📘 Quiz

Solution: Step 1: Calculate the relative speed of Vinay and Mahesh. Since they are moving towards each other, their speeds add up. Step 2: Relative Speed = 36 km/h + 54 km/h = 90 km/h. Step 3: Convert the relative speed from km/h to m/s, as the distance is in meters and time is required in seconds: - Relative Speed (m/s) = 90 * (5/18) = 5 * 5 = 25 m/s. Step 4: The initial distance between them is 250 meters. Step 5: Use the formula: Time = Distance / Speed. Step 6: Time = 250 meters / 25 m/s = 10 seconds. Step 7: Vinay and Mahesh will meet each other in 10 seconds.

Solution: Step 1: Let the distance from A to B be 3k and the distance from B to C be 5k, based on the given ratio 3:5. Step 2: The total distance for the journey is 3k + 5k = 8k. Step 3: Calculate the time taken for each segment: - Time from A to B (T_AB) = Distance / Speed = 3k / x hours. - Time from B to C (T_BC) = Distance / Speed = 5k / 50 = k / 10 hours. Step 4: Calculate the total time for the journey: - Total Time = T_AB + T_BC = (3k/x) + (k/10) = k * (3/x + 1/10) = k * ((30 + x) / (10x)) hours. Step 5: The average speed for the entire journey is given by Total Distance / Total Time. Step 6: Given average speed is 40 km/h. So, 40 = (8k) / [k * ((30 + x) / (10x))]. Step 7: Cancel 'k' from the numerator and denominator: 40 = 8 / ((30 + x) / (10x)). Step 8: Rearrange the equation: 40 = 8 * (10x / (30 + x)). Step 9: Divide both sides by 8: 5 = 10x / (30 + x). Step 10: Multiply both sides by (30 + x): 5 * (30 + x) = 10x. Step 11: Distribute 5: 150 + 5x = 10x. Step 12: Solve for x: 150 = 10x - 5x => 150 = 5x. Step 13: Therefore, x = 150 / 5 = 30 km/h. Step 14: Now, calculate the required ratio (x - 10) : (x + 1). Step 15: Substitute x = 30: (30 - 10) : (30 + 1) = 20 : 31. Step 16: The value of the ratio (x - 10) : (x + 1) is 20 : 31.

Solution: Step 1: The time interval between two consecutive buses leaving the terminal is 10 minutes. Step 2: Let the speed of the man be 'x' km/h. Step 3: The relative speed of the bus and the man (moving in opposite directions) is (20 + x) km/h. Step 4: The distance covered by the bus in the 10-minute interval (20 km/h * 10/60 h) must be equal to the distance covered by the relative speed of the bus and man in 8 minutes ((20 + x) km/h * 8/60 h). Step 5: Set up the equation: 20 * (10/60) = (20 + x) * (8/60). Step 6: Multiply both sides by 60: 20 * 10 = (20 + x) * 8. Step 7: 200 = 160 + 8x. Step 8: 8x = 200 - 160 = 40. Step 9: x = 40 / 8 = 5 km/h. Step 10: The speed of the man is 5 km/h.

Solution: Step 1: To compare the speeds, convert all given speeds to a common unit, such as kilometres per hour (km/hr). Step 2: Convert the first speed: 25 m/sec. To convert m/sec to km/hr, multiply by (18/5). Speed = 25 * (18/5) km/hr = 5 * 18 km/hr = 90 km/hr. Step 3: Convert the second speed: 1500 m/min. Convert metres to kilometres: 1500 m = 1.5 km. Convert minutes to hours: 1 min = 1/60 hr. Speed = 1.5 km / (1/60) hr = 1.5 * 60 km/hr = 90 km/hr. Step 4: The third speed is already in km/hr: 90 km/hr. Step 5: Compare all converted speeds. Speed 1 = 90 km/hr. Speed 2 = 90 km/hr. Speed 3 = 90 km/hr. Step 6: Since all three speeds are equal, none is faster than the others. They are all the same speed.

