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By a live trainer on Maths for GRE , GMAT like competive examinations and school students according to your convenience.Free sample question papers with solution are available.
Wednesday, November 30, 2011
Highlights of the Training
Monday, November 28, 2011
GRE/GMAT sample questions with solution
Substitution:
1. If n is an even integer which one of the
following is an odd integer?
a) n2 b) -2n-4
c) √(n2
+2 d)n+1/2 e)2n2-3
Sol: It is given that n is an
even integer. Let n=2
n2 = 22= 4
is an even integer. So choice a is eliminated.
-2n-4 = -2*2 -4 = -4-4 = -8 is an even integer. So choice b
is eliminated.
√(n2+2 =√(22+2
= √6 is not an integer. So choice c is
eliminated.
n+1/2 = 2+1/2 = 3/2 is not an
integer. So choice d is eliminated.
2n2-3 = 2* 22 –
3 = 8 -3 =5 is an odd integer. So choice e is the answer.
2.
If n is an integer, which of the following CANNOT be an integer?
a) n-2/2 b) √n
c) 2/n+1 d) √(n2+3 e) √(1/n2+2)
Sol : Let n =0 . So n-2/2 = 0-2/2 = -1 which is an integer. So choice a
is eliminated.
√n = √0 =0 which is an integer.
So choice b is eliminated.
2/(n+1) = 2/(0+1) =2 which is an
integer. So choice c is eliminated.
Next, √(n2+3 = √(02+3)
= √3 which is not an integer . So it may
be the answer.
√(1/n2+2) =
√(1/2) which is not an integer as well .
So we choose another number say 1.
Then √(n2+3) = √4 =2 which is
an integer .
√(1/n2+2) = √(1/4) .
Thus choice e is the answer.
3. If x, y and z are positive integers
such that x< y < z and x + y + z =6, then what is the value of
z?
a) 1 b) 2 c)3 d) 4 e) 5
Sol :
From the given inequality x<y<z, it is clear that the positive
integers x, y and z are
different
and are in increasing order of size.
Let x=1 y =2 and z= 3. So 1< 2< 3 and x+ y+z = 1+2+3 =6
So z=3. Choice c is the answer.
4. By how much is the greatest of five
consecutive even integers greater than the smallest
among them?
a) 1 b) 2 c)4 d)8 e) 10
Sol: Let the five consecutive even integers be
2, 4, 6, 8, 10
Largest no of these =10
Smallest no of these = 2
Difference =8
So answer is d.
5. Which of the following could be an
integer?
a) The average of two consecutive integers.
b) The average of three consecutive integers.
c) The average of four consecutive integers.
d) The average of six consecutive integers.
e) The average of 6 and 9
Sol: choose any 2 consecutive
integers, say, 1 and 2.
Average = 1+2/2 =1.5 is not an
integer. So choice a is eliminated.
choose any 3 consecutive
integers, say, 1 ,2 and 3.
Averge = 1+2+3/3 =6/3 =2 is an
integer. So choice b is the answer.
Friday, November 25, 2011
GRE : Sample Questions with Solution
1. In how many
different ways can the letters of the word ‘DETAILS’ be arranged in such a way
that the vowels occupy only the odd
positions?
a) 32 b) 48 c) 36
d) 60
Sol : There are 6 letters in the given word, out
of wich there are 3 vowels and 3
consonants.
3
vowels can be placed at any of the 3 places
marked 1, 3, 5.
No of ways of arranging the vowels = 3P3 = 3! =6
Also the 3 consonants can be arranged at the remaining 3 positions.
No of ways these arrangements = 3P3 = 3! = 6
Total no of ways = 6*6 =36
2. From a group of 7
men and 6 woman five persons are to be
selected to form a committee so that at least 3 men are there on te committee .
