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Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent

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Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent  [#permalink]

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New post 29 Mar 2019, 11:30
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Difficulty:

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Question Stats:

76% (03:14) correct 24% (03:09) wrong based on 29 sessions

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Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

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Re: Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent  [#permalink]

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New post 30 Mar 2019, 00:16
Noshad wrote:
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4



We can form three equations to solve ..
Let the number of 5-cent coins, 10-cent coins, and 25-cent coins be x, y and z.
I) Joe has a collection of 23 coins - x+y+z=23
II) He has 3 more 10-cent coins than 5-cent coins - y=x+3
III) the total value of his collection is 320 cents - 5x+10y+25z=320 or x+2y=5z=64...
substitute y =x+3..
I) x+x+3+z=23....2x+z=20.....z=20-2x
substitute the values in (III)
x+2(x+3)+5(20-2x)=64.......x+2x+6+100-10x=64......7x=106-64.....7x=42....x=6
therefore z=20-2*6=8..

Thus z-x=8-6=2

C
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Re: Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent  [#permalink]

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New post 30 Mar 2019, 01:41
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Noshad wrote:
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4


slightly lengthy question as it involves too much of substitution
given
x+y+x=23
x+3=y
need to find z-x value
5x+10y+25z=320
x+2y+5z=64
also 2x+z=20
z=20-2x

x+2y+100-10x=64
x+2*(x+3)+5(20-2x)=64
solve x= 6
so z= 20-12 ; 8
x-z ; 2
IMO C
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Re: Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent  [#permalink]

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New post 30 Mar 2019, 04:21
3
Noshad wrote:
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4


Simply note that options say that number of 25 cent coins will be 0/1/2/3/4 more than 5 cent coins. We already know that number of 10 cent coins is 3 more than number of 5 cent coins.

There are total 23 coins. If you remove the extra number of 25 cent and 10 cent coins, the resultant should be divisible by 3.
23 - 3 = 20. We need to remove another 2 to get 20 - 2 = 18, a number divisible by 3.
Removing 0/1/3/4 will not make a multiple of 3.

So number of 25 cent coins MUST be 2 more than number of 5 cent coins.
So only possible option is (C)
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Re: Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent   [#permalink] 30 Mar 2019, 04:21
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