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13 Apr 2026, 19:00
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D
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Difficulty:
85%
(hard)
Question Stats:
54% (03:04) correct
46%
(03:09)
wrong
based on 83
sessions
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For a trade-show booth, a crew assembled 10 display panels. Some were reinforced panels, and the rest were standard panels. Each reinforced panel required 16 mounting clips, and each standard panel required 11 mounting clips. If at least 1 panel of each type was used, were more than 4 of the 10 display panels reinforced?
(1) The total number of mounting clips used was a multiple of 9.
(2) If the crew had assembled 3 reinforced panels and 7 standard panels instead, the total number of mounting clips used would have been at least 10 fewer.
(1) The total number of mounting clips used was a multiple of 9.
(2) If the crew had assembled 3 reinforced panels and 7 standard panels instead, the total number of mounting clips used would have been at least 10 fewer.
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21 Apr 2026, 01:45
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GMAT Club Official Solution:
For a trade-show booth, a crew assembled 10 display panels. Some were reinforced panels, and the rest were standard panels. Each reinforced panel required 16 mounting clips, and each standard panel required 11 mounting clips. If at least 1 panel of each type was used, were more than 4 of the 10 display panels reinforced?
Let r be the number of reinforced panels. Then the number of standard panels was 10 - r. So the total number of mounting clips used was
16r + 11(10 - r) = 110 + 5r
We need to determine whether r > 4.
(1) The total number of mounting clips used was a multiple of 9.
So 110 + 5r is divisible by 9. Since 110 leaves remainder 2 when divided by 9, 5r must leave remainder 7 when divided by 9. The only value of r from 1 to 9 that works is r = 5. So r > 4. Sufficient.
(2) If the crew had assembled 3 reinforced panels and 7 standard panels instead, the total number of mounting clips used would have been at least 10 fewer.
That hypothetical arrangement would have used:
3(16) + 7(11) = 125 clips.
So the actual total number of clips used was at least
125 + 10 = 135
Thus:
110 + 5r >= 135
5r >= 25
r >= 5
So r > 4.
Sufficient.
Answer: D.
For a trade-show booth, a crew assembled 10 display panels. Some were reinforced panels, and the rest were standard panels. Each reinforced panel required 16 mounting clips, and each standard panel required 11 mounting clips. If at least 1 panel of each type was used, were more than 4 of the 10 display panels reinforced?
Let r be the number of reinforced panels. Then the number of standard panels was 10 - r. So the total number of mounting clips used was
16r + 11(10 - r) = 110 + 5r
We need to determine whether r > 4.
(1) The total number of mounting clips used was a multiple of 9.
So 110 + 5r is divisible by 9. Since 110 leaves remainder 2 when divided by 9, 5r must leave remainder 7 when divided by 9. The only value of r from 1 to 9 that works is r = 5. So r > 4. Sufficient.
(2) If the crew had assembled 3 reinforced panels and 7 standard panels instead, the total number of mounting clips used would have been at least 10 fewer.
That hypothetical arrangement would have used:
3(16) + 7(11) = 125 clips.
So the actual total number of clips used was at least
125 + 10 = 135
Thus:
110 + 5r >= 135
5r >= 25
r >= 5
So r > 4.
Sufficient.
Answer: D.
General Discussion
13 Apr 2026, 23:34
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Both are sufficient IMO.
First one:
This is tricky, so basically what is happening is we have 16x+11y form and x+y = 10.
Now 16x is ideally -2x and 11y is ideally 2y
So we have -2x + 2y mod 9 = 0
x+y=10 and we need to find x and y.
2(x-y) mod 9 = 0
Thus x-y must be divisible by 9 or 0.
If total is less than 10, max we can have is 9,1 only and x-y = 8 and 8 is not divisible by 9.
Thus x=y=5 is the only solution.
SUFFICIENT.
Second one:
This straightforward, since you have x+y=10 limit, thus if you increase x then y has to be reduced. Thus the net total here it says 3,7 makes 10 lesser.
OR
x,y would lead to 10 more than 3,7.
Each time you add 16 and subtract 11, then you you add a net of 5.
Since you have a net of 10 asked so you do it twice, which means u add 16 twice and subtract 11 twice to achieve 10 more
= 3+2,7-2 = 5,5.
SUFFICIENT.
Answer: Option D
First one:
This is tricky, so basically what is happening is we have 16x+11y form and x+y = 10.
Now 16x is ideally -2x and 11y is ideally 2y
So we have -2x + 2y mod 9 = 0
x+y=10 and we need to find x and y.
2(x-y) mod 9 = 0
Thus x-y must be divisible by 9 or 0.
If total is less than 10, max we can have is 9,1 only and x-y = 8 and 8 is not divisible by 9.
Thus x=y=5 is the only solution.
SUFFICIENT.
Second one:
This straightforward, since you have x+y=10 limit, thus if you increase x then y has to be reduced. Thus the net total here it says 3,7 makes 10 lesser.
OR
x,y would lead to 10 more than 3,7.
Each time you add 16 and subtract 11, then you you add a net of 5.
Since you have a net of 10 asked so you do it twice, which means u add 16 twice and subtract 11 twice to achieve 10 more
= 3+2,7-2 = 5,5.
SUFFICIENT.
Answer: Option D
Bunuel