In how many ways can it be done?
a)
564 b) 645 c)
735 d) 756
Sol : We may have
(3 men and 2 women ) or (4 men and 1 woman ) or(5 men only)
Required no of ways = 7C3 * 6C2 +
7C4 * 6C1 +7C5
= 756
3. A box contains 2 white balls , 3 black balls and 4
red balls . In how many ways can 3 balls be
drawn from
the box, if at least one black ball is to be included in the draw?
a) 32 b) 48 c) 64 d) 96
Sol: we may
have (1 black & 2 non-black ) or (2 black and 1 non black ) or 3 black
Required no of ways = 3C1 * 6C2 + 3C2 *
6C1 + 3C3
= 45+18+1=64
4. Two dice are thrown together . What is the probability that the sum of the nos on
the two
faces is
divisible by 4 or 6?
a) 7/18 b) 5/ 18
c) 5/36 d) ½
Sol : n(S) = 6 * 6 =36
Let E be the event tat the sum of the nos on the two faces is divisible
by 4 or 6 Then
E
= {(1,3) ,(1, 5) , (2, 2), (2, 4) ,(2, 6), (3,1), (3,3) , (3,5), (4,2), (4,4),
(5,1), (5,3),(6,2),
(6,6)}
n(E)
= 14
P(E) = n(E)/n(S) = 14/ 36 = 7/18
5. A box contains 5
green, 4 yellow and 3 white marbles. Three marbles are drawn at
random. What
is the probability that they are not of the same colour?
a) 3/44 b)3/55 c) 52/55 d)41/44
Sol: Let S be the sample space. Then n(S) = no of
ways of drawing 3 marbles out of 12.
=
12C3 = 220
Let E be the event of drawing 3 balls of the same colour.
Then E = events of drawing (3 balls out of 5) or (3 balls out of 4 ) or
(3 balls out of 3)
ð
n(E) = 5C3 +4C3 +3C3 = 5C2 + 4C1 + 1 = 15
ð
P(E) = n(E)/n(S) = 15/220 = 3/44
Required probability = 1-(3/44) =41/44
Sunday, November 20, 2011
GRE / GMAT : Sample Questions with Solution
Profit & Loss:
1. If CP = $
56.25 and Gain = 20% What is SP ?
a) 65.70 b) 67.50 c)76.70
d) 56.50
Sol:
SP = 120 % of $ 56.25
= 120* 56.25 / 100
= 67.50
2. If CP = $ 80 and Loss = 5% what is SP?
a) 76
b) 67 c) 77 d)68
Sol :
SP = 95% of $80
= 95 * 80 /100
= 76
3. A person
incurs 5% loss by selling an article for
$ 1140 .At what price should the article be sold to
earn 5% profit ?
a) 1200 b) 1260 c) 1270
d) 1400
Sol : Let the new SP be x. Then
(100 – Loss % ) : (1st SP ) = (100 + Gain % ) : (2nd SP )
ð
(100-5)/1140
= (100+5)/x
ð
x = 105 * 1140 /95 = 1260
So New SP = $1260.
4. A book was sold for $ 27.50 with a profit of
10% .If it were sold for $ 25.75 then what have been
the percentage of profit or loss ?
a)1 %
b)2% c) 3% d )5%
Sol :
SP = $ 27.50 profit = 10 %
So CP = 100* 27.50/110 =
25
When SP =
$25.75 profit = 25.75 – 25 = 0.75
Profit % =( 0.75 * 100 /25 )
% =
3%
5. If the cost price is 96% of the selling price
then what is the profit percentage ?
