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Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin

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Bunuel
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Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink] New post   06 May 2019, 20:52
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Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coins out of his pocket that amount to $0.70. If there are only two denominations of coins among the 6 coins in Pat’s hand,how many nickels is he holding? (quarter = 25 cents, dime = 10 cents, nickel = 5 cents)

A. 2
B. 3
C. 4
D. 5
E. 6
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C
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Re: Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink] New post   06 May 2019, 21:08
Since options don’t have nikel = 0 , there are no dimes as 6 dimes and 2 nickels are required to have 70 cents which is more than 6 coins.

So it needs to have 2 quarters and 4 nickels which amount to 70 cents.

Option C

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Re: Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink] New post   12 May 2019, 11:27
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Bunuel wrote:
Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coins out of his pocket that amount to $0.70. If there are only two denominations of coins among the 6 coins in Pat’s hand,how many nickels is he holding? (quarter = 25 cents, dime = 10 cents, nickel = 5 cents)

A. 2
B. 3
C. 4
D. 5
E. 6


Among the 6 coins that Mr. Pat takes out, one denomination has to be Nickels and the other can be Quarter or Dime.
If we don't take one denomination to be Nickel then we wouldn't be able to find the number of Nickels as we do not know the total number of coins in the pocket.

assuming x to be the number of Nickels and y to be the number of Quarter
x+y =6
5x+25y=70

Solving these we get x=4 and y = 2 , lets keep this solution and test.

assuming x to be the number of Nickels and y to be the number of Dimes
x+y =6
x+10y = 70

solving the above we get x=-2 and y=8
Since the number of Nickels cannot be negative. The first solution where x=4 is correct.

Hence number of Nickels is 4

Ans= c
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Re: Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink] New post   12 May 2019, 20:20
Bunuel wrote:
Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coins out of his pocket that amount to $0.70. If there are only two denominations of coins among the 6 coins in Pat’s hand,how many nickels is he holding? (quarter = 25 cents, dime = 10 cents, nickel = 5 cents)

A. 2
B. 3
C. 4
D. 5
E. 6


let n=number of nickels
6-n=number of quarters
.05n+.25(6-n)=.70→
n=4
C
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Re: Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink] New post   06 Jun 2019, 17:45
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Bunuel wrote:
Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coins out of his pocket that amount to $0.70. If there are only two denominations of coins among the 6 coins in Pat’s hand,how many nickels is he holding? (quarter = 25 cents, dime = 10 cents, nickel = 5 cents)

A. 2
B. 3
C. 4
D. 5
E. 6


The only way one can make 70 cents using 6 coins in two types of coins is 2 quarters and 4 nickels. Therefore, Pat is holding 4 nickels.

Answer: C
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Re: Pat has a pocket full of quarters, dimes, and nickels. He takes 6 coin [#permalink]
06 Jun 2019, 17:45
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