Target Test Prep

Example 1

hard

If x and y are positive integers, what is the units digit of 7xy+y $7^{x y + y}$ ?

1) x16=n $\frac{x}{16} = n$ , where n is a positive integer.

2) y8=m $\frac{y}{8} = m$ , where m is a positive integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

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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 7xy+y $7^{x y + y}$ . It may be helpful to simplify 7xy+y $7^{x y + y}$ to (7xy)×(7y) $(7^{x y}) \times (7^{y})$ .

Statement One Alone:

⇒ x16=n $\frac{x}{16} = n$ , 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:

⇒ y8=m $\frac{y}{8} = m$ , 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 (7multiple of 8)×(7multiple of 8) $(7^{m u l t i p l e of 8}) \times (7^{m u l t i p l e of 8})$ , and because 8 is a multiple of 4, we can further simplify our thinking to (7multiple of 4)×(7multiple of 4) $(7^{m u l t i p l e of 4}) \times (7^{m u l t i p l e of 4})$ . We know that all powers of 7 that are multiples of 4 have a units digit of 1. Thus, the units digit of x=t2 $x = t^{2}$ is 1 × 1 = 1. Statement two alone is sufficient.

Correct answer:B

Example 2

medium

In a certain 120-person 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?

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Solution:

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.

Correct answer:D

Example 3

medium

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.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

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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

⇒revenue from red winerevenue from white wine=rxwy $\Rightarrow \frac{revenue from red wine}{revenue from white wine} = \frac{r x}{w y}$

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:

⇒revenue from red winerevenue from white wine=3wxwy=3xy $\Rightarrow \frac{revenue from red wine}{revenue from white wine} = \frac{3 w x}{w y} = \frac{3 x}{y}$

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:

⇒revenue from red winerevenue from white wine=rxw2x=r2w $\Rightarrow \frac{revenue from red wine}{revenue from white wine} = \frac{r x}{w 2 x} = \frac{r}{2 w}$

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:

⇒revenue from red winerevenue from white wine=3wxw2x=32 $\Rightarrow \frac{revenue from red wine}{revenue from white wine} = \frac{3 w x}{w 2 x} = \frac{3}{2}$

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.

Correct answer:C

Example 4

medium

In the regular hexagon above, what is the value of a + b + c + d + e + f + g + h + i + j + k + l?

180°

360°

540°

720°

1,040°

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Solution:

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°.

Correct answer:D

Example 5

hard

If 5x−5x−1=500 $5^{x} - 5^{x - 1} = 500$ , what is the value of (x - 1)²?

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Solution:

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.

⇒5x−5x−1=500⇒5x−(5)x(5)−1=500 $\begin{array}{l} \Rightarrow 5^{x} - 5^{x - 1} = 500 \\ \Rightarrow 5^{x} - {(5)}^{x} {(5)}^{- 1} = 500 \end{array}$

Now we can easily see to factor out the common term of 5^x. We now have:

⇒5x−(5)x(5)−1=500⇒5x(1−5−1)=500⇒5x(1−15)=500⇒5x×45=500 $\begin{array}{l} \Rightarrow 5^{x} - {(5)}^{x} {(5)}^{- 1} = 500 \\ \Rightarrow 5^{x} (1 - 5^{- 1}) = 500 \\ \Rightarrow 5^{x} (1 - \frac{1}{5}) = 500 \\ \Rightarrow 5^{x} \times \frac{4}{5} = 500 \end{array}$

Our next step is to break down all the values into prime factors. This will make canceling out much easier.

⇒5x×45=500→5x×225=53×22⇒5x=53×22×522⇒5x=54→x=4 $\begin{array}{l} \Rightarrow 5^{x} \times \frac{4}{5} = 500 \to 5^{x} \times \frac{2^{2}}{5} = 5^{3} \times 2^{2} \\ \Rightarrow 5^{x} = \frac{5^{3} \times 2^{2} \times 5}{2^{2}} \\ \Rightarrow 5^{x} = 5^{4} \to x = 4 \end{array}$

Finally, we have to solve for (x – 1)², so (4 – 1)² = 3² = 9

Correct answer:C

Example 6

hard

How many digits are in the number 50⁸ × 8³ × 11²?

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Solution:

The first step is to prime factorize the number 508×83×112 $50^{8} \times 8^{3} \times 11^{2}$ . This becomes:

⇒508×83×112⇒(5×5×2)8×(29)×112⇒516×28×29×112⇒516×217×112 $\begin{array}{l} \Rightarrow 50^{8} \times 8^{3} \times 11^{2} \\ \Rightarrow {(5 \times 5 \times 2)}^{8} \times (2^{9}) \times 11^{2} \\ \Rightarrow 5^{16} \times 2^{8} \times 2^{9} \times 11^{2} \\ \Rightarrow 5^{16} \times 2^{17} \times 11^{2} \end{array}$

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.

Correct answer:D

Example 7

hard

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?

