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#### Page No 72:

#### Question 1:

The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.

#### Answer:

Let the tens place digit be x.

The units place digit is 3.

∴ Number = (10x + 3) ... (1)

Given:

7( x + 3) = (10 x + 3)

7 x + 21 = 10 x + 3

∴ 10 x - 7x = 21 - 3

⇒ 3 x = 18

or x = 6

Using x = 6 in equation (1):

The number is 63.

#### Page No 72:

#### Question 2:

In a two-digit number, the digit at the units place is double the digit in the tens place. The number exceeds the sum of its digits by 18. Find the number.

#### Answer:

Let the tens digit be x.

The digit in the units place is 2x.

Number = 10x + 2x

Given:

(x + 2x) + 18 = (10x + 2x)

∴ 3x + 18 = 12x

12x - 3x = 18

9x =18

x = $\frac{18}{2}$ = 2

The digit in the tens place is 2.

The digit in the units place is twice the digit in the tens place.

The digit in the units place is 4.

Therefore, the number is 24.

#### Page No 72:

#### Question 3:

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

#### Answer:

Let the tens place digit be a and the units place digit be b.

Then, number is (10a + b).

According to the question:

4(a + b) + 3 = (10 a + b)

4a + 4b + 3 = 10a + b

6a - 3b = 3

3(2a - b) = 3

2a - b =1 ... (1)

Given:

If 18 is added to the number, its digits are reversed.

The reverse of the number is (10b + a).

∴ (10a + b) + 18 = 10b + a

10a - a + b -10b = -18

9a - 9b = -18

9(a - b) = -18

a - b = -2 ... (2)

Subtracting equation (2) from equation (1):

2a - b = 1

a - b = -2

__- + + __

a = 3

Using a = 3 in equation (1):

2(3) - b = 1

6 - b = 1

∴ b = 5

Number = 10a+b = 10 $\times $ 3 + 5 = 35

#### Page No 72:

#### Question 4:

The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

#### Answer:

Let the tens place digit be a and the units place digit be b.

Then, the number is (10a + b).

Given:

a + b = 15 ... (1)

When the digits are interchanged the number will be (10 b + a).

Given:

10a + b + 9 = 10 b + a

∴ 10a - a + b - 10b = -9

9a - 9b = -9

a - b = -1 ... (2)

Adding equations (1) and (2):

a + b = 15

__a - b = -1 __

2a = 14

∴ a = 7

Using a = 7 in equation (2):

7 - b = -1

∴ b = 8

Original number = 10a+b = 10 $\times $ 7 + 8 = 78

#### Page No 72:

#### Question 5:

The difference between a 2-digit number and the number obtained by interchanging its digits is 63. What is the difference between the digits of the number?

#### Answer:

Let the tens place digit be 'x' and the units place digit be 'y'.

∴ Number = (10x + y)

Number obtained by interchanging the digits = (10y + x)

Given: (10x + y) - (10y + x) = 63

∴ 10x - x + y - 10 y = 63

9x - 9y = 63

9(x - y) = 63

x - y = 7

Therefore, the difference between the digits of the number is 7.

#### Page No 72:

#### Question 6:

In a 3-digit number, the tens digit is thrice the units digit and the hundreds digit is four times the units digit. Also, the sum of its digits is 16. Find the number.

#### Answer:

Let the units place digit be x.

Then, the tens place digit will be 3x and the hundreds place digit will be 4x.

Given:

4x + 3x + x = 16

or 8x = 16

or x =2

Units place digit = 2

Tens place digit = 3 $\times $ 2 = 6

Hundreds place digit = 4 $\times $ 2 = 8

Therefore, the number is 862.

#### Page No 77:

#### Question 1:

Test the divisibility of each of the following numers by 2:

(i) 94

(ii) 570

(iii) 285

(iv) 2398

(v) 79532

(vi) 13576

(vii) 46821

(viii) 84663

(ix) 66669

#### Answer:

A given number is divisible by 2 only when its unit digit is 0, 2, 4, 6 or 8.

(i) 94

The number 94 has '4' at its unit's place so, it is divisible by 2.

(ii) 570

The number 570 has '0' at its unit's place so, it is divisible by 2.

(iii) 285

The number 285 has '5' at its unit's place so, it is not divisible by 2.

(iv) 2398

The number 2398 has '8' at its unit's place so, it is divisible by 2.

(v) 79532

The number 79532 has '2' at its unit's place so, it is divisible by 2.

(vi) 13576

The number 13576 has '6' at its unit's place so, it is divisible by 2.

(vii) 46821

The number 46821 has '1' at its unit's place so, it is not divisible by 2.

(viii) 84663

The number 84663 has '3' at its unit's place so, it is not divisible by 2.

(ix) 66669

The number 66669 has '9' at its unit's place so, it is not divisible by 2.

#### Page No 77:

#### Question 2:

Test the divisibility of each of the following numers by 5:

(i) 95

(ii) 470

(iii) 1056

(iv) 2735

(v) 55053

(vi) 35790

(vii) 98765

(viii) 42658

(ix) 77990

#### Answer:

A given number is divisible by 5 only when its unit digit is 0 or 5.

(i) 95

The number 95 has '5' at its unit's place so, it is divisible by 5.

(ii) 470

The number 470 has '0' at its unit's place so, it is divisible by 5.

(iii) 1056

The number 1056 has '6' at its unit's place so, it is not divisible by 5.

(iv) 2735

The number 2735 has '5' at its unit's place so, it is divisible by 5.

(v) 55053

The number 55053 has '3' at its unit's place so, it is not divisible by 5.

(vi) 35790

The number 35790 has '0' at its unit's place so, it is divisible by 5.

(vii) 98765

The number 98765 has '5' at its unit's place so, it is divisible by 5.

(viii) 42658

The number 42658 has '8' at its unit's place so, it is not divisible by 5.

(ix) 77990

The number 77990 has '0' at its unit's place so, it is divisible by 5.

#### Page No 77:

#### Question 3:

Test the divisibility of each of the following numbers by 10:

(i) 205

(ii) 90

(iii) 1174

(iv) 57930

(v) 60005

#### Answer:

A given number is divisible by 10 only when its unit digit is 0.

(i) 205

The number 205 has '5' at its unit's place so, it is not divisible by 10.

(ii) 90

The number 90 has '0' at its unit's place so, it is divisible by 10.

(iii) 1174

The number 1174 has '4' at its unit's place so, it is not divisible by 10.

(iv) 57930

The number 57930 has '0' at its unit's place so, it is divisible by 10.

(v) 60005

The number 60005 has '5' at its unit's place so, it is not divisible by 10.

#### Page No 77:

#### Question 4:

Test the divisibility of each of the following numers by 3:

(i) 83

(ii) 378

(iii) 474

(iv) 1693

(v) 20345

(vi) 67035

(vii) 591282

(viii) 903164

(ix) 100002

#### Answer:

A given number is divisible by 3 only when the sum of its digits is divisible by 3.

(i) 83

The sum of the digits is 8 + 3 = 11 which is not divisible by 3. So, 83 is not divisible by 3.

(ii) 378

The sum of the digits is 3 + 7 + 8 = 18 which is divisible by 3. So, 378 is divisible by 3.

(iii) 474

The sum of the digits is 4 + 7 + 4 = 15 which is divisible by 3. So, 474 is divisible by 3.

(iv) 1693

The sum of the digits is 1 + 6 + 9 + 3 = 19 which is not divisible by 3. So, 1693 is not divisible by 3.

(v) 20345

The sum of the digits is 2 + 0 + 3 + 4 + 5 = 14 which is not divisible by 3. So, 20345 is not divisible by 3.

(vi) 67035

The sum of the digits is 6 + 7 + 0 + 3 + 5 = 21 which is divisible by 3. So, 67035 is divisible by 3.

(vii) 591282

The sum of the digits is 5 + 9 + 1 + 2 + 8 + 2 = 27 which is divisible by 3. So, 591282 is divisible by 3.

(viii) 903164

The sum of the digits is 9 + 0 + 3 + 1 + 6 + 4 = 23 which is not divisible by 3. So, 903164 is not divisible by 3.

(ix) 100002

The sum of the digits is 1 + 0 + 0 + 0 + 0 + 2 = 3 which is divisible by 3. So, 100002 is divisible by 3.

#### Page No 77:

#### Question 5:

Test the divisibility of each of the following numbers by 9:

(i) 327

(ii) 7524

(iii) 32022

(iv) 64302

(v) 89361

(vi) 14799

(vii) 66888

(viii) 30006

(ix) 33333

#### Answer:

A given number is divisible by 9 only when the sum of the digits is divisible by 9.

(i) 327

The sum of the digits is 3 + 2 + 7 = 12 which is not divisible by 9. So, 327 is not divisible by 9.

(ii) 7524

The sum of the digits is 7 + 5 + 2 + 4 = 18 which is divisible by 9. So, 7524 is divisible by 9.

(iii) 32022

The sum of the digits is 3 + 2 + 0 + 2 + 2 = 9 which is divisible by 9. So, 32022 is divisible by 9.

(iv) 64302

The sum of the digits is 6 + 4 + 3 + 0 + 2 = 15 which is not divisible by 9. So, 64302 is not divisible by 9.

(v) 89361

The sum of the digits is 8 + 9 + 3 + 6 + 1 = 27 which is divisible by 9. So, 89361 is divisible by 9.

(vi) 14799

The sum of the digits is 1 + 4 + 7 + 9 + 9 = 30 which is not divisible by 9. So, 14799 is not divisible by 9.

(vii) 66888

The sum of the digits is 6 + 6 + 8 + 8 + 8 = 36 which is divisible by 9. So, 66888 is divisible by 9.

(viii) 30006

The sum of the digits is 3 + 0 + 0 + 0 + 6 = 9 which is divisible by 9. So, 30006 is divisible by 9.

