GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #61 : Rectangular Solids & Cylinders

Each base of a right prism is a regular hexagon with sidelength 6. Its height is two thirds the perimeter of a base. Give the surface area of the prism.

Possible Answers:

$864+ 108\sqrt{3}$

972

$972\sqrt{3}$

$1,296\sqrt{3}$

$864+ 54\sqrt{3}$

Correct answer:

$864+ 108\sqrt{3}$

Explanation:

The perimeter of a regular hexagon with sidelength 6 is

$P= 6 \cdot 6 = 36$

The height of the prism is two thirds of this, so

$h = \frac{2}{3} \cdot 36 = 24$ .

The lateral area of the prism is the product of the perimeter of a base and the height of the prism, so

$L =Ph= 36 \cdot 24 = 864$

The area of each base can be calculated using the area formula for a regular hexagon:

$B = \frac{3s^{2}\sqrt{3}}{2}$

$B = \frac{3\cdot 6^{2}\sqrt{3}}{2}$

$B = \frac{108\sqrt{3}}{2}$

$B = 54\sqrt{3}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 864+ 54\sqrt{3} + 54\sqrt{3} = 864+ 108\sqrt{3}$

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Example Question #782 : Problem Solving Questions

Tetra_1

Refer to the above diagram. The perimeter of $\bigtriangleup DEG$ is 30. What is the surface area of the cube shown?

Possible Answers:

300

$600 \sqrt{2}$

Insufficient information is given to answer the question.

600

$300 \sqrt{2}$

Correct answer:

300

Explanation:

Since each side of $\bigtriangleup DEG$ is a diagonal of one of three congruent squares, each side has the same length; since the total perimeter is 30, each side measures one third of this, or 10.

Each square side, therefore, has diagonal 10, and, by the 45-45-90 Theorem, each side of each square - and each edge of the cube - measures $s= \frac{10}{\sqrt{2}}$ . We use the surface area formula:

$A= 6s^{2}$

$A= 6 \left ( \frac{10}{\sqrt{2}} \right )^{2} = 6 \cdot \frac{100}{2} = 300$

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Example Question #783 : Problem Solving Questions

A right prism has as its bases two triangles, each of which has a hypotenuse of length 25 and a leg of length 7. The height of the prism is one fourth the perimeter of a base. Give the surface area of the prism.

Possible Answers:

424

868

434

784

952

Correct answer:

952

Explanation:

The second leg of a right triangle with hypotenuse of length 25 and one leg of length 7 has length

$\sqrt{25^{2}-7^{2}}= \sqrt{625-49}= \sqrt{576}= 24$ .

The area of this right triangle is half the product of the lengths of the legs, which is

$B = \frac{1}{2} \cdot 7 \cdot 24 = 84$ .

The perimeter of each base is

P = 7+ 24 + 25 = 56 ,

and the height is one fourth this, or

$h = \frac{1}{4} \cdot P = \frac{1}{4} \cdot 56 =14$

The lateral area of the prism is the product of its height and the perimeter of a base; this is

$L = 14 \cdot 56 = 784$ .

The surface area is the sum of the lateral area and the two bases:

S= L + B + B = 784+84+84 =952 .

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Example Question #15 : Prisms

The length of a cube is increased by 20%, and the width is decreased by 20%. Which of the following must happen to the height so that the resulting rectangular prism will have the same surface area as the original cube?

Possible Answers:

The height must be increased by 4%.

The height must remain the same.

The height must be decreased by 4%.

The height must be decreased by 2%.

The height must be increased by 2%.

Correct answer:

The height must be increased by 2%.

Explanation:

To look at this more easily, assume the cube has sides of length 100; this argument generalizes to any size. The surface area of the cube is

$S= 6s^{2}= 6 \cdot 100^{2}= 6 \cdot 10,000 = 60,000$ .

After the changes, the resulting rectangular prism will have length

100 (1+0.2) = 120

and width

100 (1-0.2) = 80

The surface area of a rectangular prism is

S = 2LH + 2LW + 2WH . We can call L = 80 , W = 120 , and S= 60,000 and solve for :

2(80)H + 2(80)(120)+ 2(120)H= 60,000

160H + 19,200+240H= 60,000

400H + 19,200 = 60,000

400H = 40,800

H = 102

This means that the height of the prism must be 102% of the height of the cube - equivalently, the height must be increased by 2%.

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Example Question #61 : Rectangular Solids & Cylinders

A cube of iron has a mass of 4 kg. What is the mass of a rectangular prism of iron that is the same height but has a width and length that are twice as long as the cube?

Possible Answers:

Correct answer:

Explanation:

Let the length of the sides of the cube equal 1. The volume of the cube is then $\dpi{100} 1\times 1\times 1=1$ . Therefore, the volume of the prism is $\dpi{100} 2\times 2\times 1=4$ . Therefore, the mass of the prism must be 4 times greater than the mass of the cube.

