AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.5 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.5

Question 1.

Find the missing digits in the following additions.

Solution:

a) 1 + A + 7 = 17 ⇒ A = 17 – 8 = 9

∴ A = 9

b) 2 + 8 + B = 15 ⇒ B = 15 – 10 = 5

2 + 1 + B = 8 ⇒ B = 8 – 3 = 5

∴ B = 5

c) A + 7 + A=13 ⇒ 2A = 6 ⇒ A = 3 A + A + 1 = 7 2A = 6 ⇒ A = 3

∴ A = 3

d) From 1st column

2 + 9 + 9 + A = 26

A = 26 – 20 = 6

From 2nd column

⇒ 2 + 1 + A = 9 ⇒ A = 9 – 3 = 6

∴ A = 6

e) B + 6 + A = 11 or 21

B + A + (1 or 2) = 6

A + 1 = 4 ⇒ A = 3

⇒ From (1), B + 6 + 3 = 11 ⇒ B = 2

∴ A = 3, B = 2

Question 2.

Find the value of A in the following

(a) 7A – 16 = A9 (b) 107 – A9 = lA (e) A36 – 1A4 = 742

Solution:

a) 7A – 16 = A9

A- 6 = 9

If A = 5 it is possible.

∴ A = 5 (or)

7A- 16 = A9

⇒ 7 x 10 + (1 x A) – 16 = (A x 10 + 9 x 1)

⇒ 70 + A – 16 = 10A + 9

⇒ 9A = 45

⇒ A = 5

b) 107 – A9 = 1A

⇒ 107 – (10 x A + 9 x 1) = (1 x 10 + A x 1)

⇒ 107-10A-9= 10 +A

⇒ 11A = 88 A = 8

c) A36 – 1A4 = 742

⇒ (100 x A + 3 x 10 + 6 x 1) – (1 x 100 + A x 10 + 4 x 1) = 742

⇒ 100A + 36 – 100 – 10A – 4 = 742

⇒ 90A = 810

⇒ A = \(\frac{810}{90}\)

∴ A = 9

Question 3.

Find the numerical value of the letters given below-

Solution:

a) If E x 3 = E then E should be equal to 0 (or) 5.

5 x 3 = 15, 0 x 3 = 0

3 x D + 0 = ID [If E = 0]

⇒ 3D = 10 + D

⇒ 2D = 10

⇒ D = 5

∴ F = 1, D = 5, E = 0

b) If H x 6 = H then H should be equal to 0, 2, 6, 8.

G6 = 1G [If H = 0]

⇒ 6G + 0 = 10 + G

⇒ 5G = 10

⇒ G = 10/5 = 2

C = 1, G = 2, H = 0

Question 4.

Replace the letters with appropriate digits

(a) 73K ÷ 8 = 9L

(b) 1MN ÷ 3 = MN

Solution:

a) 73K ÷ 8 = 9L

\(\frac{73 \mathrm{~K}}{8}\) = 9L

If 73K is divisible by 8 then K = {1, 2, 3, …………. 9}

Select K = 6 from the set

∴ \(\frac{73 \mathrm{~K}}{8}\) (R = 0)

∴ \(\frac{73 \mathrm{~K}}{8}\) = 92 = 9L

⇒ 90 + 2 = (9 x 10 + L x 1)

⇒ 90 + 2 = 90 + L

∴ L = 2

∴ K = 6, L = 2

b) 1MN ÷ 3 = MN

If 1MN is divisible by 3 then sum of all the digits is divisible by 3.

⇒ 1 + M + N = 3 x {1,2,3}

Let 1 + M + N = 3 x 2 = 6 say

M + N = 5 …………..(1)

\(\frac{1 \mathrm{MN}}{3}\) = MN

⇒ 1MN = 3[MN]

⇒ 1 x 100 + M x 10 + N x 1

= 3[M x 10 + N x 1]

⇒ 100 + 10M + N = 3[10M + N]

⇒ 100 + 10M + N = 30M + 3N

⇒ 20M + 2N = 100

⇒ 10M + N = 50 …………….. (2)

From (1) & (2)

10 M + N = 50 (-) M + N = 5 9M = 45

∴ M = 5

If M = 5 then

M + N = 5

⇒ N = 0

M = 5, N = 0 [∵ \(\frac{150}{3}\) = 50 ]

Question 5.

If ABB x 999 = ABC 123 (where A, B, C are digits) find the values of A, B, C.

Solution:

From ABB x 999 = ABC 123, the product of units digit is equal to 3.

∴ B x 9 = 3 is the units digit.

If B = 7 then,

7 x 9 = 6 3

∴ A = 8, B = 7, C = 6

∴ The required product = 876123

∴ A = 8, B = 7, C = 6