Q: A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house? (S.S…

Concept & Time Difference: When a person travels the same distance at two different speeds, the distance can be found using the formula: Distance = (S_1 × S_2/S_1 \sim S_2) × Difference in Time Step-by-Step Solution: • Given: Initial speed S_1 = 2 (1/2) = 2.5 km/hr. Increased speed S_2 = 2.5 + 1 = 3.5 km/hr. • Time Difference: Late by 6 mins, then early by 6 mins. Total difference = 6 + 6 = 12 minutes = (12/60) = (1/5) hours. • Calculation: • Distance = (2.5 × 3.5/3.5 - 2.5) × (1/5) • Distance = (8.75/1) × (1/5) = 1.75 km. • 1.75 km = 1 (3/4) km. Exam Strategy & Shortcut: Use the direct formula D = (v_1 v_2/\Delta v) \Delta t. It avoids setting up algebraic equations and gives the answer in one step. Common Pitfall: A frequent error is subtracting the times (6 - 6 = 0) instead of adding them when one is late and the other is early. Final Answer: Therefore, the correct answer is 1 (3/4) km.

Q: If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 minutes only. The correct time for the train to complete its journey is

Concept & Time Equivalence: When distance is constant, Speed × Time is constant. Let the correct time be t minutes. We equate the distance for both scenarios. S_1 × T_1 = S_2 × T_2 Step-by-Step Solution: • Given: Speeds are 40 kmph and 50 kmph. • Time Equivalence: Let correct time be t minutes. • Case 1 time = (t + 11) minutes. • Case 2 time = (t + 5) minutes. • Calculation: • 40 × (t + 11/60) = 50 × (t + 5/60) • 4(t + 11) = 5(t + 5) • 4t + 44 = 5t + 25 • t = 19 Exam Strategy & Shortcut: Difference in time = 11 - 5 = 6 mins. Distance D = (40 × 50/50 - 40) × (6/60) = (2000/10) × (1/10) = 20 km. Correct time = Time taken at 50 kmph - 5 mins. Time at 50 kmph = (20/50) hrs = (2/5) × 60 = 24 mins. Correct time = 24 - 5 = 19 mins. Common Pitfall: Misinterpreting "late by 11 mins" and "late by 5 mins" and adding them. Since both are "late", you must subtract the delay from the actual time taken to find the correct time. Final Answer: Therefore, the correct answer is 19 min.

Q: Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph; he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? (D.M.R.C., 2003)

Concept & Average Speed Logic: When travelling the same distance, the relationship between speeds and times can be established using a common base time. Distance = Speed × Time Step-by-Step Solution: • Given: To reach at 2 P.M., S_1 = 10 kmph. To reach at 12 noon, S_2 = 15 kmph. We want to find the speed for 1 P.M. • Establish Times: Let t be the time in hours to reach at 1 P.M. • Time to reach at 2 P.M. = (t + 1) hours. • Time to reach at 12 noon = (t - 1) hours. • Equate Distances: • 10 × (t + 1) = 15 × (t - 1) • 10t + 10 = 15t - 15 • 5t = 25 \Rightarrow t = 5 hours. • Calculate Distance & Target Speed: • Distance = 10 × (5 + 1) = 60 km. • Speed for 1 P.M. (which takes 5 hours) = (60/5) = 12 kmph. Exam Strategy & Shortcut: Because 1 P.M. is exactly halfway between 12 noon and 2 P.M. in time, you might be tempted to average the speeds (10+15/2) = 12.5, but average speed doesn't work this way over equal distance. However, you can find distance directly: time gap is 2 hrs. D = (10 × 15/15 - 10) × 2 = (150/5) × 2 = 60 km. Time at 15kmph is 4 hrs. Start time is 8 A.M. 1 P.M. is 5 hrs from 8 A.M. Speed = 60 / 5 = 12 kmph. Common Pitfall: Taking the simple arithmetic mean of the speeds (12.5 kmph) assuming time linearity translates directly to speed linearity. Final Answer: Therefore, the correct answer is 12 kmph.

Q: A person travels 285 km in 6 hours in two stages. In the first part of the journey, he travels by bus at the speed of 40 km/hr. In the second part of the journey, he travels by train at the speed of 55 km/hr. How much distance does he travel by trai…

Concept & Linear Equations: When a total distance and total time are given for a journey split into two parts with different speeds, we can set up a linear equation based on time. Total Time = (Time on Bus) + (Time on Train) Step-by-Step Solution: • Given: Total distance = 285 km. Total time = 6 hours. Speed of bus = 40 km/hr. Speed of train = 55 km/hr. • Establish Variables: Let the time traveled by train be t hours. Therefore, the time traveled by bus is (6 - t) hours. • Formulate Equation: • Distance by bus + Distance by train = Total Distance • 40(6 - t) + 55t = 285 • 240 - 40t + 55t = 285 • 15t = 285 - 240 • 15t = 45 \Rightarrow t = 3 hours. • Calculate Distance: • Distance traveled by train = Speed of train × Time by train • Distance = 55 × 3 = 165 km. Exam Strategy & Shortcut: You can also use the method of alligation by finding the average speed of the whole journey (285/6 = 47.5 km/hr) and applying it to the two speeds (40 and 55) to find the ratio of times. Ratio of time = (55 - 47.5) : (47.5 - 40) = 7.5 : 7.5 = 1:1. Total time is 6 hours, so 3 hours each. Train distance = 3 × 55 = 165 km. Common Pitfall: A common mistake is assigning the variable x to distance rather than time, which creates a more complex fractional equation: (x/55) + (285 - x/40) = 6. While valid, it takes longer to solve. Final Answer: Therefore, the correct answer is 165 km.

