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03 Mar 2021, 15:08
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
84% (01:44) correct
16%
(01:28)
wrong
based on 150
sessions
History
Date
Time
Result
Not Attempted Yet
Bert has $1.37 of loose change in his pocket — pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). He reaches into his pocket and pulls out one coin at random. What is the probability that the coin is a nickel?
(1) There are exactly seven pennies in his pocket
(2) There are exactly three quarters in his pocket
(1) There are exactly seven pennies in his pocket
(2) There are exactly three quarters in his pocket
12 Apr 2021, 12:15
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Bunuel
Bert has $1.37 of loose change in his pocket — pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). He reaches into his pocket and pulls out one coin at random. What is the probability that the coin is a nickel?
(1) There are exactly seven pennies in his pocket
(2) There are exactly three quarters in his pocket
(1) There are exactly seven pennies in his pocket
(2) There are exactly three quarters in his pocket
MAGOOSH OFFICIAL SOLUTION:
A reminder for non-American students — $1.00 = 100¢. Thus
penny = $0.01 = 1¢
nickel = $0.05 = 5¢
dime = $0.10 = 10¢
quarter = $0.25 = 25¢
Statement #1: Bert has seven pennies, amounting to 7¢. The other coins total $1.30. This could be all nickels — 26 nickels — so that of the 7 + 26 = 33 coins in the pocket, 26 are nickel, and the probability of picking a nickel would be 26/33. Or, that $1.30 could be 4 quarters and 3 dimes, so that there were no nickels, and the probability of picking a nickel would be zero! Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient.
Statement #2: Bert has three quarters, amounting to 75¢. The other coins total 62¢ — this could be two pennies and twelve nickels, that of the 3 + 2 + 12 = 17 coins in the pocket, 12 are nickels, and the probability is 12/17. Or, that 62¢ could be entirely in pennies, without any nickels or dimes at all: then the probability of picking a nickel would be zero. Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient.
Combined statements: Now, we know Bert has seven pennies and three quarters, and these ten coins together account for 82¢. The remaining 55¢ must be composed of dimes and nickels. There could be eleven nickels and no dimes, so that of the 10 + 11 = 21 coins in the pocket, 11 are nickels, and P = 11/21. OR, there could be five dimes and one nickel in the pocket, so of the 10 + 5 + 1 = 16 coins in the pocket, only one is a nickel, and P = 1/16. Different possible choices lead to different answer to the prompt question, so both statements combined are insufficient.
Nothing is sufficient.
Answer = E
General Discussion
globaldesi
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03 Mar 2021, 22:02
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given equation:
p+5n+10d+25q= 137
using 1
5n+10d+25q=130 clearly not sufficient
using 2
p+5n+10d= 62
again N.S
combine both
5n+10d= 55
or
n+2d= 11
now various values of d = 1,2,3,4,5 solve and n will be 9,7, 5,3,1
hence combine also N.S
thus E
p+5n+10d+25q= 137
using 1
5n+10d+25q=130 clearly not sufficient
using 2
p+5n+10d= 62
again N.S
combine both
5n+10d= 55
or
n+2d= 11
now various values of d = 1,2,3,4,5 solve and n will be 9,7, 5,3,1
hence combine also N.S
thus E
