math
posted by Al .
Sasa had some 20cent and 50cent coins.
7/8 of the coins were 20cent coins and the rest were 50cent coins.After Sasa had spent $72.50 worth of 50cent coins and 5/7 of the 20cent coins, she had 2/7 of the coins left. Find the total amount of money Sasa left.

let there be:
c coins
t 20¢ coins
f 50¢ coins
7/8 c = t
1/8 c = f
so, t = 7f
72.50 is 145 of the 50¢ coins
(f + t)  145  (5/7 t) = 2/7 (f+t)
since t = 7f,
f + 7f  145  5f = 2/7 * 8f
f = 203
so, t = 1421
and 1624 coins in all
check:
1624  145  5/7 (1421)
= 1624  145  1015
= 464
464 is 2/7 of 1624 
Thank you, Steve.

The first answer is wrong. What is being asked is the amount. 464 is too much compared to $72.50 which is the combination of 50c and 5/7 of 20c.
Listing method is used to find the answer.
The total fraction for the 20c and 50c is 8/8.
7/8 were 20c
1/8 were 50c
Singaporean Math is applied using the block method.
7 units = 20c
1 unit = 50c
8 units in all
If the denominator is 8, find a multiple that will match the given amount of $72.50. The multiple is equivalent to 1 unit.
The multiple is 48.
Each unit is 48.
$72.50 is 5units of 20c and 1unit of 5c.
48 x 5 x .20 = $48
48 x .50 = $24
$48 + $24 = $72
Add .50 to make it $72.50
The question is the total amount of money left which 2/7.
So that is 48 x 2 x .20 = $19.20
48 48 48 48 48 48 48 49
20c 20c 20c 20c 20c 20c 20c 50c
There is a total of 336pcs 20c coins amounting to $67.20 and
49pcs 50c coins amounting to $24.50
=$91.70  $19.20 = $72.50
All 50c were spent. Only 2/7 of 20c were left.
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