Target Test Prep Sample GMAT Problems
Below is a mixture of sample GMAT problem solving and data sufficiency questions. Although challenging, these questions cover a wide range of topics that you could see on your GMAT, ranging from Number Properties to Statistics to Geometry.
These ten questions were handpicked from the Target Test Prep question bank, which includes over 2,400 quantitative practice GMAT questions. So don’t be shy; give it a shot and see how you do!
Example 1
If x and y are positive integers, what is the units digit of ${7}^{\mathrm{x}\mathrm{y}+\mathrm{y}}$?
1) $\frac{\mathrm{x}}{16}=\mathrm{n}$, where n is a positive integer.
2) $\frac{\mathrm{y}}{8}=\mathrm{m}$, where m is a positive integer.
Question Stem Analysis:
The units digits of powers of 7 follow the fournumber pattern 7–9–3–1. We need to determine the units digit of ${7}^{\mathrm{x}\mathrm{y}+\mathrm{y}}$. It may be helpful to simplify ${7}^{\mathrm{x}\mathrm{y}+\mathrm{y}}$ to $\left({7}^{\mathrm{x}\mathrm{y}}\right)\times \left({7}^{\mathrm{y}}\right)$.
Statement One Alone:
⇒ $\frac{\mathrm{x}}{16}=\mathrm{n}$, where n is a positive integer.
We can restate this as x = 16n. Because n is an integer, it must be true that x is a multiple of 16. This means that xy is a multiple of 16 because the product of 16 and any positive integer is a multiple of 16. Because 16 is a multiple of 4, we also know that xy is a multiple of 4. However, because we do not know anything about y, statement one alone is not sufficient.
Eliminate answer choices A and D.
Statement Two Alone:
⇒ $\frac{\mathrm{y}}{8}=\mathrm{m}$, where m is a positive integer.
We can restate this as y = 8m. Because m is an integer, it must be true that y is a multiple of 8. This also means that xy is a multiple of 8. We can now think about this as $\left({7}^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{of8}}\right)\times \left({7}^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{of8}}\right)$, and because 8 is a multiple of 4, we can further simplify our thinking to $\left({7}^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{of4}}\right)\times \left({7}^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{of4}}\right)$. We know that all powers of 7 that are multiples of 4 have a units digit of 1. Thus, the units digit of $\mathrm{x}={\mathrm{t}}^{2}$ is 1 × 1 = 1. Statement two alone is sufficient.
Example 2
In a certain 120person orchestra, each musician plays one or more of the following musical instruments: the piano, the violin, or the tuba. A total of 50 musicians play the violin, 70 musicians play the piano, and 60 musicians play the tuba. If 30 musicians play exactly two of the instruments, how many musicians play exactly all three of the instruments?
We’re looking for the number of musicians who play exactly three instruments.
$\Rightarrow $ Total # of Unique Elements = # in (Group A) + # in (Group B) + # in (Group C) – # in (Groups of Exactly Two) – 2[#in (Group of Exactly Three)] + # in (Neither)
$\Rightarrow $ Let T = # in (Group of Exactly Three)
$\Rightarrow $ 120 = 50 + 70 + 60 – 30 – 2(T) + 0
$\Rightarrow $ 120 = 150 – 2T
$\Rightarrow $ 2T = 30
$\Rightarrow $ T= 15
Thus, 15 people play exactly all three instruments. Notice that since each musician must play one or more of the three instruments, the number of people in the Neither region is zero.
Example 3
Last Sunday, the Stewart Vineyard ran a price promotion in which it sold red wine for r dollars per case and white wine for w dollars per case. What was the ratio of the revenue from red wine to the revenue from white wine?
1) Last Sunday, the Stewart Vineyard sold each case of red wine for 3 times as much as each case of white wine.
2) Last Sunday, the Stewart Vineyard sold 2 times as many cases of white wine as it sold of red wine.
Question Stem Analysis:
If we let x = the number of cases of red wine sold, the revenue from red wine can be expressed as rx. Similarly, if we let y = the number of cases of white wine sold, the revenue from white wine can be expressed as wy. Thus, the ratio we’re looking for is
$\Rightarrow \frac{\text{revenuefromredwine}}{\text{revenuefromwhitewine}}=\frac{\mathrm{r}\mathrm{x}}{\mathrm{w}\mathrm{y}}$
Remember that we’ll need to produce a constant ratio in order to produce a meaningful ratio.