a) 4% b) 3.17 %
c)4.27 % d) 4.17%
Sol : Let SP
= $100 Then CP = 96 Profit = $4
Profit % = (4 %100)/96 = 25/6 %
= 4.17 %
Friday, November 18, 2011
GRE /GMAT : Sample Questions with Solution
1. If 3/p = 6 and 3/q = 15 then p - q = ?
- 1/3
- 2/5
- 3/10
- 5/6
- None
of the above
Ans : C
Sol : 3/p = 6 so
p=3/6 =1/2
3/q = 15 so q=3/15=1/5
P – q = ½ - 1/5 =5-2/10 =3/10
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2. If x4 = 16, then 4x =
a) 2 b) 4
c) 8 d) 16
x4 = 16
2* 2*2*2=16
24 = 16
S x=2
4x = 42 = 16
3. |-5| + |3| - |-5| =
a) -2 b) -13
c) 8 d) 3 e) 2
Sol:
Absolute value of |-5|=5, |3| = 3
|-5| + |3| - |-5| = 5 + 3 – 5 = 3
4. If 2x > 24 and 3x < 48, which of
the following is a possible value for x?
a) 12 b) 14 c) 16
d) 18
Sol: 2x > 24 =>
2x/2 > 24/2 =>
x > 12
3x < 48 =>
3x/3 < 48/3
=> x < 16
So
from the given choices answers is 14
5. If a cube has a volume of 343
cubic inches, what is the length of one side?
A:7square inches B:30square inches C: 7inches D:
49 inches
Sol:
Volume of a cube = (length of one side )3 = 343
So (length of one side )3 = 73
Length of one side = 7 inches
6. If 3x - 6 = 14 - x, then x =
A: 5 B: 3
C: 4 D: 8 E: 10
Sol: It is
given that 3x - 6 = 14 – x
3x – 6 -14 + x = 0
4x –
20 =0
4x = 20
x=
5
7. Factor: a2 -3a -
18
A: (a - 3)(a - 9) B: (a - 2)(a - 9) C: (a + 3)(a + 6) D: (a + 2)(a + 9) E: (a - 6)(a + 3)
It is given a2 -3a – 18
Let (a+ x) (a + y) = a2 -3a – 18
x+ y = -3
xy = -18
So x= -6 and y = 3 factors are (a-6) and (a+ 3)
8. If a = 2, then 3a +
(a3)2 =
A: 73 B: 70
C: 7 D: 12 E: 27
Sol : It is given a =2
3a + (a3 ) 2
= 3 * 2 + (23 )2 =6 + 82 = 6+ 64 = 70
9.
If a camera purchased for $490 and sold for $460 find the loss percent?
a)
1% b) 5% c)
10% d) 2%
Sol
: CP = $490 SP = $465.50
Loss
= (490-465.50) = $24.50
Loss
% = 24.50* 100/490 = 5%
10
. A man buys an article for $ 27.50 and sells it for $ 28.60 . Find his gain
percent?
a)
3% b) 1% c) 4%
d) 5%
Sol:
CP = $ 27.50 SP= 28.60
Gain
= SP- CP = 28.60 – 27.50 = 1.10
Gain
% = Gain * 100/CP = 1.10 * 100 /27.50 =
4 %
Wednesday, November 16, 2011
GRE /GMAT : Quantitative Comparison with solution
Directions:
In this section you will be given two quantities, one in column
A and one in column B. You are to determine a relationship between
the two quantities and mark.
- If the
quantity in column A is greater than the quantity in column B.
- If the quantity
in column B is greater than the quantity in column A.
- If the
quantities are equal.
- If the
comparison cannot be determined from the information that is given.
- Quantity A: (-6)4
Quantity B: (-6)5 - if the
quantity A is greater;
- if the
quantity B is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given.
Ans : A
Soln: ( -6)4 =(-1 * 6)4 = (-1)4 * 64 = (-1)2 * (-1)2 * 64 =64
(-6)5
= (-1)5 * 65 = -1
* 65 =-65
So Ans is A
- Quantity A: Time to travel 95 miles at 50 miles per hour
Quantity B: Time to travel 125 miles at 60 miles per hour - Quantity A is
greater
- Quantity A
equals Quantity B
- Quantity B is
greater
- Relationship
Indeterminate
Ans : C
Soln : Time
= Distance/Speed
For
Case A Time= 95/ 50 = 1.