(ix) 33333

The sum of the digits is 3 + 3 + 3 + 3 + 3 = 15 which is not divisible by 9. So, 33333 is not divisible by 9.

#### Page No 77:

#### Question 6:

Test the divisibility of each of the following numbers by 8:

(i) 134

(ii) 618

(iii) 3928

(iv) 50176

(v) 39392

(vi) 56794

(vii) 86102

(viii) 66666

(ix) 99918

(x) 77736

#### Answer:

A given number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

(i) 134

The last two digits of 134 are '34' which is not divisible by 4. Hence, 134 is not divisible by 4.

(ii) 618

The last two digits of 618 are '18' which is not divisible by 4. Hence, 618 is not divisible by 4.

(iii) 3928

The last two digits of 3928 are '28' which is divisible by 4. Hence, 3928 is divisible by 4.

(iv) 50176

The last two digits of 50176 are '76' which is divisible by 4. Hence, 50176 is divisible by 4.

(v) 39392

The last two digits of 39392 are '92' which is divisible by 4. Hence, 39392 is divisible by 4.

(vi) 56794

The last two digits of 56794 are '94' which is not divisible by 4. Hence, 56794 is not divisible by 4.

(vii) 86102

The last two digits of 86102 are '02' which is not divisible by 4. Hence, 86102 is not divisible by 4.

(viii) 66666

The last two digits of 66666 are '66' which is not divisible by 4. Hence, 66666 is not divisible by 4.

(ix) 99918

The last two digits of 99918 are '18' which is not divisible by 4. Hence, 99918 is not divisible by 4.

(x) 77736

The last two digits of 77736 are '36' which is divisible by 4. Hence, 77736 is divisible by 4.

#### Page No 77:

#### Question 7:

Text the divisibility of each of the following numbers by 5:

(i) 95

(ii) 470

(iii) 1056

(iv) 2735

(v) 55053

(vi) 35790

(vii) 98765

(viii) 42658

(ix) 77990

#### Answer:

A number is divisible by 8 only when the number formed by its last three digits is divisible by 8.

(i) 6132

The last three digits of the given number are 132 which is not divisible by 8. So, 6132 is not divisible by 8.

(ii) 7304

The last three digits of the given number are 304 which is divisible by 8. So, 7304 is divisible by 8.

(iii) 59312

The last three digits of the given number are 312 which is divisible by 8. So, 59312 is divisible by 8.

(iv) 66664

The last three digits of the given number are 664 which is divisible by 8. So, 66664 is divisible by 8.

(v) 44444

The last three digits of the given number are 444 which is not divisible by 8. So, 44444 is not divisible by 8.

(vi) 154360

The last three digits of the given number are 360 which is divisible by 8. So, 154360 is divisible by 8.

(vii) 998818

The last three digits of the given number are 818 which is not divisible by 8. So, 998818 is not divisible by 8.

(viii) 265472

The last three digits of the given number are 472 which is divisible by 8. So, 265472 is divisible by 8.

(ix) 7350162

The last three digits of the given number are 162 which is not divisible by 8. So, 7350162 is not divisible by 8.

#### Page No 77:

#### Question 8:

Test the divisibility of each of the following numbers by 11:

(i) 22222

(ii) 444444

(iii) 379654

(iv) 1057982

(v) 6543207

(vi) 818532

(vii) 900163

(viii) 7531622

#### Answer:

A given number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of its digits at even places is

either 0 or a number divisible by 11.

(i) 22222

For the given number,

sum of the digits at odd places = 2 + 2 + 2 = 6

sum of digits at even places = 2 + 2 = 4

Difference of the above sums = 6 − 4 = 2

Since the difference is not 0 and neither a number divisible by 11 so, the 22222 is not divisible by 11.

(ii) 444444

For the given number,

sum of the digits at odd places = 4 + 4 + 4 = 12

sum of digits at even places = 4 + 4 + 4 = 12

Difference of the above sums = 12 − 12 = 0

Since the difference is 0 so the given number 444444 is divisible by 11.

(iii) 379654

For the given number,

sum of the digits at odd places = 4 + 6 + 7 = 17

sum of digits at even places = 5 + 9 + 3 = 17

Difference of the above sums = 17 − 17 = 0

Since the difference is 0 so the given number 379654 is divisible by 11.

(iv) 1057982

For the given number,

sum of the digits at odd places = 2 + 9 + 5 + 1 = 17

sum of digits at even places = 8 + 7 + 0 = 15

Difference of the above sums = 17 − 15= 2

Since the difference is not 0 and neither a number divisible by 11 so, the 1057982 is not divisible by 11.

(v) 6543207

For the given number,

sum of the digits at odd places = 7 + 2 + 4 + 6 = 19

sum of digits at even places = 0 + 3 + 5 = 8

Difference of the above sums = 19 − 8 = 11

Since the difference is 11 which is surely divisible by 11 so the given number 6543207 is divisible by 11.