Q: A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is

Concept & Speed-Time Variations: When speed changes cause corresponding changes in time for a constant distance, we can use the formula relating distance, speed, and time difference: Distance = (v_1 × v_2/|v_1 - v_2|) × \Delta t Step-by-Step Solution: • Establish Variables: Let actual speed be v kmph and distance be d km. Convert 40 minutes to hours: (40/60) = (2/3) hours. • Set up Equations: • Case 1 (Faster): d = (v(v + 3)/3) × (2/3) • Case 2 (Slower): d = (v(v - 2)/2) × (2/3) • Equate and Solve for v: • (v(v + 3)/3) = (v(v - 2)/2) • Divide both sides by v (since v \neq 0): • (v + 3/3) = (v - 2/2) • 2(v + 3) = 3(v - 2) • 2v + 6 = 3v - 6 • v = 12 kmph. • Calculate Distance: • Substitute v = 12 into the first equation: • d = (12(12 + 3)/3) × (2/3) • d = (12 × 15/3) × (2/3) = 4 × 15 × (2/3) = 40 km. Exam Strategy & Shortcut: Since both cases have the same time difference (40 mins), the ratio of the speed changes directly maps to the equations. You can quickly balance (v+3/3) = (v-2/2) to find original speed mentally (v=12). Then plug it into the distance formula. Common Pitfall: Forgetting to convert minutes to hours before multiplying with speeds in kmph, which throws off the final distance value entirely. Final Answer: Therefore, the correct answer is 40.

Q: A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, then it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr, then the train would have taken 6 hours more than t…

Concept & Speed-Time Variations: Just like the previous problem, use the formula for distance when variations in speed and time are given: Distance = (v × (v \pm \Delta v)/\Delta v) × \Delta t Step-by-Step Solution: • Establish Variables: Let original speed be v km/hr and distance be d km. • Set up Equations: • Case 1 (Faster): d = (v(v + 6)/6) × 4 • Case 2 (Slower): d = (v(v - 6)/6) × 6 • Equate and Solve for v: • (v(v + 6)/6) × 4 = (v(v - 6)/6) × 6 • 4(v + 6) = 6(v - 6) • 4v + 24 = 6v - 36 • 2v = 60 \Rightarrow v = 30 km/hr. • Calculate Distance: • Substitute v = 30 into the Case 1 equation: • d = (30(30 + 6)/6) × 4 • d = 5 × 36 × 4 = 180 × 4 = 720 km. Exam Strategy & Shortcut: Equate the two conditions directly by focusing on the remaining terms: (\Delta t_1/\Delta v_1)(v + \Delta v_1) = (\Delta t_2/\Delta v_2)(v - \Delta v_2). Here, (4/6)(v+6) = (6/6)(v-6), simplifying to (2/3)(v+6) = v-6, which quickly yields v=30. Common Pitfall: Misinterpreting the variables and trying to solve for total time instead of substituting the speed directly back into the distance formula. Final Answer: Therefore, the correct answer is 720 km.

Question 1

A man on tour travels 160 km by car at 64 km/hr and another 160 km by bus at 80 km/hr. The average speed for the whole journey is (L.I.C.A.D.O., 2008)

Accepted Answer

Concept & Harmonic Mean for Equal Distances: When a journey consists of two equal distances traveled at different speeds (x and y), the average speed for the entire journey is given by the harmonic mean of those two speeds. The actual distance value (160 km) is irrelevant to this specific formula. Average\ Speed = (2xy/x + y) Step-by-Step Solution: • Step 1: Identify the speeds and verify distances are equal. • Both segments are 160 km, so distances are equal. • Speed x = 64 km/hr. • Speed y = 80 km/hr. • Step 2: Apply the harmonic mean formula. • Average Speed = (2 × 64 × 80/64 + 80) • Average Speed = (10240/144) • Step 3: Simplify the fraction. • Divide by 16: (10240 \div 16/144 \div 16) = (640/9) • Calculate decimal: 640 \div 9 = 71.111... km/hr. Exam Strategy & Shortcut: You can completely ignore the 160 km figure and plug 64 and 80 directly into the (2xy/x+y) shortcut. This saves you from calculating the individual times ((160/64) and (160/80)), adding them, and dividing the total 320 km by that sum. Common Pitfall: Taking the simple arithmetic average of the two speeds: (64 + 80/2) = 72 km/hr. Average speed must account for the fact that more time is spent traveling at the slower speed. Final Answer: Therefore, the correct answer is 71.11 km/hr.

Question 2

A cyclist rides 24 km at 16 kmph and further 36 km at 15 kmph. Find his average speed for the journey. (R.R.B., 2008)

Accepted Answer

Concept & Average Speed: Average speed is calculated by dividing the total distance traveled by the total time elapsed. Average\ Speed = (Total\ Distance/Total\ Time) Step-by-Step Solution: • Step 1: Calculate Total Distance. • Total Distance = 24 km + 36 km = 60 km. • Step 2: Calculate the time taken for each segment. • Time 1 = (Distance_1/Speed_1) = (24/16) = 1.5 hours. • Time 2 = (Distance_2/Speed_2) = (36/15) = (12/5) = 2.4 hours. • Step 3: Calculate Total Time. • Total Time = 1.5 + 2.4 = 3.9 hours. • Step 4: Calculate Average Speed. • Average Speed = (60/3.9) = (600/39) • Simplify the fraction by dividing numerator and denominator by 3: (200/13) • Calculate the decimal: 200 \div 13 \approx 15.3846... kmph. Exam Strategy & Shortcut: Recognize that (24/16) and (36/15) yield simple decimals (1.5 and 2.4). Adding these directly to get 3.9 hours is much faster than cross-multiplying fractions to find a common denominator. Common Pitfall: Choosing "None of these" because 15.3846... does not terminate perfectly, failing to recognize that 15.38 is standard two-decimal place rounding for this exact value. Final Answer: Therefore, the correct answer is 15.38 kmph.

Question 3

A train running at $\frac{7}{11}$ of its own speed reached a place in 22 hours. How much time could be saved if the train would have run at its own speed?

Accepted Answer

Concept & Proportionality: When distance is constant, speed and time are inversely proportional. S_1 × T_1 = S_2 × T_2 If speed becomes (a/b) of its original, time becomes (b/a) of its original. Step-by-Step Solution: • Let the usual speed be S and usual time be T. • The new speed is (7/11)S. • Because speed and time are inversely proportional for a constant distance, the new time taken is (11/7)T. • We are given the new time is 22 hours. • (11/7) × T = 22 • T = 22 × (7/11) = 14 hours. • The usual time taken would have been 14 hours. • Time saved = New time - Usual time = 22 - 14 = 8 hours. Exam Strategy & Shortcut: Using ratios: Speed ratio is 11:7 (original to new). Therefore, the time ratio is 7:11 (original to new). If 11 units = 22 hours, 1 unit = 2 hours. The time saved is 11 - 7 = 4 units. 4 × 2 = 8 hours. Common Pitfall: Students often calculate the usual time (14 hours) and incorrectly select it as the answer, forgetting that the question asks for the "time saved". Final Answer: Therefore, the correct answer is 8 hours.

Question 4

Walking $\frac{6}{7}$ th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is

Accepted Answer

Concept & Inverse Proportionality: For a fixed distance, time and speed are inversely related. If speed becomes (a/b) of its usual value, the time taken becomes (b/a) of its usual value. Step-by-Step Solution: • Let the usual time taken be T. • The new speed is (6/7) of the usual speed. • Therefore, the new time taken is (7/6) of the usual time, which is (7/6)T. • The difference between the new time and the usual time is the amount he is late. • Late time = (7/6)T - T = (1/6)T. • We are given that he is 12 minutes late, so (1/6)T = 12 minutes. • T = 12 × 6 = 72 minutes. • Convert minutes to hours: 72 minutes = 1 hour 12 minutes. Exam Strategy & Shortcut: If new speed is (x/y) of original speed (where x < y), then usual time = (x/y - x) × (late time). Here, x=6, y=7. Usual time = (6/7 - 6) × 12 = 6 × 12 = 72 minutes, which is 1 hr 12 min. Common Pitfall: Failing to convert the final answer from minutes (72) back into the hour-minute format required by the options. Final Answer: Therefore, the correct answer is 1 hr 12 min.

Question 5

A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house? (S.S…

Accepted Answer

Concept & Time Difference: When a person travels the same distance at two different speeds, the distance can be found using the formula: Distance = (S_1 × S_2/S_1 \sim S_2) × Difference in Time Step-by-Step Solution: • Given: Initial speed S_1 = 2 (1/2) = 2.5 km/hr. Increased speed S_2 = 2.5 + 1 = 3.5 km/hr. • Time Difference: Late by 6 mins, then early by 6 mins. Total difference = 6 + 6 = 12 minutes = (12/60) = (1/5) hours. • Calculation: • Distance = (2.5 × 3.5/3.5 - 2.5) × (1/5) • Distance = (8.75/1) × (1/5) = 1.75 km. • 1.75 km = 1 (3/4) km. Exam Strategy & Shortcut: Use the direct formula D = (v_1 v_2/\Delta v) \Delta t. It avoids setting up algebraic equations and gives the answer in one step. Common Pitfall: A frequent error is subtracting the times (6 - 6 = 0) instead of adding them when one is late and the other is early. Final Answer: Therefore, the correct answer is 1 (3/4) km.

