# two cylindrical tanks have the same height, but the radius of one tank equals the diameter of the other. if the volume of the larger is k% more than the volume of the smaller, k= ?

a)50
b)100
c)200
d)300
e)400

small tank volume: pi*(r/2)^2*h
big tank volume: pi*r^2*h

factor small volume out =
pi*(r^2/4)*h

since big volume is 4 times as the small one, it's 300% more.
(4 is 3 more than 1).

## The correct answer is d) 300.

Explanation:
Given that the volume of the larger tank is k% more than the volume of the smaller tank, we can express this as:

Volume of larger tank = (100 + k)% Ã— Volume of smaller tank

Let's assume that the radius of the smaller tank is r, then the radius of the larger tank would be 2r (since it equals the diameter of the smaller tank).

The volume of the smaller tank can be given as:

Volume of smaller tank = Ï€ Ã— (r/2)^2 Ã— h

Simplifying this expression, we get:

Volume of smaller tank = Ï€ Ã— (r^2/4) Ã— h

The volume of the larger tank can be given as:

Volume of larger tank = Ï€ Ã— r^2 Ã— h

Substituting the expressions for the volumes into the equation:

Ï€ Ã— r^2 Ã— h = (100 + k)% Ã— (Ï€ Ã— (r^2/4) Ã— h)

Simplifying further by canceling out common terms:

r^2 Ã— h = (100 + k)% Ã— (r^2/4) Ã— h

Dividing both sides by r^2 Ã— h:

1 = (100 + k)% Ã— (1/4)

Multiplying both sides by 4:

4 = (100 + k)%

Dividing both sides by 100:

0.04 = k%

Thus, k = 0.04 Ã— 100 = 4 Ã— 100 = 400.

Therefore, the volume of the larger tank is 400% more than the volume of the smaller tank, so the answer is e) 400.

## To solve this problem, we need to compare the volumes of the two cylindrical tanks.

Let's say the radius of the smaller tank is "r" and the height of both tanks is "h". Therefore, the volume of the smaller tank can be calculated using the formula: V_small = Ï€ * (r/2)^2 * h.

The radius of the larger tank is equal to the diameter of the smaller tank (which is twice the radius), so the radius of the larger tank is 2r. The volume of the larger tank can be calculated using the formula: V_large = Ï€ * (2r)^2 * h.

We are given that the volume of the larger tank is k% more than the volume of the smaller tank. Mathematically, this can be expressed as:

V_large = (1 + k/100) * V_small.

Substituting the volume formulas, we have:

Ï€ * (2r)^2 * h = (1 + k/100) * Ï€ * (r/2)^2 * h.

We can simplify this equation by canceling out Ï€ * h from both sides:

(2r)^2 = (1 + k/100) * (r/2)^2.

Expanding and simplifying:

4r^2 = (1 + k/100) * (r^2 / 4).

Multiplying both sides by 4 to eliminate the denominators:

16r^2 = (1 + k/100) * r^2.

Now, we can divide both sides by r^2 to isolate k:

16 = 1 + k/100.

Subtracting 1 from both sides:

15 = k/100.

Multiplying both sides by 100 to isolate k:

k = 1500.

Therefore, k is equal to 1500. However, none of the answer choices provided match this value. This suggests that there may be a mistake in the question or the options given.