# math

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

• math -

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

• math -

Thank you, Steve.

• math -

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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