(vi) 818532

For the given number,

sum of the digits at odd places = 2 + 5 + 1 = 8

sum of digits at even places = 3 + 8 + 8 = 19

Difference of the above sums = 19 − 8 = 11

Since the difference is 11 which is surely divisible by 11 so the given number 818532 is divisible by 11.

(vii) 900163

For the given number,

sum of the digits at odd places = 3 + 1 + 0 = 4

sum of digits at even places = 6 + 0 + 9 = 15

Difference of the above sums = 15 − 4 = 11

Since the difference is 11 which is surely divisible by 11 so the given number 900163 is divisible by 11.

(viii) 7531622

For the given number,

sum of the digits at odd places = 2 + 6 + 3 + 7 = 18

sum of digits at even places = 2 + 1 + 5 = 8

Difference of the above sums = 18 − 8 = 10

Since the difference is 10 which is not divisible by 11 so the given number 7531622 is not divisible by 11.

#### Page No 77:

#### Question 9:

Test the divisibility of each of the following numbers by 7:

(i) 693

(ii) 7896

(iii) 3467

(iv) 12873

(v) 65436

(vi) 54636

(vii) 98175

(viii) 88777

#### Answer:

(i) 693

69 − (2 × 3) = 69 − 6 = 63, which is divisible by 7.

Hence, 693 is divisible by 7.

(ii) 7896

789 − (2 × 6) = 789 − 12 = 777, which is divisible by 7.

Hence, 7896 is divisible by 7.

(iii) 3467

346 − (2 × 7) = 346 − 14 = 332, which is not divisible by 7.

Hence, 3467 is not divisible by 7.

(iv) 12873

1287 − (2 × 3) = 1287 − 6 = 1281, which is divisible by 7.

Hence, 12873 is divisible by 7.

(v) 65436

6543 − (2 × 6) = 6543 − 12 = 6531, which is divisible by 7.

Hence, 65436 is divisible by 7.

(vi) 54636

5463 − (2 × 6) = 5463 − 12 = 5451, which is not divisible by 7.

Hence, 54636 is not divisible by 7.

(vii) 98175

9817 − (2 × 5) = 9817 − 10 = 9807, which is divisible by 7.

Hence, 98175 is divisible by 7.

(viii) 88777

8877 − (2 × 7) = 8877 − 14 = 8863, which is not divisible by 7.

Hence, 88777 is not divisible by 7.

#### Page No 77:

#### Question 10:

Find all possible values of *x* for which the number 7*x*3 is divisible by 3. Also, find each such number.

#### Answer:

For a number to be divisible by 3, the sum of the digits must be divisible by 3.

$\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathit{}\mathit{=}\mathit{}7\mathit{}+\mathit{}x\mathit{}+3\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{}10+\mathit{}x$

$10+x$ will be divisible by 3 in the following cases:

$10+x=12,orx=2\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{will}\mathrm{be}723.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}10+x=15,orx=5\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{will}\mathrm{be}753.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}10+x=18,orx=8\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{will}\mathrm{be}783.$

So, the numbers can be 723, 753 or 783.

#### Page No 77:

#### Question 11:

Find all possible values of *y* for which the number 53*y*1 is divisible by 3. Also, find each such number.

#### Answer:

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

Sum of the digits = $5+3+y+1=9+y$

The sum of the digits is divisible by 3 in the following cases:

$9+y=9,ory=0\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5301.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}9+y=12,ory=3\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5331.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}9+y=15,ory=6\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5361.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}9+y=18,ory=9\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5391.$

∴ y = 0, 3, 6 or 9

The possible numbers are 5301, 5331, 5361 and 5391.

#### Page No 77:

#### Question 12:

Find the value of *x* for which the number *x*806 is divisible by 9. Also, find the number.

#### Answer:

For a number to be divisible by 9, the sum of the digits must be divisible by 9.

Sum of the digits in the given number = $x+8+0+6=x+14$

The sum of the digits is divisible by 9, only in the following case:

$x=4\phantom{\rule{0ex}{0ex}}or\phantom{\rule{0ex}{0ex}}x+14=18\phantom{\rule{0ex}{0ex}}$

Thus, the number x806 is divisible by 9 if $x$ is equal to 4.

The number is 4806.

#### Page No 77:

#### Question 13:

Find the value of *z* for which the number 471*z*8 is divisible by 9. Also, find the number.

#### Answer:

If a number is divisible by 9, then the sum of the digits is also divisible by 9.

Sum of the digits of the given number = $4+7+1+z+8=20+z$

$20+z=27,forz=7$

27 is divisible by 9.

Therefore, 471z8 is divisible by 9 if $z$ is equal to 7.

The number is 47178.

#### Page No 77:

#### Question 14:

Give five examples of numbers, each one of which is divisible by 3 but not divisible by 9.

#### Answer:

For a number to be divisible by 3, the sum of the digits should be divisible by 3.

And for the number to be divisible by 9, the sum of the digits should be divisible by 9.

Let us take the number 21.

