GMAT Quantitative Reasoning

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GMAT Quantitative › GMAT Quantitative Reasoning

Questions 1 - 10

1. If the arithmetic mean of five different numbers is 50, how many of the numbers are greater than 50?

(1) None of the five numbers is greater than 100.

(2) Three of the five numbers are 24, 25 and 26, respectively.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Explanation

For statement (1), there are different combinations that satisfy the condition. For example, the five numbers can be $48,49,50,51,\ or\ 52,$ or the five numbers can be . Therefore, we cannot determine how many of the numbers are greater than by knowing the first statement.

For statement (2), even though we know three of them, the two unknown numbers can both be greater than , or one smaller and one greater. Thus statement (2) is not sufficient.

Putting the two statements together, we know that the sum of the two unknown numbers is

$50\cdot 5-\left ( 24+25+26 \right )=175$

Since none of them is greater than 100, both of them have to be greater than 50. Therefore when we combine the two statements, we know that there are two numbers that are greater than 50.

A bag contains red, yellow and green marbles. There are marbles total.

I) There are green marbles.

II) The number of yellow marbles is half of one less than the number of green marbles.

What are the odds of picking a red followed by a green followed by a yellow? Assume no replacement.

Both statements are needed to answer the question.

Neither statement is sufficient to answer the question. More information is needed.

Statement I is sufficient to answer the question, but statement II is not sufficient to answer the question.

Statement II is sufficient to answer the question, but statement I is not sufficient to answer the question.

Either statement is sufficient to answer the question.

Explanation

In order to calculate the probability the question asks for, we need to know the number of each color of marble.

I) Gives us the number of greens.

II) Gives us clues which will allow us to find the number of reds and yellows.

We need both statements to answer this question.

Sector

The circle in the above diagram has center . Give the area of the shaded sector.

Statement 1: Arc $\widehat{AB}$ has length $30 \pi$ .

Statement 2: Arc $\widehat{BD}$ has length $60 \pi$ .

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

Explanation

Assume Statement 1 alone. Since the circumference of the circle is not given, it cannot be determined what part of the circle $\widehat{AB}$ is, and therefore, the central angle of the sector cannot be determined. Also, no information about the circle can be determined. A similar argument can be given for Statement 2 being insufficient.

Now assume both statements are true. Then the length of semicircle $\widehat{ABD}$ is equal to $30 \pi + 60 \pi = 90 \pi$ . The circumference is twice this, or $180 \pi$ . The radius can be calculated as $r = C \div( 2 \pi )=180 \pi \div (2 \pi) = 90$ , and the area, $\pi r^{2}= \pi \cdot 90^{2} = 8,100 \pi$ . Also, $\widehat{AB}$ is $\frac{30 \pi}{180 \pi}= \frac{1}{6}$ of the circle, and the area of the sector can now be calculated as $\frac{1}{6} \times 8,100 \pi = 1,350 \pi$ .

Parallelogram2

Give the area of the above parallelogram if .

$25\sqrt{3}$

$25\sqrt{2}$

$50\sqrt{2}$

Explanation

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$CD= \frac{BC}{2} = \frac{10}{2}} = 5$ .

$BD =CD \sqrt{3} = 5\sqrt{3}$

The area is therefore

$BD \cdot CD = 5\sqrt{3} \cdot 5= 25\sqrt{3}$

Company B produces toy trucks for a shopping mall at a cost of \$7.00 each for the first 500 trucks and \$5.00 for each additional truck. If 600 trucks were produced by Company B and sold for \$15.00 each, what was Company B’s gross profit?

$\$5000$

$\$0$

$\$9000$

$\$4000$

$\$14,000$

Explanation

First of all, we need to know that

$Gross\ Profit=Revenue-Total\ Cost$ .

There are 600 trucks produced. According to the question, the first 500 trucks cost \$7.00 each. Therefore, the total cost of the first 500 trucks is $\$7.00\cdot 500=\$3500$ .

The other 100 trucks cost \$5.00 each for a cost of $$5.00\cdot 100=$500$ .

Add these together to find the cost of the 600 trucks: $$3500+$500=\$4000$

The total profit is easier to calculate since the selling price doesn't change: $\$15.00\cdot 600=\$9000$

At this point we have both revenue and total cost, so the answer for gross profit is $\$9000-$4000=$5000$ .

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Explanation

The total number of balls in the box will be

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right }$

Their respective probabilities are their frequencies divided by 55:

$\left { \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right }$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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

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

$75 \pi$

$50 \pi$

$200 \pi$

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

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

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 circumscribed circle has circumference $20 \pi$ , so its radius - and the length of the hypotenuse of the right triangle -

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

By the 30-60-90 Theorem, the length of the shorter leg is half this, or 5. The length of the longer leg, which is the radius of the inscribed circle, is $\sqrt{3}$ times this, or $5 \sqrt{3}$ .

The area of the inscribed circle can now be calculated:

$A= \pi r^{2} = \pi \cdot \left ( 5 \sqrt{3 }\right ) ^{2}= \pi \cdot 25 \cdot 3 = 75 \pi$

In the -plane, point $(a,b)$ lies on a circle with center at the origin. The radius of the circle is 5. What is the value of $a^{2}+b^{2}$ ?

$25$

$5$

$10$

$16$

$32$

Explanation

$a$ and $b$ are the right-angle sides of a triangle, and the radius of the circle is the hypotenuse of the triangle. From the Pythagorean Theorem we would know that $a^{2}+b^{2}=r^{2}=5^{2}=25$ .

Find the $y\textup{-intercept}$ for the following equation:

Explanation

To find the $y\textup{-intercept}$ , you must put the equation into slope intercept form:

where is the intercept.

Thus,

$y=\frac{3}2{x-3}$

Therefore, your $y\textup{-intercept}$ is

A line segment has its midpoint at and an endpoint at . What are the coordinates of the other endpoint?

Explanation

Because we are given the midpoint and one of the endpoints, we know the x coordinate of the other endpoint will be the same distance away from the midpoint in the x direction, and the y coordinate of the other endpoint will be the same distance away from the midpoint in the y direction. Given two endpoints of the form:

The midpoint of these two endpoints has the coordinates:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in values for the given midpoint and one of the endpoints, which we can see is because it lies to the right of the midpoint, we can solve for the other endpoint as follows: