If a* = (2a + 6)/4, and a = 3b + 1, then a* equals
A)0.5b + 1.5
B)1.5b + 2
C)1.5b - 2
D)6b + 7
E)b - 1/3
a*= (6b +2+6)/4 = (6b+8)/4
hence the answer is option B
Showing posts with label CAT. Show all posts
Showing posts with label CAT. Show all posts
Thursday, October 21, 2010
Tuesday, October 5, 2010
GMAT Question of the Day
In the rectangular solid , the three sides shown have areas 12, 15, and 20,respectively. What is the volume of the solid?
A. 60
B. 120
C. 450
D. 1,800
E. 3,600
Solution :
ab = 12
bc = 15
ac = 20
a=12/b
c=15/b
ac =20
(12*15)/b^2 = 20
20b^2=180
b=3 a=4 c=5
Volume = 3 * 4 * 5 = = 60
hence the answer is A
A. 60
B. 120
C. 450
D. 1,800
E. 3,600
Solution :
ab = 12
bc = 15
ac = 20
a=12/b
c=15/b
ac =20
(12*15)/b^2 = 20
20b^2=180
b=3 a=4 c=5
Volume = 3 * 4 * 5 = = 60
hence the answer is A
Friday, September 17, 2010
GMAT Question of the Day
john intended to place a mark 2 1/4 inches from one end of a narrow end 9 1/4 inches long, but he mistakenly placed the mark 2 1/4 inches from the other end. How far, in inches, is his mark from the location where he intended to place it?
A) 7
B)4 3/4
C)2 1/4
D)5
E)5 1/4
Solution
total lenght is 9 1/4 and john need to put the mark at 2 1/4 form the end.
so 9 1/4 - 2 1/4 = 7
That is actual mark should be at 7 inches.
The mark was put in 2 1/4,
so 7- 2 1/4 = 4 3/4
A) 7
B)4 3/4
C)2 1/4
D)5
E)5 1/4
Solution
total lenght is 9 1/4 and john need to put the mark at 2 1/4 form the end.
so 9 1/4 - 2 1/4 = 7
That is actual mark should be at 7 inches.
The mark was put in 2 1/4,
so 7- 2 1/4 = 4 3/4
Wednesday, September 15, 2010
GMAT Question of the Day
What is the sum of the first 50 common numbers between the series 15,19,23....... and 14,19,24.........?
A) 25550
B) 24540
C) 24450
D) 25450
E) 25540
The given series are increasing by 4 and 5 respectively..
so the LCM=20
Hence they will have common numbers every 20...so common numbers will be like 19, 39, 59...
The common numbers is also an arithmetic series starting with 19 and with a difference of 20 between consecutive numbers....
therefore a=19
d=20
n=50
sum of arithmatic series
s = n/2(a+(n-1)*d)
= 25(19+49*20)
= 25450
A) 25550
B) 24540
C) 24450
D) 25450
E) 25540
The given series are increasing by 4 and 5 respectively..
so the LCM=20
Hence they will have common numbers every 20...so common numbers will be like 19, 39, 59...
The common numbers is also an arithmetic series starting with 19 and with a difference of 20 between consecutive numbers....
therefore a=19
d=20
n=50
sum of arithmatic series
s = n/2(a+(n-1)*d)
= 25(19+49*20)
= 25450
Monday, September 6, 2010
GMAT Question of the Day
A carpenter makes 3 beds every day. A military school needs to organize beds for 143 soldiers. If there are 5 carpenters working on the job, how many whole days in advance should they receive the order and start working in order to finish the right number of beds assuming that each bed is used by two soldiers?
A)3.
B)4.
C)5.
D)6.
E)7.
143 soldiers is not divisible by 2 so we need at least 72 beds (one will sleep alone).
The carpenters together can complete 15 beds a day.
72/15 is between 4 and 5 so they will need 5 days notice in order to complete the job on time.
The best answer is C.
A)3.
B)4.
C)5.
D)6.
E)7.
143 soldiers is not divisible by 2 so we need at least 72 beds (one will sleep alone).
The carpenters together can complete 15 beds a day.
72/15 is between 4 and 5 so they will need 5 days notice in order to complete the job on time.
The best answer is C.
Friday, September 3, 2010
GMAT Question of the Day
Car A travels at three times the average speed of car B. Car A started to travel at 12:00 Hrs, car B started to travel at 16:00 Hrs. What is the speed of car B (in Km/h) if the total distance that both cars traveled until 18:00 was 1000 Km?
A)10.
B)25.
C)30.
D)38.
E)50.
Let the speed of car B be X.
Then speed of car A is 3X.
Car A traveled 3X x 6 hours = 18X Km.
Car B traveled X x 2 hours = 2X Km.
20X=1000 ---> X = 50 Km/h.
The best answer is E.
A)10.
B)25.
C)30.
D)38.
E)50.
Let the speed of car B be X.
Then speed of car A is 3X.
Car A traveled 3X x 6 hours = 18X Km.
Car B traveled X x 2 hours = 2X Km.
20X=1000 ---> X = 50 Km/h.
The best answer is E.
Monday, August 30, 2010
GMAT Question of the day
A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
A)20
B)30
C)25
D)35
E)40
The solution is
(x-2) * ( 5 + 300/ x) =420
=> 5x - 600/x - 10 + 300 = 420
=> 5x^2 - 130x - 600=0
=> x^2 - 26x - 120 = 0
Solving we get, x = 30 or x = -4
x cannot be negative as it is the number of calculators
So, there were 30 calculators.
Hence the answer is B.
A)20
B)30
C)25
D)35
E)40
The solution is
(x-2) * ( 5 + 300/ x) =420
=> 5x - 600/x - 10 + 300 = 420
=> 5x^2 - 130x - 600=0
=> x^2 - 26x - 120 = 0
Solving we get, x = 30 or x = -4
x cannot be negative as it is the number of calculators
So, there were 30 calculators.
Hence the answer is B.
Sunday, August 22, 2010
GMAT Question of the day
A, B, C, D and E are 5 consecutive points on a straight line. If BC = 2CD, DE = 4, AB = 5 and AC = 11, what is the length of AE?
(a) 21
(b) 26
(c) 30
(d) 18
(e) 16
In order to find the length of AE, find the length of CD and BC first.
BC = AC – AB = 11 – 5 = 6.
BC = 2CD
CD = 3.
AE = 5 + 6 + 3 + 4 = 18.
The answer is D.
(a) 21
(b) 26
(c) 30
(d) 18
(e) 16
In order to find the length of AE, find the length of CD and BC first.
BC = AC – AB = 11 – 5 = 6.
BC = 2CD
CD = 3.
AE = 5 + 6 + 3 + 4 = 18.
The answer is D.
Wednesday, August 18, 2010
GMAT Question of the day
How many different subsets of the set {10,14,17,24} are there that contain an odd number of elements?
(a) 3
(b) 6
(c) 8
(d) 10
(e) 12
Option D : 8 is the answer.
The different subsets are
10,
14,
17,
24,
10, 14, 17
14, 17, 24
17, 24, 10
24, 10, 14
(a) 3
(b) 6
(c) 8
(d) 10
(e) 12
Option D : 8 is the answer.
The different subsets are
10,
14,
17,
24,
10, 14, 17
14, 17, 24
17, 24, 10
24, 10, 14
Tuesday, August 17, 2010
GMAT Question of the day
A box contains 90 nuts each of 100 gms and 100 bolts each of 150 gms. If the entire box weighs 35.5 kg., then the weight of the empty box is :
A)10 kg
B)10.5 kg
C)11 kg
D)11.5 kg
E)None of the above
Total weight of nuts = 90*100 = 9 Kg
Total weight of bolts = 100*150 = 15 Kg
Total weight of box with nuts and bolts = 35.5 Kg
So the Answer is D
A)10 kg
B)10.5 kg
C)11 kg
D)11.5 kg
E)None of the above
Total weight of nuts = 90*100 = 9 Kg
Total weight of bolts = 100*150 = 15 Kg
Total weight of box with nuts and bolts = 35.5 Kg
So the Answer is D
Sunday, August 15, 2010
GMAT Question of the day
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) 12
(B) 14
(C) 16
(D) 18
(E) 20
Race 1 :
ta = tb-6 ( because A beats B by 6 sec)
Race 2 :
Ta = tb+2 ( because A looses to B by 2 sec)
We know D= S * T
Therefore
480/Sa = 432/Sb -6
and
480/Sa = 336/Sb +2
Equating these two equations we get Sb = 12
ta,Sa stand for time taken by A and speed of A resp
of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
Race 1 :
ta = tb-6 ( because A beats B by 6 sec)
Race 2 :
Ta = tb+2 ( because A looses to B by 2 sec)
We know D= S * T
Therefore
480/Sa = 432/Sb -6
and
480/Sa = 336/Sb +2
Equating these two equations we get Sb = 12
ta,Sa stand for time taken by A and speed of A resp
Thursday, August 5, 2010
GMAT Question of the day
To which of the following shapes the area can’t be calculated if the perimeter is given?
(a) Circle.
(b) An isoceles right triangle.
(c) Rectangle.
(d) A regular Hexagon.
(e) Square.
The perimeter of a rectangle is 2a + 2b.
In order to calculate the area we need to know the multiplication of a x b.
The best answer is C.
(a) Circle.
(b) An isoceles right triangle.
(c) Rectangle.
(d) A regular Hexagon.
(e) Square.
The perimeter of a rectangle is 2a + 2b.
In order to calculate the area we need to know the multiplication of a x b.
The best answer is C.
Sunday, August 1, 2010
GMAT Question of the Day
In a triangle, one side is 6 Cm and another side is 9 Cm. which of the following can be the perimeter of the triangle?
(a) 18.
(b) 25.
(c) 30.
(d) 32.
(e) 34.
The third side of the triangle is larger than 3 (The difference between the other two) and should be smaller than 15 ( The sum of the other two).
The perimeter is between (6+9+3 = 18) and (6+9+15 = 30). The only answer that is in this range is B.
The best answer is B.
(a) 18.
(b) 25.
(c) 30.
(d) 32.
(e) 34.
The third side of the triangle is larger than 3 (The difference between the other two) and should be smaller than 15 ( The sum of the other two).
The perimeter is between (6+9+3 = 18) and (6+9+15 = 30). The only answer that is in this range is B.
The best answer is B.
Saturday, July 31, 2010
GMAT Question of the Day
19. If A2 + B2 = 15 and AB = 10, what is the value of the expression
(A – B)2 + (A + B)2 ?
(a) 10.
(b) 20.
(c) 30.
(d) 60.
(e) 70.
(A – B)2 + (A + B)2
=A2 – 2AB + B2 + A2 + 2AB + B2
=2(A2 + B2)
=30.
The best answer is C.
(A – B)2 + (A + B)2 ?
(a) 10.
(b) 20.
(c) 30.
(d) 60.
(e) 70.
(A – B)2 + (A + B)2
=A2 – 2AB + B2 + A2 + 2AB + B2
=2(A2 + B2)
=30.
The best answer is C.
Friday, July 30, 2010
GMAT Question of the Day
The value of some stock X changes according to the following method:
At the end of each month her value is doubled but due to commission the stock’s value is decreases by Rs.10. If the value at the beginning of January is Rs.A, what would be her value at the end of February?
(a) 4A – 10.
(b) 4A – 20.
(c) 4A – 30.
(d) 4A – 40.
(e) 4A – 50 .
At the end of January her value is 2A – 10.
At the end of February her value is (2 x (2A – 10) – 10
= 4A – 30).
The best answer is C.
At the end of each month her value is doubled but due to commission the stock’s value is decreases by Rs.10. If the value at the beginning of January is Rs.A, what would be her value at the end of February?
(a) 4A – 10.
(b) 4A – 20.
(c) 4A – 30.
(d) 4A – 40.
(e) 4A – 50 .
At the end of January her value is 2A – 10.
At the end of February her value is (2 x (2A – 10) – 10
= 4A – 30).
The best answer is C.
Thursday, July 29, 2010
GMAT Question of the Day
A Cell divides into two each round hour. If at a certain round hour, two Cells were placed in a jar, how many Cells will be in the jar in N hours?
(a) 2N
(b) 22N
(c) 2N+1
(d) 2N-1
(e) 2N
Let’s find the number of Cells in the first hours.
After one hour (N=1) there will be 4 Cells.
After two hours (N=2) there will be 8 Cells.
After three hours (N=3) there will be 16 Cells.
Therefore the formula that fits this series is 2N+1.
The best answer is C.
(a) 2N
(b) 22N
(c) 2N+1
(d) 2N-1
(e) 2N
Let’s find the number of Cells in the first hours.
After one hour (N=1) there will be 4 Cells.
After two hours (N=2) there will be 8 Cells.
After three hours (N=3) there will be 16 Cells.
Therefore the formula that fits this series is 2N+1.
The best answer is C.
Wednesday, July 28, 2010
GMAT Question of the Day
Alfa, Beta and Gamma are inner angles in a triangle. If Alfa = Beta + Gamma, what can’t be the size of Beta?
(a) 44 degrees.
(b) 45 degrees.
(c) 89 degrees.
(d) 90 degrees.
(e) There isn’t enough data to determine.
If Beta is 90 degrees than Alfa is bigger than 90 and the sum of the angles in the triangle will be bigger than 180 degrees.
The best answer is D.
(a) 44 degrees.
(b) 45 degrees.
(c) 89 degrees.
(d) 90 degrees.
(e) There isn’t enough data to determine.
If Beta is 90 degrees than Alfa is bigger than 90 and the sum of the angles in the triangle will be bigger than 180 degrees.
The best answer is D.
Wednesday, July 21, 2010
GMAT Question of the day
What is the sum of all the even numbers bigger than (-10) and smaller than 12?
(a) 2.
(b) 10.
(c) 0.
(d) 8.
(e) 4.
This is a series of numbers with a constant difference between them.
The first number is (-8) and the last is (10), there are 10 numbers in total.
Hence its in AP .
The formula for such a series is: ((-8 + 10) x 10)/2 = 10.
The best answer is B.
(a) 2.
(b) 10.
(c) 0.
(d) 8.
(e) 4.
This is a series of numbers with a constant difference between them.
The first number is (-8) and the last is (10), there are 10 numbers in total.
Hence its in AP .
The formula for such a series is: ((-8 + 10) x 10)/2 = 10.
The best answer is B.
Monday, July 19, 2010
GMAT Question of the Day
Chandler is building a fence in the following method: He grounds 10 poles, each 10 Cm thick, in 1 meter spaces from each other. He then connects the poles with a barbed wire. What is the total length of the fence?
(a) 11.
(b) 12.
(c) 9.9.
(d) 10.
(e) 13.
The total width of the poles is (10 x 0.1 = 1) meter.
There are 9 spaces between the poles, each 1 meter, so it’s another 9 meters.
The total length is (1 + 9 = 10) meters.
The best answer is D.
(a) 11.
(b) 12.
(c) 9.9.
(d) 10.
(e) 13.
The total width of the poles is (10 x 0.1 = 1) meter.
There are 9 spaces between the poles, each 1 meter, so it’s another 9 meters.
The total length is (1 + 9 = 10) meters.
The best answer is D.
Sunday, July 18, 2010
GMAT Question of the day
There are 200 cats in Cat-City. Out of the 200, 70 are street cats and the rest are domestic cats. 110 cats are gray, 30 out of the gray cats are domestic ones. How many domestic cats are there which are not gray in Cat-City?
(a) 90.
(b) 80.
(c) 50.
(d) 40.
(e) 25.
Out of 200 cats, 130 are domestic ones.
Out of 110 gray cats, 30 are street cats therefore 80 are grey and domestic ones.
Altogether there are 130 domestic cats, 80 are grey so (130 – 80) = 50 are not grey.
The best answer is C.
(a) 90.
(b) 80.
(c) 50.
(d) 40.
(e) 25.
Out of 200 cats, 130 are domestic ones.
Out of 110 gray cats, 30 are street cats therefore 80 are grey and domestic ones.
Altogether there are 130 domestic cats, 80 are grey so (130 – 80) = 50 are not grey.
The best answer is C.
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