Solution: Step 1: Convert the speed of the train from km/hr to m/sec. Step 2: Speed = 30 km/hr = 30 × (5/18) m/sec = (5 × 5) / 3 = 25/3 m/sec. Step 3: Let the length of the bridge be 'x' meters. Step 4: When a train crosses a bridge, the total distance covered is the sum of the train's length and the bridge's length. Step 5: Total distance = Length of train + Length of bridge = (50 + x) m. Step 6: The time taken to cross the bridge is 36 seconds. Step 7: Using the formula Time = Total Distance / Speed: 36 = (50 + x) / (25/3). Step 8: Multiply both sides by (25/3): 36 × (25/3) = 50 + x. Step 9: Simplify the left side: 12 × 25 = 50 + x. Step 10: 300 = 50 + x. Step 11: Solve for x: x = 300 - 50 = 250 meters.

Solution: Step 1: Let the speed of the boat in still water be B kmph. Given: Speed of the stream (S) = 3 kmph. Speed Downstream = (B + 3) kmph. Speed Upstream = (B - 3) kmph. Step 2: The distance covered downstream and upstream is the same. Convert 1 and 1/2 hours to 1.5 hours. Distance Downstream = Speed Downstream × Time Downstream = (B + 3) × 1. Distance Upstream = Speed Upstream × Time Upstream = (B - 3) × 1.5. Step 3: Equate the distances: (B + 3) × 1 = (B - 3) × 1.5 B + 3 = 1.5B - 4.5 Step 4: Solve for B: 3 + 4.5 = 1.5B - B 7.5 = 0.5B B = 7.5 / 0.5 B = 15 kmph. The speed of the boat in still water is 15 kmph.

Solution: Step 1: Establish the relationships between the speeds of A, B, and C. * Speed of A = 2 × Speed of B (S_A = 2S_B) * Speed of B = 3 × Speed of C (S_B = 3S_C) Step 2: Express all speeds in terms of C's speed. * S_B = 3S_C * S_A = 2 × (3S_C) = 6S_C. Step 3: Determine the ratio of their speeds: * S_A : S_B : S_C = 6S_C : 3S_C : 1S_C = 6 : 3 : 1. Step 4: For a constant distance, time taken is inversely proportional to speed. So, the ratio of their times (T_A : T_B : T_C) will be the reciprocal of their speed ratios. * T_A : T_B : T_C = 1/6 : 1/3 : 1/1. Step 5: To simplify the time ratio to whole numbers, multiply each part by the LCM of the denominators (which is 6): * T_A : T_B : T_C = (1/6)*6 : (1/3)*6 : (1/1)*6 = 1 : 2 : 6. Step 6: We are given that C takes 78 minutes. From our ratio, T_C corresponds to 6 parts. * 6 parts = 78 minutes. Step 7: Calculate the value of 1 part: * 1 part = 78 / 6 = 13 minutes. Step 8: The time taken by A (T_A) corresponds to 1 part. * T_A = 1 × 13 minutes = 13 minutes.

Solution: Step 1: At 10:25 pm, the hour hand is between 10 and 11, and the minute hand is at 5. Step 2: The minute hand moves at 6° per minute, so at 25 minutes, it is at 25 * 6 = 150°. Step 3: The hour hand moves at 0.5° per minute, so at 10:25, it is at 10 * 30 + 25 * 0.5 = 300 + 12.5 = 312.5°. Step 4: Calculate the absolute difference between the two angles: |312.5 - 150| = 162.5°. Step 5: Verify: The angle between the hands at 10:25 pm is indeed 162.5°.

Solution: Step 1: Calculate the fuel consumed due to driving per hour. Speed = 40 km/hr, Efficiency = 10 km/litre. Fuel consumed (driving) = (40 km/hr) / (10 km/litre) = 4 litres/hour. Step 2: Note the fuel lost due to leakage. Fuel leakage rate = 0.5 litres/hour (half a litre per hour). Step 3: Calculate the total (net) rate of fuel consumption. Net fuel consumption rate = Fuel (driving) + Fuel (leakage) = 4 litres/hour + 0.5 litres/hour = 4.5 litres/hour. Step 4: Calculate the total time the car could run on the initial 18 litres of petrol. Total time = Initial fuel / Net fuel consumption rate = 18 litres / 4.5 litres/hour = 4 hours. Step 5: Calculate the total distance covered during this time. Distance = Car's speed * Total time = 40 km/hr * 4 hours = 160 km.

