Target Test Prep Sample GMAT Problems

Question Stem Analysis:

The units digits of powers of 7 follow the four-number 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 $\left(7^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{ of 8}}\right)\times \left(7^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{ of 8}}\right)$, and because 8 is a multiple of 4, we can further simplify our thinking to $\left(7^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{ of 4}}\right)\times \left(7^{\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\text{ of 4}}\right)$. 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.

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.

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

$\Rightarrow \frac{\text{revenue from red wine}}{\text{revenue from white wine}}=\frac{\mathrm{r}\mathrm{x}}{\mathrm{w}\mathrm{y}}$

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:

$\Rightarrow \frac{\text{revenue from red wine}}{\text{revenue from white wine}}=\frac{3\mathrm{w}\mathrm{x}}{\mathrm{w}\mathrm{y}}=\frac{3\mathrm{x}}{\mathrm{y}}$

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:

$\Rightarrow \frac{\text{revenue from red wine}}{\text{revenue from white wine}}=\frac{\mathrm{r}\mathrm{x}}{\mathrm{w}2\mathrm{x}}=\frac{\mathrm{r}}{2\mathrm{w}}$

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:

$\Rightarrow \frac{\text{revenue from red wine}}{\text{revenue from white wine}}=\frac{3\mathrm{w}\mathrm{x}}{\mathrm{w}2\mathrm{x}}=\frac{3}{2}$

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.

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°.

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 = 3² = 9

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.

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).

Now we have enough information to determine the faster and slower times:

$\begin{array}{l}\Rightarrow {\text{time}}_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\mathrm{r}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\mathrm{r}\text{ miles}}=\frac{60}{\mathrm{r}}\text{ hours}\\ \Rightarrow {\text{time}}_{\text{faster }\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\left(\mathrm{r}+2\right)\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\left(\mathrm{r}+2\right)\text{ miles}}=\frac{60}{\left(\mathrm{r}+2\right)}\text{ hours}\end{array}$

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:

$\begin{array}{l}\text{⇒ Faster Time}+\text{1 Hour}=\text{ Slower Time}\\ \text{⇒}\frac{60}{\mathrm{r}+2}+1=\frac{60}{\mathrm{r}}\\ \text{⇒}\mathrm{r}(\mathrm{r}+2)(\frac{60}{\mathrm{r}+2}+1)=\frac{60}{\mathrm{r}}\\ \text{⇒}60\mathrm{r}+\mathrm{r}(\mathrm{r}+2)=60(\mathrm{r}+2)\\ \text{⇒}60\mathrm{r}+{\mathrm{r}}^2+2\mathrm{r}=60\mathrm{r}+120\\ \text{⇒}{\mathrm{r}}^2+2\mathrm{r}-120=0\\ \text{⇒}(\mathrm{r}+12)(\mathrm{r}-10)=0\\ \text{⇒}\mathrm{r}=-12,\text{ r}=10\end{array}$

Since we cannot have a negative rate, Thomas’s actual rate was 10 mph.

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.

At this point, we’ve produced an equation in the matrix. We know that

$\begin{array}{l}\Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->40+0.50\mathrm{J}=\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->40=0.50\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->400=5\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->\mathrm{J}=80\end{array}$

Thus, 80 guests wore jackets, which means that 80 percent of the guests wore jackets.

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.

$\begin{array}{l}\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}-6\mathrm{y}\\ \Rightarrow \mathrm{x}\left(\frac{\mathrm{y}}{\mathrm{x}}=\frac{4\mathrm{y}}{\mathrm{x}}-6\mathrm{y}\right)\\ \Rightarrow \mathrm{y}=4\mathrm{y}-6\mathrm{y}\mathrm{x}\\ \Rightarrow 1=4-6\mathrm{x}\\ \Rightarrow 6\mathrm{x}=3\to \mathrm{x}=\frac{1}{2}\end{array}$

Statement one alone is sufficient. (Notice that since y $\ne \; <!--NotEqual,\; ne-->$ 0, we can divide both sides of y = 4y – 6yx by y to obtain 1 = 4 – 6y. Had we not known that y $\ne \; <!--NotEqual,\; ne-->$ 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.

Question Stem Analysis:

We must determine the average grade of the left-handed students. The average grade of all the students can be expressed as

$\Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->\mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{Grade = <!--nbsp, NonBreakingSpace-->}\frac{\left(\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}\times \text{ <!--nbsp, NonBreakingSpace-->}\mathrm{a}\mathrm{v}\mathrm{g}\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\text{ of right <!--nbsp, NonBreakingSpace-->}\right)\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\left(\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}\times \text{ <!--nbsp, NonBreakingSpace-->}\mathrm{a}\mathrm{v}\mathrm{g}\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\text{ of left <!--nbsp, NonBreakingSpace-->}\right)}{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$

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 $\frac{\mathrm{x}}{5}$ right-handed students in the class; together, they receive an average score of 95.

Since there are $\frac{\mathrm{x}}{5}$ right-handed students in the class, there are $\mathrm{x}-\frac{\mathrm{x}}{5}=\frac{5\mathrm{x}}{5}-\frac{\mathrm{x}}{5}=\frac{4\mathrm{x}}{5}$ left-handed students.

We can insert these values into the equation:

$\begin{array}{l}\Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->\mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{ Grade = <!--nbsp, NonBreakingSpace-->}\frac{\left(\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}\times \text{ <!--nbsp, NonBreakingSpace-->}\mathrm{a}\mathrm{v}\mathrm{g}\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\right)\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\left(\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}\times \text{ <!--nbsp, NonBreakingSpace-->}\mathrm{a}\mathrm{v}\mathrm{g}\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{e}\right)}{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->\mathrm{90 \; <!--nbsp,\; NonBreakingSpace-->}\text{= <!--nbsp, NonBreakingSpace-->}\frac{\left(\frac{\mathrm{x}}{5}\text{ <!--nbsp, NonBreakingSpace-->}\times \mathrm{ 95\; <!--nbsp,\; NonBreakingSpace-->}\right)\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\left(\frac{4\mathrm{x}}{5}\text{ <!--nbsp, NonBreakingSpace-->}\times \text{ <!--nbsp, NonBreakingSpace-->}\mathrm{L}\right)}{\mathrm{x}}\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->5[\; <!--lsqb,\; lbrack-->90\mathrm{x}\text{ = <!--nbsp, NonBreakingSpace-->}\left(\frac{\mathrm{x}}{5}\times 95\right)\text{ <!--nbsp, NonBreakingSpace-->}+\text{ <!--nbsp, NonBreakingSpace-->}\left(\frac{4\mathrm{x}}{5}\times \mathrm{L}\right)]\; <!--rbrack,\; rsqb-->\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->450\mathrm{x}=95\mathrm{x}+4\mathrm{x}\mathrm{L}\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->450=95+4\mathrm{L}\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->355=4\mathrm{L}\\ \Rightarrow \; <!--Rightarrow,\; DoubleRightArrow,\; rArr,\; Implies-->\mathrm{L}=88.75\end{array}$

Statement two alone is sufficient.