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Chapter 16: Playing with Numbers

Exercise 16.1

  1. In the given addition, find the value of the letters and give reason.

   4    A

+ 3    4

———-

  B    3

                    

Answer:

When we add A and 4, we are getting number 1 that is it gives number 1 in onesplace. This is possible only when A is number 9. Therefore, adding A that is 9 to 4, it gives 13. 1 will be carry for the next step.

In next step, the addition is 1 + 4 + 3 = 8. So the addition is as given below.

4   9

+ 3   4

————

8   3

Therefore, B is 8.

So, A and B are 4 and 8 respectively.

 

  1. In the given addition, find the value of the letters and also give reason.

 5   X

+ 8   7

———-

Y  Z  2

Answer:

The addition of X and 7 gives 2 that is a no. whose ones place is 2. It is possible only when digit X is 5. So, the addition of 5 and 7 gives 12. 1 will be carry for the next step.

1 + 5 + 8 = 14

Hence, the addition is as given below.

5   5

+ 8   7

—————

1   4   2

The value of X, Y and Z is 5, 1 and 4 respectively.

 

  1. In the given multiplication, find the value of the letters and give reason.

        2   Y

      ×   Y

—————–   

   1    2   Y

 

Answer:

The multiplication of a number with itself gives a number whose ones placeis that number itself. This occurs when that number is 1, 5 or 6.

Y = 1

The multiplication will be 21 × 1 = 21. Here, there is hundreds place as well. So, Y = 1 is not possible.

Y = 5

The multiplication will be 25 × 5 = 125. Here, tens as well as hundreds place match. So, Y = 5 is the correct answer.

 

  1. In the given addition, find the value of the letters and also give reason.

    X   Y

 + 4   6

————–

    7   X

Answer:

The addition of X and 3 is giving 7. There can be two cases.

(i)No carry

The value of X will be 3 so we get 7 when we add 3 and 4 that is 3 + 4 = 7. Now consider the first step, Y + 6 = 3 so the value of Y has to be 7. Then we get 3 in ones place. But the value X is single digit so it is not possible.

(ii) With carry

The value of X will be 2 as 1 + 2 + 4 = 7. Now consider the first step where Y is added to 6 to give 2 in ones place. For that the value of Y will be 6 + Y = 12. Therefore, the value of Y is 6.

2   6

+ 4   6

——————

7    2

 

Hence, the value of X and Y is 2 and 6 respectively.

 

  1. In the given multiplication, find the value of the letters and give reason.

        X   Y

 ×   3

  ————-

    Z  X   Y

 

Answer:

When 3is multiplied with Y it gives a number whose onesplace is Y again. So, Y must be 5 or 0.

 

Let Y = 5

First step: 5 × 3 = 15

1 will be carried forward. Therefore, (X × 3) + 1 = ZX. This is not possible for any number.

Therefore, value of Y has to be 0 only.

If Y = 0, then there will be no carry. So we get X × 3 = ZX.

When a number is multiplied with 3, its ones placeshould be the number itself. That is possible only for X = 0 or 5. But X cannot be 0 as it has to be two digit numbers. Therefore, the value of X is 5. Thus we get the following

 

5   0

×   3

—————

15 0

 

The value of X, Y and Z is 5, 0 and 1 respectively.

 

  1. In the given multiplication, find the value of the letters and also give reason.

        X   Y

         ×   5

   ————

     Z X   Y

 

Answer:

When 5 is multiplied with Y it gives a number whose ones placeis Y again. So, Y must be 5 or 0.

 

Let Y = 5

First step: 5 × Y = 5 × 5 = 25

2 will be carried forward. Therefore, (X × 5) + 2 = ZX. This is possible for number X = 2 or 7.

 

The multiplication is as given below.

2   5                                                     7   5

×   5                                                   ×   5

———-                                                ———-

1  2  5                                                   3  7  5

Let Y = 0

First step: 5 × Y = 5

5 × 0 = 0

There will not be any carry in this case.

In the next step, 5 × X = ZX

This can happen only when the value of X is 5 or 0.

However, X cannot be 0 as XY is two digit numbers. Therefore, the value of X is 5.

5   0

×   5

———–

2  5  0

 

Therefore, there three possible values of X, Y and Z.

(i) 2, 5 and 1 respectively

(ii) 5, 0 and 2 respectively

(iii) 7, 5 and 3 respectively

 

  1. In the given multiplication, find the value of the letters and also give reason.

      X   Y

      ×   6

     ———-

      Y  YY

Answer:

When 6 is multiplied with Y, it gives a number whose ones placeis Y. It is possible only if Y = 0, 2, 4, 6 or 8.

Y = 0;

The product will be 0 in this case so it is not possible.

 

Y = 2;

Y × 6 = 12 and 1 will be carried forward for the next step.

6X + 1 = YY = 22. Then integer value of X is not possible.

 

Y = 6;

Y × 6 = 36 and 3 will be carried forward for the next step.

6X + 3 = YY = 66. Then integer value of X is not possible.