Solution: Step 1: Let usual speed = x km/h and usual time = t hours Step 2: Distance = x * t km Step 3: New speed = (11/9)x km/h Step 4: New time = t - 2 hours Step 5: Distance = (11/9)x * (t - 2) Step 6: Set up equation: x * t = (11/9)x * (t - 2) Step 7: Cancel x: t = (11/9)(t - 2) Step 8: Multiply both sides by 9: 9t = 11t - 22 Step 9: Rearrange: 22 = 2t Step 10: Solve for t: t = 11 hours

Solution: Step 1: Calculate the relative speed of the jeep with respect to the car. Since both vehicles are moving in the same direction, subtract their speeds. - Relative Speed = Speed of Jeep - Speed of Car = 90 km/hr - 75 km/hr = 15 km/hr. Step 2: Identify the initial distance between the jeep and the car. - Initial Distance = 5 km. Step 3: Calculate the time taken for the jeep to cover this distance at the relative speed. - Time = Distance / Relative Speed = 5 km / 15 km/hr = 1/3 hour. Step 4: Convert the time from hours to minutes. - Time in minutes = (1/3) * 60 minutes = 20 minutes.

Solution: Step 1: Let vehicle length = L meters Step 2: Speed = (L + 160)/20 m/s Step 3: Same speed = (L + 110)/15 m/s Step 4: Set up equation: (L + 160)/20 = (L + 110)/15 Step 5: Cross-multiply: 15(L + 160) = 20(L + 110) Step 6: Expand: 15L + 2400 = 20L + 2200 Step 7: Rearrange: 2400 - 2200 = 20L - 15L Step 8: Solve: 200 = 5L Step 9: Length L = 40 m

Solution: Step 1: Understand that when two objects travel for the same amount of time, the ratio of the distances they cover is directly proportional to the ratio of their speeds. Step 2: Determine the distance covered by each train until they meet. Total distance between stations = 200 km. Distance covered by the first train (D1) = 110 km (from one station). Step 3: Calculate the distance covered by the second train (D2). Distance covered by the second train = Total distance - Distance by first train = 200 km - 110 km = 90 km. Step 4: Since the time taken for both trains to meet is the same, the ratio of their speeds (S1:S2) is equal to the ratio of the distances they covered (D1:D2). Ratio of speeds = D1 : D2 = 110 km : 90 km. Step 5: Simplify the ratio. Ratio of speeds = 11 : 9. Step 6: The ratio of their speeds is 11:9.

Solution: Step 1: Let time = t years Step 2: Population of first quantity: 272000 - 2700t Step 3: Population of second quantity: 190000 + 1400t Step 4: Set up equation: 272000 - 2700t = 190000 + 1400t Step 5: Combine like terms: 272000 - 190000 = 2700t + 1400t Step 6: Simplify: 82000 = 4100t Step 7: Solve for t: t = 82000 / 4100 = 20 years

Solution: Step 1: Relative speed = 50 - 20 = 30 km/hr Step 2: Convert relative speed to m/s: 30 * (1000/3600) = 8.33 m/s Step 3: Distance covered in 20 seconds = 8.33 * 20 = 166.6 meters Step 4: Length of faster vehicle = 166.6 meters

Solution: Step 1: Calculate the downstream speed (D). * Downstream Distance = 24 km * Downstream Time = 2 hours * Downstream Speed (D) = 24 / 2 = 12 km/hr. Step 2: Calculate the upstream speed (U). * Upstream Distance = 10 km * Upstream Time = 1 hour * Upstream Speed (U) = 10 / 1 = 10 km/hr. Step 3: Calculate the speed of the motorboat in still water (B). * B = (Downstream Speed + Upstream Speed) / 2 * B = (12 + 10) / 2 = 22 / 2 = 11 km/hr. Step 4: The speed of the motorboat in still water is 11 km/hr.

