Non Verbal Reasoning Questions

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 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.

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.

Similar small figures are increasing by one in each of the bigger 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.

This would not mean that K and L will always be together. It just implies that, if K is there, then L will also be there.

At the same time, it can happen that L is there but K isn't.

Remember, the condition is on K, not on L.