# Video: Pack 5 • Paper 1 • Question 21

Pack 5 • Paper 1 • Question 21

07:48

### Video Transcript

The diagram shows two solids. Solid A is made of a cylinder and a hemisphere on its top. The cylinder has a radius 𝑥 and a height ℎ. The radius of the hemisphere is 𝑥. Solid B is made of a cone and a hemisphere on its top. The cone has a radius two 𝑥 and a height of ℎ prime. The radius of the hemisphere is two 𝑥. The volume of solid B is eight times larger than the volume of solid A. Find how many times larger ℎ prime is than ℎ. Give your answer in its simplest form.

The question also gives us firstly the volume of a cone, which is one-third 𝜋𝑟 squared ℎ, where 𝑟 is the radius of the cone and ℎ is the height of the cone, and secondly the volume of a sphere, which is four-thirds 𝜋𝑟 cubed, where 𝑟 is the radius of the sphere. We’re gonna move these formula over here to give ourselves some space.

Okay, now, let’s examine the key bits of information given by the question. Both of our solids A and B are compound shapes, we’re told that solid A is made up of a cylinder and a hemisphere. We can, therefore, say that the volume of solid A, which we’ll call 𝑉 A, is equal to the volume of the cylinder add the volume of the hemisphere that makes it up. We’ll call this hemisphere A to avoid confusion.

Next, we’re told that solid B is made up of a cone and a hemisphere. Again, we can say that the volume of solid B is equal to the volume of the cone add the volume of the hemisphere that makes it up. It’s worth noting that the two hemispheres are different since one has a radius of 𝑥 and one has a radius of two 𝑥. They’ll therefore have different volumes.

Lastly, we are told that the volume of solid B is eight times larger than the volume of solid A. Using annotation, we can write that 𝑉 B, which is the volume of solid B, is eight times 𝑉 A, which is the volume of solid A.

Now that we’ve condensed the question to these key statements, again, let’s give ourselves some room. So given this information, we should be able to find how many times larger ℎ prime is than ℎ. To begin, let’s work on finding the volume of solid A by examining the first term, which is the volume of the cylinder.

Now, the question hasn’t given us a formula for the volume of a cylinder. However, another way to think of a cylinder is a circular prism. We should be able to recall that the volume of any prism is equal to the area of the cross section, which in our case is a circle with radius 𝑥, multiplied by the length of the prism, which in our case is ℎ.

The area of a circle is 𝜋 times the radius squared. Since this circle’s radius is given as 𝑥, we can replace this. We multiply this by the length of the prism ℎ and we have the volume of our cylinder.

The next term for the volume of solid A is the volume of hemisphere A. We can recall the definition that a hemisphere is a sphere that has been cut exactly in half. Since the question has given us a formula for the volume of a sphere, we can multiply this by exactly half to give the volume of a hemisphere. To simplify this, we can divide the four in the numerator by the two in the denominator. And we can recall that this 𝑟 is the radius of the original sphere, which we cut in half, which has been given as 𝑥.

Replacing this, we get that the volume of hemisphere A is two-thirds times 𝜋𝑥 cubed. Putting this together, we get that the volume of solid A is equal to 𝜋𝑥 squared ℎ add two-thirds of 𝜋𝑥 cubed.

Let’s now work on the volume of solid B. The first term of this is the volume of a cone. And a formula for this has been given by the question. In this formula, the 𝑟 represents the radius of the circle which forms the base of the cone. For our cone, this radius is two 𝑥. Next, the ℎ represents the height of the cone. For our cone, this has been given as ℎ prime.

We can replace these two terms to find that the volume of our cone is given by one-third 𝜋 times two 𝑥 squared times ℎ prime. And we should be extra careful not to be confused by the ℎ given in the formula and the ℎ used for the cylinder in solid A since they are not the same. Another misconception that we wanna clear up is not to just stick a squared sign on this two 𝑥. Remember we’re squaring the entire radius. So put brackets around your two 𝑥 to make it clear.

When we simplify these brackets, we find that this term is in fact four 𝑥 squared. And we can move this four over to the numerator of our fraction. We now have that the volume of our cone is four over three times 𝜋𝑥 squared ℎ prime.

Now, let’s look at the volume of hemisphere B. For this, we can use the same method as the volume of hemisphere A. But, now, instead, we’ll replace our radius with two 𝑥 instead of 𝑥. Again, we remember to put our brackets around two 𝑥 since we’re cubing the entire radius. And this is not the same as two times 𝑥 cubed. Multiplying out these brackets, we in fact get eight 𝑥 cubed.

We can simplify this by moving the eight to the numerator of our fraction to find that the volume of hemisphere B is 16 over three times 𝜋𝑥 cubed. We can now add these together. And we find that the volume of solid B is equal to four over three times 𝜋𝑥 squared ℎ prime add 16 over three times 𝜋𝑥 cubed.

Let’s now use the third bit of key information given by the question. Since we know that the volume of B is equal to eight times the volume of A and we have an equation for both of these, we’re able to substitute in the following, which is the volume of B on the left-hand side and eight times the volume of A on the right-hand side.

We can now work on simplifying the right-hand side of the equation by multiplying both of the terms by the eight which is outside of the brackets. Examining this new line, we should be able to see that we have the same term of plus 16 over three 𝜋𝑥 cubed on both sides of the equation. By subtracting 16 over three 𝜋𝑥 cubed from both sides of the equation, we can eliminate this term.

Next, we can see that both the left- and right-hand side involve a factor of 𝜋𝑥 squared. We can, therefore, simplify again by dividing both sides by 𝜋𝑥 squared eliminating this factor. Now, we see that we’re very close to finding ℎ prime in terms of ℎ, which is what our question is asking for.

To give our answer in its simplest form, we want ℎ prime on its own on the left-hand side of the equation. We, therefore, want to get rid of this factor of four over three. To do so, we can multiply both sides of our equation by three over four since three over four times four over three is equal to one.

Finally, we simplify the numerator on the right-hand side of our equation. Three times eight is equal to 24. And the fraction 24 over four is just six. We have, therefore, found that ℎ prime is equal to six ℎ.

This line alone should satisfy the question. But for completeness, we can write down the answer six or put it in words ℎ prime is six times larger than ℎ.