The Lirr River runs from Rosedale in the west to Oceanside in the east with the current moving at an average of 10 miles per hour. Sasha is traveling by motorboat from Oceanside to Rosedale and back. If the water were not moving, Sasha’s motorboat would travel at an average speed of 20 miles per hour. Given the current, what was Sasha’s actual average speed, in miles per hour, for the round trip?

Solution

Answer. 15

To answer this question, Plug In. Here, the unknown is the distance traveled; pick a value that works well with the numbers in this question, such as 60 miles. When Sasha is traveling from Oceanside to Rosedale, the current adds 10 miles per hour to her boat’s “still water” speed, and thus she travels at 30 miles per hour.Therefore, it takes her 60 ÷ 30 = 2 hours. When Sasha is traveling back, she is moving “against” the current, so her speed is 10 miles per hour less, and she thus travels at 10 miles per hour.Therefore, it takes her 60 ÷ 10 = 6 hours. Altogether, she travels \(\frac{60miles+60miles}{2hours+6hours}=\frac{120miles}{8hours}\) = 15 miles per hour.

There are x red marbles, y blue marbles, and z yellow marbles in a sack. Three marbles are chosen at random, without replacements. If x, y, and z are all at least equal to 3, then which of the following must be true?
Indicate all possible values.

The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+z}\cdot \frac{x-1}{x+y+z-1}\cdot \frac{x-2}{x+y+z-2}\)

The probability of drawing at least 1 red marble is
\(1-\left (\frac{y+z}{x+y+z}\cdot \frac{y+z-1}{x+y+z-1}\cdot \frac{y+z-2}{x+y+z-2} \right )\)

The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z}\cdot \frac{y-x}{x+y+z-1}\cdot \frac{y-z}{x+y+z-2}\)

Solution

A and B

As soon as you see variables in the answer choices, set up your scratch paper to Plug In. Start with x = 6, y = 5, and z = 4. The probability of drawing exactly 3 red marbles is \(\frac{6}{15}\times \frac{5}{14}\times \frac{4}{13}=\frac{1\times 1\times 4}{1\times 7\times 13}=\frac{4}{91}\)
. Plugging the values into choice (A) gives an answer that matches our target number, so choice (A) works.The probability of drawing at least 1 red marble is best calculated bySnding the probability of drawing no red marbles and subtracting that result from 1: 1 - \(\left ( \frac{9}{15}\times \frac{8}{14}\times \frac{7}{13} \right )=1-\left ( \frac{3}{5}\times \frac{4}{1}\times \frac{1}{13}\right )=\frac{65}{65}-\frac{12}{65}=\frac{53}{65}\).Plugging the values into choice (B) gives an answer that matches our target number, so choice (B) works. The probability of drawing exactly 3 blue marbles is:\(\frac{5}{15}\times \frac{4}{14}\times \frac{3}{13}=\frac{1\times 2\times 1}{1\times 7\times 13}=\frac{2}{91}\). Plugging the values into choice (C) gives \(\frac{5}{15}\times \frac{5-4}{14}\times \frac{6+4}{13}=\frac{1\times 1\times 5}{3\times 7\times 13}=\frac{5}{273}\), which does not match our target.Choices (A) and (B) are correct.

x and y are both integers

Quantity A	Quantity B
xy	6

Quantity A is greater.
Quantity B is greater
The two quantities are equal
The relationship cannot be determined from the information give.

Solution

As soon as you see variables and Quant Comp, make your set-up. Plugging In is tricky here until you realize that one number must be negative.The 6 in Quantity B is also a clue.Try x = 3 and y = –2. y to the x power will be –8 and x to the y power will be ¹⁄₉, so these numbers work. xy = –6 so cross off choices (A) and (C). No matter what you plug in, and there are very few options, y must be negative and x must be odd.Therefore xy will always be negative and 6 will always be greater.The answer is choice (B).

If A = q − r,B = r − s, and C = q − s, what is the value of A − (B − C)?

