Target Test Prep Sample GMAT Problems

Solution:

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.

Solution:

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.

Solution:

Question Stem Analysis:

We must determine the value of x.

Statement One Alone:

$$\Rightarrow \frac{y}{x}=\frac{4y}{x}-6y$$

Most students will see this equation and conclude that it cannot yield a unique value for x because it contains another 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.

$$\Rightarrow \frac{y}{x}=\frac{4y}{x}-6y$$

To simplify the equation, we can multiply the entire equation by x:

$$\begin{array}{l}\Rightarrow x\left(\frac{y}{x}=\frac{4y}{x}-6y\right)\\ \Rightarrow \frac{y\cancel{x}}{\cancel{x}}=\frac{4y\cancel{x}}{\cancel{x}}-6yx\\ \Rightarrow y=4y-6yx\\ \Rightarrow y=y\left(4-6x\right)\end{array}$$

Dividing both sides by y, which we can do since we are given that y ≠ 0, we have:

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

Statement one alone is sufficient. (Notice that since y ≠ 0, we can divide both sides of y = 4y – 6yx by y to obtain 1 = 4 – 6x. Had we not known that y ≠ 0, we could not have.)

Eliminate answer choices B, C, and E.

Statement Two Alone:

$\Rightarrow$ y = 2

With no information given about x, statement two alone is not sufficient.

Solution:

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

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

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.