 # Question

Bala has some 5cents, 10cent and 20cents coins. The number of 5cents coins is three times that of 10cents coins. There are 2 more 20cents coins than 10cents coins. If the total value of the coins is \$10.35, how many 5cent coins does Bala have?

let number of 10- cent coins be x.

no of 5-cent coins = 3x

no of 10-cent coins = x

no of 20-cent coins = 2+x

Total value: 3x(0.05) + x(0.10) + (2+x)(0.20) = 10.35

0.15x+ 0.10x + 0.40 + 0.2x = 10.35

0.45x = 9.95

x= 22.1111 round off to 22

no of 5 cent coins = 22×3= 66

… …

No. of 5-cent coins: 66
No. of 10-cent coins: 66÷3 = 22
No. of 20-cent coins: 22 + 2 = 24

 Quantity Value No. of 5-cent coins 66 66 × 0.05 = \$3.30 No. of 10-cent coins 66 ÷ 3 = 22 22 × 0.10 = \$2.20 No. of 20-cent coins 22 + 2 = 24 24 × 0.20 = \$4.80 Total value: \$10.30

This still does not give the total value of \$10.35.

If you have defined the value of x to be the NUMBER of coins, how can it output a decimal? And how can it simply be rounded off?

This is not a fluid algebraic quantity; it refers to an actual quantity of a physical object.

This question is NOT solvable. There has to be an error or typo in the question.