 

Y = 8;

Y × 6 = 48 and 4 will be carried forward for the next step.

6X + 4 = YY = 88.

6X = 84.

X = 14

But X is single digit number.

Then value of X is not possible.

Y = 4;

Y × 6 = 24 and 2 will be carried forward for the next step.

6X + 2 = YY = 44.

6X = 42.

X = 7

The multiplication is given below

 

7   4

×   6

———-

4  4  4

 

Thus integer value of X and Y is 7 and 4 respectively.

 

  1. In the given addition, find the value of the letters and also give reason.

 

               X   1

            + 1   Y

         ————

             Y    0

 

Answer:

 

When 1 is added to Y, it gives 0 that is a number whose ones place is 0. This is possible when digit Y is 9.

So the addition of 1 and Y will be 10 so 1 will be carried forward for the next step.

 

In the next step,

1+X+1=9

Therefore, X is 7.

1 + 7 + 1 = 9 = Y

 

Hence, the addition is as given below.

 

7   1

+ 1   9

———–

9   0

Thus value of X and Y is 7 and 9 respectively.

 

  1. In the given addition, find the value of the letters and also give reason.

               2   X   Y

            + X   Y   1

         —————-

             Y   1   8

Answer:

When 1 is added to Y, it gives 8 that is a number whose ones place is 8. This is possible when digit Y is 7.

 

In the next step, X + Y = 1. Therefore, the value of X is 4.

4 + 7 = 11 and 1 will be carried forward for the next step.

 

In the next step,

1+ 2 + X = Y

1+ 2 + 4 = 7

 

Hence, the addition is as given below.

2   4   7

+ 4   7   1

————–

7   1   8

 

Thus value of X and Y is 4 and 7 respectively.

 

  1. In the given addition, find the value of the letters and also give reason.

 

             1   2   X

           + 6X   Y

         —————-

            X   0   9

Answer:

When X is added to Y, it gives 9 that is a number whose ones place is 9.Sum can be 9 only as summation of two single digits cannot be 19. So no carry generated.

In the next step, X + 2 = 0

It is possible if X = 8.

Therefore, 2 + 8 = 10 and 1 will be carried forward for the next step.

1 + 1 + 6 = 8. Therefore, value of X = 8.

When X is added to Y, it gives 9.

X + Y = 9

8 + Y = 9

Therefore, value of Y = 1

1   2   8

+ 6   8   1

—————–

8   0   9

Thus value of X and Y is 8 and 1 respectively.

 

 

Exercise 16.2 

 

 

  1. What would be the value of m to make 23m3 to be a multiple of 9?

Answer:

As, 23m3 is a multiple of 9

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

∴ 2 +3 +m + 3 = 8 + m

m + 8 = 9

m = 1

 

  1. If 31p5 is a multiple of 9, where p is a digit, find the value of p?

In this problem you will get two results. Explain why??

Answer:

Since 31p5 is a multiple of 9

 

So now, if a number is a multiple of 9, then the sum of the digits will be divisible by 9.

 

∴ \(3+1+p+5=9+p\)

9+p = 9      [Since, 9+p should be multiple of 9]

p = 0

 

If \(3+1+p+5=9+p\)

9 +p = 18

p = 9

However, since p is a single digit number, the sum can be either 9 or 18.

But in this case, 0 and 9 are two possible answers.

 

  1. If 24p is a multiple of 3, where p is a digit, find the value of p? (Since 24p is a multiple of 3, its sum of digits 6 + p is a multiple of 3; so 6 + p is one of these numbers: 0, 3, 6, 9, 12, 15, 18 … But since p is a digit, it can only be that 6 + p = 6 or 9 or 12 or 15. Therefore, p = 0 or 3 or 6 or 9. Thus, p can have any of four different values.)

Answer:

Since 24p is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Here, the sum of digits of 24p is = 2 + 4+ p

∴  2 + 4+ p=6 + p

Since ‘p’ is a digit.

6 +p = 6

p = 0

 

6 + p = 9

p = 3

 

6 + p = 12

p = 6

 

6 + p = 15

p = 9

 

Since p is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of p comes to 0 or 3 0r 6 or 9 respectively.

Therefore, p can have any of the four different values.

 

  1. If 31p5 is a multiple of 3, where p is a digit, find out the values of p.

Answer:

Since 31p5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Here p is a digit.

∴ 3+1+p+5=9+p

9 + p = 9

p = 0

 

∴ 3+1+p+5=9+p

9 + p = 12

p = 3

 

∴ 3+1+p+5=9+p

9 + p = 15

p = 6

 

∴ 3+1+p+5=9+p

9 + p = 18

p = 9

 

Since p is a single digit number, the sum of the digits can be 9 or 12 or 15 or 18 and thus, the value of p comes to 0 or 3 0r 6 or 9 respectively.

Therefore, p can have any of the four different values.



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1 Comment

  1. harshini harshini
    April 24, 2018    

    so nice to join the group of byjus

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