Question

A, B and C has few coins. Eight times the number of coins that A has equals to five times the number of coins B has, while seven times the number of coins B has is equal to eleven times the number of coins C has. Find the minimum number of coins with A, B and C put together.
1. 200
2. 199
3. 196
4. 197

Answer 1

Correct Answer - Option 2 : 199

Given:

8A = 5B

7B = 11C

Calculation:

Eight times the number of coins that A has equals to five times the number of coins B has

8A = 5B

A : B = 5 : 8      ----- (i)

seven times the number of coins B has is equal to eleven times the number of coins C has

7B = 11C

B : C = 11 : 7      ----- (ii)

Solving equation i and ii

Multiply equation (i) with 11 and equation (ii) with 8

A : B : C = 55 : 88 : 56

Minimum number of coins

⇒ 55 + 88 + 56

∴ Minimum number of coins with A, B and C put together is 199 coins