Gruber's Complete GRE Guide 2015   Gruber, Gary R
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Gruber's Complete GRE Guide 2015 Gruber, Gary R


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with Choice E and 
Work Backward
If you must check each choice for the correct answer, start with Choice E and work backward. 
The reason for this is that the test maker of a question in which each choice must be tested often 
puts the correct answer as Choice D or E in order to trick the careless student who starts 
testing with choice A. So if you\u2019re trying all the choices, start with the last choice, then the 
next to last choice, etc. See example 8 for an instance of when this strategy should not be used.
Math 
StratEg
y 8
example 1
If p is a positive integer, which could be an odd integer?
(A) 2p 1 2
(B) p3 2 p
(C) p2 1 p
(D) p2 2 p
(E) 7p 2 3
Choice E is correct. Start with Choice E first, since you 
have to test out the choices.
Method 1: Try a number for p. Let p 5 1. Then (starting 
with Choice E):
7p 2 3 5 7(1) 2 3 5 4. 4 is even, so try another number 
for p to see whether 7p 2 3 is odd. Let p 5 2.
7p 2 3 5 7(2) 2 3 5 11. 11 is odd. Therefore, Choice E 
is correct.
Method 2: Look at Choice E. 7p could be even or odd, 
depending on what p is. If p is even, 7p is even. If p is 
odd, 7p is odd. Accordingly, 7p 2 3 is either even or odd. 
Therefore, Choice E is correct.
Note: By using either Method 1 or Method 2, it is not 
necessary to test the other choices.
example 2
If y 5 x2 1 3, then for which value of x is y divisible by 7?
(A) 10
(B) 8
(C) 7
(D) 6
(E) 5
Choice E is correct. Since you must check all of the 
choices, start with Choice E:
y 5 52 1 3 5 25 1 3 5 28
28 is divisible by 7
If you had started with Choice A, you would have had 
to test four choices instead of one choice before finding 
the correct answer.
example 3
Which fraction is greater than 2
1 ?
(A) 9
4
(B) 
35
17
(C) 
13
6
(D) 
25
12
(E) 15
8
Choice E is correct.
Look at Choice E first.
Is 
15
8
2
12 ?
Use the cross-multiplication method.
 
2
1 15
8
 15 16
 15 , 16
So, 2
1
15
81 .
You also could have looked at Choice E, said 16
8 5 2
1, 
and realized that 15
8
2
12 because 15
8 has a smaller 
denominator than 16
8 .
GRE2015_P04.indd 94 4/18/14 11:29 AM
STRATEGY SECTION \u2022 95
example 4
If n is an even integer, which of the following is an odd 
integer?
(A) n2 2 2
(B) n 2 4
(C) (n 2 4)2
(D) n3
(E) n2 2 n 2 1
Choice E is correct.
Look at Choice E first.
 n2 2 n 2 1
 If n is even,
 n2 is even
 n is even
 1 is odd
So, n2 2 n 2 1 5 even 2 even 2 odd 5 odd.
example 5
Which of the following is an odd number?
(A) 7 3 22
(B) 59 2 15
(C) 55 1 35
(D) 75 ÷ 15
(E) 47
Choice D is correct.
Look at Choice E first.
47 is even, since 4 3 4 3 4 \u2026 is even.
So now look at Choice D: 15
75 5= , which is odd.
example 6
 
3 # 2
8
28 6
#
*
If # and * are different digits in the correctly calculated 
multiplication problem above, then # could be
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
Choice E is correct.
Try Choice E first.
 3 # 2 3 2
 3 8 3 8
 
 28 * 6 28 6
9 and 6 are different numbers, so Choice E is correct.
6
9
example 7
Which choice describes a pair of numbers that are 
unequal?
(A) ,6
1
66
11
(B) . ,3 4 10
34
(C) ,75
15
5
1
(D) , .8
3 0 375
(E) ,24
86
10
42
Choice E is correct.
Look at Choice E first.
 
