(1b) (5g)(xp)(11r)= 88000=2^6 x 5^3 x 11, where b, g, p and r are the number of respective colored chips chosen.pg850 wrote: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?
OA: 2
- 1
2
3
4
5
5<x<11
(1 . 2^3) (5 . 5^2) x (2^3 . p) x (11 . 1) Tentative.
You found x to be 2^3 by using 5<x<11.
Now p can take on values.1, 2 , 4, 5, 5^2 and possibly a few more. There is nothing in the problem that restricts p from taking on any of these values. The only restriction is that 2^3 is fixed as the value of purple chips. Can’t be 25, 5, 4, 11 or 2. Besides that you can play musical chairs with the rest. Some respective distributions are:
(1 . 2^3) (5 . 5^2) x (2^3 . 1) (11 . 1) P=1 8 blue, 25g, 1 P, 1R chips chosen
(1 . 2^2) (5 . 5^2) (2^3 . 2) (11 . 1) P=2 4b, 25g, 2p, 1R chips chosen
(1 . 2) (5 .5^2) (2^3 . 2^2) (11 . 1) P=4, 2b, 25g, 4p, 1 R chips chosen
(1 . 2^3) (5 .5) (2^3 . 5) (11 . 1) P=5, 8b, 5g, 5p, 1 R chips chosen
I don’t see any mathematical justification consistent with the wording of the problem that makes 2 the only valid answer in the absence of information on the number of blue or green chips chosen. All these distributions will yield 88000 in total value.
I am looking for valid mathematical reasons why I should rule out the 1st, 3rd and 4th distributions above.
