Friday, April 26, 2013

GRE Practice Questions : 2013

9:00 AM

1. Question :
The average age of the 5 members of a family is 25 years. The age of the youngest member is 5 years. What was the average age of the members of the family just before the youngest member was born?
Explanation:
Sum of ages of 5 members = 25*5
= 125
The sum of ages of five members increased by 25 years in the last five years.
Five years ago the sum of the ages was = 125 – 25 = 100.
Five years back there were 4 members in the family.
Hence, average age five years back = 100/4
= 25 years.

2. Question :
The odds in favour of an event are 4:5. In how many chances out of 36 shall the event occur?
Explanation:
Odds n favour of the event are 4:5
Probability of occurrence of the event = 4/(4+5)
= 4/9
If the chances are 36, the number of chances of the occurrence of the event are 36/9*4
= 4*4 = 16

3. Question:
If two pipes work simultaneously, the reservoir fills in 12 hours. One pipe fills the reservoir 10 hours faster than the other does. How many hours does the faster pipe take to fill the reserviour alone?
Explanation:
Let the two pipes be F and S where F is the faster pipe and S is the slower pipe.
Let S take x hours to fill the tank and F will take (x-10) hours.
Work done by S in one hour = 1/x
Work done by F n one hour = 1/(x-10)
Work done by F and S working together in one hour = 1/12
1/x+1/(x-10) = 1/12
(x-10+x)/[x(x-10)]=1/12
12(2x-10) = x(x-10)
24x-120=x^2-10x
x^2-34x+120=0
x^2-30x-4x+120=0
x(x-30)-4(x-30)=0
x=4,30
x cannot be 4 since x<10
x=30
The faster pipe takes 30-10 = 20 hours to fill the reserviour.

4. Question :
The product of the positive numbers is 300 and the sum of their squares is 25 more than two times their product. Find the sum of the two numbers.
Explanation:
Let the numbers be x and y.
According to the given conditions, we have
xy=300
x^2+y^2= 2xy+25
=2*300+25
=600+25=625
(x+y)^2=x^2+y^2+2xy
=625+2*300
=625+600=1225
x+ysqrt(1225)
= 35
The sum of the numbers is 35

5. Question :
The product of the digits of the number is 15 and the sum of the digits is 8. The digits interchange their places when 18 is added to the number.
Find the number.
Explanation:
Let the digit in the units place be x and the one in the tens place be y.
The number will be 10y+x.
xy=15 and x+y=8
Putting x = 8-y in xy=15, we get
(8-y)y=15
8y-y^2-15=0
y^2-8y+15=0
y^2-5y-3y+15=0
y(y-5)-3(y-5)=0
y=3,5
When y = 3, x = 5 and when y = 5, x = 3
The number is either 35 or 53.
When 18 is added to the number the digits interchange their places.
35+!8=53
Hence, the number is 35.

6. Question :
The price of wheat has fallen by 10%. How many quintals can be purchased for the same price for which 27 quintals could be bought before?
Explanation:
Let the price per quintal before be x.
Total cost of 27 quintals = 27x
Reduction in price is 10%
New price = x-10x/100
=90x/100 = 0.9x
Number of quintals that can be purchased now = 27x/0.9x
= 30
30 quintals can be bought now.

7. Question :
Find the probability of the occurrence of exactly one of A and B when P(AUB) = 0.59 and the probability of the occurrence of both A and B is 0.01.[AUB = A union B]
Explanation:
Probability of occurrence of exactly one of A and B = P(AUB) - probability of occurrence of both A and B
= 0.59 - 0.01
= 0.58

