## RS Aggarwal Class 8 solutions Chapter 5 Playing with Numbers Ex 5C in pdf free download

Contents

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

**Question 1.**

**Solution:**

Here A can be as 6 + 7 = 13

Now 1 + 5 + 8 = 14

∴C = 1, B = 4, A = 6

**Question 2.**

**Solution:**

Here A can be 7, as 6+7 = 13

1 + B + 9 = 10 + B

∴B can be 7

∴10 + 7 = 17

1 + C + 6 = 7 + C

∴C can be 4

∴1 + 4 + 6 = 11

and 1 + 4 + 3 = 8

∴A = 7, B = 7, C = 4

**Question 3.**

**Solution:**

Here A + A + A = A

∴A can = 5

∴5 + 5 + 5 = 15

∴B = 1

Hence A = 5, B = 1

**Question 4.**

**Solution:**

6 – A = 3

1 + 5 – A = 3

5 – A = 3

∴A = 5 – 3 = 2

Now 2 – B = 7

=>12 – B = 7

∴B = 5

Hence A = 2, B = 5

**Question 5.**

**Solution:**

– 5 – A = 9 =>A = 5 – 9 or 15 – 9

= 6

∴A = 6

Now B – 1 – 8 = 5 =>B – 9 = 5

=>B = 5 + 9 = 14

∴B = 4

Now C – 1 – 2 = 2 =>C – 3 = 2

C = 2 + 3 = 5

∴A = 6, B = 4, C = 5

**Question 6.**

**Solution:**

B x 3 = B

∴B can be 5 or 0

∴5 x 3 = 15 => B = 5 or 3 x 0 =0

If B = 0, then A can be 5

∴3 x 5 = 15

∴A = 5 and C = 1

Hence A = 5, B = 0, C = 1

**Question 7.**

**Solution:**

∴B x A = B

∴A can be 1

∴AB = B

=>A = 1

and A² + B² – 1 + B² + C

∴B² +1 = C

∴B² in one digit

If B = 3

∴3² + 1 = 9 + 1 = 10 = C

∴C = 0

B x 1 + 1 = B + 1 = 3 + 1

Hence A = 1, B = 3, C = 0

**Question 8.**

**Solution:**

Here we see that 6 x 9 = 54

∴A – 4 = 3 => A = 3 + 4 = 7

and 6 x 6 = 36

3B = 36 => B = 6

and C = 6

Hence A = 7, B = 6, C = 6

**Question 9.**

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

**Solution:**

Product of two numbers = 1 -digit number

and sum = 2-digit numbers

Let first number = x

and second number = y

∴x X y = 1-digit number

x + y = 2-digit number

By hit and hail, we sec that

1 x 9 = 9 which is I-digit number

and 1 + 9 = 10 which is 2-digit number

**Question 10.**

**Find three whole numbers whose product and sum are equal.**

**Solution:**

By hit and trail method, we see that

1 + 2 + 3 = 6 and 1 x 2 x 3 = 6

1, 2 and 3 are the required whole numbers

whose sum and product is same

**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.**

**Solution:**

In the given square, we have to interest the numbers from 1 to 9, such that the sum in each raw, column on diagonal to be 15

So, we complete it as given here

**Question 12.**

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

**Solution:**

We shall complete the triangle by intersecting the numbers from 1 to 6 without repetition so that the sum in each side be 12

**Question 13.**

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

**a, b (a + b), (a +2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21 a + 34b). Sum of all these numbers = 11 (5a + 8b) =11 x 7th number. Taking a = 8, b = 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 x 7th number.**

**Solution:**

The given numbers are

a, b (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b)

Sum of there numbers = 11 (5a + 8b)

= 11 x 7th number

Now taking a = 8, b = 13, then the 10 number be 8, 13, 21, 34, 55, 89, 144, 233, 377, 610

Whose 7th number = 144

By adding these 10 numbers, we get the

sum

= 8+ 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 + 610 = 1584 and 11 x 7th number =11 x 144 , = 1584

Which is same in each case

**Question 14.**

**Complete the magic square**

**Solution:**

We see that in the magic box sum of 0 + 11 + 7 + 12 = 30

Now we shall complete this magic square, to get 30 as the sum in each row and column and also diagonal wise

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