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For bringing each copper coin from the bottom of a river, a coin-diver

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For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 01 Dec 2015, 04:24
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Question Stats:

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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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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 01 Dec 2015, 06:26
Let number of copper coins = c
number of bronze coins = b
.20c + .25 b = 2.80
=> 20c + 25b = 280
=> 20c = 280 - 25b
=> c = 14 - (25/20)b = 14 - (5/4)b

To minimize c , we need to maximze b . b needs to be multiple of 4
b = 4 , c = 14 - 5 = 9
b = 8 , c = 14 - 10 = 4
b = 12 , c = 14 - 15 = -1 , Number of coins can be negative .
Therefore , minimum number of copper coins = 4
Answer A

Alternatively ,
we use backsolving here . It will be much quicker .
if b= 0 , 2.8 is not a multiple of .25 . Not possible
b = 1 , 2.6 is not a multiple of .25 . Not possible
b = 2 , 2.4 is not a multiple of .25 . Not possible
b= 3 , 2.2 is not a multiple of .25 . Not possible
b=4 , 2.0 is a multiple of .25 .

Answer A
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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 01 Dec 2015, 09:52
0.2C + 0.25B = 2.8
For C to be minimum, we have to maximize B such that 2.8 - 0.25B is divisible by 0.2.
Max value of B can be 8
C = (2.8 - 2)/0.2 = 4

Answer: A
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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 01 Dec 2015, 10:24
Bunuel 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


Copper = 20 c

Brass = 25 c

Now, 280 = 20 (x) + 25 (y)

We are required to find the minimum number of copper coins so we maximize the number of brass coins.

280 = 20x + 25(8)

280 = 20x + 200

x = 4

Hence answer is definitely 4
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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 02 Dec 2015, 04:03
Cents earned for copper coin = 20
Cents earned for brass coin = 25
Cents earned in total = 280

we need to find minimum number of copper coins. shortest way IMO here would be to keep reducing the amount from 280 in multiple of 20 and checking whether the reduced number is multiple of 25 or not.

is 260 multiple of 25 - NO
is 240 multiple of 25 - NO
is 220 multiple of 25 - NO
is 200 multiple of 25 - Yes

280-200 = 80

80/20 = 4

4 copper coins and 8 brass coins.

option A is the correct answer.
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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 03 Dec 2015, 23:39
Hi All,

There's a great 'brute-force' solution to this question - you don't need any excessive equations or math work - you just have to use basic arithmetic and logic. From the answer choices, we know that the minimum number of copper coins is an integer from 0 to 4.

The prompt tells us:
1) Copper coins get us 20 cents
2) Brass coins get us 25 cents
3) Total revenue = $2.80

Since we CANNOT get to $2.80 in 25 cent increments, there must be at least 1 copper coin.

IF...there was just 1 copper coin, then Brass coins would account for $2.60. But that's not possible, so there must be at least 2 copper coins...

IF...there were just 2 copper coins, then Brass coins would account for $2.40. But that's not possible, so there must be at least 3 copper coins...

IF...there was just 3 copper coins, then Brass coins would account for $2.20. But that's not possible, so there must be at least 4 copper coins...

And then we're done.

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Re: For bringing each copper coin from the bottom of a river, a coin-diver [#permalink]

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New post 02 May 2017, 16:48
Bunuel 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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Re: For bringing each copper coin from the bottom of a river, a coin-diver   [#permalink] 02 May 2017, 16:48
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