Question 6

If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 minutes only. The correct time for the train to complete its journey is

Accepted Answer

Concept & Time Equivalence: When distance is constant, Speed × Time is constant. Let the correct time be t minutes. We equate the distance for both scenarios. S_1 × T_1 = S_2 × T_2 Step-by-Step Solution: • Given: Speeds are 40 kmph and 50 kmph. • Time Equivalence: Let correct time be t minutes. • Case 1 time = (t + 11) minutes. • Case 2 time = (t + 5) minutes. • Calculation: • 40 × (t + 11/60) = 50 × (t + 5/60) • 4(t + 11) = 5(t + 5) • 4t + 44 = 5t + 25 • t = 19 Exam Strategy & Shortcut: Difference in time = 11 - 5 = 6 mins. Distance D = (40 × 50/50 - 40) × (6/60) = (2000/10) × (1/10) = 20 km. Correct time = Time taken at 50 kmph - 5 mins. Time at 50 kmph = (20/50) hrs = (2/5) × 60 = 24 mins. Correct time = 24 - 5 = 19 mins. Common Pitfall: Misinterpreting "late by 11 mins" and "late by 5 mins" and adding them. Since both are "late", you must subtract the delay from the actual time taken to find the correct time. Final Answer: Therefore, the correct answer is 19 min.

Question 7

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph; he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? (D.M.R.C., 2003)

Accepted Answer

Concept & Average Speed Logic: When travelling the same distance, the relationship between speeds and times can be established using a common base time. Distance = Speed × Time Step-by-Step Solution: • Given: To reach at 2 P.M., S_1 = 10 kmph. To reach at 12 noon, S_2 = 15 kmph. We want to find the speed for 1 P.M. • Establish Times: Let t be the time in hours to reach at 1 P.M. • Time to reach at 2 P.M. = (t + 1) hours. • Time to reach at 12 noon = (t - 1) hours. • Equate Distances: • 10 × (t + 1) = 15 × (t - 1) • 10t + 10 = 15t - 15 • 5t = 25 \Rightarrow t = 5 hours. • Calculate Distance & Target Speed: • Distance = 10 × (5 + 1) = 60 km. • Speed for 1 P.M. (which takes 5 hours) = (60/5) = 12 kmph. Exam Strategy & Shortcut: Because 1 P.M. is exactly halfway between 12 noon and 2 P.M. in time, you might be tempted to average the speeds (10+15/2) = 12.5, but average speed doesn't work this way over equal distance. However, you can find distance directly: time gap is 2 hrs. D = (10 × 15/15 - 10) × 2 = (150/5) × 2 = 60 km. Time at 15kmph is 4 hrs. Start time is 8 A.M. 1 P.M. is 5 hrs from 8 A.M. Speed = 60 / 5 = 12 kmph. Common Pitfall: Taking the simple arithmetic mean of the speeds (12.5 kmph) assuming time linearity translates directly to speed linearity. Final Answer: Therefore, the correct answer is 12 kmph.

Question 8

A person travels 285 km in 6 hours in two stages. In the first part of the journey, he travels by bus at the speed of 40 km/hr. In the second part of the journey, he travels by train at the speed of 55 km/hr. How much distance does he travel by trai…

Accepted Answer

Concept & Linear Equations: When a total distance and total time are given for a journey split into two parts with different speeds, we can set up a linear equation based on time. Total Time = (Time on Bus) + (Time on Train) Step-by-Step Solution: • Given: Total distance = 285 km. Total time = 6 hours. Speed of bus = 40 km/hr. Speed of train = 55 km/hr. • Establish Variables: Let the time traveled by train be t hours. Therefore, the time traveled by bus is (6 - t) hours. • Formulate Equation: • Distance by bus + Distance by train = Total Distance • 40(6 - t) + 55t = 285 • 240 - 40t + 55t = 285 • 15t = 285 - 240 • 15t = 45 \Rightarrow t = 3 hours. • Calculate Distance: • Distance traveled by train = Speed of train × Time by train • Distance = 55 × 3 = 165 km. Exam Strategy & Shortcut: You can also use the method of alligation by finding the average speed of the whole journey (285/6 = 47.5 km/hr) and applying it to the two speeds (40 and 55) to find the ratio of times. Ratio of time = (55 - 47.5) : (47.5 - 40) = 7.5 : 7.5 = 1:1. Total time is 6 hours, so 3 hours each. Train distance = 3 × 55 = 165 km. Common Pitfall: A common mistake is assigning the variable x to distance rather than time, which creates a more complex fractional equation: (x/55) + (285 - x/40) = 6. While valid, it takes longer to solve. Final Answer: Therefore, the correct answer is 165 km.

