Problem-Solving Questions

Help Questions

GMAT Quantitative › Problem-Solving Questions

Questions 1 - 10

Sheryl is competing in an archery tournament. She gets to shoot three arrows at a target, and her best one counts.

Sheryl hits the bullseye 42% of the time. What is the probability (two decimal places) that she will hit the bullseye at least once in her three tries?

Explanation

This is most easily solved by finding the probability that she will not hit the bullseye at all in her three tries. If she hits 42% of the time, she misses 58% of the time, and the probability she misses three times will be

$0.58 ^{3} = 0.195112 \approx 0.20$ .

The probability of hitting the bullseye at least once in three tries is the complement of this, or .

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 the figure with vertices ?

Explanation

This figure can be seen as a composite of two simple shapes: the rectangle with vertices , and the triangle with vertices .

The rectangle has length and height , so its area is the product of these dimensions, or $9 \times 10 = 90$ .

The triangle has as its base the length of the horizontal segment connecting and , which is ; its height is the vertical distance from the other vertex to this segment, which is . The area of this triangle is half the product of the base and the height, which is $\frac{1}{2} \times 10 \times 2 = 10$ .

Add the areas of the rectangle and the triangle to get the total area:

Export-png

Triangle is a right triangle with . What is the length of its height ?

$\frac{12}{5}$

$\frac{5}{2}$

$\frac{36}{20}$

$\frac{52}{20}$

Explanation

The height AE, divides the triangle ABC, in two triangles AEC and AEB with same proportions as the original triangle ABC, this property holds true for any right triangle.

In other words, $\frac{AE}{AB}=\frac{AC}{BC}$ .

Therefore, we can calculate, the length of AE:

$AE= \frac{AC\cdot AB}{BC}= \frac{12}{5}$ .

What is the mean of the following data set in terms of and ?

$\left \{a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right \}$

$\frac{3}{2} a +\frac{1}{2} x$

$\frac{3}{2} a -\frac{1}{2} x$

$\frac{3}{2} a$

Explanation

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

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

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

What is the area of the quadrilateral on the coordinate plane with vertices ?

Explanation

The quadrilateral is a parallelogram with two vertical bases, each with length . Its height is the distance between the bases, which is the difference of the -coordinates: . The area of the parallelogram is the product of its base and its height: