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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

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All SSAT Upper Level Math Resources

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Example Questions

Example Question #164 : Geometry

The height of a cylinder is 3 inches and the radius of the circular end of the cylinder is 3 inches. Give the volume and surface area of the cylinder.

Possible Answers:

V=74.78 in^3, A=103.04in^2

V=84.78 in^3, A=103.04in^2

V=84.78 in^3, A=123.04in^2

V=74.78 in^3, A=113.10in^2

V=84.82 in^3, A=113.10in^2

Correct answer:

V=84.82 in^3, A=113.10in^2

Explanation:

The volume of a cylinder is found by multiplying the area of one end of the cylinder (base) by its height or:

$Volume=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder. So we can write:

$Volume=\pi r^2h=\pi \times 3^2\times 3=84.82in^3$

The surface area of the cylinder is given by:

$A=2\pi r^2+2\pi rh$

where is the surface area of the cylinder, is the radius of the cylinder and is the height of the cylinder. So we can write:

$A=2\pi r^2+2\pi rh$

$A=2\pi (3)^2+2\pi \times 3\times 3$

$A=18\pi+18\pi$

$A=36\pi$

A=113.10

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Example Question #171 : Geometry

The height of a cylinder is two times the length of the radius of the circular end of a cylinder. If the volume of the cylinder is $16\pi$ , what is the height of the cylinder?

Possible Answers:

$2\sqrt{2}$

Correct answer:

Explanation:

The volume of a cylinder is:

$Volume=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder.

Since h=2r , we can substitute that into the volume formula. So we can write:

$Volume=\pi r^2h$

$16\pi=\pi r^2\cdot (2r)$

$16\pi =2\pi r^{3}$

$16=2r^{3}$

$8=r^{3}$

r=2

So we get:

$h=2r=2\times 2=4$

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Example Question #951 : Ssat Upper Level Quantitative (Math)

The end (base) of a cylinder has an area of $16\pi$ square inches. If the height of the cylinder is half of the radius of the base of the cylinder, give the volume of the cylinder.

Possible Answers:

$16\pi$

$8\pi$

$32\pi$

Correct answer:

$32\pi$

Explanation:

The area of the end (base) of a cylinder is $\pi r^2$ , so we can write:

$\pi r^2=16\pi\Rightarrow r^2=16\Rightarrow r=4\ inches$

The height of the cylinder is half of the radius of the base of the cylinder, that means:

$h=\frac{r}{2}=\frac{4}{2}=2\ inches$

The volume of a cylinder is found by multiplying the area of one end of the cylinder (base) by its height:

$Volume=Area\times h=16\pi\times 2=32\pi$

$Volume=\pi r^2h=\pi\times 4^2\times 2=32\pi$

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Example Question #173 : Geometry

We have two right cylinders. The radius of the base Cylinder 1 is $\sqrt{3}$ times more than that of Cylinder 2, and the height of Cylinder 2 is 4 times more than the height of Cylinder 1. The volume of Cylinder 1 is what fraction of the volume of Cylinder 2?

Possible Answers:

$V_{1}=\frac{4}{3}V_{2}$

$V_{1}=V_{2}$

$V_{1}=\frac{3}{4}V_{2}$

$V_{1}=\frac{2}{3}V_{2}$

$V_{1}=\frac{3}{2}V_{2}$

Correct answer:

$V_{1}=\frac{3}{4}V_{2}$

Explanation:

The volume of a cylinder is:

$V=\pi r^2h$

where is the volume of the cylinder, is the radius of the circular end of the cylinder, and is the height of the cylinder.

So we can write:

$V_{1}=\pi (r_{1})^2h_{1}$

and

$V_{2}=\pi (r_{2})^2h_{2}$

Now we can summarize the given information:

$r_{1}=\sqrt{3}\cdot r_{2}$

$h_{2}=4\cdot h_{1}\Rightarrow h_{1}=\frac{h_{2}}{4}$

Now substitute them in the $V_{1}$ formula:

$V_{1}=\pi (\sqrt{3}r_{2})^2\times \frac{h_{2}}{4}\Rightarrow V_{1}=\frac{3}{4}\pi (r_{2})^2\times h_{2}\Rightarrow V_{1}=\frac{3}{4}V_{2}$

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Example Question #174 : Geometry

Two right cylinders have the same height. The radius of the base of the first cylinder is two times more than that of the second cylinder. Compare the volume of the two cylinders.

