# ICSE Class 8 Maths Selina Solutions Chapter 5 Playing with Number

The Class 8 Maths Chapter “Playing with Number” revolves around numbers which makes the chapter more interesting. Mathematics is that world of numbers where one can simply play with the numbers and can easily discover various analogies and relations of the numbers from the number line. A number is said to be in a generalized form if it is expressed as the sum of the product of its digits with their respective place values. For eg: 56 = 10 x 5 + 6, 37 = 10 x 3 + 7, 90 = 10 x 9 = 0, etc. The chapter also explains about deducing the divisibility test rules for a two or three-digit number expressed in the general form.

The Selina Solutions of Class 8 Maths Chapter 5 Playing with Numbers are provided here to help the students to clear all their doubts easily and can work on the weak points. These solutions are comprehensive and given in a step by step format.

## Download ICSE Class 8 Maths Selina Solutions PDF for Chapter 5 – Playing with Number

### ICSE Class 8 Maths Selina Solutions Chapter 5 Playing with Number – Exercise 5 (A)

Question 1. Write the quotient when the sum of 73 and 37 is divided by

(i) 11

Solution:

Sum of 73 and 37 is to be divided by

Let ab = 73

and ba = 37

∴a=7

and b=3

The quotient of ab+bc i.e. (73+37)when

Divided by 11 is  a+b=7+3=10

$\left(\frac{a b + b a}{11}=a+b\right)$

(ii) 10

Solution:

Sum of 73 and 37 is to be divided by

Let ab=73

and ba=37

∴a=7

And  b=3

The quotient of ab+ba i.e. (73 + 37) when

Divided by 10 ( i.e.  a + b is  11) ,

$\left(\frac{a b+b a}{a+b}=11\right)$

Question 2.Write the quotient when the sum of 94 and 49 is divided by

(i) 11

Solution:

Sum of 94 and 49 is to be divided by

Let ab=94

and  ba=49

∴a=9 and b=4

The quotient of 94+49 (i.e. ab + ba)

When divided by

11 is a+b i.e. 9 + 4 = 13

$\left(\frac{a b+b a}{11}=a+b\right)$

(ii) 13

Solution:

Sum of 94 and 49 is to be divided by

Let ab = 94

and ba = 49

∴ a = 9 and b = 4

The quotient of 94+49 (i.e. ab+ba)

When divided by i.e. (a+b) is 11

$\left(\frac{a b+b a}{a+b}=11\right)$

Question 3. Find the quotient when 73 – 37 is divided by

(i) 9

Solution:

(i) Difference of 73 – 37 is to be divided by 9

Let ab=73 and ba=37

a=7 and b=3

The quotient of 73-37(i.e. ab-bc) when

Divided by 7 is a-b i.e. 7-3=4

$\left(\frac{a b-b a}{9}=a-b\right) [\latex] (ii) 4 Solution: Let ab=73 and ba=37 (a=7 and b=3) The quotient of 73-37 (i.e. ab – ba) when Divided by 4 i.e. (a-b) is 9) \(\left(\frac{a b-b a}{a-b}=9\right)$

Question 4.

Find the quotient when 94-49 is divided by

(i) 9

Solution:

Difference of 94 and 49 is to be divided by

ab = 94 and ba = 49

The quotient of 94 – 49 i.e. (ab – ba) when

Divided by 9 is (a-b) i.e. 9 – 4 = 5

$\left(\frac{a b-b a}{9}=a-b\right)$

(ii) 5

Solution:

The quotient of 94-49 i.e. (ab-ba) when

Divided by 5 i.e. (a-b) is 9

$\left(\frac{a b-b a}{a-b}=9\right)$

Question 5. Show that 527 + 752 + 275 is exactly divisible by 14.