8 mph

10 mph

12 mph

15 mph

18 mph

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Solution:

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	rmileshour $r \frac{m i l e s}{h o u r}$		60 miles
Faster Speed	(r+2)mileshour $(r + 2) \frac{m i l e s}{h o u r}$		60 miles

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

⇒timeactual speed=distancerate=60 milesr mileshour=60 miles1×1 hourr miles=60r hours⇒timefaster speed=distancerate=60 miles(r+2) mileshour=60 miles1×1 hour(r+2) miles=60(r+2) hours $\begin{array}{l} \Rightarrow {time}_{a c t u a l s p e e d} = \frac{distance}{r a t e} = \frac{60 miles}{r \frac{m i l e s}{h o u r}} = \frac{60 miles}{1} \times \frac{1 h o u r}{r miles} = \frac{60}{r} hours \\ \Rightarrow {time}_{faster s p e e d} = \frac{distance}{r a t e} = \frac{60 miles}{(r + 2) \frac{m i l e s}{h o u r}} = \frac{60 miles}{1} \times \frac{1 h o u r}{(r + 2) miles} = \frac{60}{(r + 2)} hours \end{array}$

	Rate	Time	Distance
Actual Speed	r mileshour $r \frac{m i l e s}{h o u r}$	60rhours $\frac{60}{r} h o u r s$	60 miles
Faster Speed	(r+2) mileshour $(r + 2) \frac{m i l e s}{h o u r}$	60(r+2)hours $\frac{60}{(r + 2)} h o u r s$	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:

Faster Time + 1 Hour = Slower Time ⇒60r+2+1=60r $\begin{array}{l} Faster Time +1 Hour = Slower Time \\ \Rightarrow \frac{60}{r + 2} + 1 = \frac{60}{r} \end{array}$

To cancel out the denominators of r + 2 and r, we can multiply the entire equation by r(r + 2), which is the Least Common Multiple of r + 2 and r:

⇒r(r+2)×(60r+2+1=60r)⇒60r+r(r+2)=60(r+2)⇒60r+r2+2r=60r+120⇒r2+2r−120=0⇒(r+12)(r−10)=0⇒r=−12,r=10 $\begin{array}{l} \Rightarrow r (r + 2) \times (\frac{60}{r + 2} + 1 = \frac{60}{r}) \\ \Rightarrow 60 r + r (r + 2) = 60 (r + 2) \\ \Rightarrow 60 r + r^{2} + 2 r = 60 r + 120 \\ \Rightarrow r^{2} + 2 r - 120 = 0 \\ \Rightarrow (r + 12) (r - 10) = 0 \\ \Rightarrow r = - 12, r = 10 \end{array}$

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

(Note: If you thought that factoring this quadratic was difficult, keep in mind that once you have r² + 2r - 120 = 0, you can look at the answer choices to help you determine r. Notice that 10 and 12 are 2 units apart and also multiply to 120. Thus, we can quickly see that r² + 2r - 120 = 0= 0 will factor down to (r + 12)(r - 10). This is a useful method to use anytime you have a tricky quadratic in a problem solving question.)

Correct answer:B

Example 8

hard

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?

20 percent

40 percent

60 percent

70 percent

80 percent

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Solution:

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

⇒40+0.50J=J⇒40=0.50J⇒400=5J⇒J=80 $\begin{array}{l} \Rightarrow 40 + 0.50 J = J \\ \Rightarrow 40 = 0.50 J \\ \Rightarrow 400 = 5 J \\ \Rightarrow J = 80 \end{array}$

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

Correct answer:E

Example 9

medium

If y ≠ $\neq$ 0, what is the value of x?

1) yx=4yx−6y $\frac{y}{x} = \frac{4 y}{x} - 6 y$

2) y = 2

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

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Solution:

Question Stem Analysis:

We must determine the value of x.

Statement One Alone:

\Rightarrow y x = 4 y x - 6 y

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

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 y x = 4 y x - 6 y

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

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

\Rightarrow x (y x = 4 y x - 6 y) \Rightarrow y x x = 4 y x x - 6 y x \Rightarrow y = 4 y - 6 y x \Rightarrow y = y (4 - 6 x)

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

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

\Rightarrow y y = y ( 4 - 6 x ) y \Rightarrow 1 = 4 - 6 x \Rightarrow 6 x = 3 \to x = 1 2

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

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.

Correct answer:A

Example 10

medium

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 left-handed students?

1) There are a total of 20 students in the class.

2) There are x5 $\frac{x}{5}$ right-handed students in the class; together, they receive an average score of 95.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

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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

⇒Average Grade = (right × avg grade of right) + (left × avg grade of left)left + right $\Rightarrow Average Grade = \frac{(r i g h t \times a v g g r a d e of right) + (l e f t \times a v g g r a d e of left)}{l e f t + r i g h t}$

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 x5 $\frac{x}{5}$ right-handed students in the class; together, they receive an average score of 95.

Since there are x5 $\frac{x}{5}$ right-handed students in the class, there are x−x5=5x5−x5=4x5 $x - \frac{x}{5} = \frac{5 x}{5} - \frac{x}{5} = \frac{4 x}{5}$ left-handed students.

We can insert these values into the equation:

⇒Average Grade = (right × avg grade) + (left × avg grade)left + right⇒90 = (x5 × 95) + (4x5 × L)x⇒5[90x = (x5×95) + (4x5×L)]⇒450x=95x+4xL⇒450=95+4L⇒355=4L⇒L=88.75 $\begin{array}{l} \Rightarrow A v e r a g e Grade = \frac{(r i g h t \times a v g g r a d e) + (l e f t \times a v g g r a d e)}{l e f t + r i g h t} \\ \Rightarrow 90 = \frac{(\frac{x}{5} \times 95) + (\frac{4 x}{5} \times L)}{x} \\ \Rightarrow 5 [90 x = (\frac{x}{5} \times 95) + (\frac{4 x}{5} \times L)] \\ \Rightarrow 450 x = 95 x + 4 x L \\ \Rightarrow 450 = 95 + 4 L \\ \Rightarrow 355 = 4 L \\ \Rightarrow L = 88.75 \end{array}$

Statement two alone is sufficient.

Correct answer:B

Target Test Prep Sample GMAT Problems

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10