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A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

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by BTGModeratorVI » Fri May 29, 2020 6:41 am

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Answers

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E

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A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7

Answer: D
Source: Magoosh
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Re: A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

by Brent@GMATPrepNow » Mon Jun 01, 2020 8:34 am
BTGModeratorVI wrote: ↑
Fri May 29, 2020 6:41 am
 A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7

Answer: D
Source: Magoosh
We can quickly solve this question by testing values that satisfy the given information

GIVEN: The probability of selecting a red chip at random is 3/7
So, it COULD be the case that there are 3 red chips, and a TOTAL of 7 chips
If there are 7 chips and 3 are red, then the other 4 chips must be blue
In other words, y = 3 and x = 4

This means x/y = 4/3

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

by Scott@TargetTestPrep » Fri Jun 05, 2020 10:56 am
BTGModeratorVI wrote: ↑
Fri May 29, 2020 6:41 am
 A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7

Answer: D
Solution:

Based on the given information, we can create the equation:

y/(x + y) = 3/7

3(x + y) = 7y

3x + 3y = 7y

3x = 4y

x/y = 4/3

Answer: D

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