# maths

A coin box contained some twenty-cent and fifty-cent coins in the ratio 4 : 3. After 20 twenty-cent coins were taken out to exchange for fifty-cents coins of the same value and put back in the box, the ratio of the number of twenty-cent coins to the number of fifty-cents coins became 7 : 11. Find the sum of money in the box.

I posted this question before and based on formula below my final calculation did not match the given answer which is \$27.60
(4x-20)/(3x+8) = 7/11
x = 12

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1. number of 20 cent coins ---- 4x
number of 50 cent coins ---- 3x
(notice 4x : 3x = 4 : 3)

20 of the twenty cents coins were taken out
so we have 4x - 20 left

value of those 20 coins = 20(20) = 400 cents
which would make 400/50 or 8 fifty cent pieces
So number of 50 cent coins = 3x + 8

(4x-20)/(3x+8) = 7/11
44x - 220 = 21x + 56
23x = 276
x = 12

So originally there were 4(12) or 48 twenty cent pieces
and 2(12) or 36 fifty cent pieces.
total amount = 48(20cents) + 36(50cents) = 2760 cents or \$27.60

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posted by Reiny

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