Statement One Alone:
$\Rightarrow $ Last Sunday, the Stewart Vineyard sold each case of red wine for 3 times as much as each case of white wine.
This means that r = 3w, which we can substitute for r in our ratio:
$\Rightarrow \frac{\text{revenuefromredwine}}{\text{revenuefromwhitewine}}=\frac{3\mathrm{w}\mathrm{x}}{\mathrm{w}\mathrm{y}}=\frac{3\mathrm{x}}{\mathrm{y}}$
This is not sufficient because there are still two different variables in the fraction. Different values of x and y would change the value of the fraction. Statement one alone is not sufficient.
Eliminate answer choices A and D.
Statement Two Alone:
$\Rightarrow $ Last Sunday, the Stewart Vineyard sold 2 times as many cases of white wine as it sold of red wine.
This means that y = 2x, which we can substitute for y in our ratio:
$\Rightarrow \frac{\text{revenuefromredwine}}{\text{revenuefromwhitewine}}=\frac{\mathrm{r}\mathrm{x}}{\mathrm{w}2\mathrm{x}}=\frac{\mathrm{r}}{2\mathrm{w}}$
This ratio is not sufficient because there are two different variables in the fraction. Different values of r and w would change the value of the fraction. Statement two alone is not sufficient.
Eliminate answer choice B.
Statements One and Two Together:
Substituting the information from both statements for variables in our original ratio, we can produce the following fraction:
$\Rightarrow \frac{\text{revenuefromredwine}}{\text{revenuefromwhitewine}}=\frac{3\mathrm{w}\mathrm{x}}{\mathrm{w}2\mathrm{x}}=\frac{3}{2}$
Thus, the ratio of revenue from red wine to revenue from white wine is 3:2. Statements one and two together are sufficient to answer the question.
Example 4
In the regular hexagon above, what is the value of a + b + c + d + e + f + g + h + i + j + k + l?
The sum of the exterior angles of any polygon is 360°. However, this statement is only true when we take only one exterior angle per vertex and add up only the measures of each of those angles. Here there are actually two exterior angles per vertex, a situation that will yield two sets of exterior angles. We can consider angles a, c, e, g, i, k as one set and angles b, d, f, h, j, and l as the other. Since each of these two sets of exterior angles adds up to 360°, the two sets together give us a total of 2 × 360° = 720°.
Example 5
If ${5}^{\mathrm{x}}{5}^{\mathrm{x}1}=500$, what is the value of (x  1)^{2}?
We should recognize that we have subtraction of bases with exponents. This means before we can combine the equation’s terms, we need to factor out common factors. However, to help see what we can factor out, we can rewrite the equation.
$\begin{array}{l}\Rightarrow {5}^{\mathrm{x}}{5}^{\mathrm{x}1}=500\\ \Rightarrow {5}^{\mathrm{x}}{\left(5\right)}^{\mathrm{x}}{\left(5\right)}^{1}=500\end{array}$
Now we can easily see to factor out the common term of 5^{x}. We now have:
$\begin{array}{l}\Rightarrow {5}^{\mathrm{x}}{\left(5\right)}^{\mathrm{x}}{\left(5\right)}^{1}=500\\ \Rightarrow {5}^{\mathrm{x}}\left(1{5}^{1}\right)=500\\ \Rightarrow {5}^{\mathrm{x}}\left(1\frac{1}{5}\right)=500\\ \Rightarrow {5}^{\mathrm{x}}\times \frac{4}{5}=500\end{array}$
Our next step is to break down all the values into prime factors. This will make canceling out much easier.
$\begin{array}{l}\Rightarrow {5}^{\mathrm{x}}\times \frac{4}{5}=500\to {5}^{\mathrm{x}}\times \frac{{2}^{2}}{5}={5}^{3}\times {2}^{2}\\ \Rightarrow {5}^{\mathrm{x}}=\frac{{5}^{3}\times {2}^{2}\times 5}{{2}^{2}}\\ \Rightarrow {5}^{\mathrm{x}}={5}^{4}\to \mathrm{x}=4\end{array}$
Finally, we have to solve for (x – 1)^{2}, so (4 – 1)^{2} = 3^{2} = 9
Example 6
How many digits are in the number 50^{8} × 8^{3} × 11^{2}?