Question 9

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is

Accepted Answer

Concept & Speed-Time Variations: When speed changes cause corresponding changes in time for a constant distance, we can use the formula relating distance, speed, and time difference: Distance = (v_1 × v_2/|v_1 - v_2|) × \Delta t Step-by-Step Solution: • Establish Variables: Let actual speed be v kmph and distance be d km. Convert 40 minutes to hours: (40/60) = (2/3) hours. • Set up Equations: • Case 1 (Faster): d = (v(v + 3)/3) × (2/3) • Case 2 (Slower): d = (v(v - 2)/2) × (2/3) • Equate and Solve for v: • (v(v + 3)/3) = (v(v - 2)/2) • Divide both sides by v (since v 
eq 0): • (v + 3/3) = (v - 2/2) • 2(v + 3) = 3(v - 2) • 2v + 6 = 3v - 6 • v = 12 kmph. • Calculate Distance: • Substitute v = 12 into the first equation: • d = (12(12 + 3)/3) × (2/3) • d = (12 × 15/3) × (2/3) = 4 × 15 × (2/3) = 40 km. Exam Strategy & Shortcut: Since both cases have the same time difference (40 mins), the ratio of the speed changes directly maps to the equations. You can quickly balance (v+3/3) = (v-2/2) to find original speed mentally (v=12). Then plug it into the distance formula. Common Pitfall: Forgetting to convert minutes to hours before multiplying with speeds in kmph, which throws off the final distance value entirely. Final Answer: Therefore, the correct answer is 40.

Question 10

A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, then it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr, then the train would have taken 6 hours more than t…

Accepted Answer

Concept & Speed-Time Variations: Just like the previous problem, use the formula for distance when variations in speed and time are given: Distance = (v × (v \pm \Delta v)/\Delta v) × \Delta t Step-by-Step Solution: • Establish Variables: Let original speed be v km/hr and distance be d km. • Set up Equations: • Case 1 (Faster): d = (v(v + 6)/6) × 4 • Case 2 (Slower): d = (v(v - 6)/6) × 6 • Equate and Solve for v: • (v(v + 6)/6) × 4 = (v(v - 6)/6) × 6 • 4(v + 6) = 6(v - 6) • 4v + 24 = 6v - 36 • 2v = 60 \Rightarrow v = 30 km/hr. • Calculate Distance: • Substitute v = 30 into the Case 1 equation: • d = (30(30 + 6)/6) × 4 • d = 5 × 36 × 4 = 180 × 4 = 720 km. Exam Strategy & Shortcut: Equate the two conditions directly by focusing on the remaining terms: (\Delta t_1/\Delta v_1)(v + \Delta v_1) = (\Delta t_2/\Delta v_2)(v - \Delta v_2). Here, (4/6)(v+6) = (6/6)(v-6), simplifying to (2/3)(v+6) = v-6, which quickly yields v=30. Common Pitfall: Misinterpreting the variables and trying to solve for total time instead of substituting the speed directly back into the distance formula. Final Answer: Therefore, the correct answer is 720 km.

Question 11

A train can travel 50% faster than a car. Both start from point $A$ at the same time and reach point $B$ 75 kms away from $A$ at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the c…

Accepted Answer

Concept & Speed Ratio: When two vehicles cover the same distance, the ratio of the time taken is inversely proportional to the ratio of their speeds. Time \propto (1/Speed) Step-by-Step Solution: • Given: Distance = 75 km. Train is 50% faster than car. Train stoppage time = 12.5 mins = (12.5/60) hours. • Speed Ratio: Let speed of car be v. Speed of train = v + 0.5v = 1.5v. Ratio of speeds (Car : Train) = v : 1.5v = 2 : 3. • Time Ratio: Ratio of time taken (without stoppages) = 3 : 2. • Formulate Equation: The difference in actual travel time is exactly the stoppage time (since they start and arrive at the same time). • Time taken by Car - Travel time of Train = Stoppage time • (75/v) - (75/1.5v) = (12.5/60) • (75/v) - (50/v) = (12.5/60) • (25/v) = (12.5/60) • v = (25 × 60/12.5) = 2 × 60 = 120 kmph. Exam Strategy & Shortcut: Using ratios: Time ratio (Car:Train) = 3:2. Difference is 1 unit. This 1 unit is the time the train stopped to let the car catch up: 1 unit = 12.5 mins. Time taken by car = 3 units = 3 × 12.5 = 37.5 mins = (37.5/60) hours. Speed of car = \frac{Distance}{Time} = (75/\frac{37.5){60}} = (75 × 60/37.5) = 120 kmph. Common Pitfall: A common error is equating the train's total journey time to the car's, forgetting to account for the stoppage time algebraically. Always remember: Train moving time = Car time - Stoppage time. Final Answer: Therefore, the correct answer is 120 kmph.

