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Option A

:Size of box - L =11, W = 10 & H =7. To be filled with wooden cubes of side 3 (l = w = h 3).

: Can start with putting cubes one over another - max number of cubes which can be put together in this pattern = 2 (total h = 3 + 3 = 6, which is constrained by the height H, of box = 7)

: Now, this set of 2 cubes (one over other) can be placed together, max number of such sets which can be put together along the length = 3 (total l of cubes = =3*3 = 9, which is constrained by the length L, of box = 11)

: Similarly, these sets can be placed along the width, max number of such sets which can be put together along the width = 3 (total w of cubes = =3*3 = 9, which is constrained by the width W, of box = 10)

Total number wooden cubes = 2*3*3 = 18
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Answer: A

Box = 11 by 10 by 7
Cubes = 3 by 3 by 3
Number of Cubes = 3 * 3 * 2 [ (11/3) * (10/3) * (7/3) ] Only integer wholes
Number of Cubes =18
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Answer: A

Box = 11 by 10 by 7
Cubes = 3 by 3 by 3
Number of Cubes = 3 * 3 * 2 [ (11/3) * (10/3) * (7/3) ] Only integer wholes
Number of Cubes =18
How did u arrive at ds??
Any formulae

Sent from my SM-G600FY using GMAT Club Forum mobile app
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No formula. One cannot fit fractional cubes at the end (you would need to visualize). If it would have been a fluid, then Box volume divided by volume would have sufficed.
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Bunuel
A box measuring 11 inches by 10 inches by 7 inches is filled with solid wooden cubical blocks, each measuring 3 inches per side. How many wooden blocks can fit into the box?

A. 18
B. 27
C. 28
D. 29
E. 54

Official solution from Veritas Prep.

This problem is tricky because you must think about it logically BEFORE you do the math. Mathematically, you can quickly calculate \(11∗10∗7=770\). Each cube has a volume of \(3∗3∗3=27\). This problem invites you to do the math \(\frac{770}{27}=28.5\). And if you've done that, choices C and D are waiting for you to determine whether to round up or down.

HOWEVER, you should not do the math before logically setting up the problem. Since you have solid wooden blocks, you will not be able to fit blocks across all levels. For instance, if one side measures 7 inches, only two blocks can fit. There will be 1 inch of wasted space. You will get only three blocks on each of the two remaining sides. Consequently, you will be able to fit \(2∗3∗3=18\) blocks inside the box.
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The key here is to realize that we can't have fractional cubes.

Therefore in a box measuring 11 inches by 10 inches by 7 inches,

the side measuring 11 inches / 3 = 3
the side measuring 10 / 3 = 3
the side measuring 7 / 3 = 2

3 * 3 * 2 = 18

Answer is A.
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Bunuel
A box measuring 11 inches by 10 inches by 7 inches is filled with solid wooden cubical blocks, each measuring 3 inches per side. How many wooden blocks can fit into the box?

A. 18
B. 27
C. 28
D. 29
E. 54

Number of boxes =floor(11/3) * floor(10/3)*floor(7/3) = 3*3*2 = 19 OA is A
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