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Cathy has equal numbers of nickels and quarters worth a total of $7.50
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31 Mar 2021, 05:27
31 Mar 2021, 05:27
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Cathy has equal numbers of nickels and quarters worth a total of $7.50. How many coins does she have?
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: Cathy has equal numbers of nickels and quarters worth a total of $7.50
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31 Mar 2021, 05:42
31 Mar 2021, 05:42
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Carcass wrote:
Cathy has equal numbers of nickels and quarters worth a total of $7.50. How many coins does she have?
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
One approach is to test the answer choices.
C) 50
If there are 50 coins in total, and half are nickels and half are quarters, then we have 25 nickels and 25 quarters.
25 nickels is worth $1.25 (since 25 x $0.05 = $1.25)
25 quarters is worth $6.25 (since 25 x $0.65 = $6.25)
TOTAL value = #1.25 + $6.25 = $7.50 (PERFECT!)
Answer: C
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Status:Founder & Quant Trainer
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Re: Cathy has equal numbers of nickels and quarters worth a total of $7.50
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31 Mar 2021, 07:00
31 Mar 2021, 07:00
1
Carcass wrote:
Cathy has equal numbers of nickels and quarters worth a total of $7.50. How many coins does she have?
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
(nickel = 5 cents, quarter = 25 cents)
(A) 20
(B) 25
(C) 50
(D) 62
(E) 70
Given, \(n = d\)
\(5n + 25d = 750\)
\(30d = 750\)
\(d = 25\)
Total coins \(= 2d = 50\)
hence, option C
