GRE Questions

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

Hence, Planet : Orbit : : Projectile : Trajectory.

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.

Let, Hema attempted 'k' sum correctly, then

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

so 26 correct sums.

5 L from A are V --- 5 Letters from Alphabets are Vowels.

Given P = 54 and M = 80

According to the question,

Let ‘A’ years ago, M = 3P

=> M - A/P - A = 3

=> 80 - A = 3(54 - A)

=> 80 - A = 162 - 3A

=> 2A = 82

=> A = 41

Therefore, A = 41 years ago Sweety’s mother was 3 times of Sweety’s age.

Hence, at the age S = 13 and M = 39.

The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1

Let the distance in one direction = k kms

Speed in still water = 4.5 kmph

Speed of river = 1.5

Hence, speed in upstream = 4.5 - 1.5 = 3 kmph

Speed in downstream = 4.5 + 1.5 = 6 kmph

Time taken by Rajesh to row upwards = k/3 hrs

Time taken by Rajesh to row downwards = k/6 hrs

Now, required Average speed =$$\begin{gathered} Total distanceTotal speed = 2kk3 + k6 \\ = 2k x 186k + 3k = 4 kmph \end{gathered}$$ 

Therefore, the average speed of the whole journey = 4kmph.