New GRE Math Question Types - Pitfalls to avoid
Numeric Entry Questions

Another type of Questions, that are relatively new to the GRE are Numeric Entry Questions.

To answer these Questions, you will either need to enter a number in a single answer box or as a fraction in
two boxes, one corresponding to the numerator & one for the denominator.


Things to watch out for these Question Types

• In case the answer is a  decimal, round the answer as directed in the Question.

• Round your answer if a question so indicates; otherwise, enter the exact answer.

• Since there are no answer choices, make sure that you answer the exact question asked.

• Pay attention to unit of measures like feet, kilometers, miles etc.

• Check your answer to make sure it seems reasonable with reference to the question.
Maths questions (Numbers 1-12 are from test 5 )

New GRE Math questions  Practice

 
1.  The ratio of days worked to days not worked for a certain employee was 8 to 7 in June. If the employee worked 5 more days in July than in June, then what per cent of the total number of days in July were days that the employee worked?
 
Give your answer to the nearest percent.

 

------------------------%

 

 
 
 
2.

 

n

1

2

3

4

5

6

7

8

…

r

1

2

2

3

2

4

2

4

…


In the table above, the numbers in the row labeled n form the sequence of positive integers and each number in the row labeled r is the number of different positive divisors of the corresponding integer n. What is the value of r that corresponds to the integer 36?

 


r      =     -----------------------

 
 
 
3. A manager has to write performance appraisals for all of her staff. She has written 20% of the total number of appraisals. After she writes 2 more, she will still have 75% of the total number of them to write. How many appraisals will she still have to write?

-----------------------------

 

 
 

4. S is a series of natural numbers. S1 represents the 'i' th number of the series. Si (for i > 2) is equal to the sum of Si – 1 and Si – 2 . Also S2 = 5. What is the difference between the square of S100 and the product of S99 and S101 if S1 = S2?
 

--------------------------

 
 
 5. A total of $2,030 was spent for a 30-day vacation. If transportation expenses were $230, lodging was $750, and exactly $20 was spent per day for meals and tips, what was the average (arithmetic mean) amount per day that was spent on other expenses?


 -------------------- dollars
 
 
 

 
6. A pentagonal prism is to be painted. No two Adjacent faces are painted. In how many ways can the prism be painted?

 

--------------------
 
 
 
 
7. COST OF A TELEPHONE CALL AT DAY AND NIGHT RATES

 

Cost for First Minute

Cost for Each Additional Minute

Day rate

$ 0.50

$ 0.34

Night Rate

$ 0.20

$ 0.14

 
 
According to the table above, a night-rate call that costs the same as a 3-minute day rate call would last a total of how many minutes?
 
  -------------------- minutes

 


8. The calorie content of five fruits is measured. The sum of calorie content of three fruits taken at a time are 150, 160, 120, 180, 140, 150, 170, 130, 160 and 140. What is the average calorie content of a fruit?
 
 ---------------------- calories
 
 
 

9. If a total of x identical disks can be arranged in 8 stacks of equal height or in 12 stacks of equal height, what is the least possible value of x?
 
x =   ----------------------
 
 
 
 
10.  Forty percent of Mr. Johnson’s seventh-grade class are boys. If on a certain day 10 per cent of the boys and 20 per cent of the girls are absent, what proportion of Mr. Johnson’s class is absent?
 
Give your answer as a fraction.
 
---------/-------

 
 

 

 

 

Answer Key:

 

 Question

Correct Answer

1

68

2

9

3

30

4

25

5

15

6

13

7

8

8

50

9

24

10

4 / 25

 

 

 

Detailed Explanations

 

 
1. There are 30 days in June.
working: nonworking days = 8:7 = 16 days : 14 days.
In July = 3: days. He worked 16 + 5 = 21 days.
 
% of working days =  

 

 

2. 1 is divisible by only 1 T = 1
2 is divisible by 1 and 2 T = 2
3 is divisible by 1 and 3 T = 2
 
36 is divisible by 1, 2, 3, 4, 6, 9, 12, 18, 36, T = 9.
 
