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 =

[BTGModeratorVI]

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

[terminator12]

Probability of selecting red chip = \(\frac{y}{x+\ y}\) = \(\frac{3}{7}\)

Solving this, we get \(\frac{x}{y}\) = \(\frac{4}{3}\)

[Brent@GMATPrepNow]

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