Aptitude and Reasoning Questions

Volume of rod = πr2h = πr2×r = πr3  

Volume of one spherical ball =r/2 

Volume of the spherical ball = 43πr38  

Number of balls=Volumeof rod/volumeof one ball

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.

Taking the 2 investments to be 3x and 5x respectively
Total income of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25
Therefore, Gain% = 0.25/8 x 100 = 3.125 %.

Justification:      

In the first and second statements, the common digit is '2' and the common word is 'from'.

So, '2' is the code for 'from'.

In the first and third statements, the common  code digit is '8' and the common word is 'paper'.

So, '8' is the code for 'paper'.

As shown above, '8' is the code for 'paper'.

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

Given 45 kg for painting 

And 6% 0f 45 for wasting => 6x45/100 = 2.7 kg

Now total paint required = 45+2.7 = 47.7 kgs

But tins are available in only 4kgs

Then, no. of tins req = 47.7/4 = 11.4 => 12 tins