Number and Logic Puzzles

Letting the weight of the brick be x, the equation is x = 1 + 0.5x. Solving for x yields 0.5x = 1, which means x = 2 kg. Distractors like 1.5 kg or 3 kg fail this basic algebraic balance.

The key insight is that the fly keeps flying continuously until the trains collide. The trains start 200 km apart and approach each other at 50 km/h each, so their relative speed is 100 km/h. The collision time is 200 km ÷ 100 km/h = 2 hours. The fly travels at 75 km/h for those 2 hours, covering 75 × 2 = 150 km. The complex back-and-forth path is irrelevant - we only need total time and speed.

To average 60 km/h over a 60 km trip, the total travel time must be exactly 1 hour. Traveling the first 30 km at 30 km/h takes 1 hour, leaving 0 seconds for the remaining half. Thus, it is physically impossible, making 'No Answer' correct.

Let current time be T. If it's two hours later (T+2), time to midnight = M-(T+2). If it's one hour later (T+1), time to midnight = M-(T+1). Given: M-(T+2) = 1/2[M-(T+1)]. Solving: 2(M-T-2) = M-T-1, so 2M-2T-4 = M-T-1, giving M-T = 3. At 9pm, time to midnight is 3 hours. Check: At 11pm (2hrs later), 1 hour to midnight. At 10pm (1hr later), 2 hours to midnight. 1 = 1/2(2) ✓

The minute and hour hands coincide at intervals of exactly 12/11 hours, which is 1 hour, 5 minutes, and 27.273 seconds. At this exact time, the second hand is at 27.273 seconds, which corresponds to an angle of approximately 131 degrees from the other two hands.

If the number is x, then 100000 + x is the number with 1 in front, and 10x + 1 is the number with 1 behind. The equation is 3(100000 + x) = 10x + 1. Solving gives 7x = 299999, so x = 42857.

If I have no siblings, 'my father's son' must be me. Therefore, the father of the man in the photo is me. This means the man in the photo is my son.

The pattern follows consecutive and alternating letters. First letters are S, T, U, V, W (consecutive alphabetically). Second letters are C, E, G, I, K (every other letter). Third letters are D, F, H, J, L (every other letter starting from D). This gives VIJ.

The pattern cycles the number position through the second, third, and fourth letter positions sequentially. The numbers 2, 3, 4, 5, 6 appear in positions 2, 3, 4, 2, 3 respectively. So BC3D places 3 in the third position, continuing the cycle.

The first and third letters of each triplet follow an alphabetical sequence (Q-R, R-S, S-T, T-U for the first; R-S, S-T, T-U, U-V for the third). The middle letter 'A' remains constant. Thus, the next is UAV.

The pattern consists of groups of three consecutive letters in reverse alphabetical order, moving backward through the alphabet: QPO, NML, KJI, (HGF), EDC. HGF follows KJI as the next set of three letters.