

So, we can say the value of ‘n’ will be 1, 4, or 7

Question no – (7)

Solution :

If 597n8 is divisible by 9, where n is a digit, The value of the n will be  7

If 597n8 be divisible by 9, then (5 + 9 + 7 + n + 8)

(29 + n) must be divisible by 9, if n = 7

We know, a number is divisible by 9 if the sum of all digits is divisible by 9.

Question no – (8)

Solution :

If 814y is divisible by 6, where y is a digit, The value of y will be 2 or 8

As we know a number is divisible by 3 if the sum of digits is divisible by 3.

If 814y be divisible by 6, then y must be an even number and (8 + 1 + 4 + y) must be divisible by 3.

Now, (13 + y) will be divisible by 3 when y = 2 or 8

Therefore, the value of y will be 2 or 8

Question number – (10)

Solution :

If 36×2 is divisible by 4, where x is a digit, then the value of x will be 1, 3, 5, 7, 9

If 36×2 be divisible by 4, then the number (x2) must be divisible by 4

And this will happen if x = 1, 3, 5, 7, 9

We know, A number is divisible by 4 if the last two digit of the number is divisible by 4 or the whole number is divisible by 4.

Question no – (11)

Solution :

If 148y is divisible 4, where y is a digit, then the value of will be 0, 4 or 8

If 148y be divisible by 4 the number (8y) must be divisible by 4.

And, it will happen if y = 0, 4, 8

Because we know, a number is divisible by 4 if the last two digit of the number is divisible by 4 or the whole number is divisible by 4.

Playing with Numbers Exercise 5.3 Solution :

Question no – (1)

Solution :

(a) A B A

+ 1 B 2
——————-
6 6 6

(b) 3 A 5

+ 6 2 B
——————-
B 7 A

(c) 1 A 8

+ B B B
——————-
A B 5

(d) 2 A B

+ 5 9 B
——————-
A 7 A

(e) 5 A

×      A
——————-
B B A

(f) 7 A

×      3
——————-
B B A

(g) A B

×       9
——————-
6 A B

(h) A 2 B

+   2 4 3
——————-
B A 2

Revision Exercise Questions Solution :

Question no – (1)

Solution :

If 32a8 is a multiple of 9, then the value of a will be 5

32a8 is a multiple of 9

(3 + 2 + a + 8) = (13 + a) is a multiple of 9

Hence, it means a = 5

Question no – (2)

Solution :

If 543x is a multiple of 8, then the possible value of x will be 2

543x is a multiple of 8 ….(according to the question)

It means, the number 43x is divisible by 8

It will only happen if x = 2

Because we know a number is divisible by 8 if its last three digits are divisible by 8 then the number is divisible by 8

Thus, we can say the possible value of x will be 2.

Question no – (3)

Solution :

Here, at the unit place (C + C + C) = C

It means, either C = 0 or C = 5

But the summation of three positive numbers can’t be 000.

That’s why, C = 5

Now, the summation of three equal numbers is 555.

Each number,

= 555 ÷ 3

= 185

Question no – (4)

Solution :

The number will be 194

Suppose, the number is ABC

Now, B – A = 8.

So, if B = 8, Then A = 0

and if B = 9, then A = 1

If we take B = 8, then the number becomes of two digit.

So, we must take B = 9, A = 1

And then, we know, B – C = 5, 9 – C = 5 => C = 4

Therefore, the required number is 194

Next Chapter Solution :

Updated: June 24, 2023 — 7:32 am