Question 12

Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was 17 km, and another 15 minutes later, one sea trawler was 10.5 km farther from the origin than the ot…

Accepted Answer

Concept & Pythagorean Theorem: When two objects travel in mutually perpendicular directions from the same starting point, the shortest distance between them forms the hypotenuse of a right-angled triangle, where the distances traveled by each form the perpendicular sides. Distance = Speed × Time c^2 = a^2 + b^2 Step-by-Step Solution: 1. Let the speeds of the two trawlers be x km/hr and y km/hr respectively. 2. Distance traveled in half an hour (0.5 hr): 0.5x and 0.5y. 3. The shortest distance between them is the hypotenuse: \sqrt{(0.5x)^2 + (0.5y)^2} = 17 0.25x^2 + 0.25y^2 = 289 x^2 + y^2 = 1156 4. Another 15 minutes later, total time = 30 + 15 = 45 minutes = 0.75 hours. 5. Distance traveled in 0.75 hours: 0.75x and 0.75y. 6. One is 10.5 km farther than the other: 0.75x - 0.75y = 10.5 \quad (Assuming x > y ) x - y = (10.5/0.75) = 14 7. We have a system of two equations: x^2 + y^2 = 1156 and x - y = 14. 8. Rather than solving quadratics, substitute the options into these simplified equations. - Option (a): 30^2 + 16^2 = 900 + 256 = 1156. And 30 - 16 = 14. This fits perfectly. Exam Strategy & Shortcut: Instead of solving the quadratic equation (y + 14)^2 + y^2 = 1156, use option elimination based on the derived condition x - y = 14. - (a) 30 - 16 = 14 (Matches) - (b) 24 - 18 = 6 (Fails) - (c) 22 - 20 = 2 (Fails) - (d) 36 - 18 = 18 (Fails) Only option (a) satisfies the difference in speeds. Common Pitfall: A common mistake is incorrectly converting 45 minutes into hours (e.g., using 0.45 instead of 0.75), which completely ruins the second equation. Final Answer: Therefore, the correct answer is 16 km/hr, 30 km/hr.

Question 13

$A$ runs twice as fast as $B$ and $B$ runs thrice as fast as $C$. The distance covered by $C$ in 72 minutes, will be covered by $A$ in (C.P.O., 2007; R.R.B., 2006)

Accepted Answer

Concept & Speed-Time Inversely Proportional Relation: Speed and Time are inversely proportional when the distance is constant. If the speed of an object is n times that of another, the time taken by it will be 1/n of the time taken by the other to cover the same distance. Speed Ratio = a : b \implies Time Ratio = (1/a) : (1/b) Step-by-Step Solution: 1. Let the speed of C be v_c. 2. Since B runs thrice as fast as C, speed of B (v_b) = 3v_c. 3. Since A runs twice as fast as B, speed of A (v_a) = 2 × v_b = 2 × 3v_c = 6v_c. 4. The speed of A is 6 times the speed of C. 5. Because time is inversely proportional to speed for a fixed distance, A will take (1/6)th of the time C takes. 6. Time taken by C = 72 minutes. 7. Time taken by A = (72/6) = 12 minutes. Exam Strategy & Shortcut: Write down the speed ratios directly: A : B : C = (2 × 3) : 3 : 1 = 6 : 3 : 1. The time ratio for the same distance is the inverse of the speed ratio: T_a : T_b : T_c = (1/6) : (1/3) : (1/1) = 1 : 2 : 6. If 6 units of time corresponds to 72 minutes, 1 unit of time corresponds to (72/6) = 12 minutes. Common Pitfall: A common error is confusing speed and time relationships, accidentally multiplying the time by 6 (resulting in 72 × 6) instead of dividing it. Final Answer: Therefore, the correct answer is 12 minutes.

Question 14

A ship, 40 kilometres from the shore, springs a leak which admits $3 \frac{3}{4}$ tonnes of water in 12 minutes.  60 tonnes would suffice to sink her, but the ship's pumps can throw out 12 tonnes of water in one hour. Find the average rate of sailin…

Accepted Answer

Concept & Net Rate of Accumulation: This problem requires determining the net rate at which water is accumulating in the ship. We find the difference between the inflow (leak) and outflow (pump) rates, then calculate the time it takes to reach the sinking threshold. Net Inflow Rate = Leak Rate - Pump Rate Required Speed = \frac{Distance to Shore}{Time to Sink} Step-by-Step Solution: 1. Find the water intake rate per hour: The leak admits 3 (3/4) = (15/4) tonnes in 12 minutes. Since there are 60 / 12 = 5 twelve-minute periods in an hour, the hourly leak rate is: (15/4) × 5 = (75/4) = 18.75 tonnes/hour 2. Find the net water accumulation rate per hour: The pump throws out 12 tonnes per hour. Net inflow = 18.75 - 12 = 6.75 tonnes/hour = (27/4) tonnes/hour. 3. Calculate the total time before the ship sinks: The ship sinks at 60 tonnes. Time to sink = (60/27/4) = (60 × 4/27) = (240/27) = (80/9) hours. 4. Calculate the required speed: The ship is 40 km from the shore. It must cover this in (80/9) hours. Speed = \frac{Distance}{Time} = (40/80/9) = (40 × 9/80) = 4.5 km/hour. 5. 4.5 km/hour is equivalent to 4 (1/2) km/hour. Exam Strategy & Shortcut: Work exclusively with fractions to avoid messy decimal division. Leak = 15/4 tonnes in 1/5 hr \rightarrow 75/4 tonnes/hr. Net = 75/4 - 48/4 = 27/4. Time = 60 / (27/4) = 80/9. Speed = 40 / (80/9) = 9/2 = 4.5. Common Pitfall: A common pitfall is forgetting to standardize the time units. Students might compare the 12-minute leak rate directly with the 1-hour pump rate without aligning them first. Final Answer: Therefore, the correct answer is 4 (1/2) km per hour.

