GRE Questions

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 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

216004989600=0.043

Let’s assume John bought the laptop for Rs. 100
He then sells it to Mark at a loss of 20% => Rs. 80,
Who consequently sells it to Kevin at a profit of 25% =>  Rs. 100
Then, Kevin sells it back to Mark for Rs. 90 (Rs.4.50 for every Rs. 5 that he had paid initially), so he suffers a loss of Rs. 10
Therefore, for Rs. 100 there is a loss of Rs. 10,
Then for Rs. 1,75,000 there will be a loss of Rs. 17,500.

This question concerns a committee’s decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that K and N are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.

The fourth condition given in the passage on which this question is based requires that exactly two of K, N, and J are reduced. Since the question asks us to suppose that both K and N are reduced, we know that J must not be reduced:

Reduced         ::      K, N
Not reduced   ::      J

The second condition requires that if L is reduced, neither N nor O is reduced. So L and N cannot both be reduced. Here, since N is reduced, we know that L cannot be. Thus, adding this to what we’ve determined so far, we know that J and L are a pair of areas that cannot both be reduced if both K and N are reduced:

Reduced        ::      K, N
Not reduced  ::      J, L

Answer choice (B) is therefore the correct answer.

Here the letters follow the rule (AxZ)+(A+Z)

The number corresponding to the letters are used in the rule

A=1, Z=26 => 26x1 + 1 + 26 = 26+27 = 53 =  053

(LxP)+(L+P) = (12 x16) + 12 + 16 = 192+28 = 220

Similarly (11x12) + 11 + 12 => 132 + 23 = 155 

To solve a pair of simultaneous equations such as those given we can add or subtract them.

Adding we get 4x + 4y = 20

Therefore 2x + 2y = 10

1. Answer (b)

Santosh , Ujjwal and Varun must each receive at least three pieces, but Tarun must receive at least four pieces. The maximum number of pieces that any person may receive is six ( 6 + 3+ 3+3 = 15 ), and only Tarun may receive six pieces. Ujjwal and Varun each may receive as many as five pieces, while Santosh may receive only four pieces at maximum.
At least one pear.

2. Answer (c)

Varun may receive no more than five pieces and Tarun may receive no fewer than four pieces. As a result, if Varun receives more pieces than Tarun, then Varun must receive five pieces, and Tarun must receive four pieces. Thus, statement (a) must be true. Since Varun will receive five pieces, none of the other three friends will receive any apple, accordingly, each of the other three friends must receive at least one orange and at least one pear. Thus, statement (b) and (d) must be true.

They walked and did this.

It was never specified that the boat was at sea. Since this logic is observed by them to tackle situation.

Error is in the first part of the sentence. 'Whole' is used with singular nouns.
The correct sentence is - 'All Northern states in India are reeling under the devastating of the drought.