- The clock is set right at 5 a.m. on Monday. - The clock loses 16 minutes in 24 hours. This means for every 24 hours of true time, the clock shows 24 hours - 16 minutes. - 24 hours true time = (24 60) minutes true time = 1440 minutes true time. - Clock shows: 1440 - 16 = 1424 minutes. - The ratio of (Time shown by clock) / (True time) is constant. - Ratio = 1424 / 1440. Simplify the fraction: 1424/1440 = 712/720 = 356/360 = 89/90. - So, for every 90 units of true time that pass, the faulty clock shows 89 units of time have passed. - The clock indicates 10 p.m. on Thursday. We need to find the total time elapsed on the faulty clock from the starting point (Monday 5 a.m.) to Thursday 10 p.m. - From Monday 5 a.m. to Thursday 5 a.m. is 3 full days, which is 3 24 = 72 hours. - From Thursday 5 a.m. to Thursday 10 p.m. is 17 hours (5 a.m. to 12 p.m. is 7 hours, 12 p.m. to 10 p.m. is 10 hours; 7+10=17). - Total time shown by the faulty clock = 72 hours + 17 hours = 89 hours. - Let T_clock be the time shown by the clock and T_true be the true time elapsed. - T_clock / T_true = 89 /
A clock is set right at 5 a.m. on Monday. The clock loses 16 minutes in 24 hours. What will be true time when clock indicates 10 p.m on Thursday of the same week?
Asked in GPSC Prelims CSP 1 2023 · Topic: Mechanics (Newton's Laws of Motion and their applications)
More Physics questions
- - 89 hours / T_true = 89 / 90. - T_true = 90 hours. - Now, calculate the true time that is 90 hours after Monday 5 a.m. - Monday 5 a.m. + 24 hours = Tuesday 5 a.m. - Monday 5 a.m. + 48 hours = Wednesday 5 a.m. - Monday 5 a.m. + 72 hours = Thursday 5 a.m. - Monday 5 a.m. + 90 hours = (Monday 5 a.m. + 72 hours) + 18 hours = Thursday 5 a.m. + 18 hours. - 18 hours after 5 a.m. is 5 a.m. + 12 hours (to 5 p.m.) + 6 hours = 5 p.m. + 6 hours = 11 p.m. - The true time is 11 p.m. on Thursday. 276. The wheels of bicycles P and Q have radii 30 cm and 40 cm, respectively. While travelling a certain distance, wheel of P required 5000 more revolutions than wheel of Q. If bicycle Q travelled this distance in 45 minutes, then how much was its speed, in km per hour?
- If a journey of 120 km is done by train and the rest by car, then it takes 8 hours for total 600 km journey. And if 200 km is done by train and the rest by car, then it takes 20 minutes more. Then what would be the ratio of the speed of the train to that of the car?
- $\frac{120}{v_T} + \frac{480}{v_C} = 8$. (Equation 1)Case 2: 200 km by train, remaining by car. Remaining distance $= 600 - 200 = 400$ km by car. Total time = 8 hours 20 minutes. 8 hours 20 minutes $= 8 + \frac{20}{60}$ hours $= 8 + \frac{1}{3}$ hours $= \frac{25}{3}$ hours. Time by train + Time by car = $\frac{25}{3}$. $ \frac{200}{v_T} + \frac{400}{v_C} = \frac{25} {3}$. (Equation 2)Let $x = \frac{1}{v_T}$ and $y = \frac{1}{v_C}$. The equations become: $120x + 480y = 8$ (Equation 1') $200x + 400y = \frac{25} {3}$ (Equation 2')Divide Equation 1' by 8: $15x + 60y = 1$. (Equation 1'')From Equation 1'', $15x = 1 - 60y$.Substitute this into Equation 2': $200 \left(\frac{1 - 60y}{15}\right) + 400y = \frac{25} {3}$. $\frac{40}{3} (1 - 60y) + 400y = \frac{25} {3}$. Multiply the entire equation by 3: $40(1 - 60y) + 1200y = 25$. $40 - 2400y + 1200y = 25$. $40 - 1200y = 25$. $1200y = 40 - 25 = 15$. $y = \frac{15}{1200} = \frac{1}{80}$. Since $y = \frac{1}{v_C}$, $v_C = 80$ km/h.Substitute the value of $y$ back into Equation 1'': $15x + 60\left(\frac{1}{80}\right) = 1$. $15x + \frac{60} {80} = 1$. $15x + \frac{3}{4} = 1$. $15x = 1 - \frac{3}{4} = \frac{1}{4}$. $x = \frac{1}{4 \times 15} = \frac{1}{60}$. Since $x = \frac{1}{v_T}$, $v_T = 60$ km/h.The ratio of the speed of the train to that of the car is $v_T : v_C = 60 : 80$. Divide both parts by 20: $60 : 80 = \frac{60}{20} : \frac{80}{20} = 3 : 4$. 278. A speedboat runs 6 km upstream in a river and comes back to the starting point in 33 minutes. The stream of the river is running at 2 km/hr. What is the speed of speedboat in still water?
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