GRE Questions

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Let Cost Price(C.P) = P
gain% = {(S.P-C.P)/C.P} x 100
25 = {(1540-P)/P} x 100
25/100 = (1540-P)/P
=> P = 4(1540)-4P
=> 5P = 4(1540)
=> P = 1232
So, Cost Price = Rs. 1232

From the given data,

18 snakes can eat 18 bananas in 18 hrs.

=> 1 snake can eat 1 banana in 18 hrs

=> 79 snakes can eat 79 bananas in 18 hrs.

Hence, 18 hrs.

From the given information :

D O W N A M E == 5 @ 9 # 6 % 3
MODE = %@53

0.127 expressed as a fraction = 127/1000

1/8 can also be expressed as (1 x 125) / (8 x 125) = 125/1000

The difference is 2/1000, which is 1/500

7th stop only one was left and in each preceding stop twice the nos... there were so 6 stops before 7th.

so 26=64

Total number of female members = 115

Let the number of males = x

Number of females = y

=> x + y = 115  ......(1)

According to the given data,

x + y(0.50) = 69 ......(2)

From (1) & (2), we get

0.5y = 46

=> y = 46 x 2 = 92

=> x = 115 - 92 = 23

Hence, required number of gentlemen in the family = x = 23.

Let the the age of the elder boy = E

Let the the age of the younger boy = Y

Given that Y = cube root of EY

=> Y3 = EY => E = Y2 .....(1)
By the condition of number replacement the age of the father is YE

The Mother's age = EY/2

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

Then now from the given options we can identify which satisfies the all the conditions.

Here Y =2 and E = 4 satisfies all the conditions.

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.