GMAT Quantitative - GMAT Quantitative Reasoning

The arc $\widehat{AB}$ of a circle measures $90^{\circ }$ . The chord of the arc, $\overline{AB}$ , has length . Give the length of the arc $\widehat{AB}$ .

$\frac{ \pi \sqrt{2}}{4} x + \frac{ \pi \sqrt{2}}{2}$

$\frac{ \pi \sqrt{2}}{2} x + \pi \sqrt{2}$

$\frac{ \sqrt{2}}{\pi}x + \frac{ \sqrt{2}}{\pi}$

$\frac{2 \sqrt{2}}{\pi}x + \frac{4 \sqrt{2}}{\pi}$

$\pi x \sqrt{2}+ 2 \pi \sqrt{2}$

Explanation

Examine the figure below, which shows the arc and chord in question.

Circle x

If we extend the figure to depict the circle as the composite of four quarter-circles, each a $90^{\circ }$ arc, we see that is also the side of an inscribed square. A diagonal of this square, which measures $\sqrt{2}$ times this sidelength, or

$(x+2) \sqrt{2} = x\sqrt{2} + 2\sqrt{2}$ ,

is a diameter of this circle. The circumference is $\pi$ times the diameter, or

$\pi ( x\sqrt{2} + 2\sqrt{2}) = x \pi \sqrt{2} + 2 \pi \sqrt{2}$ .

Since a $90^{\circ }$ arc is one fourth of a circle, the length of arc $\widehat{AB}$ is

$\frac{1}{4}( x \pi \sqrt{2} + 2 \pi \sqrt{2})$

$= \frac{ \pi }{4} x\sqrt{2} + \frac{ \pi }{2} \sqrt{2}$

$= \frac{ \pi \sqrt{2}}{4} x + \frac{ \pi \sqrt{2}}{2}$

What is the diagonal of a rectangular prism?

Its surface area is
Its height = twice its width = thrice its length.

Together, the two statements are sufficient.

Statement 1 alone is sufficient.

Statement 2 alone is sufficient.

Either of the statements is sufficient.

Neither of the statements, separate or together, is sufficient.

Explanation

The diagonal of a rectangular prism is found via the formula $\sqrt{w^2+l^2+h^2}$

The second statement reduces this to $\sqrt{(3l)^2+l^2+(6l)^2}=\sqrt{46l^2}$

However, the actual length is unknown. Statement 1 allows the calculation of a numerical value:

$l^2=\frac{48}{54}in^2$

$l=\sqrt{\frac{8}{9}}in$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

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:

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

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:

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

Let be a positive integer.

True or false: $i^{N} = 1$

Statement 1: is a prime number.

Statement 2: is a two-digit number ending in a 7.

EITHER 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.

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

BOTH statements TOGETHER are insufficient to answer the question.

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

Explanation

If is a positive integer, then $i^{N} = 1$ if and only if is a multiple of 4.

It follows that if $i^{N} = 1$ , cannot be a prime number. Also, every multiple of 4 is even, so as an even number, cannot end in 7. Contrapositively, if Statement 1 is true and is prime, or if Statement 2 is true and if ends in 7, it follows that is not a multiple of 4, and $i^{N} e 1$ .

Sector

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

Statement 1: The circle has circumference $40 \pi$ .

Statement 2: $m \angle ACB = 30 ^{\circ }$

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.

BOTH statements TOGETHER are insufficient 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.

Explanation

To find the area of a sector of a circle, we need a way to find the area of the circle and a way to find the central angle of the sector.

Statement 1 alone gives us the circumference; this can be divided by $2 \pi$ to yield the radius, and that can be substituted for in the formula $A = \pi r^{2}$ to find the area. However, it provides no clue that might yield $m \angle AOB$ .

Statement 2 alone asserts that $m \angle ACB = 30 ^{\circ }$ . This is an inscribed angle that intercepts the arc $\widehat{AB}$ ; therefore, the arc - and the central angle that intercepts it - has twice this measure, or $60 ^{\circ }$ . Therefore, Statement 2 alone gives the central angle, but does not yield any clues about the area.

Assume both statements are true. The radius is $40 \pi \div 2 \pi = 20$ and the area is $A = \pi \cdot 20^{2 }= 400 \pi$ . The shaded sector is $\frac{60^{\circ }}{360^{\circ }}= \frac{1}{6}$ of the circle, so the area can be calculated to be $\frac{1}{6} \times 400 \pi = \frac{200 \pi}{3}$ .

$\bigtriangleup ABC$ is a right triangle with right angle $\angle B$ . Evaluate .

Statement 1: $\bigtriangleup ABC$ has area 24.

Statement 2: $\bigtriangleup ABC$ can be circumscribed by a circle with area $25 \pi$ .

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

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

BOTH statements TOGETHER are insufficient to answer the question.

Explanation

Since $\angle B$ is given as the right angle of the triangle $\bigtriangleup ABC$ , we are being asked to evaluate the length of hypotenuse $\overline{AC}$ .

Statement 1 alone gives insufficient information. We note that the area of a right triangle is half the product of the lengths of its legs, and we examine two scenarios:

Case 1:

The area is $\frac{1}{2} \cdot 6 \cdot 8 = 24$

By the Pythagorean Theorem, hypotenuse $\overline{AC}$ has length

$\sqrt{6^{2}+8^{2}} = \sqrt{36+64} = \sqrt{100}= 10$

Case 2:

The area is $\frac{1}{2} \cdot 4 \cdot 12 = 24$

By the Pythagorean Theorem, hypotenuse $\overline{AC}$ has length

$\sqrt{4^{2}+12^{2}} = \sqrt{16+144} = \sqrt{160}= \sqrt{16} \cdot \sqrt{10} = 4\sqrt{10}$

Both triangles have area 24 but the hypotenuses have different lengths.

Assume Statement 2 alone. A circle that circumscribes a right triangle has the hypotenuse of the triangle as one of its diameters, so the length of the hypotenuse is the diameter - or, twice the radius - of the circle. Since the area of the circumsctibed circle is $25 \pi$ , its radius can be determined using the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 25 \pi$

$r^{2}= 25$

The diameter - and the length of hypotenuse $\overline{AC}$ - is twice this, or 10.

Which of the following is the area of a triangle on the coordinate plane with its vertices on the points , where ?

Explanation

We can view the horizontal segment connecting , and as the base; its length wiill be . The height will be the perpendicular (vertical) distance to this segment from the opposite point , which is , the -coordinate; therefore, the area of the triangle will be half the product of these two numbers, or