These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions

NCERT Intext Question Page No. 250

Question 1.

Write the following numbers in generalised form.;

(i) 25

(ii) 73

(iii) 129

(iv) 302

Answer:

(i) 25 = 20 + 5 = 2 × 10 + 5 × 1 = 10 × 2 + 5

(ii) 73 = 70 + 3 = 7 × 10 + 3 × 1 = 10 × 7 + 3

(iii) 129 = 100 + 20 + 9

= 1 × 100 + 2 × 10 + 9 × 1

= 100 × 1 + 10 × 2 + 9

(iv) 302 = 3 × 100 + 0 × 10 + 2 × 1

= 100 × 3 + 10 × 0 + 2

Question 2.

Write the following in the usual form.

(i) 10 × 5 + 6

(ii) 100 × 7 + 10 × 1 + 8

(iii) 100 × a + 10 × c + b

Answer:

(i) 10 × 5 + 6 = 50 + 6 = 56

(ii) 100 × 7 + 10 × 1 + 8

= 700 + 10 + 8 = 718

(iii) 100 × a + 10 × c + b

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

NCERT Intext Question Page No. 257

Question 1.

If the division N + 5 leave a remainder of 3, what might be the one’s digit of N?

Answer:

The unit digit when divided by 5 must leave a remainder of 3.50, the one’s digit must be either 3 or 8.

Question 2.

If the division N + 5 leaves a remainder of 1, what might be the ones digit of N?

Answer:

The one’s digit when divided by 5 must have a remainder of 1. So the one’s digit must be either 1 or 6.

Question 3.

If the division N + 5 leaves a remainder of 4. What might be the one’s digit of N?

Answer:

The one’s digit, when divided by 5 must leave a remainder of 4. So the one’s digit must be either 4 or 9.

NCERT Intext Question Page No. 259

Question 1.

Check the divisiblity of the following number by 9.

(i) 108

(ii) 616

(iii) 294

(iv) 432

(v) 927

Answer:

(i) 108

Sum of the digits = 1 + 0 + 8 = 9 which is divisible by 9.

∴ 108 is divisible by 9.

(ii) 616

Sum of the digits = 6 + 1 + 6

= 13 which is not divisible by 9.

∴ 616 is not divisible by 9.

(iii) 294

Sum of the digits = 2 + 9 + 4 = 15

which is not divisible by 9.

∴ 294 is not divisible by 9.

(iv) 432

Sum of the digit = 4 + 3 + 2 = 9

which is divisible by 9.

∴ 432 is divisible by 9.

(v) 927

Sum of the digits = 9 + 2 + 7 = 18

which is divisible by 9.

∴ 927 is divisible by 9.

Question 9.

Check the divisibility of the following number by 3.

(i) 108

(ii) 616

(iii) 294

(iv) 432

(v) 927

Answer:

(i) 108

Sum of the digits =1 + 0 + 8 = 9

which is divisible by 3.

∴ 108 is divisible by 3.

(ii) 616

Sum of the digits = 6 + 1 + 6 = 13

which is not divisible by 3.

∴ 613 is also not divisible by 3.

(iii) 294

Sum of the digits = 2 + 9 + 4 = 15

which is divisible by 3.

∴ 294 is also divisible by 3.

(iv) 432

Sum of the digits = 4 + 3 + 2 = 9

which is divisible by 3.

∴ 432 is also divisible by 3.

(v) 927

Sum of the digits = 9 + 2 + 7 = 18 which is divisible by 3.

∴ Thus 927 is also divisible by 3.

NCERT Intext Question Page No. 251

Question 1.

Check what the result would have been if Sundaram had chosen the numbers shown below.

1. 27

2. 39

3. 64

4. 17

Answer:

1. Chosen number = 27

Number with reversed digits = 72

Sum of the two numbers = 27 + 72 = 99

Now, 99 = 11 [9] = 11 [2 + 7]

= 11 [Sum of the digits of the chosen number]

2. Chose number = 39

Number with reversed digits = 93

Sum of the two numbers = 39 + 93 = 132

Now, 132 ÷ 11 = 12

i. e„ 132 = 11 [12] = 11 [3 + 9]

= 11 [Sum of the digits of the chose number]

3. Chosen number = 17

Number with reversed digits = 71

Sum = 17 + 71 = 88

Now, 88 = 11 [8] = 11 [1 + 7]

= 11 [Sum of the digits of the chosen number]

Question 2.

Check what the result would have been if Sundaram had chosen the numbers shown.

1. 17

2. 21

3. 96

4. 37

Answer:

1. Chosen number =17

Number with reversed digits = 71

Difference = 71 – 17 = 54 = 9 × [6]

= 9 x [Difference of the digits of the chosen number (7 – 1 = 6)]

2. Chosen number = 21

Number with reversed digits = 12

Difference = 21 – 12 = 9 = 9 × [1]

= 9 x [Difference of the digits of the chosen number (12 -1 = 1)]

3. Chosen number = 96

Number with reversed digits = 69

Difference = 96 – 69 = 27 = 9 × [3]

=9 x [Difference of the digits of the chosen number (9 – 1 = 3)]

4. Chosen number = 37

Number with reversed digits = 73

Difference = 73 – 37 = 36 = 9 × [4]

=9 x [Difference of the digits of the chosen number (7-3 = 4)]

NCERT Intext Question Page No. 252

Question 1.

Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.

1. 132

2. 469

3. 737

4. 901

Answer:

1. 132 Chosen number = 132

Reversed number = 231

Difference = 231 – 132 = 99

We have 99 ÷ 99 = 1, remainder = 0

2. 469 Chosen number = 469

Reversed number = 964

Difference = 964 – 469 = 495

We have 99 ÷ 99 = 5, remainder = 0

3. Chosen number = 737

Reversed number = 737

We have Difference = 737 – 737 = 0

0 ÷ 99 =0, remainder = 0

4. Chosen number = 901

Reversed number = 109

Difference = 901 – 109 = 792

We have 792 ÷ 99 = 8, remainder = 0

Forming three-digit number with given three digits

NCERT Intext Question Page No. 253

Question 1.

Check what the result would have been if Sundaram had chosen the numbers shown below.

1. 417

2. 632

3. 117

4. 937

Answer:

1. Chosen number = 417

Two other numbers with the same digits are 741 and 174

Sum of the three numbers

We have 1332 ÷ 37 = 36, remainder = 0

2. Chosen number = 632

Two other numbers are 263 and 326

Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

3. Chosen number =117

Other numbers are 711 and 171

Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

4. Chosen number = 937

Other two numbers are 793 and 379

Sum of the three numbers

We have 2109 + 37 = 57, remainder = 0