 
3. Let total no. of staff be ‘x’,

therefore 

 

 

 

 

 

 

 

 

 

 x = 40
 
 
 Appraisal still to be written =  
 
 
 
4. S1 = S2 = 5
therefore, S3 = 5 + 5 = 10, S4 = 10 + 5, S5 = 25
S22 - S1S3 = 25 - 10 x 5 = 25
S32 - S2S4 = 100 - 5 x 15 = 25
S42 - S3S5 = 225 - 250 = -25
 
Therefore, in general (Sn2 - Sn = 1 Sn-1) = (-1) (n-1), 25
 
S1002 - S101 S99 = -25 But the difference means only the magnitude
 
Hence |S1002 – S101 S99| = |-25|
 
 
 
5. Transportation expenses $ 230
Lodging $750
Meals and tips @ $20 per day $600
Total $1580
 
Therefore other expenses = 2030 - 1580 = $450
Therefore average daily other expenses = 450/30 = $15
 
6. If only one face is painted then there are seven ways of painting the
prism.
Two faces can be painted as follows:
The base and top are painted (1 way)
Two vertical faces (not adjacent) can be painted in 5 ways.
 
Hence total number of ways = 7 + 1 + 5 = 13
 
 
7. A 3 minute day call costs 50 + (2 x 34) cents = $1.18
 
At night the first minute is $0.20 and the next n minutes cost $1.18
- $0.20 = $0.98
rate per minute after the first minute is $0.14
n = .98 divided by .14 = 7
Total number of minutes = 1 + 7 = 8
 
 
8. The sum of calories contents of fruits taken 3 at a time = 1500
 
Now average calories content for 3 fruits taken at a time =   
 
Average calories content of a fruit =   
 
 Alternatively, the sum of five fruits taken 3 at a time = 1500
 
When total amount of 1500 is considered, it will have 10 x 3 = 30 fruits.
 
But there are only 5 different kinds of fruits.
 
Thus, each of the 5 fruit is counted     =   6 times.
 
To find sum of five fruits, we have to divided total amount by 6,
 
which is =   .
 
 
Now, to find average of each of five fruit, we get   = 50.
 
 
 
9. The least possible value of x is the LCM of 8 and 12 , which is 24.


 
10.  Let x be total no. of students in the class


Then, boys = .4x absent boys = .1 x .4x = 0.4x
 
Girls = .6x absent girls = .2 x .6x = 0.12x
 
Total of absent students = 0.16x
 

16% of the class is absent. 16% =

Multiple Choice Questions - Select one or more answer choices

These are a new type of questions introduced in the New GRE.

The question may provide you with the number of correct answers, or may simply ask you to mark all the
correct answers.

After you have got the solution, you need to check it against each answer choice.

Here are some sample questions to give you an introduction:



Select ALL answer choices that apply:


1.                Find the value of 1.1! + 2.2! + 3.3! + ......+n.n!              (Note: n! = n.(n-1).(n-2)....2.1)
(a)         n! +1                (b)        (n+1)!                (c)        (n+1)!-1        
(d)        (n+1)(n)! -1      (e)        none of these


2.                From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the         
probability that equal numbers of boys and girls will be selected?
(a)        <0.1        (b)         >0.5        (c)         <0.8        (d)        1        (e)         0


3.                How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21?
(a)         <2000            (b)           >2000          (c)  <4000        (d)     >4000         
(e)        all of the above


4.                If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?
(a)        >35000        (b)        >40000        (c)   <50000        (d)        >60000        
(e)        >100000


5.                Two numbers are less than a third number by 30% and 37 % respectively. How much percent is
the second number less than the first?
(a)         >10%        (b)        <15%        (c)         >20%        (d)  <12%        (e)        13%


6.                How many different subsets of the set {10, 14, 17, 24} are there that contain an odd number of
elements?
(a)         >8        (b)         <8        (c)         8        (d)        <10        (e)         >10


7.                If P represents the product of the first 15 positive integers, then P is not a multiple of:
(a)        99        (b)         84        (c)         72        (d)         68        (e)         57


8.                A group of people participate in different activities; 20 of them practice Yoga, 10 study         
cooking, 12 study weaving, 3 of them study cooking only, 4 of them study both cooking and yoga, 2 of them
participate all activities. How many people study both cooking and weaving?
(a)        >1        (b)        <2        (c)        >3        (d)        <4        (e)        >5


9.                In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3.
50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the
survey liked all three products, what percentage of the survey participants liked more than one of the three
products?
(a)        <5        (b)         >10        (c)         <15        (d)         20        (e)        > 25




10.                A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats  him by
1/10th of a minute. In the second heat, A gives B a head start of 144 m and is  beaten by 1/30th of a minute.
What is B’s speed in m/s?
(a)        <5        (b)         >5        (c)        <15        (d)         >15        (e)         none of the above




Answers & Solutions

1.               Ans:        c and d

Sol:
1.1! + 2.2! + 3.3! + ......+n.n!
=1.1! + (3-1)2! + (4-1)3! +......+ ((n+1)-1) n!
=1.1! + 3! - 2! + 4! - 3! +.......+ (n+1)! - n!  (The common terms like + 3! and  - 3!  etc. cancel out)
So it is (n+1)! -1 which can also be written as (n+1)(n)! -1
Hence (c) and (d) is the answer.