Possible Answers:

$V_{1}=4\cdot V_{2}$

$V_{1}=\sqrt{2}\cdot V_{2}$

$V_{1}=V_{2}$

$V_{1}=3\cdot V_{2}$

$V_{1}=2\cdot V_{2}$

Correct answer:

$V_{1}=4\cdot V_{2}$

Explanation:

The volume of a cylinder is:

$V=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder. So we can write:

$V_{1}=\pi (r_{1})^2h_{1}$

$V_{2}=\pi (r_{2})^2h_{2}$

We know that

$h_{1}=h_{2}$

and

$r_{1}=2r_{2}$ .

So we can write:

$V_{1}=\pi (r_{1})^2h_{1}=\pi (2r_{2})^2\cdot h_{2}=4\pi (r_{2})^2h_{2}\Rightarrow V_{1}=4V_{2}$

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Example Question #113 : Volume Of A Three Dimensional Figure

A cylinder has a diameter of inches and a height of inches. Find the volume, in cubic inches, of this cylinder.

Possible Answers:

98.3 in^3

100.5 in^3

402.1 in^3

308.9 in^3

Correct answer:

100.5 in^3

Explanation:

$\text{Volume of cylinder}=\pi \times \text{radius}^2 \times \text{height}$

Since we are given the diameter, divide that value in half to find the radius.

$\text{radius}=8\div2=4$

Now plug this value into the equation for the volume of a cylinder.

$\text{Volume}=\pi \times 4^4 \times 2$

$\text{Volume}=\pi \times 32=100.5$ in^3

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Example Question #1993 : Hspt Mathematics

Find the volume, in cubic inches, of a cylinder that has a radius of inches and a height of inches.

Possible Answers:

2695.3 in^3

737.5 in^3

4456.1 in^3

3308.1 in^3

Correct answer:

3308.1 in^3

Explanation:

The formula to find the volume of a cylinder is $\pi r^2\times height$ .

Now, plug in the given numbers into this equation.

$\text{Volume}=\pi \times 9^2 \times 13$

$\text{Volume}=\pi \times 1053=3308.1$ in^3

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Example Question #731 : Geometry

Find the volume of the cylinder if the circular base has an area of $\pi$ , and the height of the cylinder is also $\pi$ .

Possible Answers:

$\frac{\pi^3}{4}$

$\frac{\pi^3}{2}$

$\pi^3$

$\frac{\pi^2}{2}$

$\pi^2$

Correct answer:

$\pi^2$

Explanation:

Write the formula for volume of a cylinder.

$V=\pi r^2 h$

Remember that area of a circle is $A=\pi r^2$

Since the area of the circle is known, substitute the area into the formula.

$V=\pi(\pi)= \pi^2$

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Example Question #952 : Ssat Upper Level Quantitative (Math)

A regular hexagon has perimeter 9 meters. Give the length of one side in millimeters.

Possible Answers:

$1,800\textrm{ mm }$

$1,200\textrm{ mm }$

$1,500 \textrm{ mm }$

$2,400 \textrm{ mm }$

$2,000\textrm{ mm }$

Correct answer:

$1,500 \textrm{ mm }$

Explanation:

One meter is equal to 1,000 millimeters, so the perimeter of 9 meters can be expressed as:

9 meters = $9 \times 1,000 = 9,000$ millimeters.

Since the six sides of a regular hexagon are congruent, divide by 6:

$9,000 \div 6 = 1,500$ millimeters.

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Example Question #1 : How To Find The Perimeter Of A Hexagon

A hexagon with perimeter 60 has four congruent sides of length t+1 . Its other two sides are congruent to each other. Give the length of each of those other sides in terms of .

Possible Answers:

28-2t

56+2t

56-2t

28+2t

28-t

Correct answer:

28-2t

Explanation:

The perimeter of a polygon is the sum of the lengths of its sides. Let:

Length of one of those other two sides

Now we can set up an equation and solve it for in terms of :

2x+4(t+1)=60

$\Rightarrow 2x+4t+4=60$

$\Rightarrow 2x=60-4-4t$

$\Rightarrow 2x=56-4t$

$\Rightarrow x=28-2t$

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #164 : Geometry

Example Question #171 : Geometry

Example Question #951 : Ssat Upper Level Quantitative (Math)

Example Question #173 : Geometry

Example Question #174 : Geometry

Example Question #113 : Volume Of A Three Dimensional Figure

Example Question #1993 : Hspt Mathematics

Example Question #731 : Geometry

Example Question #952 : Ssat Upper Level Quantitative (Math)

Example Question #1 : How To Find The Perimeter Of A Hexagon

All SSAT Upper Level Math Resources

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