GRE Questions

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The given puzzle follows a logic that,

2 + 10 = 24 => 2 + 10 = 12 x 2 = 24

3 + 6 = 27 => 3 + 6 = 9 x 3 = 27

7 + 2 = 63 = 7 + 2 = 9 x 7 = 63

Similarly, 5 + 3 = 40 => 5 + 3 = 8 x 5 = 40.

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

Here the given number analogy follows two patterns that,

6 

18 --> 6 x 3

4

12 --> 4 x 3

OR

6

18 ----> 6x6/2 = 18

4

8 ----> 4x4/2 = 8

As 12 is not available in the options, 8 is correct answer.

S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 . 

S.I. for 2 years = Rs (156 x 23 x 2)= Rs 208. 

Therefore, Principal = Rs (1008 - 208) = Rs 800. 

Now, P = 800, T= 2 and S.I. = 208.

Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%

Here the given words represent synonyms.

Exhibit and Display are synonyms for each other. Similarly, Send and Emit are synonyms for each other.

Hence, Exhibit : display :: send : Emit.

Let each instalment be Rs.x .

1st year =  [x + (x * 12 * 2)/100] 

2nd year = [ x + (x *12 * 1)/100] 

3rd year = x 

Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092 

3x + ( 24x/100 ) + ( 12x/100 )  = 1092 

336x =109200

Therefore, x = 325 

Each instalment = Rs. 325

This is a quadratic equation. To find n we multiply the last terms in each bracket together (2 times -5) = -10

To find ‘a’ we multiply the first terms in each bracket together (3x times 2x) = 6x2, So a = 6.

Thus a - n = 16

To find k we multiply the first term in the first bracket by the second term in the second bracket and add the result to the product of the second term in the first bracket and the first term in the second. (-15x + 4x) = -11x. So k = -11

16 - 11 = 5