Problem Solving

For bringing each copper coin from the bottom of a river, a coin-diver gets 20 cents, and for each brass coin he gets 25 cents. If after one dive, he got $2.80, what is the minimum number of copper coins that he brought?

(A) 4

(B) 3

(C) 2

(D) 1

(E) 0

In general, to find the minimum that one value can be, maximize all the other values. Conversely to find the maximum that one value can be, minimize the other values. In this case, the minimum number of copper coins will be paired with the maximum number of brass coins.

To exchange copper for brass, we must consider the least common multiple of their values (20 cents and 25 cents). This is $1.00. This amount is equivalent to 5 copper or 4 brass coins. This means that to replace as many copper coins as we can with brass coins, we must add brass coins in groups of 4.

4 brass --> $1.00
4 more brass --> $1.00

Adding another 4 brass coins would take us over the sum of $2.80. If we have 1, 2, or 3 brass coins it would take us to $2.25, 2.50 or 2.75; there would be no copper amount that gets us to exactly $2.80. Therefore, we can only add copper coins once we get to $2.80

4 copper and 8 brass = $0.80 + $2.00

4 is the minimum number of copper coins

0.2C+0.25B=2.8

2C+2.5B=28



B=(28-2C)/2.5=(280-20C)/25

Since B should be positive intger as it represent number of brass coins.

Sub the options in the above equation.Only 4 satisifies.

So 4 copper and 8 Brass