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genxer123
Bunuel
If the areas of three of the faces of a rectangular solid are 6, 10 and 15, what is the volume of the solid?

(A) 30
(B) 90
(C) 150
(D) 300
(E) 450
Kudos to Sasindran - I used the approach for similar questions that do not involve volume.
Next time I will notice if volume is involved. Fortunately the numbers are easy.

Three surface areas of 6, 10, and 15

6 = 2 * 3
10 = 2 * 5
15 = 3 * 5

6 and 10 must share side with length 2
6 and 15 must share side with length 3
10 and 15 must share side with length 5

Volume = (L * H * W) = (5 * 2 * 3) = 30

Answer A

Thank you.

But I feel that your method is a little “trial and error” method.
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Kudos to Sasindran - I used the approach for similar questions that do not involve volume. . . .
Next time I will notice if volume is involved.

Thank you.

But I feel that your method is a little “trial and error” method.
Sasindran , perhaps I was not clear. Or perhaps you misunderstood what I meant.

Your approach to this problem is vastly superior to mine. You were shrewd. I did not catch the algebraic connection. (Not good!)

And that meant I had to resort to finding the common side. I got lucky. As I mentioned, the numbers were easy. I will choose shrewd over lucky every time.

I put my method there for one reason only: I have seen questions where edge length must be found, and the only way to find edge length involves finding common sides. Those questions look a lot like this one.

That was my mistake. Your good work taught me that I got complacent. Here, yours is by far the better way, IMO.

Nice work.

(And next time, at least from me, please assume the best. If I go out of my way to name you in a compliment, that means I am thrilled to see someone do well and grateful to have learned something.) Cheers.
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Kudos to Sasindran - I used the approach for similar questions that do not involve volume. . . .
Next time I will notice if volume is involved.

Thank you.

But I feel that your method is a little “trial and error” method.
Sasindran , perhaps I was not clear. Or perhaps you misunderstood what I meant.

Your approach to this problem is vastly superior to mine. You were shrewd. I did not catch the algebraic connection. (Not good!)

And that meant I had to resort to finding the common side. I got lucky. As I mentioned, the numbers were easy. I will choose shrewd over lucky every time.

I put my method there for one reason only: I have seen questions where edge length must be found, and the only way to find edge length involves finding common sides. Those questions look a lot like this one.

That was my mistake. Your good work taught me that I got complacent. Here, yours is by far the better way, IMO.

Nice work.

(And next time, at least from me, please assume the best. If I go out of my way to name you in a compliment, that means I am thrilled to see someone do well and grateful to have learned something.) Cheers.

Thanks and good luck to you too

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