8. Question :
Find the value of (3/y-1/x) if 3(2x+y)=7xy and 3(x+3y)=11xy.
Explanation:
Let 1/x = u and 1/y = v
The two equations can be rewritten as
6x+3y = 7xy and 3x+9y=11xy
6/y+3/x=7 and 3/y+9/x = 11
6v+3u=7 and 3v+9u=11
The equations now are
3u+6v=7 ...(1)
9u+3v=11 ...(2)
Multiplying (1) by 3 and subtracting from (2), we get
9u+3v-9u-18v=11-21
-15v=-10
v=2/3
u = 1/3(7-6*2/3)
=1/3(21-12)/3
=9/9=1
Hence, x = 1/u = 1 and y = 1/v = 3/2
3/y-1/x = 3/(3/2)-1/1
=2-1=1

9. Question :
The simple interest on a sum of money is 1/6 of the principal. The number of months is equal to twice the rate percent. Find the time in months of the investment.
Explanation:
Let the principal be Rs.x.
The simple interest will be Rs.x/6
Let the rate of interest be y%.
The time in months will be 2y
Simple interest = P*R*T/100, where P, R and T are the principal, rate and interest.
x/6 = (x*y*2y/12)/100
y^2=100
y=10
Hence, the time in months is 2*10=20 months
[y^2=y*y]

10. Question :
Find the value of (sinA+sinA^2) if cosA^2+cosA^4=1
[sinA^2=sinA*sinA]
Explanation:
We know that sinA^2+cosA^2=1
cosA^2+cosA^4=1
1-cosA^2=cosA^4
sinA^2=cosA^4
Taking square root of both sides, we get
sinA=cosA^2
sinA=1-sinA^2
sinA+sinA^2=1

11. Question :
How many straight lines can be drawn by joining 12 points in a plane when no three points are collinear?
Explanation:
The number of straight lines is given by C(12,2)
C(12,2)=12!/[(12-2)!2!]
=12!/(10!2!)
=12*11/2
= 6*11=66

12. Question :
How many three digit even numbers exist with distinct digits?
Explanation:
The numbers shall contain the digits 0,2,4,6 and 8 only.
The hundreds place can be filled with any of 2,4,6 and 8. It can be filled in 4 ways.
The tens place can be occupied by any of the remaining digits. It can be filled in 4 ways.
The units place can be occupied by any of the remaining digits. Hence, it can be filled in 3 ways.
Total numbers = 4*4*3
= 48

13. Question :
Find the value of (n-r), when 7/2<(r+5)/2<=4 and P(n,r)=60
Explanation:
7/2<(r+5)/2<=4
7
7-5
2
r=3
P(n,r)=60
n!/(n-r)! = 60
n!(n-3)!=60
n(n-1)(n-2)(n-3)!/(n-3)!=60
n(n-1)(n-2)=60
5*4*3=60
n=5
Hence, (n-r)=5-3
=2

14. Question :
Find the least possible value of x, (5x-2)/3-(7x-3)/5>=x/4
Explanation:
(5x-2)/3-(7x-3)/5>=x/4
(25x-10-21x+9)/15>=x/4
(4x-1)/15>=x/4
4(4x-1)>=15x
16x-4>=15x
x>=4
Hence, the least possible value of x is 4.

15. Question :
Find the value of k for which the equations have infinite solutions.
(k-3)x+3y=k and kx+ky=12
Explanation:
For infinite solutions, we have the condition
(k-3)/k=3/k=k/12
(k-3)/k=3/k and 3/k=k/12
k(k-3)=3k and 3*12=k*k
k^2-3k=3k and k^2=36
k^2=6k and k=-6,6
k=0, 6 and k = -6, 6
k = 6

16. Question :
Find the value of a for which a+5, 3a-10 and 2a+a/2 are in AP
Explanation:
Since a+5, 3a-10 and 2a+a/2 are in AP, we have
2(3a-10)=(a+5)+(2a+a/2)
6a-20=a+5+2a+a/2
6a-20=(6a+a)/2+5
12a-40=7a+10
12a-7a=10+40
5a=50
a=10

