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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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Example Questions

Example Question #841 : Ssat Upper Level Quantitative (Math)

A car dealership wants to fill a large spherical advertising ballon with helium. It can afford to buy 1,000 cubic yards of helium to fill this balloon. What is the greatest possible diameter of that balloon (nearest tenth of a yard)?

Possible Answers:

$15.4 \;\textrm{yd}$

$7.7 \;\textrm{yd}$

$12.4 \;\textrm{yd}$

$9.3 \;\textrm{yd}$

$6.2 \;\textrm{yd}$

Correct answer:

$12.4 \;\textrm{yd}$

Explanation:

The volume of a sphere, given its radius, is

$V = \frac{4\pi r^{3}}{3}$

Set V = 1,000 , solve for , and double that to get the diameter.

$1,000= \frac{4\pi r^{3}}{3}$

$r^{3} = \frac{1,000 \cdot 3 }{4\pi} \approx 238.7$

$r \approx \sqrt[3]{238.7} \approx 6.2$

The diameter is twice this, or 12.4 yards.

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Example Question #842 : Ssat Upper Level Quantitative (Math)

The diameter of a sphere is . Give the volume of the sphere in terms of .

Possible Answers:

$30\pi t^3$

$24\pi t^3$

$36\pi t^3$

$72\pi t^3$

$10\pi t^3$

Correct answer:

$36\pi t^3$

Explanation:

The diameter of a sphere is so the radius of the sphere would be $6t\div 2=3t$

The volume enclosed by a sphere is given by the formula:

$Volume=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times (3t)^3=\frac{4}{3}\pi\times 27t^3\Rightarrow Volume=36\pi t^3$

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Example Question #843 : Ssat Upper Level Quantitative (Math)

A spherical balloon has a diameter of 10 meters. Give the volume of the balloon.

Possible Answers:

723.3m^3

500m^3

1023.3m^3

523.6m^3

1000m^3

Correct answer:

523.6m^3

Explanation:

The volume enclosed by a sphere is given by the formula:

$Volume=\frac{4}{3}\pi r^3$

where is the radius of the sphere. The diameter of the balloon is 10 meters so the radius of the sphere would be $10\div 2=5$ meters. Now we can get:

$Volume=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\times 5^3=\frac{4}{3}\times \pi \times 125\approx 523.6 m^3$

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Example Question #844 : Ssat Upper Level Quantitative (Math)

The volume of a sphere is 1000 cubic inches. What is the diameter of the sphere.

Possible Answers:

$10\ inches$

$16.4\ inches$

$12\ inches$

$12.4\ inches$

$10.4\ inches$

Correct answer:

$12.4\ inches$

Explanation:

The volume of a sphere is:

$Volume=\frac{4}{3}\pi r^3$

Where is the radius of the sphere. We know the volume and can solve the formula for :

$Volume=\frac{4}{3}\pi r^3=1000\Rightarrow r^3=\frac{3\times1000 }{4\pi}\approx 238.85\Rightarrow r\approx 6.2$ inches

So we can get:

$Diameter=2r=2\times 6.2=12.4inches$

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Example Question #845 : Ssat Upper Level Quantitative (Math)

A sphere has a diameter of inches. What is the volume of this sphere?

Possible Answers:

523.6 in^3

632.9 in^3

458.3 in^3

104.7 in^3

Correct answer:

523.6 in^3

Explanation:

To find the volume of a sphere, use the following formula:

$\text{Volume}=\frac{4}{3}\pi\times r^3$ , where is the radius of the sphere.

Now, because we are given the diameter of the sphere, divide that value in half to find the radius.

$r=10\div2=5$

Now, plug this value into the volume equation.

$\text{Volume}=\frac{4}{3}\pi\times 5^3$

$\text{Volume}=\frac{4}{3}\pi\times 125$

$\text{Volume}=\frac{500}{3}\pi=523.6$ in^3

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Example Question #3 : How To Find The Volume Of A Sphere

What is the volume of a sphere with a diameter of ?

Possible Answers:

$\frac{500}{3}\pi$

$100\pi$

$200\pi$

$150\pi$

$\frac{100}{3}\pi$

Correct answer:

$\frac{500}{3}\pi$

Explanation:

Write the formula for the volume of the sphere.

$V=\frac{4}{3}\pi r^3$

The radius is half the diameter, which is five. Substitute the value.

