Aptitude and Reasoning Questions

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.

100 cm is read as 102 cm

a = 100 x 100 sq.cm and b = 102 x 102 sq.cm

then, Increase in area = [ (102 x 102) - (100 x 100) ] sq.cm

percentage error =404100*100*100= 4.04%

In the figure, there are 9 columns containing 5 cubes each, 7 columns containing 4 cubes each, 5 columns containing 3 cubes each and 1 column containing 1 cube.

∴Total number of cubes = (9 x 5) + (7 x 4) + (5 x 3) + (1 x 1) = 89

First comments on the second. Appraiser comments on buildings and a Critic comments on books.

Hence, Appraiser : Building :: Critic : Book.

The statement doesn't give any clue to the use of language. Hence a is not implicit.
b is also not implicit because in the sentence 'only' indicates false meaning.

Let the sum be Rs. 100. Then,

S.I. for first 6 months = Rs.[ (100 x 10 x 1)/(100 x 2) ]= Rs.5

S.I. for last 6 months =Rs.[(102 x 10 x 1)/(100 x 2) ] = Rs.5.25

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25

Effective rate = (110.25 - 100) = 10.25%

4/3 π x 10 x 10 x 10 = 8 x 4/3 π x r x r x r
r = 5
4π x 5 x 5 = 100π