For Case B Time= 125/60=2.
So B>A
- Quantity A: (9/13)2
Quantity B: (9/13)1/2 - Quantity A
equals Quantity B
- Relationship
Indeterminate
- Quantity B is
greater
- Quantity A is
greater
Ans : C
Soln : a1/m = m√a
For Case A it is (9/13)2
For Case B it is (9/13)1/2 = √(9/13)
So B>A
- Quantity A: 4 / 100
Quantity B: 0.012 / 3 - Quantity B is
greater
- Quantity A
equals Quantity B
- Quantity A is
greater
- Relationship
Indeterminate
Ans : C
For Case A
: 4/100=0.04
For Case B :
0.012 / 3 = 0.04
So A=B
- x = 2y + 3
y = -2
Quantity A: x
Quantity B: -1
Quantity B: -1
- if the
quantity in Column A is greater;
- if the
quantity in Column B is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given
Ans : C
x=2y+3
= 2(-2)+3 = -1
So
both quantities are equal.
- x + 2y > 8
Quantity A: 2x + 4y
Quantity B: 20
Quantity B: 20
- if the
quantity in Column A is greater;
- if the
quantity in Column B is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given.
Ans : D
x+2y
> 8 => 2(x+2y) > 2(8)
2x+4y > 16
So
the relationship cannot be determined from the information given
- Quantity A: The number of
months in 7 years
Quantity B: The number of days in 12 weeks - if the
quantity in Column A is greater;
- if the
quantity in Column B is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given
Ans : C
The no of months in 7 years = 7* 12 months = 84 months
The
no of days in 12 weeks = 12* 7 days = 84 days
So
two quantities are equal.
- Quantity A: 1-1/27
Quantity B: 8/9 + 1/81 - if the
quantity in is greater;
- if the
quantity in is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given.
Ans : A
Case A 1-1/27 = 27-1/27 = 26/27=0.96
Case B 8/9 + 1/81
LCM of 9 and 81 is 81
So 8/9
+ 1/81 = (8*9 +1)/81 = 73/81 =0.90
So A>B
- r>s>0
Quantity A: rs/r
Quantity B: rs/s
Quantity B: rs/s
- if the
quantity A is greater;
- if the
quantity B is greater;
- if the two
quantities are equal;
- if the
relationship cannot be determined from the information given.
Ans : B
Case A rs/r and case B is rs/s
It is given that r>s
If
you divide the same quantity by a smaller no the result will be greater.
- Quantity A: 0.83
Quantity B: 0.81/3 - Quantity B is
greater
- Relationship
Indeterminate
- Quantity A is
greater
- Quantity A
equals Quantity B
Ans : C
Case A 0.83
Case B is 0.81/3 = 0.27
So
A>B
Tuesday, November 15, 2011
Sample questions of GRE with solution
1
1 Questions Related to Chain
rule , Time & Work
1) If 6 workers can build 4 cars in 2 days, then
how many days would it take 8 workers to
build 6 cars?
(A) 5/3
(B) 9/4
(C) 8/3
(D) 11/4
(E) 10/3
Ans: more workers less no of days (indirect proportion)
More cars more
days (direct
proportion)
Ratio of workers= 8:6
Ratio of cars =4:6
Ratio of days = 2:x
6*6*2=8*4*x
X=(6*6*2)/8*4
= 9/4
22) If 7 workers can build 7 cars in 7 days, then
how many days would it take 5 workers to
build 5 cars?
Ans : more workers less no of days (indirect proportion)
More cars more days (direct proportion)
Ratio of workers= 5:7
Ratio
of cars =7:5
Ratio
of days = 7:x
5*7*x=7*5*7
X=7*5*7/5*7
=7
33 If 5 men and 9 women can do a piece of work in
19 days how many days will take 3men and 6 women to do the same work?
