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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
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
y 5 52 1 3 5 25 1 3 5 28
28 is divisible by 7
to test four choices instead of one choice before finding
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
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
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```