Non Verbal Reasoning Questions

From figures (i) and (ii) we conclude that the number 1, 2, 3 and 4 appear adjacent to 6. Thus, the number 5 will appear opposite 6. Therefore, when six is at the bottom, then 5 will be at the top.

From positions X and Y we conclude that 1, 5, 6 and 3 lie adjacent to 4. Therefore, 2 must lie opposite 4. From positions Y and Z we conclude that 4, 3, 2 and 5 lie adjacent to 6. Therefore, 1 must lie opposite 6. Thus, 2 lies opposite 4, 1 lies opposite 6 and consequently 5 lies opposite 3.

As analysed above, 1 lies opposite 6.

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.

1, 6, 9 are figures which are half shaded by slanting lines.
2, 4, 7 are all divided into equal parts (either three or four parts) by straight lines and also have a black circle at the centre.
3, 5, 8 have similar designs and have their four corners shaded black.

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.