There are 125 chips on a table. If as many of the chips as possible are to be arranged into an equal number 3-chip and 4-chip stacks and the remaining chips are to removed, how many of the chips are to be removed
A. one
B. two
C. five
D. six
E. seven
The OA is D.
No of 3 chip stacks = No of 3 chip stacks = n
So, 3x + 4x = 125
Or, x = 17 and remainder 6.
Please, can anyone explain another way to solve this PS question? Thanks in advance!
There are 125 chips on a table. If as many of the chips as
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You have done it correctly except the equality must be inequality.BTGmoderatorLU wrote:There are 125 chips on a table. If as many of the chips as possible are to be arranged into an equal number 3-chip and 4-chip stacks and the remaining chips are to removed, how many of the chips are to be removed
A. one
B. two
C. five
D. six
E. seven
The OA is D.
No of 3 chip stacks = No of 3 chip stacks = n
So, 3x + 4x = 125
Or, x = 17 and remainder 6.
Please, can anyone explain another way to solve this PS question? Thanks in advance!
Instead of writing 3x + 4x = 125, you should write 3x + 4x ≤ 125; where x is a positive integer and has a maximum possible value.
Hope this helps!
-Jay
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Notice that we are told that the number of 3-chip and 4-chip stacks must be equal.
3x + 4x <= 125 --> 7x <= 25 --> the greatest multiple of which is less than 125 is 119, so 125 - 119 = 6 chips to be removed.
Option D.
3x + 4x <= 125 --> 7x <= 25 --> the greatest multiple of which is less than 125 is 119, so 125 - 119 = 6 chips to be removed.
Option D.
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Since the total number of chips in one 3-chip stack and one 4-chip stack is 7 and 125/7 = 17 R 6, we see that the number of 3-chip (or 4-chip) stacks we can have is 17, leaving 6 chips unstacked and thus requiring removal.BTGmoderatorLU wrote:There are 125 chips on a table. If as many of the chips as possible are to be arranged into an equal number 3-chip and 4-chip stacks and the remaining chips are to removed, how many of the chips are to be removed
A. one
B. two
C. five
D. six
E. seven
Answer: D
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