17. Question :
A retailer buys 25 items from a wholesaler. The total price for which the flawless items were sold was equal to the price that he had paid for buying the 25 items and 20% of the items were defected. Find the gain percent of the retailer.
Explanation:
Number of items purchased = 25
Defected items = 20/100*25
= 500/100 = 5
Let the cost price of each item be Rs.x
Total cost price = 25x
Number of items bought = 25
Total selling price = 25x
Number of items sold = 25-5=20
Gain% = [(25x/20-25x/25)/25x/25]*100
= (25x-20x)/20x*100
=25%

18. Question :
Find the sum of all two-digit natural numbers that are divisible by 4.
Explanation:
The smallest and thelargest such numbers are 12 and 96
The numbers form an AP with common difference d = 4 and the first term a = 12
The last term is 96 and let there be n terms in the AP
nth term is given by
Tn= a+(n-1)d
96=12+(n-1)4
96=12+4n-4
88=4n
n=88/4=22
Sum of n terms is given Sn=(n/2)[2a+(n-1)d]
=(22/2)[2*12+(22-1)4]
=11(24+21*4)
=11(24+84)
=11*108=1188
The sum of the required numbers is 1188

19. Question :
A's expenses are 20% less than B's expenses. How much percent more are B's expenses than A's expenses?
Explanation:
Let the expenses of A and B be x and y respectively.
x=y-20% of y
x=y-20y/100
x=(100y-20y)/100
x=80y/100
y=100/80x
Required percentage = (y-x)/x*100
= (100x/80-x)/x*100
= (100x-80x)/(80x)*100
=20/80*100
= 25%

20. Question :
The breadth of a rectangle becomes one and a half times itself and the length increases by 40%. What is the percentage change in the area of the rectangle?
Explanation:
Let the original length and breadth be l and b respectively
Area = l*b
New length = l+40% of l
= l+40l/100
=140l/100
= 3b/2
New area= 140l/100*3b/2
= 140*3lb/200
= 21lb/10
Percentage increase=increase/original area*100
= (21lb/10 - lb)/lb*100
= (21-10)/10*100
= 1100/10=110%

21. Question :
The volume of a cylindrical pipe is 17600 cc. Its height is 14cm and the external radius is 25cm. Find the thickness of the cylinder.
Explanation:
Volume of the cylindrical pipe = pi(R^2-r^2)h, where R and r are the external and internal radii and h is the height of the cylinder.
pi(R^2-r^2)h=17600
22/7(25^2-r^2)14=17600
22*2*(625-r^2) = 17600
r^2=625-400
r^2=225
r=sqrt(225)=15
R=25
r=15
Thickness = R-r
=25-15=10 cm
[x^2=x*x]

22. Question :
The height and radius of the base of a right circular cone are in the ratio 3:4. The ratio of the total surface area to the curved surface area is 9:5. The sum of the total surface area and the curved surface area of the cone is 224*pi sq.cm. Find the radius of the base.
Explanation:
Let the height and radius of the base be 3x and 4x respectively.
Let the total surface area and curved surface area be 9y and 5y respectively.
9y+5y=224*pi
14y=224pi
y=224pi/12=16pi
Curved surface area = 5y = 5*16pi = 80pi ...(1)
Curved surface area =pi*r*l, where r is the radius of the base and l is the slant height
l=sqrt[(3x)^2+(4x)^2]
=sqrt(9x^2+16x^2)
=sqrt(25x) = 5x
Hence, curved surface area = pi*r*l = pi*4x*5x
=20pi*x^2
Equating with (1), we get
20pi*x^2=80pi
x^2=80/20=4
x=2
[x^2=x*x]

23. Question :
The distance between the points (x,-1) and (3,2) is 5. Find the value of x if the point lies in the fourth quadrant.
The distance between the points is given by
Sqrt[(x-3)^2+(-1-2)^2]=5
Squaring both sides, we get
(x-3)^2+(-3)^2=25
x^2-6x+9+9=25
x^2-6x-7=0
x^2-7x+x-7=0
x(x-7)+1(x-7)=0
x=7,-1
x=7 since (x,-1) lies in the fourth quadrant.

