GRE Questions

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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

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.

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.

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%

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

3, 7, 6, 5, 9, 3, 12, 1, 15, ... has two different series

1) 3, 6, 9, 12, 15.....

2) 7, 5, 3, 1, .....

-1 is the next number.

Let the sum be Rs. x

a. gives, S.I = Rs. 9000 and time = 9 years.

b. gives, Sum + S.I for 6 years = 2 x Sum

--> Sum = S.I for 6 years.

Now, S.I for 9 years = Rs. 9000

S.I for 1 year = Rs. 9000/9 = Rs. 1000.

S.I for 6 years = Rs. (1000 x 6)= Rs. 6000.

--> x = Rs. 6000

Thus, both a and b are necessary to answer the question.

I.  (a - 3)(a - 4) = 0

=> a = 3, 4

II.   (b - 2)(b - 1) = 0

=> b = 1, 2

=> a > b