GMAT Quantitative - Problem-Solving Questions

Define a function as follows:

$f(x) = 2 ^{x-3}+ 5$

Give the vertical aysmptote of the graph of .

The graph of does not have a vertical asymptote.

$x= - \frac{5}{2}$

Explanation

Since any number, positive or negative, can appear as an exponent, the domain of the function $f(x) = 2 ^{x-3}+ 5$ is the set of all real numbers; in other words, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

Two circles are constructed; one is inscribed inside a given square, and the other is circumscribed about the same square.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

$200 \pi$

$400 \pi$

$300 \pi$

$50 \pi$

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

Explanation

Examine the diagram below, which shows the square, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

Note that the segment from the center of the square to the midpoint of a side is a radius of the inscribed circle, and the segment from the center to a vertex is a radius of the circumscribed circle. The two segments and half a side of the square form a 45-45-90 triangle, so by the 45-45-90 Theorem, the radius of the circumscribed circle is $\sqrt{2}$ times that of the inscribed circle.

The inscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The circumscribed circle has radius $\sqrt{2}$ times this, or $10 \sqrt{2}$ , so its area is

$A= \pi r^{2} = \pi \cdot \left (10 \sqrt{2} \right ) ^{2}= \pi \cdot 100 \cdot 2 = 200 \pi$

Two circles are constructed; one is inscribed inside a given regular hexagon, and the other is circumscribed about the same hexagon.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

$\frac{400 \pi }{3}$

$200 \pi$

$50 \pi$

$75 \pi$

$\frac{200 \pi }{3}$

Explanation

Examine the diagram below, which shows the hexagon, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

Note that the segment from the center of the hexagon to the midpoint of a side is a radius of the inscribed circle, and the segment from the center to a vertex is a radius of the circumscribed circle. The two segments and half a side of the hexagon can be proved to form a 30-60-90 triangle.

The inscribed circle has circumference $20 \pi$ , so its radius - and the length of the longer leg of the right triangle - is

$20 \pi \div 2 \pi = 10$

By the 30-60-90 Theorem, the length of the shorter leg is this length divided by $\sqrt{3}$ , or $\frac{10}{\sqrt{3}}$ ; the length of the hypotenuse, which is the radius of the circumscribed circle, is twice this, or $\frac{20}{\sqrt{3}}$ .

The area of the circumscribed circle can now be calculated:

$A= \pi r^{2} = \pi \cdot \left (\frac{20}{\sqrt{3}}\right ) ^{2}= \pi \cdot \frac{400}{3} = \frac{400 \pi }{3}$

A rectangle twice as long as it is wide has perimeter . Write its area in terms of .

$2N^{2} -8N + 8$

$N^{2} -4N + 4$

$4N^{2} -16N + 16$

$2N^{2} -6N + 4$

$N^{2} -3N + 2$

Explanation

Let be the width of the rectangle; then its length is , and its perimeter is

Set this equal to and solve for :

$\frac{6W}{6} =\frac{ 6N-12}{6}$

The width is and the length is $L = 2W = 2 \left (N - 2 \right ) = 2N - 4$ , so multiply these expressions to get the area:

$A = LW = (N-2) \cdot \left (2N-4 \right ) = 2N^{2} -8N + 8$

On January 1, Gary borrows $10,000 to purchase an automobile at 12% annual interest, compounded quarterly beginning on April 1. He agrees to pay $800 per month on the last day of the month, beginning on January 31, over twelve months; his thirteenth payment, on the following January 31, will be the unpaid balance. How much will that thirteenth payment be?

Explanation

12% annual interest compounded quarterly is, effectively, 3% interest per quarter.

Over the course of one quarter, Gary pays off $$800 \times 3 = $2,400$ , and the remainder of the loan accruses 3% interest. This happens four times, so we will subtract $2,400 and subsequently multiply by 1.03 (adding 3% interest) four times.

First quarter:

$$7,600 \times 1.03 = $7,828$

Second quarter:

$$ 5,428 \times 1.03 = $5,590.84$

Third quarter:

$$ 3,190.84 \times 1.03 = $3,286.57$

Fourth quarter:

$$ 886.57 \times 1.03 = $913.16$

The thriteenth payment, with which Gary will pay off the loan, will be $913.16.

A large group of students is given a standardized test. The following information is given about the scores:

Mean: 73.8

Standard deviation: 6.3

Median: 71

25th percentile: 61

75th percentile: 86

Highest score: 100

Lowest score: 12

What is the interquartile range of the tests?

More information about the scores is needed.

Explanation

The interquartile range of a data set is the difference between the 75th and 25th percentiles:

$\textrm{IQR }= 86-61=25$

All other given information is extraneous to the problem.

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

is a cube and face has an area of . What is the length of diagonal of the cube ?

$3\sqrt{3}$

$2\sqrt{3}$

$3\sqrt{2}$

Explanation

To find the diagonal of a cube we can apply the formula $d=e\sqrt{3}$ , where is the length of the diagonal and where is the length of an edge of the cube.

Since we are given an area of a face of the cube, we can find the length of an edge simply by taking its square root.

$9=s^2 \rightarrow \sqrt 9 =s \rightarrow 3=s$

Here the length of an edge is 3.

Thefore the final andwer is $d=e\sqrt3=3\sqrt{3}$ .

Given: $\bigtriangleup ABC$ with $m \angle A = m \angle C = 30 ^{\circ}$ and .

Construct the altitude of $\bigtriangleup ABC$ from to a point on $\overline{AC}$ . What is the length of $\overline{BM}$ ?

$7\sqrt{3}$

$7\sqrt{2}$

$14\sqrt{3}$

Explanation

$\bigtriangleup ABC$ is shown below, along with altitude $\overline{BM}$ .

Isosceles

By the Isosceles Triangle Theorem, since $\angle A \cong \angle C$ , $\bigtriangleup ABC$ is isosceles with $\overline{AB } \cong \overline {CB}$ . By the Hypotenuse-Leg Theorem, the altitude cuts $\bigtriangleup ABC$ into congruent triangles $\bigtriangleup AMB$ and $\bigtriangleup CMB$ , so $\overline{AM } \cong \overline {CM}$ ; this makes the midpoint of $\overline{AC}$ . $\overline{AC }$ has length 42, so $\overline{AM}$ measures half this, or 21.

Also, since $m \angle A = 30^{\circ }$ , and $\overline{BM}$ , by definition, is perpendicular to $\overline{AC}$ , $\bigtriangleup AMB$ is a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, $\overline{BM}$ , as the shorter leg of $\bigtriangleup AMB$ , has length equal to that of longer leg $\overline{AM}$ divided by $\sqrt{3}$ ; that is,

$BM = \frac{AM}{\sqrt{3}} = \frac{21}{\sqrt{3}} = \frac{21\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}= \frac{21\cdot \sqrt{3}}{3} = 7 \sqrt{3}$

What is the area of a circle with a diameter of ?

$324\pi cm^2$

$81\pi cm^2$

$54\pi cm^2$

$36\pi cm^2$

$90\pi cm^2$

Explanation

The area of a circle is defined by $A=\pi r^{2}$ , where is the radius of the circle. We are provided with the diameter of the circle, which is twice the length of .