GRE Questions

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Let the sum be Rs. x.

I  gives, S.I. = Rs. 7000 and T = 7 years.

II gives, Sum + S.I. for 5 years = 2 x Sum Sum = S.I. for 5 years.

Now, S.I. for 7 years = Rs. 7000.

therefore, S.I. for 1 year =  Rs. 1000.

Thus, I and II both are needed to get the answer.

Tadpole is the larval stage in all the amphibians whereas larva in all the insects.

Let the radii of the smaller and the larger circles be 's' m and 'l' m respectively.  

 2πs = 264 and 2πl = 352  

 s = 2642π and l = 3522π   

Difference between the areas =πl2-πs2   

  = π1762π2-1322π2 

  = 1762π-1322π  

  = (176 - 132)(176 + 132) / π  

  = (44 x 308) / (22/7) = 4312 sq.m

 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.

P = 68000, R =503%  & T = 9 months (3/4 years)  

 S.I=p×r×t100=(6800×503×34)100=Rs.8500

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.

When ice melts it changes its physical state to water. Similarly, when rock melts it changes into lava or magma.

Hence, ice is to water as rock is to lava.