Question 15

The speeds of John and Max are 30 km/h and 40 km/h. Initially Max is at a place L and John is at a place M. The distance between L and M is 650 km. John started his journey 3 hours earlier than Max to meet each other. If they meet each other at a pl…

Accepted Answer

Concept & Relative Speed: When two objects move towards each other, their relative speed is the sum of their individual speeds. Before applying relative speed, we must first determine the remaining distance between them at the exact moment both are moving simultaneously. Relative Speed = S_1 + S_2 Step-by-Step Solution: 1. John's speed is 30 km/h, and Max's speed is 40 km/h. Initial distance is 650 km. 2. John starts from M, 3 hours earlier than Max. 3. Distance covered by John in these 3 hours: Distance = 30 × 3 = 90 km 4. Remaining distance between John and Max when Max starts moving: Remaining Distance = 650 - 90 = 560 km 5. Now both are moving towards each other. Their relative speed: Relative Speed = 30 + 40 = 70 km/h 6. Time taken for them to meet after Max starts: Time = (560/70) = 8 hours 7. The meeting point P's distance from M is exactly the total distance John traveled. 8. Total time John traveled = 3 hours (alone) + 8 hours (with Max moving) = 11 hours. 9. Distance from M to P = John's total distance: Distance = 30 × 11 = 330 km Exam Strategy & Shortcut: Separate the head start from the simultaneous movement. Calculate the head start distance (90 km). The remaining 560 km is closed at a combined speed of 70 km/h, which takes 8 hours. The question asks for distance from M (John's start). John travelled for 3 + 8 = 11 hours at 30 km/h. 11 × 30 = 330 km. Common Pitfall: Calculating the distance from L (Max's start point) instead of M. Max travels for 8 hours at 40 km/h = 320 km. Selecting 320 km would be incorrect as the prompt asks for distance between P and M. Final Answer: Therefore, the correct answer is 330 km.

Question 16

A man performs 2/15 of the total journey by rail, 9/20 by bus and the remaining 10 km, on the cycle. His total journey is

Accepted Answer

Concept & Fractional Parts of a Whole Journey: To find the total distance, we add the given fractions of the journey and subtract the sum from 1 (the whole). This remaining fraction corresponds exactly to the absolute remaining distance provided in the problem. Step-by-Step Solution: 1. Let the total distance of the journey be D. 2. The fractions of the journey covered by rail and bus are (2/15) and (9/20), respectively. 3. Add these fractions together. The least common multiple (LCM) of 15 and 20 is 60. (2/15) + (9/20) = (2 × 4/60) + (9 × 3/60) = (8/60) + (27/60) = (35/60) 4. Simplify the combined fraction: (35/60) = (7/12) 5. Determine the fraction representing the remaining journey on the cycle: 1 - (7/12) = (5/12) 6. We are given that this remaining fraction equals 10 km. Therefore: (5/12) of D = 10 km 7. Solve for D: D = 10 × (12/5) = 2 × 12 = 24 km Exam Strategy & Shortcut: Use the LCM of the denominators (15 and 20) to assume a hypothetical total distance. Let the total distance be 60 units. Rail journey = (2/15) of 60 = 8 units. Bus journey = (9/20) of 60 = 27 units. Total covered = 8 + 27 = 35 units. Remaining distance = 60 - 35 = 25 units. The problem states the remaining distance is 10 km. So, 25 units = 10 km. We need the value of 60 units. Total Distance = (10/25) × 60 = (2/5) × 60 = 24 km. Common Pitfall: Incorrectly calculating the least common multiple of the fractions or making a basic arithmetic error during addition, leading to a wrong remaining fraction. Final Answer: Therefore, the correct answer is 24 km.

Question 17

Two men start together to walk to a certain destination, one at 3 kmph and another at 3.75 kmph. The latter arrives half an hour before the former. The distance is

Accepted Answer

Concept & Time Difference for Constant Distance: When the distance (d) is constant, the difference in time taken at two different speeds (S_1 and S_2) can be expressed algebraically. (d/S_1) - (d/S_2) = \Delta T Step-by-Step Solution: 1. Let the total distance to the destination be d km. 2. The speed of the first man is S_1 = 3 kmph. Time taken by him is T_1 = (d/3) hours. 3. The speed of the second man is S_2 = 3.75 kmph (which is (15/4) kmph). Time taken by him is T_2 = (d/3.75) = (4d/15) hours. 4. We are given that the second man arrives half an hour ((1/2) hour) before the first. T_1 - T_2 = (1/2) (d/3) - (4d/15) = (1/2) 5. Find a common denominator (15) to solve for d: (5d - 4d/15) = (1/2) (d/15) = (1/2) d = (15/2) = 7.5 km Exam Strategy & Shortcut: Use the specific distance formula based on speeds and time difference: Distance = (S_1 × S_2/S_1 \sim S_2) × \Delta T Distance = (3 × 3.75/3.75 - 3) × (1/2) = (11.25/0.75) × (1/2) = 15 × (1/2) = 7.5 km This yields the answer directly without algebraic setup. Common Pitfall: A common error is keeping the time difference as 30 minutes instead of converting it to (1/2) an hour. Always ensure your units (km/h and hours) are consistent. Final Answer: Therefore, the correct answer is 7.5 km.

