We can simply form an equation like:
5*x+7*y=135 or, 5*x+7*15=135 or, 5*x=30 or, x=6 [Here, we count for the largest value of 7-inch seg­ments so that we can get the minimum number of 5-inch segments]
So A is the answer.

7x + 5y = 135. We need to find minimum number of 5's that can go in. To do this, we need to MAXIMIZE the NUMBER OF 7's that can go inside while also making sure that the remaining amount is divisible by 5. We start with MAX and we keep going down till we find a suitable number that fits both 7 and 5.

135/7 = 19.7 (not divisible)
So we find the closest number to 135 divisible by 7. We can do this by 7 * quotient value (19) = 133.

For 133, the remaining value from 135 is 2, which is not divisible. From 133, we go behind 7 steps at a time and see if the remainder is divisible by 5.

133
126 (Remaining 9)
119 (Remaining 16)
112 (Remaining 22)
105 (Remaining 30, which is a multiple of 5)

So the minimum amount of 5's that can go is 30/5 = 6.
Quantity A is greater.

A is greater.

LCM of 5 and 7 is 35. So maximum possible number of 7 inches sheet will give us minimum possible number of 5 inches sheet.

Just ask yourself: What should be a number that is both divisible by 5, and 7, and when subtracted from 135 yield a number which is also divided by 5. Hence 105 is that number. Thus minimum possible number of 5inches sheet = (135-105)/5=6.

Ans A

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.