The first step is to prime factorize the number ${50}^{8}\times {8}^{3}\times {11}^{2}$. This becomes:
$\begin{array}{l}\Rightarrow {50}^{8}\times {8}^{3}\times {11}^{2}\\ \Rightarrow {\left(5\times 5\times 2\right)}^{8}\times \left({2}^{9}\right)\times {11}^{2}\\ \Rightarrow {5}^{16}\times {2}^{8}\times {2}^{9}\times {11}^{2}\\ \Rightarrow {5}^{16}\times {2}^{17}\times {11}^{2}\end{array}$
The sixteen (5 × 2) pairs contribute a total of sixteen trailing zeros to the number. One 2 and two 11’s remain. The product of (2 × 11 × 11) = 242, which is 3 digits. Thus, the number has (16 + 3) = 19 total digits.
Example 7
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?
This is an if/then rate question. We’re given a hypothetical “if” scenario, and we need to use this to determine the actual scenario. We are told that the distance traveled was 60 miles. Most importantly, we are told that the hypothetical speed was 2 mph faster than the actual speed. Since we do not have any values for the actual speed, we can express the actual speed as r and the faster speed as (r+2).
Rate 
Time 
Distance 

Actual Speed 
$\mathrm{r}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}$ 
60 miles 

Faster Speed 
$\left(\mathrm{r}+2\right)\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}$ 
60 miles 
Now we have enough information to determine the faster and slower times:
$\begin{array}{l}\Rightarrow {\text{time}}_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\text{}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{miles}}{\mathrm{r}\text{}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{miles}}{1}\times \frac{\text{1}\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\mathrm{r}\text{miles}}=\frac{60}{\mathrm{r}}\text{hours}\\ \Rightarrow {\text{time}}_{\text{faster}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{miles}}{\left(\mathrm{r}+2\right)\text{}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{miles}}{1}\times \frac{\text{1}\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\left(\mathrm{r}+2\right)\text{miles}}=\frac{60}{\left(\mathrm{r}+2\right)}\text{hours}\end{array}$
Rate 
Time 
Distance 

Actual Speed 
$\mathrm{r}\text{}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}$ 
$\frac{60}{\mathrm{r}}\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}$ 
60 miles 
Faster Speed 
$\left(\mathrm{r}+2\right)\text{}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}$ 
$\frac{60}{\left(\mathrm{r}+2\right)}\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}$ 
60 miles 
We can now set up an equation using the travel time. We know that if Thomas had ridden at the faster speed, he would have arrived 1 hour earlier. Hence:
$\begin{array}{l}\text{\u21d2FasterTime}+\text{1Hour}=\text{SlowerTime}\\ \text{\u21d2}\frac{60}{\mathrm{r}+2}+1=\frac{60}{\mathrm{r}}\\ \text{\u21d2}\mathrm{r}(\mathrm{r}+2)(\frac{60}{\mathrm{r}+2}+1)=\frac{60}{\mathrm{r}}\\ \text{\u21d2}60\mathrm{r}+\mathrm{r}(\mathrm{r}+2)=60(\mathrm{r}+2)\\ \text{\u21d2}60\mathrm{r}+{\mathrm{r}}^{2}+2\mathrm{r}=60\mathrm{r}+120\\ \text{\u21d2}{\text{}\mathrm{r}}^{2}+2\mathrm{r}120=0\\ \text{\u21d2}(\mathrm{r}+12)(\mathrm{r}10)=0\\ \text{\u21d2}\mathrm{r}=12,\text{r}=10\end{array}$
Since we cannot have a negative rate, Thomas’s actual rate was 10 mph.
Example 8
At a dinner party, 40 percent of the guests wore both jackets and ties. If 50 percent of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?
We can assume that there are 100 guests at the dinner party since no specific number of guests is defined. This assumption results in having 40 people wear both jackets and ties. The second sentence takes some care. Since we don’t know the number of guests who wore jackets, we can let the variable J represent that number. This means that the guests who did wear a jacket but not a tie can be represented as 0.50J.
