Exam-level Analytical Puzzles: XAT practice questions

Solution

G is in seat 4 and A is to its right. If A is in 5 or 6, B must be in the other of 5 and 6 (B cannot be in 4 or 7). Then C, D, E and F take seats 1, 2, 3 and 7: D two right of E forces E = 1 and D = 3, so C and F are in 2 and 7; C cannot be in 7, so F = 7, which is not to the left of C. So A = 7 and B = 6.

That leaves seats 1, 2, 3 and 5 for C, D, E and F, with D two seats right of E (E in 1 and D in 3, or E in 3 and D in 5) and F not in 3 or 5 (next to G).

E = 1, D = 3: F = 2 and C = 5 (C not at an end, right of F). E = 3, D = 5: F = 1 and C = 2. So exactly 2 arrangements: E F D G C B A and F C E G D B A.

Solution

In both possible arrangements (E F D G C B A and F C E G D B A), B is in seat 6.

Solution

Only E F D G C B A has C (seat 5) to the right of G (seat 4). Seat 1 is E.

Solution

In E F D G C B A, D (3) is next to G (4); in F C E G D B A, D (5) is next to G (4). The other statements hold in only one of the two arrangements.

Solution

The condition 'if B then H' already forces H. Checking every valid committee with B (BCEH, BCFH, BDEH, BDFH, BFGH), only H appears in all of them.

Solution

Listing all valid committees with one man: AFGH, BFGH, CEFG, CEFH, DEFG, DEFH, that is 6.

Solution

A and B are both in Finance, so the other two members must be Marketing women G and H. But G needs F, which would make three Finance members. So A and B can never be chosen together.

The other pairs all appear in valid committees, for example ACFH, BCEH, CEFH and AFGH.

Solution

Choose the two Finance members from A, B, E, F and two Marketing members from C, D, G, H, then apply the rules. The valid committees are: ACFG, ACFH, ADFG, ADFH, AFGH, BCEH, BCFH, BDEH, BDFH, BFGH, CEFG, CEFH, DEFG, DEFH, EFGH: 15 in all.