GRE Questions

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Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

The path traced by planet is called Orbit. Similarly, the path traced by projectile is called Trajectory.

Hence, Planet : Orbit : : Projectile : Trajectory.

Let the age of ruler is x so that of son = x/2 (given)
Now according to the given condition
(x/6) + (x/12) + (x/7) + 5 + (x/2) + 4 = x
=> x = 84

The letters move two places backward each in the alphabet as given below:

 M→-2K

 P→-2N

Therefore,  D→-2B and G→-2E => BE

Any person can have only 1 birthday i.e, the day he was born. And every year we celebrate it's anniversaries.

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

Therefore, the radius of the circle implies the length of the rope = 30 mts.