GRE Questions

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ADDRESS is what the kind of dress can never be worn by anyone.

The given series follows the rule,
+(11x1), +(11x3), +(11x5), +(11x7), +(11x9),....

So starting from 7, 7 + 11 = 18, 18 + 33 = 51, 51 + 55 = 106, ...
So wrong number is 40 which is to be 51.

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

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

100 P2 = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

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

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

 M→-2K

 P→-2N

Therefore,  D→-2B and G→-2E => BE

Let, Hema attempted 'k' sum correctly, then

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

so 26 correct sums.

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.