Quiz

Yes, 44, but your Ment solution is too complicated for me. I'll try to understand it now, but I figured it out much simpler myself. My solution: The 120° angle will occur when the minute hand is 120° (2π/3 radians) or 240° (4π/3 radians) ahead of the hour hand. The angular velocity of the minute hand is w_m = 2π rad/hour. The angular velocity of the hour hand is w_h = π/6 rad/hour. First, I calculated how many times the hour hand would be ahead of the minute hand by 2π/3 radians. It will be ahead when the angle between them is 2π/3 + 2πn, where n is the number of revolutions it has already overtaken. But since we already have one instance when 2πn = 0 (n = 0), and we need to find the number of times, I wrote the expression for the angle as 2π/3 + 2π(n₁₂₀ - 1). The angle that the minute hand travels in time t hours is w_m*t. The angle that the hour hand travels in time t hours is w_h*t. The angle by which the minute hand will be ahead of the hour hand is: w_m*t - w_h*t = (w_m - w_h)*t Thus, we get the following to calculate the number of times the minute hand will overtake the hour hand: 2π/3 + 2π(n₁₂₀ - 1) = (w_m - w_h)*t t = 24 hours. 2π/3 + 2π(n₁₂₀ - 1) = (2π - π/6)*24 n₁₂₀ = 68/3, which is 22 whole times. Then, I found how many times the minute hand overtakes the hour hand by 240°. Here, the angle is 4π/3 + 2π(n₂₄₀ - 1). The expression is: 4π/3 + 2π(n₂₄₀ - 1) = (2π - π/6)*24 n₂₄₀ = 67/3, which is also 22 whole times. Total: n₁₂₀ + n₂₄₀ = 22 + 22 = 44 times. Added 7 minutes later Since Heroist was the first to write the correct answer and has a different explanation, I think it's right to say that he was the first to solve it. Now, it's your turn, Heroist!