Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Intex Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Intex Questions

Try These (Page 250)

Question 1.

Write the following numbers in generalised form?

- 25
- 73
- 129
- 302

Solution:

1. 25 = 20 + 5 = 10 Ć 2 + 5 Ć 1

= 10 Ć 2 + 5

2. 73 = 70 + 3

= 10 Ć 7 + 3 Ć 1

= 10 Ć 7 + 3

3. 129 = 100 + 20 + 9

= 100 Ć 1 + 10 Ć 2 + 1 Ć 9

= 100 Ć 1 + 10 Ć 2 + 9

4. 302 = 100 Ć 3 + 10 Ć 0 + 1 Ć 2 = 300 + 0 + 2

Question 2.

Write the following in the usual form?

- 10 Ć 5 + 6
- 100 Ć 7 + 10 Ć 1 + 8
- 100 Ć a + 10 Ć c + b

Solution:

- 10 Ć 5 + 6 = 50 + 6 = 56
- 100 Ć 7 + 10 Ć 1 + 8 = 700 + 10 + 8 = 718
- 100 Ć a + 10 Ć c + b = 100a + 10c + b

=

Try These (Page 251)

[Adding the number with reversed digits to the chosen numbers.]

Question 1.

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

- 27
- 39
- 64
- 17

Solution:

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. Chosen 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 chosen number]

3. Chosen number = 64

Number with reversed digits = 46

Sum = 64 + 46 = 110

Now, 110 = 11[10] = 11 [6 + 4]

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

4. 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]

Try These (Page 251)

[Finding the difference of the chosen number and the number obtained by reversing the digits.]

Question 1.

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

- 17
- 21
- 96
- 37

Solution:

1. Chosen number =17

Number with reversed digits = 71

Difference = 71 – 17 = 54 = 9 Ć [6]

= 9 Ć [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 Ć [1]

= 9 Ć [Difference between the digits of the chosen number (2 – 1 = 1)]

3. Chosen number = 96

Number with reversed digits = 69

Difference = 96 – 69 = 27 = 9 Ć [3]

= 9 Ć [Difference between the digits of the chosen number (9 – 6 = 3)]

4. Chosen number = 37

Number with reversed digits = 73

Difference = 73 – 37 = 36 = 9 Ć [4]

= 9 Ć [Difference between the digits of the chosen number (7 – 3 = 4)]

Try These (Page 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?

- 132
- 469
- 737
- 901

Solution:

1. 132 Chosen number = 132

Reversed number = 231

Difference = 231 – 132 = 99

We have 99 Ć· 99 = 1, remainder = 0

2. 469 Chosen number = 469

Reversed number = 964

Difference = 964 – 469 = 495

We have 495 Ć· 99 = 5, remainder = 0

3. Chosen number = 737

Reversed number = 737

We have Difference = 737 – 737 = 0

0 Ć· 99 = 0, remainder = 0

4. Chosen number = 901

Reversed number = 109

Difference = 901 – 109 = 792

We have 792 Ć· 99 = 8, remainder = 0

Forming three-digit number with given three digits

Try These (Page 253)

Question 1.

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

- 417
- 632
- 117
- 937

Solution:

1. Chosen number = 417

Two other numbers with the same digits are 741 and 174

Sum of the three numbers

We have 1332 Ć· 37 = 36, remainder

2. Chosen number = 632

Two other numbers are 263 and 326

Sum of the three numbers

We have 1221 Ć· 37 = 33, remainder = 0

3. Chosen number =117

Other numbers are 711 and 171

Sum of the three numbers

We have 999 Ć· 37 = 27, remainder = 0

4. Chosen number = 937

Other two numbers are 793 and 379

ā“ Sum of the three numbers

We have 2109 Ć· 37 = 57, remainder = 0

Try These (Page 257)

Question 1.

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

Solution:

The ones digit, when divided by 5, must leave a remainder of 3. So the ones digit must be either 3 or 8.

Question 2.

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

Solution:

If remainder = 1, then the ones digit of āNā must be either 1 or 6.

Question 3.

If the division N Ć· 5 leaves a remainder of 4, what might be the ones digit of N?

Solution:

If remainder = 4, then the ones digit of āNā must be either 9 or 4.

Try These (Page 257)

Question 1.

If the division N Ć· 2 leaves a remainder of 1, what might be the ones digit of N?

Solution:

N is odd; so its ones digit is odd. Therefore. the ones digit must be 1, 3, 5, 7 or 9.

Question 2.

If the division N Ć· 2 leaves no remainder (i.e.. zero remainder), what might be the one digit of N?

Solution:

āµ Remainder = 0

ā“ Ones digit can be 0, 2, 4, 6 or 8.

Question 3.

Suppose that the division N Ć· 5 leaves a remainder of 4 and the division N Ć· 2 leaves a remainder of I. What must be the ones digit of N?

Solution:

āµ N + 5 and remainder = 4

ā“ Ones digit can be 4 or 9.

Again N Ć· 2 and remainder = 1

ā“ N must be an odd number.

Thus, ones digit can be 9 only.

Try These (Page 259)

Question 1.

Check the divisibility of the following numbers by 9?

1. 108

2. 616

3. 294

4. 432

5. 927

Solution:

1. 108

āµ 1 + 0 + 8 = 9

and 9 is divisible by 9.

[āµ 9 + 9 = 1 and remainder = 0]

ā“ 108 is divisible by 9.

2. 616

We have 6 + 1 + 6 = 13

and 13 Ć· 9 = 1, remainder = 4

i.e., 13 is not divisible 9.

ā“ 616 is also not divisible by 9.

3. 294

We have 2 + 9 + 4 = 15

and 15 + 9 = 1, remainder = 6

i.e., 15 is not divisible by 9.

ā“ 294 is also not divisible by 9.

4. 432

We have 4 + 3 + 2 = 9

9 + 9 = 1, remainder = 0

ā“ 432 is divisible by 9.

5. 927

We have 9 + 2 + 7 = 18

and 18 + 9 = 2, remainder = 0

i.e., 18 is divisible by 9.

ā“ 927 is also divisible by 9.

Try These (Page 260)

Question 1.

Check the divisibility of the following numbers by 3?

1. 108

2. 616

3. 294

4. 432

5. 927

Solution:

1. 108

We have 1 + 0 + 8 = 9

and 9 + 3 = 3, remainder = 0

ā“ 108 is divisible by 3.

[āµ 9 is divisible by 3]

2. 616

We have 6 + 1 + 6 = 13

and 13 Ć· 3 = 4, remainder = 1

ā“ 13 is not divisible by 3.

Thus 616 is also not divisible by 3.

3. 294

We have 2 + 9 + 4 = 15

and 15 Ć· 3 = 5, remainder = 0

ā“ 15 is divisible by 3.

Thus, 294 is also divisible by 3.

4. 432

We have 4 + 3 + 2 = 9

and 9 Ć· 3 = 3, remainder = 0

i.e; 9 is divisible by 3.

Thus, 294 is also divisible by 3.

5. 927

we have 9 + 2 + 7 = 18

and 18 Ć· 3 = 6, remainder = 0

i.e; 18 is divisible by 3.

Thus, 927 is also divisible by 3.