Let required no be X
5 men = 9 women => 1man =9/5
women
3 men and 6 women = 3*9/5 +6
=27+30/5=57/5
More women less no of days (indirect proportion)
57/5:9=19:x
X=9*19*5/57=15
44 A
man completes 5/8 of a job in 10 days. At this rate how many
more days will it take him to finish the job?
a)
5 b)
6 c) 7
d) 71/2
Ans Work done
=5/8 Balance work = (1- 5/8)=3/8
Less work Less
days (Direct Proportion )
Let the required no of days be x.
Then 5/8:3/8=10:x
5/8*x=3/8*10
X=3*8*10/5*8
=6 days
5) 3 pumps , working 8 hours a day can empty a tank in 2 days , how many hours a day must 4
5) 3 pumps , working 8 hours a day can empty a tank in 2 days , how many hours a day must 4
pu mps work to empty the tank in 1 day?
a) 9
b) 10 c) 11 d) 12
Ans: Let the required number of working hours per
day be x
More pumps Less working hours per
day (Indirect Proportion )
Less days, More Working Hours per
day (Indirect
Proportion )
Pumps 4:3
::
8: x
Days 1:2
4*1*x=3*2*8
X=3*2*8/4
=12
56) If
the cost of x metres of wire is d dollars , then what is the cost
of y metres of wire at the same rate?
a)
xy/d
b) xd c) yd d)yd/x
Cost of x metres
= d
Cost of 1 metre
= d/x
Cost of y metres
= (d/x)y=dy/x
67) In a camp, 95 men had provision for 200 days , After 5 days 30 men left the camp. For how many days will the
remaining food last now?
a)
180 b)
285 c) 139 18/19
d) none of these
Ans: Let the remaining food will last for x days.
95 men had provisions for 195 days . 65 men had
provision for x days.
Less men
,More days (Indirect Proportion )
65 : 95 ::
195 : x => 65*x=95*195
X =95*195/65
=285
78) A rope makes 70 rounds of the circumference of a cylinder whose radius of
the base is 14cm. How many times can it go round a cylinder with radius 20 cm ?
a)
40 b)49
c) 100 d)None of these
Ans : Let the required number of rounds be x
More
radius , Less rounds (Indirectly
Proportional)
20:14 ::
70:x => 20*x=14*70
X =
14*70/20
=49
89) If 3/5 of
a cistern is filled in 1 minute, how much more time will be required to fill
the rest of it?
a)
30 sec
b) 40 sec c) 36 sec
d) 24 sec
Let the required time be x seconds.
Part Filled = 3/5 ,
Remaining part = (1-3/5) = 2/5
Less part , Less time (Direct proportion)
3/5 : 2/5 :: 60 : x
=> 3/5*x =2/5*60
= > x=40
910) If 7
spiders make 7 webs, then 1 spider will make 1 web in how many days?
a)
1
b)7/2 c) 7 d) 49
Ans : Let the required no
of days be x. Then
Less spiders, More days
(Indirect Proportion)
Less webs , Less days
(Direct Proportion)
Spiders 1 : 7
:: 7:x
Webs 7 : 1
1*7*x= 7*1*7
x = 7*7*1 / 1*7
= 7
111) 400
persons working 9 hours per day
complete ¼ th of the work in 10 days. The no of
additional persons working 8 hours per day , required to complete the remaining
work in 20 days, is :
a)
675
b) 275 c) 250 d) 225
Ans: Let the no of persons completing te work in
20 days be x.
Work done
= ¼ Remaining work = (1- ¼ ) = ¾
Less hours
per day , More men required ( Indirect Proportion)
More work
, More men required (
Direct proportion )
More days
, Less men required
( Indirect Proportion)
Hours per
day 8 : 9
Work ¼ : ¾
: : 400:x
Days 20 : 10
8 * ¼ * 20 * x = 9 * ¾ * 10 * 400
x = 9 * 3*
10* 400*4 / 8 * 20 * 4
=675
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