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5)^3=\frac{4}{3}\pi (125)= \frac{500}{3}\pi$

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Example Question #621 : Geometry

What is the volume of a sphere with diameter 12 feet?

Possible Answers:

None of the other answers are correct

$3,619.1\;ft^{2}$

$904.8 ft^{3}$

$452.4\;ft^{3}$

$7,238.2\;ft^{3}$

Correct answer:

$904.8 ft^{3}$

Explanation:

The radius of the sphere is half the diameter, or 6 feet; use the formula

$V= \frac{4}{3} \pi r^{3}$ .

Setting r=6 :

$V= \frac{4}{3} \pi \cdot 6^{3} \approx904.8 ft^{3}$

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Example Question #846 : Ssat Upper Level Quantitative (Math)

Chestnut wood has a density of about $0.45\frac{g}{cm^3}$ . A right circular cone made out of chestnut wood has a height of three meters, and a base with a radius of two meters. What is its mass in kilograms (nearest whole kilogram)?

Possible Answers:

$5,855 \; \textrm{kg}$

$5,755 \; \textrm{kg}$

$5,955 \; \textrm{kg}$

$6,055 \; \textrm{kg}$

$5,655 \; \textrm{kg}$

Correct answer:

$5,655 \; \textrm{kg}$

Explanation:

First, convert the dimensions to cubic centimeters by multiplying by 100 : the cone has height 300cm , and its base has radius 200cm .

$3m*\frac{100cm}{1m}=300cm\ \ \2m*\frac{100cm}{1m}=200cm$

Its volume is found by using the formula and the converted height and radius.

$V=\frac{1}{3} \pi r^{2}h$

$V=\frac{1}{3} \pi (200cm)^{2} ( 300cm )\approx 12,566,371 \; \textrm{cm}^{3}$

Now multiply this by $0.45\frac{g}{cm^3}$ to get the mass.

$m=V*\rho$

$12,566,371cm^3*0.45\frac{g}{cm^3} \approx 5,654,867g$

Finally, convert the answer to kilograms.

$5654867g *\frac{1kg}{1,000g} \approx 5,655kg$

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Example Question #1 : How To Find The Volume Of A Cone

A cone has the height of 4 meters and the circular base area of 4 square meters. If we want to fill out the cone with water (density = $1000\frac{Kg}{m^3}$ ), what is the mass of required water (nearest whole kilogram)?

Possible Answers:

10,000

5333

6333

4333

2133

Correct answer:

5333

Explanation:

The volume of a cone is:

$Volume=\frac{1}{3}\pi r^2h$

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

As the circular base area is $\pi r^2$ , so we can rewrite the volume formula as follows:

$Volume =\frac{A\times h}{3}$

where is the circular base area and known in this problem. So we can write:

$Volume =\frac{A\times h}{3}=\frac{4\times 4}{3}=\frac{16}{3}\approx 5.333 m^3$

We know that density is defined as mass per unit volume or:

$\rho =\frac{m}{v}$

Where $\rho$ is the density; is the mass and is the volume. So we get:

$m=\rho v=1000\times 5.333=5333 Kg$

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Example Question #847 : Ssat Upper Level Quantitative (Math)

The vertical height (or altitude) of a right cone is . The radius of the circular base of the cone is . Find the volume of the cone in terms of .

Possible Answers:

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

$\frac{1}{3}\pi t^2$

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

$\pi t^3$

$\frac{1}{3}\pi t^3$

Correct answer:

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

Explanation:

The volume of a cone is:

$Volume=\frac{1}3{}\pi r^2h$

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

$Volume=\frac{1}3{}\pi r^2h =\frac{1}3{}\pi t^2\times 2t=\frac{2}{3}\pi t^3$

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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 #841 : Ssat Upper Level Quantitative (Math)

Example Question #842 : Ssat Upper Level Quantitative (Math)

Example Question #843 : Ssat Upper Level Quantitative (Math)

Example Question #844 : Ssat Upper Level Quantitative (Math)

Example Question #845 : Ssat Upper Level Quantitative (Math)

Example Question #3 : How To Find The Volume Of A Sphere

Example Question #621 : Geometry

Example Question #846 : Ssat Upper Level Quantitative (Math)

Example Question #1 : How To Find The Volume Of A Cone

Example Question #847 : Ssat Upper Level Quantitative (Math)

All SSAT Upper Level Math Resources

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