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Stem of the problem W * L = 60
A) H * L = 60
B) H * W = 60
Three variables, Three equations
You can determine W, H and L from both of these. Answer is C indeed.
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Cube Area = side\(3\)
Cube Area = 60*6 =360
side\(3\) = 360
side=3\sqrt{360}
360 = 2*2*2*3*3*5,
so we have 2\(3\) and 3\(2\) and 5\(1\)
Is there any number that can satisfy the conditions?
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

Cube Area = side\(3\)
Cube Area = 60*6 =360
side\(3\) = 360
side=3\sqrt{360}
360 = 2*2*2*3*3*5,
so we have 2\(3\) and 3\(2\) and 5\(1\)

Is there any number that can satisfy the conditions?

Can you please elaborate what you mean?

The volume of a cube = \(side^3\);
The surface area of a cube = \(6*side^2\) (since a cube has 6 faces).
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Bunuel,

Thx for the correction.
Now I realized where I made a mistake.

1. Cube volume = side\(3\)
2. Cube's side Area is equal to side\(2\)
hence side of the Cube =\(\sqrt[]{60}\) ,
or 2\(\sqrt[]{15}\)
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Bunuel,

Thx for the correction.
Now I realized where I made a mistake.

1. Cube volume = side\(3\)
2. Cube's side Area is equal to side\(2\)
hence side of the Cube = \sqrt{60},
or 2\sqrt{15}

Please check the following post.

Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube.

For #1) w*h = 60. w*d = 60 --> h=d but that isn't enough...we don't know what the height is.
For #2) d*h = 60. w*d = 60 --> w=h but that isn't enough...we don't know what the depth is.

1+2) w=h=d Sufficient (a cube is w*h*d where all sides are equal)
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Thank you for that. For some reason I was thinking that the cube needed to have a whole number on each side which is obviously untrue.

Bunuel
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = \(\sqrt{60}\), then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Answer: C.

Hope it helps.
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = \(\sqrt{60}\), then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Answer: C.

Hope it helps.

I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could anyone please elaborate on this please?

Thanks
Cheers!
J :)
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = \(\sqrt{60}\), then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Answer: C.

Hope it helps.

I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could anyone please elaborate on this please?

Thanks
Cheers!
J :)

Which statement gives the area of FOUR faces?
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It says that the base of the rectangular block is 60 and then in each of the statements you are given the area of the side face and the front face separately. Therefore if you know the bottom face then the top is also the same and likewise if you know the side face is 60 then the side just in front of that one is also 60. Thats why I thought that we needed the remaining to faces to be 60 otherwise the figure would not fit

Let me know if you understand my reasoning. I'm trying to figure out where exactly am I going wrong

Thanks!
J

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+1 on @jlgdr's statement....A alone says that 4 faces are of same surface area....which implies all 6 faces should be of the same surface area...which implies all dimensions should be same?

B alone says the same thing. The answer choice should be "D"? I am not convinced that we need A+B to answer & the answer choice be "C". Can someone clarify?
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+1 on @jlgdr's statement....A alone says that 4 faces are of same surface area....which implies all 6 faces should be of the same surface area...which implies all dimensions should be same?

B alone says the same thing. The answer choice should be "D"? I am not convinced that we need A+B to answer & the answer choice be "C". Can someone clarify?

In my post here: the-base-of-a-rectangular-block-has-an-area-of-60-square-4397.html#p1302020 specif examples are given where the block is NOT a cube:
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