$\small (4)(4)=\ 16$

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Example Question #2 : Calculating The Volume Of A Prism

What is the volume of a cube whose diagonal measures 10 inches?

Possible Answers:

$353.6\;in^{3}$

$192.5\; in ^{3}$

$333.3 \;in^{3}$

$100\; in ^{3}$

$1,000\; in ^{3}$

Correct answer:

$192.5\; in ^{3}$

Explanation:

By an extension of the Pythagorean Theorem, if is the length of an edge of the cube and is its diagonal length,

$d^{2}= s^{2}+s^{2}+s^{2} = 3s^{2}$

$\sqrt{3s^{2} }= \sqrt{d^{2}}$

$s\sqrt{3 }= d$

$s= \frac{d}{\sqrt{3 }}$

The volume is therefore

$V = s^{3} = \left (\frac{d}{\sqrt{3}} \right )^{3} = \left (\frac{10}{\sqrt{3}} \right )^{3}\approx 192.5$

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Example Question #62 : Rectangular Solids & Cylinders

The length, width, and height of a rectangular prism, in inches, are three different prime numbers. All three dimensions are between one yard and four feet. What is the volume of the prism?

Possible Answers:

$71,299\textrm{ in}^{3}$

$74,777\textrm{ in}^{3}$

$65,231\textrm{ in}^{3}$

$82,861\textrm{ in}^{3}$

It is impossible to tell from the information given.

Correct answer:

It is impossible to tell from the information given.

Explanation:

One yard is equal to 36 inches; four feet are equal to 48 inches. There are four prime numbers between 36 and 48 - 37, 41, 43, and 47. Since we are only given that the dimensions are three different prime numbers between 36 and 48, we have no way of knowing which three they are.

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Example Question #63 : Rectangular Solids & Cylinders

A company makes cubic metal blocks whose edges measure 5 inches. The blocks are supposed to be packed in crates 40 inches long, 30 inches wide, and 32 inches high. How many blocks can be packed in one of these crates?

Possible Answers:

336

240

288

308

307

Correct answer:

288

Explanation:

Each layer of blocks will be $40 \div 5 = 8$ blocks long and $30 \div 5 = 6$ blocks wide, making a layer of $8 \times 6 = 48$ blocks. The height is 32 feet, which divided by 5 is

$32 \div 5 = 6\textrm{ R }2$

so there will be 6 layers of blocks.

Therefore, $48 \times 6 = 288$ blocks can be packed.

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Example Question #5 : Calculating The Volume Of A Prism

The sum of the length, the width, and the height of a rectangular prism is one yard. The length of the prism is eleven inches greater than its width, and the width is twice its height. What is the volume of the prism?

Possible Answers:

$3,150 \textrm{ in}^{3}$

$1,050 \textrm{ in}^{3}$

$2,450 \textrm{ in}^{3}$

$1,750 \textrm{ in}^{3}$

$2,150 \textrm{ in}^{3}$

Correct answer:

$1,050 \textrm{ in}^{3}$

Explanation:

Let be the height of the prism. Then the width is , and the length is 2h+ 11 . Since the sum of the three dimensions is one yard, or 36 inches, we solve for in this equation:

h + 2h + (2h+11) = 36

5h+11 = 36

5h+11 -11 = 36-11

5h = 2 5

$5h \div 5 = 25 \div 5$

h = 5

The height is 5 inches; the width is twice this, or 10 inches; the length is eleven inches greater than the width, or 21 inches. The volume is the product of the three dimensions:

$V = lwh = 5 \cdot10\cdot21 = 1,050$ cubic inches.

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Example Question #71 : Rectangular Solids & Cylinders

A given cylinder has circular bases of radius and height . What is the volume of the cylinder?

Possible Answers:

$36\pi$

$72\pi$

$18\pi$

$432\pi$

$144\pi$

Correct answer:

$432\pi$

Explanation:

The volume of a given cylinder is the product of its base area (A) and height (h) , or V=Ah . Since the circular base area $A=\pi r^{2}$ , we can substitute and plug in the values provided:

V=Ah

$V=(\pi r^{2})h$

$V=(\pi (6)^{2})(12)$

$V=432\pi$

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #61 : Rectangular Solids & Cylinders

Example Question #782 : Problem Solving Questions

Example Question #783 : Problem Solving Questions

Example Question #15 : Prisms

Example Question #61 : Rectangular Solids & Cylinders

Example Question #2 : Calculating The Volume Of A Prism

Example Question #62 : Rectangular Solids & Cylinders

Example Question #63 : Rectangular Solids & Cylinders

Example Question #5 : Calculating The Volume Of A Prism

Example Question #71 : Rectangular Solids & Cylinders

All GMAT Math Resources

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