Question

Three cube-shaped aquariums that are five inches on each side are filled with water to capacity. All of the water from those three aquariums is to be transferred into a larger cube aquarium so that it must be filled to at least 50% of its total capacity without overflowing. Which of the following could be the length, in inches, of a side of the larger aquarium?

(A) 6.9
(B) 8.4
(C) 9.5

Answer 1

Answer: (B) 8.4

Explanation:

Given : Dimension of each cube = 5 inches

Then, volume of each cube =
`(5)^3=125\text{ in.}^3`

Then, volume of 3 such cubes =
`3*125=375\text{ in.}^3`

Let x be the dimension of larger cube , then its volume =
`x^3`

Now, by considering the given question, we have

`0.50 x^3\leq375\\\\\Rightarrow\ x^2\leq(375)/(0.5)\\\\\Rightarrow\ x^3\leq750`

Taking cube-root on both sides , we get

`x\leq(750)^(1/3)\\\\\Rightarrow\ x\leq9.086`

The most nearest value from the options less than 9.086 is 8.4.

Hence, the the length of a side of the larger aquarium could be 8.4 inches.

Answer 2

The correct answer is B.

Explanation:

We need to start calculating the total volume of water. This is, the sum of the three cube-shaped aquariums, that are 5in on each side. As the three aquariums are equal, the total volume of water is V=3V', where V' is the volume of one of those three aquariums.

We know that the volume of a cube is the third power of its side: V'=(5in)^3 = 125 in^3. Then, the total volume of water in the three aquariums is

V=3*125 in^3=325 in^3.

Now we only need to compare with the volumes of the given aquariums. Recall that the 50% is just the half of the volume of the larger aquarium.

(A) If the side of the aquarium is 6.9 in, its volume is 328.509 in^3. This volume is less that the amount of water we have from the initial three aquariums.

(B) If the side of the aquarium is 8.4 in, its volume is 592.704 in^3. It is not difficult to see that 325>592.704/2.

(C) If the side of the aquarium is 9.5 in, its volume is 857.375 in^3. Here it is not hard to check that 325<857.375/2.