Non Verbal Reasoning Questions

The simplest triangles are AML, LRK, KWD, DWJ, JXI, IYC, CYH, HTG, GOB, BOF, FNE and EMA i.e. 12 in number.

The triangles composed of two components each are AEL, KDJ, HIC and FBG i.e. 4 in number.

The triangles composed of three components each are APF, EQB, BQH, GVC, CVJ, IUD, DUL and KPA i.e. 8 in number.

The triangles composed of six components each are ASB, BSG, CSD, DSA, AKF, EBH, GGJ and IDL i.e. 8 in number.

The triangles composed of twelve components each are ADB, ABC, BCD and CDA i.e. 4 in number.

Total number of triangles in the figure = 12 + 4 + 8 + 8 + 4 = 36.

In the figure, there are

5 columns containing 4 cubes each;

33 columns containing 3 cubes each;

9 columns containing 2 cubes each and 3 columns containing 1 cube each.

∴ Total Number of cubes = ( 5 x 4) + (33 x 3) + (9 x 2) + (3 x 1) = 20 + 99 + 18 + 3 = 140

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.

Answer : C

Explanation : 

Since the examiner likes D, it means he must like C also, since the question states that those who like D like C also.

Since last year a question was asked on F, who is a modern poet, this year a question will be asked on an ancient poet which means A,B,C or D

Now the examiner is not likely to ask question on D because he has written an article on him. So he will ask the question on C since he likes C as well

If 1 is adjacent to 2, 4 and 6 then either 3 or 5 lies opposite to 1. So, the numbers 3 and 5 cannot lie opposite to each other. Hence, 3 is adjacent to 5 (necessarily).

If 1 is adjacent to 2, 3 and 5, then either 4 or 6 lies opposite to 1. So, the numbers 4 and 6 cannot lie opposite to each other. Hence, 4 necessarily lies adjacent to 6.

When the sheet in fig. (A) is folded, then one of the faces of the cube formed will be of the form   and this face will lie opposite the face bearing a square. Also, one of the blank faces lies opposite another blank face and the third blank face lies opposite the face bearing an '=' sign. Clearly, all the three blank faces cannot appear adjacent to each other. So, the cube shown in fig. (2) which has all the three blank faces adjacent to each other cannot be formed. Hence, only the cubes shown in figures A, C and D can be formed.