Question 18

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is

Accepted Answer

Concept & Equating Time: When a person travels for the same unknown duration (t) but at two different speeds, resulting in two different distances, we can equate the time to find the actual distance. Time = (Distance/Speed) Step-by-Step Solution: 1. Let the actual distance travelled by the person be x km. 2. The actual speed of the person is 10 km/hr. The time taken is (x/10) hours. 3. If he walks at 14 km/hr, the new distance covered is (x + 20) km. 4. The time taken in this hypothetical scenario is (x + 20/14) hours. 5. Since the time spent walking in both cases is the same, we equate the two time expressions: (x/10) = (x + 20/14) 6. Cross-multiply to solve for x: 14x = 10(x + 20) 14x = 10x + 200 14x - 10x = 200 4x = 200 x = 50 km Exam Strategy & Shortcut: Think about relative change. The increase in speed is 4 km/hr (14 - 10). This 4 km/hr extra speed causes an extra 20 km to be covered. Therefore, the time he walked is \frac{20 km}{4 km/hr} = 5 hours. The actual distance travelled at the actual speed (10 km/hr) in those 5 hours is 10 × 5 = 50 km. Common Pitfall: Students often calculate the time (5 hours) and then mistakenly multiply it by the hypothetical speed (14 km/hr) yielding 70 km, forgetting the question asks for the *actual* distance travelled. Final Answer: Therefore, the correct answer is 50 km.

Question 19

In covering a certain distance, the speeds of $A$ and $B$ are in the ratio of 3 : 4. $A$ takes 30 minutes more than $B$ to reach the destination. The time taken by $A$ to reach the destination is

Accepted Answer

Concept & Inverse Proportion of Speed and Time: For a constant distance, the time taken is inversely proportional to the speed. If the ratio of speeds of two moving bodies is a : b, the ratio of the time taken by them to cover the same distance is b : a. Step-by-Step Solution: 1. We are given the ratio of speeds of A and B as S_A : S_B = 3 : 4. 2. Therefore, the ratio of time taken by them will be T_A : T_B = 4 : 3. 3. Let the time taken by A be 4x minutes and the time taken by B be 3x minutes. 4. The problem states that A takes 30 minutes more than B: 4x - 3x = 30 x = 30 minutes 5. We need to find the time taken by A, which is 4x: T_A = 4 × 30 = 120 minutes 6. Convert minutes to hours: 120 minutes = 2 hours Exam Strategy & Shortcut: The time ratio is 4 : 3. The difference in ratio units is 4 - 3 = 1 unit. This 1 unit difference corresponds to the 30 minutes given in the problem. A takes 4 units of time, so A's time is 4 × 30 = 120 minutes = 2 hours. Very fast and requires no complex algebra! Common Pitfall: A common mistake is applying the ratio directly without taking its inverse, incorrectly assuming the faster person (B) takes more time. Final Answer: Therefore, the correct answer is 2 hours.

Question 20

Three persons are walking from a place $A$ to another place $B$. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach $B$ by these persons will be

Accepted Answer

Concept & Inverse Proportionality of Ratios: Because the distance is identical for all three persons, the time taken is inversely proportional to their respective speeds. If speeds are in ratio x : y : z, times are in ratio (1/x) : (1/y) : (1/z). Step-by-Step Solution: 1. The given ratio of speeds for the three persons is 4 : 3 : 5. 2. The distance is constant (from place A to place B). Therefore, the ratio of the time taken will be the reciprocal of the ratio of their speeds. Time \ Ratio = (1/4) : (1/3) : (1/5) 3. To convert these fractions into a simplified integer ratio, find the Least Common Multiple (LCM) of the denominators 4, 3, and 5. LCM(4, 3, 5) = 60 4. Multiply each term of the fractional ratio by the LCM: Time \ Ratio = ( (1/4) × 60 ) : ( (1/3) × 60 ) : ( (1/5) × 60 ) 5. Simplify the terms: Time \ Ratio = 15 : 20 : 12 Exam Strategy & Shortcut: For an inverse ratio of three numbers a:b:c, a quick shortcut to find the new ratio without writing fractions is to multiply the "other two" numbers for each spot: First term: b × c = 3 × 5 = 15 Second term: a × c = 4 × 5 = 20 Third term: a × b = 4 × 3 = 12 Resulting ratio: 15 : 20 : 12. This avoids LCM calculation entirely! Common Pitfall: A frequent error is simply reversing the order of the numbers from 4 : 3 : 5 to 5 : 3 : 4, incorrectly assuming a direct reverse works the same way for three items as it does for two. Final Answer: Therefore, the correct answer is 15 : 20 : 12.

Time and Distance Questions

Questions

Practice smarter