Solution: Step 1: Let the speed of the boat in still water be B km/hr and the speed of the stream be S km/hr. Step 2: Upstream speed = (B - S) km/hr. Downstream speed = (B + S) km/hr. Step 3: Given: Time to row 'd' km upstream and 'd' km downstream is 5 hours 15 minutes. Step 4: Convert 5 hours 15 minutes to hours: 5 + 15/60 = 5 + 1/4 = 21/4 hours. Step 5: So, the equation for the first scenario is: (d / (B + S)) + (d / (B - S)) = 21/4. (Equation 1) Step 6: Given: Time to row '2d' km upstream is 7 hours. Step 7: So, the equation for the second scenario is: (2d / (B - S)) = 7. (Equation 2) Step 8: From Equation 2, divide by 2 to find the time to row 'd' km upstream: d / (B - S) = 7/2 hours. Step 9: Substitute the value of d / (B - S) into Equation 1: (d / (B + S)) + 7/2 = 21/4. Step 10: Solve for d / (B + S): d / (B + S) = 21/4 - 7/2. Step 11: Find a common denominator: d / (B + S) = 21/4 - 14/4 = 7/4 hours. Step 12: We need to find the time to row '2d' km downstream. This is 2 times the value of d / (B + S). Step 13: Time for 2d km downstream = 2 * (d / (B + S)) = 2 * (7/4) hours. Step 14: Time for 2d km downstream = 14/4 hours = 7/2 hours. Step 15: It will take 7/2 hours to row 2d km downstream.

Solution: Step 1: Understand the Goal. The goal is to find the speed of the boat in still water (let's call it 'B'). Step 2: Analyze the Main Problem Statement. Total time for round trip (P to Q downstream, Q to P upstream) = 3 hours. Let distance PQ = D. Let speed of boat in still water = B. Let speed of current = S. Time_downstream = D / (B + S). Time_upstream = D / (B - S). Total Time = D / (B + S) + D / (B - S) = 3. Step 3: Evaluate Statement I alone. Statement I: S = 1 km/hr. The equation becomes: D / (B + 1) + D / (B - 1) = 3. We have two unknowns (D and B) and only one equation. We cannot solve for B uniquely. So, Statement I alone is not sufficient. Step 4: Evaluate Statement II alone. Statement II: D = 4 km. The equation becomes: 4 / (B + S) + 4 / (B - S) = 3. We have two unknowns (B and S) and only one equation. We cannot solve for B uniquely. So, Statement II alone is not sufficient. Step 5: Evaluate Statements I and II together. Using both statements: D = 4 km. S = 1 km/hr. Substitute these values into the total time equation: 4 / (B + 1) + 4 / (B - 1) = 3. This is a single equation with one unknown (B). It can be solved for B. (4(B - 1) + 4(B + 1)) / ((B + 1)(B - 1)) = 3. (4B - 4 + 4B + 4) / (B^2 - 1) = 3. 8B / (B^2 - 1) = 3. 8B = 3B^2 - 3. 3B^2 - 8B - 3 = 0. This is a quadratic equation, which will yield a specific value(s) for B. Therefore, both statements together are necessary and sufficient. Step 6: Conclusion. Both I and II are necessary to answer the question.

Solution: Step 1: Calculate the relative speed of the faster train with respect to the slower train. Since they are moving in the same direction, subtract their speeds. Step 2: Relative speed = Speed of faster train - Speed of slower train = 45 km/hr - 40 km/hr = 5 km/hr. Step 3: Convert the relative speed from km/hr to m/sec. Step 4: Relative speed = 5 × (5/18) m/sec = 25/18 m/sec. Step 5: When one train crosses another, the total distance covered is the sum of their lengths. Step 6: Total distance = Length of Train 1 + Length of Train 2 = 200 m + 150 m = 350 m. Step 7: Calculate the time taken using the formula Time = Total Distance / Relative Speed. Step 8: Time = 350 m / (25/18) m/sec = 350 × (18/25) seconds. Step 9: Simplify the expression: Time = (14 × 18) seconds = 252 seconds.