NCERT solutions for Class 8 Maths Chapter 16 Playing with Numbers

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.2 numbers in general form

Question: 1(i) Write the following numbers in generalised form.

Answer:

any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b

hence,

25 = 10*2 + 5 = 20 + 5

Question: 1(ii) Write the following numbers in generalised form.

Answer:

any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b

hence generalized form of the number

73 = 10*7 + 3

Question:1(iii) Write the following numbers in generalised form.

129

Answer:

A 3 digit number madeup of digits a, b, c will be written as,

abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c

hence generalised form of number

129 = 100*1 + 10*2 + 1*9

Question: 1(iv) Write the following numbers in generalised form.

302

Answer:

A 3-digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c

hence generalised form of

302 = 100*3 + 10*0 + 1*2

Question:2(i) Write the following in the usual form.

$10\times 5 + 6$

Answer:

As we know ab = 10 × a + b = 10a + b

usual form of number

10*5 + 6 = 50 + 6 = 56

Question: 2(ii) Write the following in the usual form.

$100\times 7 + 10\times 1 + 8$

Answer:

As we know abc = 100 × a + 10 × b + 1 × c

the usual form of number

100*7 + 10*1 + 8 = 700 + 10 + 8 = 718.

Question:2(iii) Write the following in the usual form.

$100\times a + 10\times c + b$

Answer:

As we know abc = 100 × a + 10 × b + 1 × c

casual form of 100 × a + 10 × b + 1 × c = abc.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.3 games with numbers

Question:1 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

As we know

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 2 and b = 7

27 + 72 = 99= 11 *9 = a multiple of 11

Question:2 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

39 + 93 = 133 = 11*12 = a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

Here a = 3 and b = 9

Question:3 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

64 + 46 = 110 = 11*10= a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 6 and b = 4

Question:4 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

17 + 71 = 88 = 11*8 = a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 1 and b = 7

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.3 games with numbers

Question:1 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

If the tens digit is larger than the ones digit (that is, a > b), then
(10a + b) – (10b + a) = 10a + b – 10b – a
= 9a – 9b = 9(a – b).
If the unit digit is larger than the tens digit (that is, b > a), he does:
(10b + a) – (10a + b) = 9(b – a).

here a = 1 and b = 7

71 - 17 = 54 = 9*6 = multiple of 9

Question:2 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

here a = 2 and b = 1

21 - 12 = 9 = 9*1 = multiple of 9

Question:3 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

here a = 9 and b = 6

96 - 69 = 27= 9*3= multiple of 9

Question:4 Check what the result would have been if Sundaram had chosen the numbers shown below.

Answer:

here a = 3 and b = 7

73 - 37 = 36 = 9*4 = multiple of 9

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.3 games with numbers

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

Answer:

Let's assume the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• If a & c are equal,then the difference is 0.

here a = 1, b = 3 and c = 2

231 - 132 = 99 = multiple of 99

Question:2 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.

469

Answer:

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original numbedr & reversed numbers is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• if a & c are equal, then the difference is 0.

here a,b and c are 4, 6 & 9 respectively.

964 - 469 = 495 = 99*5 = multiple of 99

Question:3 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.

737

Answer:

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• If A 7 C are equal, the difference is 0.

here a = 7, b = 3 and c = 7

737- 737 = 0= multiple of 99

Question:4 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.

901

Answer:

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• If a & c are equal, the difference is 0.

here a = 9, b = 0 and c = 1

901- 109= 792= 99*8 = multiple of 99

quotient in each case = c - a.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.3 games with numbers

Question:1 Check what the result would have been if Sundaram had chosen the numbers shown below.

417

Answer:

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

After adding all the above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c),

It will be divisible by 37 becuase 37 is present in the equation.

here a = 4, b = 1, and c = 7

417 + 741 + 147 = 1332 = 37*36 i.e. divisible by 37.

Question:2 Check what the result would have been if Sundaram had chosen the numbers shown below.

632

Answer:

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of all the above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 6, b = 3, and c = 2

632+ 263+ 362= 1221= 37*33 i.e. divisible by 37.

Question:3 Check what the result would have been if Sundaram had chosen the numbers shown below.

117

Answer:

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of the above all three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 1, b = 1, and c = 7

117 + 711 + 117 = 999 = 37*27 i.e. divisible by 37.

Question:4 Check what the result would have been if Sundaram had chosen the numbers shown below.

937

Answer:

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of all above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 9, b = 3, and c = 7

937 + 793 + 397 = 2109 = 37*57 i.e. divisible by 37.

NCERT solutions for class 8 maths chapter 16 playing with numbers-Exercise: 16.1

Question:1 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

Here we are adding two numbers and unit place of the first number and the second number is A and 5 respectively. unit place of the answer is 2 so the way we can get this result is when we get 12 on adding unit places of both number i.e.

A + 5 = 12

which implies A = 12 - 5 = 7.

Ten's digit of both numbers are 3 and 2.remainder=1

so ten's digit of the answer(B) = 3 + 2 + 1 = 6

Hence A= 7 and B = 6.

Question:2 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

Here

answer's unit place = 3 , possible addition of unit places digit = 13

A + 8 = 13

A = 13 - 8 = 5

remainder = 1

Ten's place of answer = 4 + 9 + 1 = 14

B = 4

remainder = 1

100's place = C = 1

Hence value of A = 5 , B = 4 and C = 1.

Question:3 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

Here

first clue :

we have A = a number which when multiplied by itself gives the same number in the unit digit.

possible numbers = 1 and 6

Second Clue:

number when multiplied with 1 and added with the reminder of previous multiplication( A*A) = 9

both first and second clue implies that A = 6.