Solution:

abc = 100a+106+c……(i)

bca = 1006+10c+a…….(ii)

And cab = 100c+10a+b…….(iii)

Adding,(i),( II) and (iii), we get abc + bca + cab = 111a + 111c + 111c = 111(a + b + c) = 3 x 37 (a + b + c)

Now, let us try this method on

527 + 752 + 275 to check is it exactly divisible by 14

Here, a = 5, b = 2,c = 7

$527+752+275=3 \times 37(5+2+7)=3 \times 37 \times 14$

Hence, it shown that 527 + 752 + 275 is exactly divisible by 14.

Question 6. If a = 6, show that abc = bac.

Solution:

Given: a = 6

To show: abc  = bac

Proof: abc = 100a + 106 + c……(i)

(By using property 3)

Bac = 1006 + 10a + c…..(ii)

(By using property 3)

Since, a = 6

Substitute the value of a=6 in equation (i) and (ii), we get

abc = 1006 + 106 + c…. (iii)

bac = 1006 + 106 + c…..(iv)

Subtracting (iv) from (iii) abc  – bac=0

abc  = bac

Hence proved.

Question 7. If a>c; show that abc – cba = 99 (a – c).

Solution:

Given, a>c

To show: abc  –  cba = 99 (a – c)

Proof:abc = 100a + 10b + c….(i)

(By using property 3)

cba = 100c + 10b + a…..(ii)

(By using property 3)

Subtracting, equation (ii) from (i), we get

abc – cba = 100a + c – 100c – a

abc  – cba = 99a – 99c

abc  – cba = 99(a – c)

Hence proved.

Question 8. If c>a; show that  cba – abc = 99(c – a).

Solution:

Given: c>a

To show: cba – abc = 99 (c – a)

Proof:

cba = 100c + 106 + a…..(i)

(By using property 3)

abc = 100a + 106 + a….(ii)

(By using property 3)

cba-abc=100c+106+a-100a-106-c

⇒ cba-abc=99c-99a

⇒ cba-abc=99(c-a)

Hence proved.

Question 9. If a = c, show that cba – abc = 0

Solution:

Given: a=c

To show : cba  – abc = 0

Proof:

cba = 100c + 106 + a….(i)

(By using property 3)

Since, a = c,

Substitute the value of a = c in equation (i) and (ii), we get

cba =100c +10b + c…..(iii)

abc = 100c + 10b +c…..(iv)

Subtracting (iv) from (iii), we get

cba-abc-100c+106+c-100c-106-c

⇒ cba-abc = 0

⇒ cba = abc

Hence proved

Question 10. Show that 954 – 459 is exactly divisible by 99.

Solution:

To show: 954 – 459 is exactly divisible by 399, where a = 9, b = 5, c = 4

abc = 100a + 10b + c

=>954=100×9+10×5+4

=>954=900+50+4…… (i)

and 459=100×4+10×5+9

459=400+50+9…… (ii)

Hence, 954 – 459 is exactly divisible by 99

Hence proved.

### ICSE Class 8 Maths Selina Solutions Chapter 5 Playing with Number – Exercise 5 (B)

Question 1.

Solution:

A=7 as 7+5=12. We want 2 at units place

and 1 is carry over. Now 3+2+1=6

B=6

Hence A=7 and B=6

Question: 2

Solution:

A=5 as 8+5=13. We want 3 at units place

and 1 is carry over. Now 9+4+1=14.

B=4

and C=1 Hence A=5 and B=4 and C=1

Question: 3

Solution:

B=9 as 9+1=10. We want 0 at units place

and 1 is carry over. Now B-1-1=A.

∴A=9-2=7

Hence A=7 and B=9

Question: 4

Solution:

B=7 as 7+1=8. We want 8 at unit place.

Now

7+A=11

∴A=11-7=4

Hence A=4 and B=7

Question: 5

Solution:

A+B=9

and 2+A=10

∴A=10-2=8

and 8+B=9

∴ B=9-8=1

Hence A=8 and B=1

#### Practise This Question

When 7xy + 5yz - 3zx, 4yz + 9zx - 4y and - 3xz + 5x - 2xy are added, we get