Analytical 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.

The given figure can be labelled as :

Straight lines :

The number of straight lines are 19

i.e. BC, CD, BD, AF, FE, AE, AB, GH, IJ, KL, DE, AG, BH, HI, GJ, IL, JK, KE and DL.

The given figure can be labelled as shown :

The horizontal lines are AK, BJ, CI, DH and EG i.e. 5 in number.
The vertical lines are AE, LF and KG i.e. 3 in number.
The slanting lines are LC, CF, FI, LI, EK and AG i.e. 6 in number.
Thus, there are 5 + 3 + 6 = 14 straight lines in the figure.

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.

The horizontal lines are DE, FH, IL and BC i.e. 4 in number.

The slanting lines are AC, DO, FN, IM, AB, EM and HN i.e. 7 in number.

Thus, there are 4 + 7 = 11 straight lines in the figure.

The simplest triangles are IJO, BCJ, CDK, KQL, MLQ, GFM, GHN and NIO i.e. 8 in number.

The triangles composed of two components each are ABO, AHO, NIJ, IGP, ICP, DEQ, FEQ, KLM, LCP and LGP i.e.10 in number.

The triangles composed of four components each are HAB, DEF, LGI, GIC, ICL and GLC i.e. 6 in number.

Total number of triangles in the figure = 8 + 10 + 6 = 24.

The given figure can be labelled as shown :

The spaces P, Q and R have to be shaded by three different colours definitely (since each of these three spaces lies adjacent to the other two).
Now, in order that no two adjacent spaces be shaded by the same colour, the spaces T, U and S must be shaded with the colours of the spaces P, Q and R respectively.
Also the spaces X, V and W must be shaded with the colours of the spaces S, T and U respectively i.e. with the colours of the spaces R, P and Q respectively. Thus, minimum three colour pencils are required.