−r
1
q + r
2(q − r)

Solution

It’s very easy to make a careless error on this one. Make sure to Plug In and write your work down in an organized manner.Try q = 10, r = 5, and s = 2. So, A = 10 − 5 = 5,B = 5 − 2 = 3, and C = 10 − 2 = 8. So, A − (B − C) = 5 − (3 − 8) = 5 − (−5) = 10. Only choice (E) yields 10 when you plug in 10 for q and 5 for r.

If ¹⁄_y < x < 0, then which one of the following must be true?

1 < x²
x² < x
−1 < x³ < 0
¹⁄_x > -1
x³ < x

Solution

Choose a value for x that satisfies the conditions of the question: x = –¹⁄₂, for example. Substituting this value into the answer choices, you see that all of the choices are false, except for choice (C).

Quantity A	Quantity B
(x + y)² − 2xy	x² + y²

Quantity A is greater.
Quantity B is greater
The two quantities are equal.
The relationship cannot be determined from the information given.

Solution

If you recognize the common quadratics, you know that (x + y)2 = x² + 2xy + y²; hence, x² + 2xy + y² − 2xy = x² + y².Thus, the two quantities are equal. Alternately, you could plug in values for x and y: If x = 2 and y = 3, then Quantity A equals 25 − 12 = 13, and Quantity B equals 4 + 9 = 13. Any set of values gives the same outcome, so select choice (C).

¹⁄_r of a circular pizza has been eaten. If the rest of the pizza is divided into m equal slices, then each of these slices is what fraction of the whole pizza?

\(\frac{r}{rm}\)
\(\frac{r-1}{rm}\)
¹⁄_m
\(\frac{m-1}{rm}\)
\(\frac{m-r}{rm}\)

Solution

To solve this one, Plug In for r and m:Try r = 2 and m = 4. If ¹⁄₂ of the pizza has been eaten, and the remaining ¹⁄₂ is divided into 4 equal slices, then each of those remaining pieces is ¹⁄₈ of the whole pizza. Now plug in 2 for r and 4 for m in the answer choices; only choice (B) hits your target answer of ¹⁄₈.

If \(b=\frac{c+d^{2}}{c}\) and \(a=\frac{c}{d^{2}}\), what is b in terms of a ?

1 + ¹⁄_a
1 + a
\(\frac{1}{1 + a}\)
a² + 1
\(\frac{a}{a+1}\)

Solution

Plug In for c and d, in both equations, and solve for a and b. If c = 8 and d = 4, then a = ¹⁄₂ and b = 3. Now plug in 8 for c and 4 for d in the answer choices; only choice (A) hits your target answer of 3.

Let x, y, and z be non-zero numbers such that the average (arithmetic mean) of x and twice y is equal to the average (arithmetic mean) of y and twice z. What is the average (arithmetic mean) of x and y?

^z⁄₂
z
2z
z − x
z − y

Solution

Variables in the question and the answer choices tell you this is a Plug In question. Plug in, but make sure you follow the rules set up in the question. If you start with something simple, like x = 2 and y = 3 then the average of x + 2y = 4, which is equal to the average of y + 2z, so 3 + 2z = 4. z must therefore equal 2.5.You’re asked to find the average of x and y, which is also 2.5. Write down 2.5 and circle it.This is your target number. Go to the answer choices and plug in a 2.5 whereever you see a z.The answer choice that equals 2.5 is the correct one.The answer is choice (B).

\(\frac{y^{2}}{x^{^{12}}}\) is an integer, y > x > 1.

Quantity A	Quantity B
x²	√y

Quantity A is greater.
Quantity B is greater
The two quantities are equal
The relationship cannot be determined from the information given.

Solution

The minute you see variables, make your set-up.Because x¹² is such a large number use a small value for x, such as 2.You now have a fraction with twelve 2’s on the bottom. For this fraction to be an integer, y must be a number that contains at least six 2’s.To keep it simple, try y = 2⁶. Quantity A is 4 and Quantity B is 2³ or 8.Eliminate choices (A) and (C).You can add more 2’s or any other number to the top of this fraction as long as y contains six 2’s.Therefore, y can only get bigger; it cannot get smaller.The answer is choice (B).

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