Sum of the digits is 2 + 1 = 3, which is divisible by 3 but not by 9. Hence, 21 is divisible by 3 not by 9.

Similarly, lets check the number 24. Here, 2 + 4 = 6. This is divisible by 3 not by 9.

The number 30 will be divisible by 3 not by 9 as 3 + 0 = 3.

The number 33 will give 3 + 3 = 6, which is divisible by 3 not by 9.

Also 39 has the digits 3 + 9 = 12, which is divisible by 3 not by 9.

Hence, the numbers 21, 24, 30, 33 and 39 are divisible by 3 not by 9.

#### Page No 77:

#### Question 15:

Give five examples of numbers, each one of which is divisible by 4 but not divisible by 8.

#### Answer:

For a number to be divisible by 4, the number formed by its last two digits should be divisible by 4.

And for the number to be divisible by 8, the number formed by its last three digits should be divisible by 8.

So, the numbers divisible by 4 and not by 8 will be 28, 36, 44, 52, 60.

#### Page No 79:

#### Question 1:

**Replace A, B, C by suitable numerals.
$\begin{array}{ccc}& 5& A\\ +& 8& 7\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}C& B& 3\end{array}}}$**

#### Answer:

$A=6\phantom{\rule{0ex}{0ex}}\therefore A+7=6+7=13$

1 is carried over.

$(1+5+8)=14$

1 is carried over.

∴ $B=4$

and $C=1$

∴ $A=6,B=4andC=1$

#### Page No 79:

#### Question 2:

**Replace A, B, C by suitable numerals.
$\begin{array}{ccccc}& 4& C& B& 6\\ +& 3& 6& 9& A\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\overline{)\overline{)\begin{array}{ccccc}& 8& 1& 7& 3\end{array}}}}}\phantom{\rule{0ex}{0ex}}$**

#### Answer:

$A=7,A+6=7+6=13$ (1 is carried over)

$(1+B+9)=17,orB=7$ (1 is carried over)

$A=7,B=7andC=4$ (1 is carried over)

∴ $A=7,B=7andC=4$

#### Page No 79:

#### Question 3:

**Replace A, B, by suitable numerals.
$\begin{array}{cc}& A\\ +& A\\ +& A\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{cc}B& A\end{array}}}$**

#### Answer:

$A+A+A=A$ (with 1 being carried over)

This is satisfied if $A$ is equal to $5$.

When $A=5$:

$A+A+A=15$ (1 is carried over)

Or $B=1$

∴ $A=5andB=1$

#### Page No 79:

#### Question 4:

**Replace A, B by suitable numerals.
$\begin{array}{ccc}& 6& A\\ -& A& B\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}& 3& 7\end{array}}}$**

#### Answer:

First look at the left column, which is:

$6-A=3$

This implies that the maximum value of A can be 3.

$A\le 3$ ... (1)

The next column has the following:

$A-B=7$

To reconcile this with equation (1), borrowing is involved.

We know:

$12-5=7$

∴ $A=2andB=5$

#### Page No 80:

#### Question 5:

**Replace A, B, C by suitable numerals.
$\begin{array}{cccc}& C& B& 5\\ -& 2& 8& A\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{cccc}& 2& 5& 9\end{array}}}$**

#### Answer:

$5-A=9$

This implies that 1 is borrowed.

We know:

$15-6=9$

∴ $A=6$

$B-5=8$

This implies that 1 is borrowed.

$13-5=8$

But 1 has also been lent

∴$B=4$

$C-2=2$

This implies that 1 has been lent.

∴ $C=5$

∴ $A=6,B=4andC=5$

#### Page No 80:

#### Question 6:

**Replace A, B, C by suitable numerals.
$\begin{array}{ccc}& A& B\\ & \times & 3\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}C& A& B\end{array}}}$**

#### Answer:

$(\mathrm{B}\times 3)=\mathrm{B}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\mathrm{B}\mathrm{can}\mathrm{either}\mathrm{be}0\mathrm{or}5.\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{B}\mathrm{is}5,\mathrm{then}1\mathrm{will}\mathrm{be}\mathrm{carried}.\phantom{\rule{0ex}{0ex}}\mathrm{Then},\mathrm{A}\times 3+1=\mathrm{A}\mathrm{will}\mathrm{not}\mathrm{be}\mathrm{possible}\mathrm{for}\mathrm{any}\mathrm{number}.\phantom{\rule{0ex}{0ex}}\therefore \mathrm{B}=0\phantom{\rule{0ex}{0ex}}\mathrm{A}\times 3=\mathrm{A}\mathrm{is}\mathrm{possible}\mathrm{for}\mathrm{either}0\mathrm{or}5.\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{we}\mathrm{take}\mathrm{A}=0,\mathrm{then}\mathrm{all}\mathrm{number}\mathrm{will}\mathrm{become}0.\mathrm{However},\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{possible}.\phantom{\rule{0ex}{0ex}}\therefore \mathrm{A}=5\phantom{\rule{0ex}{0ex}}\mathrm{Then},1\mathrm{will}\mathrm{be}\mathrm{carried}.\phantom{\rule{0ex}{0ex}}\therefore \mathrm{C}=1\phantom{\rule{0ex}{0ex}}\therefore \mathrm{A}=5,\mathrm{B}=0\mathrm{and}\mathrm{C}=1$

#### Page No 80:

#### Question 7:

**Replace A, B, C by suitable numerals.
$\begin{array}{ccc}& A& B\\ \times & B& A\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}(B+1)& C& B\end{array}}}$**

#### Answer:

$A\times B=B\Rightarrow A=1$

In the question:

First digit = B+1

Thus, 1 will be carried from 1+B^{2} and becomes (B+1) (B^{2} -9) B.

