Each of P, Q, R, S, T, U and V has an exam on a different day of a week, starting from Monday and ending on Sunday of the same week. T has an exam on Thursday. U has an exam immediately before T and P has an exam immediately after T. Only 3 people have exams between Q and V. R has an exam immediately after V. How many people have exams before Q?

Solution:

Days of the week: Monday to Sunday (7 days)

Let’s assign positions 1 to 7:
1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday, 7 = Sunday

Step 1: Given

• T is on Thursday → position 4
• U is immediately before T → position 3
• P is immediately after T → position 5

So:

• U = 3, T = 4, P = 5

Step 2: Remaining people: Q, R, S, V

We are told:

• Only 3 people have exams between Q and V
  → positions differ by 4

Possible pairs:

• (1,5), (2,6), (3,7)

But positions 3, 4, 5 are already occupied → eliminate pairs involving them.

So valid pair: Q and V at positions 1 and 5 or 2 and 6 or 3 and 7

But 5 is already P, so Q and V cannot include 5.

Thus possible: (1,5) invalid, (2,6) valid, (3,7) invalid (3 is U)

So:

• Q and V must be at positions 2 and 6

Step 3: R is immediately after V

If V = 2 → R = 3 (but 3 is U ❌)
So V = 6 → R = 7

Thus:

• V = 6, R = 7
• Q = 2

Step 4: Final arrangement:

1 = S (remaining)
2 = Q
3 = U
4 = T
5 = P
6 = V
7 = R

Step 5: People before Q

Q is at position 2 → only 1 person before it (position 1)

Final Answer: 1