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?
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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
Assuming your answer is correct,
No. of 5-cent coins: 66
No. of 10-cent coins: 66÷3 = 22
No. of 20-cent coins: 22 + 2 = 24
|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|
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.
Good point. The total sum should be $10.30 instead of $10.35. Maybe the original poster can clarify.
This is an interesting problem. So sorry it slipped unnoticed. Any solutions to this?
Doesn’t seem solvable.
Lol. It looks like I’m not so stupid after all 🙂
But it does seem solvable, maybe by some devious method. Probably some brain teaser or something.