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The least common multiple of two numbers is 44 and the highest common factor of the numbers is 264. How many such pairs of numbers exist?

Correct Answer: 2

Explanation:

Product of two numbers = LCM of the numbers/HCF of the numbers
= 264/44
= 6
Pairs of numbers whose product is 6 are (1,6) and (2,3).
Two such pairs of numbers exist.

Question: 2

The average of the sacks lifted by each of 70 workers in a day is 30. Ten workers are sick on a paricular day and others have to share their work. On an average, how many additional sacks will be lifted by each worker on that day?

Correct Answer: 5

Explanation:

Total number of sacks lifted by each worker = Total workers * Average of sacks lifed
= 70 * 30
= 2100
Since 10 workers fall ill, there are 70-10=60 left to lift 2100 sacks.
Average sacks lifted by each worker = 2100/60
= 35
Each worker has to lift 35-30=5 additional sacks on an average.

Question: 3

In a game show, each participant has to pay $10 as entry fee. The game is played in sets of 7. There are first, second and third prizes of $20, $15 and $10 respectively, two consolation prizes of $5 each and two participants do not get any prize at all. The game is played 17 times, out of which the results were inconclusive on 3 occasions and hence no prizes were given. How much did the organizers profit in dollars?

Correct Answer: 420

Explanation:

Money earned by the organizers when the game was conducted 17 times = 17*7*10
= $1190

Money given as prizes in (17-3) = 14 successful games = 14 *($20 + $15 + $10 + $5*2)
= 14*$55 = $770

Profit of organizers = 1190-770
= $420
The organizers gained $420

Question: 4

A man weighs 70 kgs at the age of 65. He steadily loses weight in the next few years at the rate of 5%. Approximately how many kgs (rounded to the nearest whole number) does he weigh when he is 67 years old?

Correct Answer: 63

Explanation:

This problem can be compared to a compound interest problem where the rate of interest is negative.
Here the original weight is 70 kgs.
The time to be considered is (67-65) = 2 years
Weight at end of 2 years = P(1-r/100)^t, where P, r and t are the principle, rate and time.
= 70(1-5/100)^2
= 70(19/20)^2
= 70*19*19/(20*20)
= 63.175 kg
= 63 kgs

[r^t = r*r*r*...t times]

Question: 5

There are three teams A, B and C containing 5, 4 and 5 equally skilled members respectively. If teams A, B and C can complete a set of tasks in 2, 3 and 3 days respectively, then how many days will it take a fourth team to complete the work when it comprises of three members, one from each of the three teams?

Correct Answer: 4

Explanation:

Time taken by one member of team A to complete the work = 5*2 days
Work done by one member of team B in one day = 1/(5*2)
Similarly, work done by one member of team B in one day = 1/(4*3)
Work done by one member of team C in one day = 1/(5*3)
Work done by the three members of the fourth team in one day = 1/(5*2) + 1/(4*3) + 1/(5*3)
= (6+5+4)/(3*4*5) = 15/(3*4*5)
= 1/4
Time taken by the fourth team to complete the task = 4 days

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