karthikpandian19 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
Let x = number of copper coins; y = number of brass coins
By translation, we have: 0.20x + 0.25y = 2.80
Now, let's make the equation easier to handle (remember: time management!)
100[0.20x + 0.25y]=2.80(100) ---> multiply 100 to both sides of the equation so we could move the decimal point to places to the right; we have: 20x + 25y = 280
Let's isolate x: 20x = 280 - 25y
Divide both sides by 1/20; we have: x = 14 - 5y/4
Now here's the critical points:
1. Since you want to minimize x, we have to maximize y
2. But remember, 5y/4 should still be an integer (or y for that matter)
5y/4 could only lead to integers when y is equal to:
y = 4 (test this, 5*4/4); so x = 14 - 5 = 9 coins
y = 8 (test this, 5*8/4); so x = 14 - 10 = 4 coins
y = 12 (test this, 5*12/4); so x = 14 - 15 = -1 coin (no such thing)
The minimum is x = 4 coins
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Found another way:
from x = 14 - 5y/4
By Algebra:
5y/4 < 14 (I didn't include the "=" sign since x couldn't be 0, check the equation again)
by manipulation we have: y < 11.2
This should satisfy:
1. y is an integer, y = 11, 10, 9, 8, 7, 6,...
2. 5y/4 should be an integer
so y could by 8 or 4
Again, maximize y = 8, we get x = 14 - 10
x = 4