Jacket 
No Jacket 
Total 

Tie 
40 

No Tie 
0.50J 

Total 
J 
100 
At this point, we’ve produced an equation in the matrix. We know that
$\begin{array}{l}\Rightarrow 40+0.50\mathrm{J}=\mathrm{J}\\ \Rightarrow 40=0.50\mathrm{J}\\ \Rightarrow 400=5\mathrm{J}\\ \Rightarrow \mathrm{J}=80\end{array}$
Thus, 80 guests wore jackets, which means that 80 percent of the guests wore jackets.
Example 9
If y $\ne $ 0, what is the value of x?
1) $\frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}6\mathrm{y}$
2) y = 2
Question Stem Analysis:
We must determine the value of x.
Statement One Alone:
$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}6\mathrm{y}$
Most students will see this equation and conclude that it cannot be used to determine a unique value for x because it contains another unknown variable, y. However, if y can be removed from the equation, x will be the only unknown. Indeed, this one equation is sufficient to determine the value of x.
$\begin{array}{l}\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}6\mathrm{y}\\ \Rightarrow \mathrm{x}\left(\frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}6\mathrm{y}\right)\\ \Rightarrow \mathrm{y}=4\mathrm{y}6\mathrm{y}\mathrm{x}\\ \Rightarrow 1=46\mathrm{x}\\ \Rightarrow 6\mathrm{x}=3\to \mathrm{x}=\frac{1}{2}\end{array}$
Statement one alone is sufficient. (Notice that since y $\ne $ 0, we can divide both sides of y = 4y – 6yx by y to obtain 1 = 4 – 6y. Had we not known that y $\ne $ 0, we could not have.)
Eliminate answer choices B, C, and D.
Statement Two Alone:
$\Rightarrow $ y = 2
With no information given about x, statement two alone is not sufficient.
Example 10
If the average (arithmetic mean) grade of x students in a particular class is 90, what is the average (arithmetic mean) grade received by the lefthanded students?
1) There are a total of 20 students in the class.
2) There are $\frac{\mathrm{x}}{5}$ righthanded students in the class; together, they receive an average score of 95.
Question Stem Analysis:
We must determine the average grade of the lefthanded students. The average grade of all the students can be expressed as
$\Rightarrow \mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{Grade=}\frac{\left(\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\text{}\times \text{}\mathrm{a}\mathrm{v}\mathrm{g}\text{}\text{}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\text{of right}\right)\text{}+\text{}\left(\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{}\times \text{}\mathrm{a}\mathrm{v}\mathrm{g}\text{}\text{}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\text{of left}\right)}{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{}+\text{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$
Statement One Alone:
$\Rightarrow $ There are a total of 20 students in the class.
Merely knowing the total number of students doesn’t allow us to weight either group of students. Statement one alone is not sufficient.
Eliminate answer choices A and D.
Statement Two Alone:
$\Rightarrow $ There are $\frac{\mathrm{x}}{5}$ righthanded students in the class; together, they receive an average score of 95.
Since there are $\frac{\mathrm{x}}{5}$ righthanded students in the class, there are $\mathrm{x}\frac{\mathrm{x}}{5}=\frac{5\mathrm{x}}{5}\frac{\mathrm{x}}{5}=\frac{4\mathrm{x}}{5}$ lefthanded students.
We can insert these values into the equation:
$\begin{array}{l}\Rightarrow \mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{Grade=}\frac{\left(\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\text{}\times \text{}\mathrm{a}\mathrm{v}\mathrm{g}\text{}\text{}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\right)\text{}+\text{}\left(\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{}\times \text{}\mathrm{a}\mathrm{v}\mathrm{g}\text{}\text{}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\right)}{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{}+\text{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}\\ \Rightarrow 90\text{=}\frac{\left(\frac{\mathrm{x}}{5}\text{}\times 95\right)\text{}+\text{}\left(\frac{4\mathrm{x}}{5}\text{}\times \text{}\mathrm{L}\right)}{\mathrm{x}}\\ \Rightarrow 5\left[90\mathrm{x}\text{=}\left(\frac{\mathrm{x}}{5}\times 95\right)\text{}+\text{}\left(\frac{4\mathrm{x}}{5}\times \mathrm{L}\right)\right]\\ \Rightarrow 450\mathrm{x}=95\mathrm{x}+4\mathrm{x}\mathrm{L}\\ \Rightarrow 450=95+4\mathrm{L}\\ \Rightarrow 355=4\mathrm{L}\\ \Rightarrow \mathrm{L}=88.75\end{array}$
Statement two alone is sufficient.