Non Verbal Reasoning Questions

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.

In the figure, there are 9 columns containing 5 cubes each, 7 columns containing 4 cubes each, 5 columns containing 3 cubes each and 1 column containing 1 cube.

∴Total number of cubes = (9 x 5) + (7 x 4) + (5 x 3) + (1 x 1) = 89

Draw out the diagram and you will see that the width of the picture becomes 8.5 - 3 = 5.5

The length becomes 10 - 3 = 7. The area is now 5.5 x 7 = 38.5

From figures X and Y, we conclude that dot, circle, square and cross lie adjacent to the triangle. Therefore, the arrow must lie opposite the triangle. From figures X and Z, we conclude that dot, triangle, arrow and cross lie adjacent to the circle. Therefore, the square must lie opposite the circle. Thus, the arrow lies opposite the triangle, the square lies opposite the circle and consequently, the cross lies opposite the dot.

As analysed above, the cross lies opposite the dot.

One line is increasing in each subsequent figue.

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.