Note: If you are stuck and do not not how to proceed, simply assume n = 2  and eliminate some of the answer
choices.


2.                Ans:        b and c

Sol:
Total number of ways of selecting 4 children = 6C4 =  6!/4!.(6-4)! = 6.5/2 = 15
with equal boys and girls. => 2 boys and 2 girls. => 3C2 × 3C2 = 9.
Hence p = 9/15  = 3/5   or 0.6
Hence (b) and (c) is the answer.


3.                Ans:        b and d

Sol:         
We first determine the number of integers less than 5,000 that are evenly divisible by 15. This can be found by
dividing 4,999 by 15:
= 4,999/15        
= 333 integers

Now we will determine the number of integers evenly divisible by 21:
= 4,999/21  
= 238 integers

Some numbers will be evenly divisible by BOTH 15 and 21. The least common multiple of 15 and 21 is 105.
This means that every number that is evenly divisible by 105 will be divisible by BOTH 15 and 21. Now we will
determine the number of integers evenly divisible by 105:
= 4,999/105        
= 47 integers

Therefore the positive integers less than 5000 that are not evenly divisible by 15 or 21 are 4999-(333+238-
47) =4475  
b and d is the answer.


4.                Ans:        a and c

Sol:
This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the
number of ways independent events can occur together can be determined by multiplying together the number
of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).
There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)
Using the Multiplication Principle:
= 8 × 10 × 10 × 10 × 5
= 40,000
Hence, (a) and (c) is the answer.


5.               Ans:        b and d

Sol:

Assuming the biggest number of the three is 1, then the other numbers are:
.7 & .63
diff in %= ((.7- .63))/(.7)  ×100=  (.07)/(.7)  ×100=10%                
Hence, (b) and (d) is the answer.


6.                Ans:        c and d

Sol:
The different subsets are
10,
14,
17,
24,
10, 14, 17
14 17 24
17 24 10
24 10 14
Hence, (c) and (d) is the answer

Note: This problem could have been attempted by using Combination/Permutations as well.


7.               Ans:        d and e

Sol:
If P represents the product of the first 15 integers, P would consist of the prime numbers that are below 15.
2, 3,5,7,11,13
Any value that has a prime higher than 13 would not be a value of P.
57 =  19 x 3
19 is a prime greater than 13, hence P is not a multiple of 57

Similarly 68 = 17 x 4, hence P is not a multiple of 68

Hence, (d) and (e) is the answer


8.                Ans:        b and d

Sol:         
We know there are 10 people who do cooking as an activity.
3 -> people who do cooking only
4 -> do cooking and Yoga
2 -> do all of the activities
x -> number of people doing cooking and weaving
When you sum all this up, we should have 10. So 3+4+2+x=10 --> x=10-9=1
 Hence, (b) and (d) is the answer.

Note: This problem could have been attempted by drawing a standard Venn diagram we well.


9.                Ans:        b and d

Sol:
n(1U2U3) = n(1) + n(2) + n(3) - n(1n2) - n(2n3) - n(1n3) + n (1n2n3)    

(Note: n(1n2)  indicates the number of elements in a set made up of Intersection of set 1 and set 2)

85 = 50 + 30 + 20 - [n(1n2) - n(2n3) - n(1n3)] + 5

[n(1n2) - n(2n3) - n(1n3)] = 20

Hence, (b) and (d) is the answer




10.                Ans:        b and c

Sol:
race 1:- ta = tb-6 (because A beats B by 6 sec)
race 2:- ta = tb+2 (because A loses to B by 2 sec)

By the formula D= S × T
we get two equations
480/Sa= 432/Sb- 6        ------------1)
480/Sa= 336/Sb+2        ------------2)
Equating these two equations we get Sb = 12
ta, Sa stand for time taken by A and speed of A resp.
Hence, (b) and (c)  is the answer.