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Coins issue

by francoisph » Wed Jun 23, 2010 6:16 am
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
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by Patrick_GMATFix » Wed Jun 23, 2010 6:33 am
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
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by selango » Wed Jun 23, 2010 6:59 am
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
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by Scott@TargetTestPrep » Thu Dec 14, 2017 6:05 am
francoisph wrote: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
We can let the number of copper coins found = c and the number of brass coins = b. Since the diver gets 20 cents per copper coin and 25 cents per brass coin, and makes 280 cents, we can create the following equation:

20c + 25b = 280

4c + 5b = 56

4c = 56 - 5b

c = (56 - 5b)/4

We see that 56 - 5b must be a multiple of 4. Furthermore, to minimize the value of c, we want to minimize the value of 56 - 5b. Thus, c will be smallest when b = 8, so we have:

c = (56 - 5(8))/4

c = 16/4

c = 4

Answer: A

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