∴ C = B^{2} -1

Now, all B, B+1 and B^{2} -9 are one digit number.

This condition is satisfied for B=3 or B=4.

For B< 3, B^{2} -9 will be negative.

For B>3, B^{2} -9 will become a two digit number.

For B=3 , C = 3^{2} - 9 = 9-9 = 0

For B = 4, C = 4^{2} -9 = 16-9 = 7

Required answer:

A=1, B=3, C = 0

or

A=1, B=4, C = 7

#### Page No 80:

#### Question 8:

**Replace A, B, C by suitable numerals.
$\begin{array}{cccccc}6)& 5& A& B& (9& C\\ -& 5& 4& & & \end{array}\phantom{\rule{0ex}{0ex}}\overline{)\begin{array}{cccc}& & 3& B\\ & -& 3& 6\end{array}}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}& \times & \end{array}}}$**

#### Answer:

$(A-4)=3\Rightarrow A=7$

Also, $6\times 6=36\Rightarrow C=6$

$36-36=0\Rightarrow B=6$

∴ $A=7\phantom{\rule{0ex}{0ex}}B=C=6$

#### Page No 80:

#### Question 9:

Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.

#### Answer:

$1and9$ are two numbers, whose product is a single digit number.

∴ $1\times 9=9$

Sum of the numbers is a two digit number.

∴ $1+9=10$

#### Page No 80:

#### Question 10:

Find three whole numbers whose product and sum are equal.

#### Answer:

The three whole numbers are 1, 2 and 3.

$1+2+3=6=1\times 2\times 3$

#### Page No 80:

#### Question 11:

complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15.

6 | 1 | |

5 | ||

#### Answer:

Taking the diagonal that starts with 6:

$6+5+x=15\Rightarrow x=4$

6 | 1 | |

5 | ||

4 |

Now, taking the first row:

$6+1+x=15\Rightarrow x=8$

6 | 1 | 8 |

5 | ||

4 |

Taking the last column:

$8+x+4=15\Rightarrow x=3$

6 | 1 | 8 |

5 | 3 | |

4 |

Taking the second column:

$1+5+x=15\Rightarrow x=9$

6 | 1 | 8 |

5 | 3 | |

9 | 4 |

Taking the second row:

$x+5+3=15\Rightarrow x=7$

6 | 1 | 8 |

7 | 5 | 3 |

9 | 4 |

Taking the diagonal that begins with 8:

$8+5+x=15\Rightarrow x=2$

6 | 1 | 8 |

7 | 5 | 3 |

2 | 9 | 4 |

#### Page No 80:

#### Question 12:

Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12.

#### Answer:

6+2+4 = 12

4+3+5 = 12

6+1+5 = 12

#### Page No 80:

#### Question 13:

**Fibonacci numbers** Take 10 numbers as shown below:

*a*, *b*, (*a* + *b*), (*a* + 2*b*), (2*a* + 3*b*), (3*a* + 5*b*), (5*a* + 8*b*), (8*a* + 13*b*), (13*a* + 21*b*), and (21*a* + 34*b*). Sum of all these numbers = 11(5*a* + 8*b*) = 11 × 7th number.

Taking *a* = 8, *b* = 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 × 7th number.

#### Answer:

Given:

$a=8andb=13$

The numbers in the Fibonnaci sequence are arranged in the following manner:

$1st,2nd,(1st+2nd),(2nd+3th),(3th+4th),(4th+5th),(5th+6th),(6th+7th),(7th+8th),(8th+9th),(9th+10th)$

The numbers are $8,13,21,34,55,89,144,233,377\mathrm{and}610$.

Sum of the numbers = $8+13+21+34+55+89+144+233+377+610\phantom{\rule{0ex}{0ex}}$

$=1584$

$11\times 7thnumber=11\times 144=1584$

#### Page No 80:

#### Question 14:

Complete the magic square:

14 | 0 | ||

8 | 6 | 11 | |

4 | . | 7 | |

2 | 1 | 12 |

#### Answer:

The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30.

3 |
14 | 13 |
0 |

8 | 5 |
6 | 11 |

4 | 9 |
10 |
7 |

15 |
2 | 1 | 12 |

#### Page No 80:

#### Question 1:

**Tick (✓) the correct answer**

If 5*x*6 is exactly divisible by 3, then the least value of *x* is

(a) 0

(b) 1

(c) 2

(d) 3

#### Answer:

(b) 1

If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3.

