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

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

----

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

OA is A

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

Hi!

We can solve this question super-duper quickly via backsolving. Since we want the minimum number of copper coins, let's start with the smallest answer, (E).

(E) 0 copper means $2.80 from brass - since 280 is not a multiple of 25, wrong!

(D) 1 copper means $2.60 from brass - since 260 is not a multiple of 25, wrong!

(C) 2 copper means $2.40 from brass - since 240 is not a multiple of 25, wrong!

(B) 3 copper means $2.20 from brass - since 220 is not a multiple of 25, wrong!

Only (A) remains, so no reason to test it - choose (A)!

(If you had tested it: 4 copper means $2.00 from brass - since 200 IS a multiple of 25, ding ding ding!)

* * *

We could also have just worked our way down from 280 in 20 cent increments until we found a multiple of 25:

280
260
240
220
200... ding ding ding! That's 4 copper coins, choose A!

Remember: there's no bonus points for beautiful proofs on test day; quick and dirty is always better than long and elegant.