Non Verbal Reasoning Questions

The figure may be labelled as shown

Rectangles :

The simplest rectangles are CVSR, VETS, RSWM and STKW i.e 4 in number.

The rectangles composed of two components each are CETR, VEKW, RTKM and CVWM i.e 4 in number.

The rectangles composed of three components each are ACRP, PRMO, EGHT and THIK i.e 4 in number.

The rectangles composed of four components each are CEKM, AVSP, PSWO,VGHS and SHIW i.e 5 in number.

The rectangles composed of five components each are AETP, PTKO, CGHR and RHIM i.e 4 in number.

The rectangles composed of six components each are ACMO and EGIK i.e 2 in number.

The rectangles composed of eight components each are AGHP, PHIO, AVWO and VGIW i.e 4 in number.

The rectangles composed of ten components each are AEKO and CGIM i.e 2 in number.

AGIO is the only rectangle having sixteen components

Total number of rectangles in the given figure = 4 + 4 + 4 + 5 + 4 + 2 + 4 + 2 + 1 = 30.

Hexagons :

The hexagons in the given figure are CDEKLM, CEUKMQ, CFHJMQ, BEUKNP and BFHJNP. So, there are 5 hexagons in the given figure.

Draw out the diagram and you will see that the width of the picture becomes 8.5 - 3 = 5.5

The length becomes 10 - 3 = 7. The area is now 5.5 x 7 = 38.5

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

(Total numbers of cubes in a line  x  Number of stack / tower) + ...

= (6x1)+(5x2)+(4x3)+(3x4)+(5x2)+(6x1)

= 6+10+12+12+10+6 = 56

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.