Question:4 Find the values of the letters in each of the following and give reasons for the steps involved

Answer:

There can be two cases

1. when the addition of unit place digit doesn't produce Carry

A + 3 = 6

A = 3

However, to get 3 in unit place of our answer our B has to be 6 and that would produce carry hence this case is not possible.

2. when the addition of unit place digit produces Carry

A + 3 +1 = 6

A = 2

for getting 2 in unit place of answer we need the sum of unit digit of numbers = 12

B + 7 = 12

B = 5

Hence A = 2 and B = 5

Question:5 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

Here multiplication of 3 and B gives a number whose unit place digit is B .

Possible value of B = 0 and 5

let B = 5

3 * A + 1 = CA

this is not possible for any value of A.

Hence B = 0

now A * 3 = CA ( a number whose unit place digit is A itself when multiplied by 3) hence

possible value of A = 5 and 0

since AB is a two digit number A can not equal to 0.

hence A = 5

A * 3 + 1 = CA

5 * 3 + 1 = 15

hence C = 1

A = 5, B = 5 and C = 1.

Question:6 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

Here , multiplication of B and 5 gives a number whose ones digit is B. this is possible when B = 0 or 5

let B = 5

B * 5 = 5 * 5 = 25

Carry = 2

5*A + 2 = CA , This os possible only when A = 2 or 7

when A = 2 ,

5*2 + 2 = 12 which implies C = 1

when A = 7

5*7 + 2 = 37 which implies C = 3

now B = 0

B*5 = 0*5 = 0

Carry = 0 so

5 * A = CA which is possible when A = 0 or 5 Howerver A cannot equal to 0 since AB is a two digit number

so A = 5

5*5 = 25 which implies C = 2

hence possible values of A B and C are

A = 5, 2, 7 , B = 0, 5, 5 and C = 2, 1, 3

Question:7 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

The product of 6 & B gives a number whose unit digit is B again.

possible value of B = 0. 2 , 4, 6 or 8

If B = o then our product will be zero. hence this value of B is not possible.

If B = 2, then B x 6 = 12 . Carry for next step = 1.

6A + 1 = BB = 22

implies A = 21/6 = not any integer value hence this case is also not possible.

If B = 6 then B *6 = 36 and 3 will be carry for next step.

6A + 3 = BB = 66

implies A = 63/6 = not an integer value hence this case is also not possible.

If B = 8 then B * 6 = 48 and 4 will be carry for next step.

6A + 4 = BB = 88

implies A = 14 however A is a single digit number hence this case is also not possible.

If B = 4 then B*6 = 24 and 2 will be carry for next step.

6A + 2 = BB = 22

implies A = 7 .

Hence A = 7 and B = 4 is the correct answer.

Question:8 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

The addition of 1 and B gives a number whose ones digit is 0. this is possible when digit B = 9 .

1 + B = 10 and 1 is the Carry for the next step

Now, A + 1 + 1 = B => 9

Implies A = 7.

Hence A = 7 and B = 9 is Correct answer

Question:9 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

The additiion of B and 1 is 8 is giving a number whose ones digit is 8. this means digit B is 7 .

B + 1 = 8 and no carry for next step.

next step :

Now, A + B = 1 => A + 7

which implies A = 4

A + B = 11 and 1 is carry for next step

1 + 2 + A = B

1 + 2 + 4 = 7

Hence A = 4 and B = 7 is correct answer.

Question:10 Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:

The addition of A and B is giving a number whose ones digit is 9. The sum can only be 9 not 10 as a sum of two single digits cannot exceed 18. hence there will not be any carry for the next step

2 + A = 0

implies A = 8

2 + 8 = 10 and 1 is the carry for next step.

1 + 1 + 6 = A = 8

it satisfies hence A = 8 and B = 1 is the correct answer.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.5.2 divisibility by 5

Question:1 If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N? (The one’s digit, when divided by 5, must leave a remainder of 3. So the one’s digit must be either 3 or 8.)

Answer:

The detailed solution for the above-written problem is as follows,

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

Question:2 If the division $N\div 5$ leaves a remainder of 1, what might be the one’s digit of ?

Answer:

The detailed solution for the above-written question is as follows

If a number is divisible by 5 then it's unit digit must be 0 or 5. so if we need the remainder of 1 when divided by 5 then the numbers unit digit must be 1 or 6.

Question: 3 If the division $N\div 5$ leaves a remainder of 4, what might be the one’s digit of N?

Answer:

The detailed solution for the above-written question is as follows

If the unit digit of a number is 0 or 5, then it is divisible by 5. hence if we need the remainder of 4 then unit digit of number should be 4 or 9.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.5.3 divisibility by 2

Question: 1 If the division N ÷ 2 leaves a remainder of 1, what might be the one’s digit of N? (N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7 or 9.)

Answer:

The detailed solution for the above-written question is as follows

N is odd; so it's unit digit is odd. Therefore, the unit digit must be 1, 3, 5, 7 or 9.

Question:2 If the division $N \div 2$ leaves no remainder (i.e., zero remainders), what might be the one’s digit of N?

Answer:

The detailed solution for the above-written question is as follows

N is Even; so it's unit digit is even. Therefore, the unit digit must be 2, 4, 6, 8 or 0.

Question: 3 Suppose that the division $N\div 5$ leaves a remainder of 4, and the division $N\div 2$ leaves a remainder of 1. What must be the one’s digit of N?

Answer:

Since N leaves the remainder of 4 when divided by 5. the possible values in ones place of number N are 4 or 9.

now, since it leaves a remainder of 1 when divided by 2, the N would be an odd number. hence ones digit of N is also an odd number. which means ones digit of our number N is 9.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.5.4 divisibility by 9 and 3

Question:1 Check the divisibility of the following numbers by 9.

108

Answer: