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
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
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