Select a suitable figure from the Answer Figures that would replace the question mark (?).

Problem Figures:                            Answer Figures:

    (A)     (B)      (C)     (D)                  (1)      (2)      (3)      (4)      (5)

The figure may be labelled as shown.

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.

Two and three half-leaves are added to the figure alternately. The addition of half-leaves takes place in an ACW direction.

The number 6 is common to both the positions of the dice. We assume the dice in fig. (ii) to be rotated so that 6 remains on the FR-LH face (i. e. on face IV as per activity 1) and the numbers 1 and 3 move to the faces hidden behind 4 and 5 respectively [in fig. (i)]. Then, 5 lies on FR-RH face (i.e. face I), 4 lies on Top face {i.e. face V), 3 lies on RR-LH face (i.e. face III) and 1 lies on Bottom face (i.e. face VI). Thus, 3 lies opposite 5. Hence, when the top is 3, then the number at the bottom is 5.

The outer element rotates I35 ^oCW and the inner element rotates 135 ^oACW.

The figure is rotated through 90 ^oCW; the triangle is converted to an arrowhead and the zig-zag or the lines inside the main figure turn outwards.

In all the figures except fig. (4), the two elements in the vertically opposite corners are laterally inverted images of each other.