# Let E = 3, B = 7 and A = 4. Find the other digits in the sum

B A S E

__+B A L L__

G AM E S

**Solution:**

Since E = 3, B = 7 and A = 4 we can write

74 S 3

__+7 4 L L__

G4 M 3 S

3 + L = S

S + L = 3

L can assume the value of 5 because

3 + 5 = 8 = S

S = 8

8 + 5 = 13 (one carry over)

M = 4 + 4 + 1(from Carry over)

M = 9

G = 1 from the carry over of 7 + 7 = 14=

L = 5; S = 8; M = 9; G = 1

**✦ Try This:** Find the other digits in the sum

4 A B

__+B A 4__

1 C 1 0

B + 4 = number ending with 0

Therefore

B = 6

1 + A + A = number ending with 1

That is possible if A = 5

1 + 5 + 5 = 11 (number ending with 1)

4 + B + 1 = D

B = 6

4 + 6 + 1 = 11( and one carries over)

4 5 6

__+6 5 4 __

1 1 1 0

A = 5; B = 6 and C = 1

**☛ Also Check: **NCERT Solutions for Class 8 Maths Chapter 16

**NCERT Exemplar Class 8 Maths Chapter 13 Problem 67**

## Let E = 3, B = 7 and A = 4. Find the other digits in the sum B A S E + B A L L = G AM E S

**Summary:**

The other digits are as follows L = 5, S = 8, M = 9, G = 1

**☛ Related Questions:**

- Let D = 3, L = 7 and A = 8. Find the other digits in the sum M A D + A S + A = B U L L
- If from a two-digit number, we subtract the number formed by reversing its digits then the result so . . . .
- Work out the following multiplication. 12345679 × 9 = ___. Use the result to answer the following qu . . . .

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