$5+x+6=11+x$ must be divisible by 3.

The smallest value of $x$ is 1.

$x=1$

$\Rightarrow x+11=12$ is divisible by 3.

#### Page No 81:

#### Question 2:

**Tick (✓) the correct answer**

If 64*y*8 is exactly divisible by 3, then the least value of *y* is

(a) 0

(b) 1

(c) 2

(d) 3

#### Answer:

(a) 0

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

$6+4+y+8=18+y$

This is divisible by 3 as *y* is equal to 0.

#### Page No 81:

#### Question 3:

**Tick (✓) the correct answer**

If 7*x*8 is exactly divisible by 9, then the least value of *x* is

(a) 0

(b) 2

(c) 3

(d) 5

#### Answer:

(c) 3

If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9.

$7+x+8=15+x$

18 is divisible by 9.

∴ $15+x=18\Rightarrow x=3$

#### Page No 81:

#### Question 4:

**Tick (✓) the correct answer**

If 37*y*4 is exactly divisible by 9, then the least value of *y* is

(a) 2

(b) 3

(c) 1

(d) 4

#### Answer:

(d) 4

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

$3+7+y+4=14+y$

For this sum to be divisible by 9:

$14+y=18\Rightarrow y=4$

#### Page No 81:

#### Question 5:

**Tick (✓) the correct answer**

If 4*xy*7 is exactly divisible by 3, then the least value of (*x* + *y*) is

(a) 1

(b) 4

(c) 5

(d) 7

#### Answer:

(a) 1

If a number is divisible by 3, the sum of the digits is also divisible by 3.

$4+x+y+7=11+(x+y)$

For the sum to be divisible by 3:

$11+(x+y)=12\Rightarrow (x+y)=1$

#### Page No 81:

#### Question 6:

**Tick (✓) the correct answer**

If *x*7*y*5 is exactly divisible by 3, then the least value of (*x* + *y*) is

(a) 6

(b) 0

(c) 4

(d) 3

#### Answer:

(d) 3

When a number is divisible by 3, the sum of the digits must also be divisible by 3.

$x+7+y+5=(x+y)+12$

This sum is divisible by 3 if x+y+12 is 12 or 15.

For x+y+12 = 12:

x+y=0

But x+y cannot be 0 because then x and y both will have to be 0.

Since x is the first digit, it cannot be 0.

∴ x+y+12 = 15

or x+y = 15-12=3

#### Page No 81:

#### Question 7:

**Tick (✓) the correct answer**

If *x*4*y*5*z* is exactly divisible by 9, then the least value of (*x* + *y* + *z*) is

(a) 3

(b) 6

(c) 9

(d) 0

#### Answer:

(c) 9

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

$x+4+y+5+z=9+(x+y+z)$

The lowest value of $(x+y+z)\mathrm{is}\mathrm{equal}\mathrm{to}0$ for the number x4y5z to be divisible by 9.

In this case, all x, y and z will be 0.

But x is the first digit, so it cannot be 0.

∴ x+4+y+5+z = 18

or x+y+z+9 = 18

or x+y+z = 9

#### Page No 81:

#### Question 8:

**Tick (✓) the correct answer**

If 1*A*2*B*5 is exactly divisible by 9, then the least value of (*A* + *B*) is

(a) 0

(b) 1

(c) 2

(d) 10

#### Answer:

(b) 1

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

$1+A+2+B+5=(A+B)+8$

The number will be divisible by 9 if $(A+B)=1$.

#### Page No 81:

#### Question 9:

**Tick (✓) the correct answer**

If the 4-digit number *x*27*y* is exactly divisible by 9, then the least value of (*x* + *y*) is

(a) 0

(b) 3

(c) 6

(d) 9

#### Answer:

(d) 9

If a number is divisible by 9, then the sum of the digits is divisible by 9.

$x+2+7+y=(x+y)+9$

For this to be divisible by 9, the least value of $(x+y)\mathrm{is}0$.

But for x+y = 0, x and y both will be zero.

Since x is the first digit, it can never be 0.

∴ x + y + 9 = 18

or x + y = 9

#### Page No 82:

#### Question 1:

Find all possible values of *x* for which the 4-digit number 320*x* is divisible by 3. Also, find the numbers.

#### Answer:

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

$3+2+0+x=5+x$ must be divisible by 3.

This is possible in the following cases:

$i)x=1\phantom{\rule{0ex}{0ex}}\therefore 5+x=6\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{is}3201.$

$ii)x=4\phantom{\rule{0ex}{0ex}}\therefore 5+x=9\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{is}3204.$

$iii)x=7\phantom{\rule{0ex}{0ex}}\therefore 5+x=12\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{number}\mathrm{is}3207.$

#### Page No 82:

#### Question 2:

Find all possible values of *y* for which the 4-digit number 64*y*3 is divisible by 9. Also, find the numbers.

