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Darion has at least one quarter, one dime, one nickel, and one penny
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09 Oct 2020, 04:00
09 Oct 2020, 04:00
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Darion has at least one quarter, one dime, one nickel, and one penny in his pocket. He has three times as many pennies as nickels, three times as many nickels as dimes, and three times as many dimes as quarters. If he has 120 coins, how much money does he have?
A. $1.20
B. $3.81
C. $4.06
D. $4.07
E. $7.75
A. $1.20
B. $3.81
C. $4.06
D. $4.07
E. $7.75
Joined: 24 Mar 2019
Posts: 193
Given Kudos: 196
Location: India
Concentration: Marketing, Operations
Schools: IIMA PGPX'23 IIM
WE:Operations (Aerospace and Defense)
Re: Darion has at least one quarter, one dime, one nickel, and one penny
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09 Oct 2020, 09:15
09 Oct 2020, 09:15
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Darion has at least one quarter, one dime, one nickel, and one penny in his pocket. He has three times as many pennies as nickels, three times as many nickels as dimes, and three times as many dimes as quarters. If he has 120 coins, how much money does he have?
A. $1.20
B. $3.81
C. $4.06
D. $4.07
E. $7.75
Explanation:
let us assume that Darion has Q quarters, D dimes, N nickels, and P pennies
then given P=3N, N=3D, D=3Q and also given
P+N+D+Q=120
i.e 27Q+9Q+3Q+Q=120
Q=3
means P=81, N=27, D=9 (as P=3N, N=3D, D=3Q)
also we know that one penny=1 cent, one nickel=5 cent ,one dime=10 cent, one quarter=25 cent
So overall money =81*1 + 27*5 + 9*10 + 3*25
=381 cents OR $3.81
Hence B is the correct answer.
A. $1.20
B. $3.81
C. $4.06
D. $4.07
E. $7.75
Explanation:
let us assume that Darion has Q quarters, D dimes, N nickels, and P pennies
then given P=3N, N=3D, D=3Q and also given
P+N+D+Q=120
i.e 27Q+9Q+3Q+Q=120
Q=3
means P=81, N=27, D=9 (as P=3N, N=3D, D=3Q)
also we know that one penny=1 cent, one nickel=5 cent ,one dime=10 cent, one quarter=25 cent
So overall money =81*1 + 27*5 + 9*10 + 3*25
=381 cents OR $3.81
Hence B is the correct answer.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4127
Given Kudos: 97
Q51 V47
Re: Darion has at least one quarter, one dime, one nickel, and one penny
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09 Oct 2020, 10:29
09 Oct 2020, 10:29
1
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Expert Reply
If he had 1 quarter, he'd have 3 dimes, 9 nickels and 27 pennies. So he has coins in a 1 to 3 to 9 to 27 ratio. Those four numbers sum to 40, but he actually has 120 coins in total, which is (3)(40), so he must have 3 quarters, 9 dimes, 27 nickels and 81 pennies. He clearly has more than $1.20 in total, and when we compute the total amount of money he has, the result in cents must be a multiple of 3 (because the number of coins of each type is a multiple of 3). Summing the digits of each answer choice from B through E, only B can be right.
