## In a certain game, a large bag is filled with blue, green, purple, and red chips worth $$1, 5, x$$ and $$11$$ points eac

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### In a certain game, a large bag is filled with blue, green, purple, and red chips worth $$1, 5, x$$ and $$11$$ points eac

by M7MBA » Sun Apr 11, 2021 12:14 pm

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In a certain game, a large bag is filled with blue, green, purple, and red chips worth $$1, 5, x$$ and $$11$$ points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is $$88,000,$$ how many purple chips were selected?

A. 1
B. 2
C. 3
D. 4
E. 5

Source: Manhattan GMAT

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### Re: In a certain game, a large bag is filled with blue, green, purple, and red chips worth $$1, 5, x$$ and $$11$$ points

by [email protected] » Tue Apr 13, 2021 7:19 am
M7MBA wrote:
Sun Apr 11, 2021 12:14 pm
In a certain game, a large bag is filled with blue, green, purple, and red chips worth $$1, 5, x$$ and $$11$$ points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is $$88,000,$$ how many purple chips were selected?

A. 1
B. 2
C. 3
D. 4
E. 5

Source: Manhattan GMAT
This question begs for some prime factorization.

88,000 = (2)(2)(2)(2)(2)(2)(5)(5)(5)(11)

First, we can see that there must be one (11-point) red chip.
Now, what role do these 2's play? Since there are no 2's hiding among the 5-point chips or the 11-point chips, the 2's must be associated with the x-point chips.
Since we know that each purple chip is worth 6,7,8,9 or 10 points, we know that x must equal 6, 8 or 10.

x cannot equal 6, because we don't have any 3's in the prime factorization.
If x were to equal 10, we'd need six 5's to go with our six 2's. Since we don't have six 5's in the prime factorization of 88,000, we can rule out the possibility that x equals 10.

By the process of elimination, x MUST equal 8.
Since 8 = (2)(2)(2), we can see that the six 2's can be used to create two products of 8.