#### Answer:

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

$6+4+y+3=13+y$

For this to be divisible by 9:

$y=5$

The number will be 6453.

#### Page No 82:

#### Question 3:

The sum of the digits of a 2-digit number is 6. The number obtained by interchanging its digits is 18 more than the original number. Find the original number.

#### Answer:

Let the two numbers of the two-digit number be 'a' and 'b'.

$a+b=6$ ... (1)

The number can be written as $(10a+b)$.

After interchanging the digits, the number becomes $(10b+a)$.

$(10a+b)+18=(10b+a)\phantom{\rule{0ex}{0ex}}9a-9b=-18$

$a-b=-2$ ... (2)

Adding equations (1) and (2):

$2a=4\Rightarrow a=2$

Using $a=2$ in equation (1):

$b=6-a=6-2=4$

Therefore, the original number is 24.

#### Page No 82:

#### Question 4:

Which of the following numbers are divisible by 9?

(i) 524618

(ii) 7345845

(iii) 8987148

#### Answer:

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

Number |
Sum of the digits |
Divisible by 9? |

524618 | 26 | No |

7345845 | 36 | Yes |

8987148 | 45 | Yes |

#### Page No 82:

#### Question 5:

Replace *A*, *B*, *C* by suitable numerals:

$\begin{array}{cccc}& 5& 7& A\\ -& C& B& 8\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{cccc}& 2& 9& 3\end{array}}}$

#### Answer:

$A-8=3$

This implies that 1 is borrowed.

$11-8=3\phantom{\rule{0ex}{0ex}}\Rightarrow A=1$

Then, $7-B=9$

1 is borrowed from 7.

∴ $16-B=9\phantom{\rule{0ex}{0ex}}\Rightarrow B=7$

Further, $5-C=2$

But 1 has been borrowed from 5.

∴ 4 - C = 2

$\Rightarrow C=2$

∴ $A=1,B=7andC=2$

#### Page No 82:

#### Question 6:

Replace *A*, *B*, *C* by suitable numerals:

$\begin{array}{cccccc}7)& 6& A& B& (8& C\\ -& 5& 6& & & \end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}& 6& B\\ -& 6& 3\end{array}}}\phantom{\rule{0ex}{0ex}}\times $

#### Answer:

Here, $A-6=6\phantom{\rule{0ex}{0ex}}\Rightarrow A=2(\mathrm{with}1\mathrm{being}\mathrm{borrowed})$

$B=3$

Since $7\times 9=63$, $C=9$

∴ $A=2,B=3andC=9$

#### Page No 82:

#### Question 7:

Find the values of *A*, *B*, *C* when

$\begin{array}{ccc}& A& B\\ \times & B& A\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\begin{array}{ccc}B& C& B\end{array}}$

#### Answer:

$A\times B=B\Rightarrow A=1$

$\begin{array}{ccc}& 1& B\\ \times & B& 1\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}& 1& B\\ B& {B}^{2}& \times \end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}B& (1+{B}^{2})& B\end{array}$

Now, $B\ne A=1$ and $(1+{B}^{2})$ is a single digit number.

∴ $B=2$

$C=(1+{B}^{2})=(1+4)=5$

∴ $A=1,B=2andC=5$

#### Page No 82:

#### Question 8:

**Mark (✓) against the correct answer**

If 7*x*8 is exactly divisible by 3, then the least value of *x* is

(a) 3

(b) 0

(c) 6

(d) 9

#### Answer:

(b) 0

If a number is exactly divisible by 3, the sum of its digits is also divisible by 3.

$7+x+8=15+x$

$15+x$ can be divisible by 3 even if *x* is equal to 0.

#### Page No 82:

#### Question 9:

**Mark (✓) against the correct answer**

If 6*x*5 is exactly divisible by 9, then the least value of *x* is

(a) 1

(b) 4

(c) 7

(d) 0

#### Answer:

(c) 7

When a number is divisible by 9, the sum of the digits is also divisible by 9.

$6+x+5=11+x$

To be divisible by 9:

$11+x=18\Rightarrow x=7$

#### Page No 82:

#### Question 10:

**Mark (✓) against the correct answer**

If *x*48*y* is exactly divisible by 9, then the least value of (*x* + *y*) is

(a) 4

(b) 0

(c) 6

(d) 7

#### Answer:

(c) 6

When a number is divisible by 9, the sum of its digits is also divisible by 9.

$x+4+8+y=12+(x+y)$

For $12+(x+y)$ to be divisible by 9:

$12+(x+y)=18\Rightarrow (x+y)=6$

#### Page No 82:

#### Question 11:

If 486*7 is divisible by 9, then the least value of * is

(a) 0

(b) 1

(c) 3

(d) 2

#### Answer:

(d) 2

For a number to be divisible by 9, the sum of its digits must be divisible by 9.

$4+8+6+*+7=25+*$

Now, $25+*=27(\mathrm{if}*=2$ and 27 is divisible by 9)

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