24
86 ? 
10
42
Cross-multiply:
 
24
86 
10
42
 860 ends in 0 24 3 42 ends in 8
Thus, the numbers must be different and unequal.
When Not to Use This Strategy:
If you can spot something in the question that shows 
you how to solve the problem readily without having to 
test each choice, there\u2019s no need to go through every 
answer by working backwards.
example 8
If |6 2 5y| . 20, which of the following is a possible 
value of y?
(A) 23
(B) 21
(C) 1
(D) 3
(E) 5
Choice A is correct.
Instead of plugging in values for y, starting with Choice 
E, you should realize there will only be one answer 
listed for which 6 2 5y . 20. So which choice gives you 
the largest product for 25y? Start by checking the most 
negative choice, or y 5 23.
This gives you |6 2 5(23)| 5 |6 1 15| 5 |21|, which is 
greater than 20.
GRE2015_P04.indd 95 4/18/14 11:29 AM
96 \u2022 GRUBER\u2019S COMPLETE GRE GUIDE 2015
Know how to Solve Problems Using the Formula R 3 T 5 D
Almost every problem involving motion can be solved using the formula
R 3 T 5 D
or
rate 3 elapsed time 5 distance
Math 
StratEg
y 9
example 1
The diagram below shows two paths: Path 1 is 10 miles 
long, and Path 2 is 12 miles long. If Person X runs along 
Path 1 at 5 miles per hour and Person Y runs along Path 
2 at y miles per hour, and if it takes exactly the same 
amount of time for both runners to run their whole path, 
then what is the value of y?
 A Path 1 B 
 C Path 2 D 
(A) 2
(B) 
6
14
(C) 6
(D) 20
(E) 24
Choice C is correct. Let T 5 Time (in hours) for either 
runner to run the whole path.
Using R 3 T 5 D, for Person X, we have
(5 mi/hr)(T hours) 5 10 miles
or 5T 5 10, or 1
 T 5 2
For Person Y, we have
( y mi/hr)(T hours) 5 12 miles
or yT 5 12
Using 1 , y(2) 5 12, or y 5 6
example 2
A car traveling at 50 miles per hour for two hours travels 
the same distance as a car traveling at 20 miles per hour 
for x hours. What is x?
(A) 
5
4
(B) 
4
5
(C) 5
(D) 2
(E) 
2
1
Choice C is correct.
Use R 3 T 5 D. Call the distance both cars travel D 
(since the distance is the same for both cars).
So we get:
 50 3 2 5 D 5 100 1
 20 3 x 5 D 5 100 2
Solving 2 , you can see that x 5 5.
example 3
John walks at a rate of 4 miles per hour. Sally walks at a 
rate of 5 miles per hour. If John and Sally both start at 
the same point, how many miles is one person from the 
other after t hours of walking? (Note: Both are walking 
on the same road in the same direction.)
(A) t2
(B) t
(C) 2t
(D) t5
4
(E) t4
5
Choice B is correct.
GRE2015_P04.indd 96 4/18/14 11:29 AM
STRATEGY SECTION \u2022 97
Draw a diagram:
John (4 mph) 
 
 
Sally (5 mph)
Let DJ be distance that John walks in t hours.
Let DS be distance that Sally walks in t hours.
Then, using R 3 t 5 D,
for John: 4 3 t 5 DJ
for Sally: 5 3 t 5 DS
The distance between Sally and John after t hours of 
walking is:
DS 2 DJ 5 5t 2 4t 5 t
example 4
A man rode a bicycle a straight distance at a speed of 
10 miles per hour and came back the same distance at 
a speed of 20 miles per hour. What was the man\u2019s total 
number of miles for the trip back and forth, if his total 
traveling time was 1 hour?
(A) 15
(B) 72
1
(C) 6
3
1
(D) 63
2
(E) 133
1
Choice E is correct.
Always use R 3 T 5 D (Rate 3 Time 5 Distance) in 
problems like this. Call the first distance D and the time 
for the first part T1. Since he rode at 10 mph:
 10 3 T1 5 D 1
Now for the trip back. He rode at 20 mph. Call the time 
it took to go back T2. Since he came back the same 
distance, we can call that distance D also. So for the trip 
back using R 3 T 5 D, we get:
 20 3 T2 5 D 2
Since it was given that the total traveling time was 1 
hour, the total traveling time is:
 T1 1 T2 5 1 
Now here\u2019s the trick: Let\u2019s make use of the fact that T1 1 
T2 5 1. Dividing Equation 1 by 10, we get:
 T D101 =
(t hours)
DJ
(t hours)
DS
Dividing 2 by 20, we get:
T D202 =
Now add T1 1 T2 and we get:
1T T D D10 201 2+ = = +
Factor D:
1 D 10
1
20
1
= +c m
Add 10
1
20
1
+ . Remember the fast way of adding 
fractions?
 
20
1
20 10
20 10
200
30
#
= + =+
1
So:
1 ( )D 200
30
=
Multiply by 200 and divide by 30 and we get:
; 6D D30
200
3
2
= =
Don\u2019t forget, we\u2019re looking for 2D: 2D 5 13 3
1
example 5
What