24. Question :
The sum of the squares of two positive numbers is 340. The sum of the numbers is 22. Find the difference between the numbers.
Explanation:
Let the numbers be x and y
x+y=22 ...(1)
x^2+y^2=340 ...(2)
Put x = 22-y from (1) in (2)
(22-y)^2+y^2=340
484-44y+y^2+y^2=340
2y^2-44y+484-340
y^2-22y+72=0
y^2-18y-4y+72=0
y(y-18)-4(y-18)=0
y=4, 18
When y = 4, x = 18
When y = 18, x = 4
Difference of the numbers = 18-4=14
[x^2=x*x]

25. Question :
Sam deposited Rs. 30000 in a bank at 10% per annum and Rs.40000 in another bank at a certain rate of intertest per annum. What was this rate of interest if the rate of interest on the whole amount was (78/7)%?
Explanation:
Simple interest is given by
SI= P*R*T/100, where P, R and T are the principle, rate and time.
Let the time be 1 year
SI for 30000= 30000*10*1/100 = 3000
Let the interest on Rs.40000 be x%
SI for 40000= 40000*x*1/100 = 400x
Total SI = 3000 + 400x
Total principal = 30000 + 40000 = 70000
From the given conditions
3000 + 400x = 70000*78/7*1/100
3000 + 400x = 7800
400x = 7800-3000
x = 4800/400 = 12
The required rate is 12%

26. Question
What is the sum of all integers between 83 and 718 which are multiples of 5?
Explanation:
The numbers form an AP with first term as 85 and last term 715.
Let the number of terms be n and the common difference will be 5.
The last term is given by a+(n-1)d, where a is the first term and d is the common difference
715 = 85 + (n-1)5
n=126+1=127
Sum of n terms of an AP with a as the first term and l as the last term is given by
Sn = (n/2)(a+l)
Sum of all multiples of 5 lying between 83 and 718 = 127(85+715)/2
= 127*400 = 50800

27. Question :
A car moves 16 meters in the first minute, 48 meters in the second minute, 80 meters in the third, 112 meters in the fourth and so on. How far in meters from its starting point is it after 11 minutes?
Explanation:
The distances moved in first, second, third and further minutes are 16, 48, 80, 112…
These form an AP with first term 16 and common difference 32.
We have to find the sum of 11 terms
The sum of n terms of an AP with n terms with a as its first term and d as its common difference is givne by
Sum = (n/2)[2a+(n-1)d]
= (11/2)[2*16+(11-1)*32]
= (11/2)(32+320)
=1936m

28. Question :
A light bulb blinks at 12 noon. It then blinks after 4 seconds, then after 8 seconds, then after 12 seconds and so on. How many times shall it have already blinked until it blinks at12.08pm?
Explanation:
The seconds after which the light bulb blinks are 0, 4, 8, 12 …These form an AP with common difference 4 and first term 0.
Let the number of times it blinks in 8 minutes be n.
Sum of n terms = 8*60 = 480
Sum of n terms in an AP where a is the first term and d is the common difference is given by
Sn = (n/2)[2a+(n-1)d]
(n/2)[2*0+4(n-1)] = 480
2n^2-2n-480=0
n^2-n-240=0
n^2-16n+15n-240=0
n(n-16)+15(n-16)=0
n=16

29. Question :
The weight of a woman reduces by 10% every year. If she weighs 100kg today, then after how many years will her weight be 72.9 kg?
Explanation:
This problem is similar to a compound interest problem where the rate is negative. Let the number of years be n.
Weight = 100(1-10/100)^n = 72.9
100(0.9)^n = 72.9
(0.9)^n = 0.729
n = 3
Her weight will be 72.9 kg afer 3 years

30. Question :
The LCM of 3/4, 5/6 and 7/8 is ..../2. Fill in the blank with the appropriate number.