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Time, Work, and Distance
Work Rate Fundamentals
Every work problem boils down to a single relationship: the total work done equals the work rate multiplied by the time spent. If a job is treated as one whole unit, then the rate of a person is the fraction of the job they complete per unit time.
$$W = R \times T$$
Total work equals rate multiplied by time — the foundation of every work problem.
$W$=Total work done (often 1 for one complete job)(jobs or units)
$R$=Work rate — fraction of the job completed per unit time(jobs per unit time)
$T$=Time spent working(hours, days, etc.)
$W = 1$
→The rate is $R = \frac{1}{T}$, meaning if someone finishes in $T$ days, their daily rate is $\frac{1}{T}$.
Whole Job Convention: Set $W = 1$ for "one complete job". A person who finishes in 10 days has rate $R = \frac{1}{10}$ jobs/day.
Rate from Time: Given completion time $T$, the rate is always $R = \frac{1}{T}$. Shorter time means faster rate.
Work from Rate and Time: If you know the rate and time, multiply: $W = R \times T$. The answer is a fraction of the total job.
Combined Work and Efficiency
When two or more people work together, their rates add. This is because each person independently contributes their share of work per unit time.
$$R_{\text{total}} = R_1 + R_2 + R_3 + \cdots$$
The combined rate of multiple workers is the sum of their individual rates.
$R_{\text{total}}$=Combined work rate of all workers(jobs per unit time)
$R_1, R_2, \ldots$=Individual work rates(jobs per unit time)
Two workers with times $t_1$ and $t_2$
→Combined time = $\frac{t_1 \cdot t_2}{t_1 + t_2}$
A works for $a$ days, then B joins for $b$ days
→$\frac{a}{t_1} + \frac{b}{t_2} = 1$ (total work = 1 job)
Rates Add, Not Times: If A takes 6 days and B takes 12 days, the combined rate is $\frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$, so they finish in 4 days together.
LCM Shortcut: To avoid fractions, set total work = LCM of individual times. For A (6 days) and B (12 days), total work = 12 units. A's rate = 2 units/day, B's rate = 1 unit/day. Combined = 3 units/day → 4 days.
Sequential Work: If A works alone for some days and then B joins, split the work into two parts: A's solo contribution plus the combined contribution.
Efficiency compares how fast two workers are relative to each other. If A is twice as efficient as B, A's rate is double B's rate, which means A takes half the time B takes for the same job.
$$\frac{\text{Efficiency of A}}{\text{Efficiency of B}} = \frac{R_A}{R_B} = \frac{T_B}{T_A}$$
Efficiency is directly proportional to rate and inversely proportional to time.
$R_A, R_B$=Work rates of A and B(jobs per unit time)
$T_A, T_B$=Individual completion times(days, hours, etc.)
Efficiency Ratio = Inverse Time Ratio: If A takes 10 days and B takes 30 days, A is 3 times as efficient as B ($\frac{30}{10} = 3$).
Wages Split by Efficiency: If A and B work together, wages are divided in the ratio of their efficiencies (or their work contributions).
Man-Days Concept: Total work = number of workers × days. If 5 workers finish in 12 days, total work = 60 man-days. If 3 workers are assigned, they take 20 days.
Pipes and Cisterns
Pipe and cistern problems are identical to work problems, with one twist: a drain pipe has a negative rate. A fill pipe adds water (positive rate), while a drain pipe removes water (negative rate).
$$R_{\text{net}} = R_{\text{fill}} - R_{\text{drain}}$$
The net rate is the sum of all fill rates minus all drain rates.
$R_{\text{net}}$=Net filling rate of the system(tanks per unit time)
$R_{\text{fill}}$=Rate of fill pipe(s)(tanks per unit time)
$R_{\text{drain}}$=Rate of drain pipe(s)(tanks per unit time)
$R_{\text{net}} < 0$
→The tank will never fill — it drains faster than it fills.
$R_{\text{net}} = 0$
→Fill rate equals drain rate — the water level stays constant.
Fill Pipe as Worker: A pipe that fills a tank in $t$ minutes has rate $\frac{1}{t}$ tanks/minute — exactly like a worker.
Drain Pipe is Negative Rate: A drain that empties in $d$ minutes has rate $-\frac{1}{d}$ tanks/minute. Subtract it from fill rates.
Alternating Pipes: If pipes open/close in rotation, calculate work done per cycle (e.g., Pipe A open 3 min + Pipe B open 2 min = one cycle) and track the tank level.
Speed, Distance, and Time
The relationship between speed, distance, and time is the backbone of all motion problems. It's a simple three-variable equation that you rearrange depending on what you need to find.
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$
Speed equals distance divided by time. Rearrange to find distance ($d = s \times t$) or time ($t = \frac{d}{s}$).
$s$=Speed (rate of motion)(km/h, m/s, mph)
$d$=Distance travelled(km, m, miles)
$t$=Time taken(hours, seconds, minutes)
Three Forms: Memorize all three: $s = \frac{d}{t}$, $d = s \times t$, $t = \frac{d}{s}$. The triangle (S on top, D and T below) helps remember which to divide.
Constant Speed Assumption: These formulas assume constant speed throughout the journey. If speed changes, split the journey into segments.
Key Unit Conversions
•
$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$
•
$1 \text{ m/s} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$
•
$1 \text{ km} = 1000 \text{ m}$
•
$1 \text{ hour} = 60 \text{ min} = 3600 \text{ s}$
Average Speed
Average speed is NOT the average of two speeds. It is always total distance divided by total time — this is the single most important fact to remember.
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
Average speed weights each speed by the time spent at that speed — longer times have more influence.
$\text{Total Distance}$=Sum of all distances covered in each segment(km, m)
$\text{Total Time}$=Sum of all times for each segment(hours, seconds)
The Universal Rule: No matter how many segments, always compute $\frac{\sum d_i}{\sum t_i}$. Never just average the speeds.
Equal Distances Shortcut: When two equal distances are covered at speeds $a$ and $b$, the average speed is $\frac{2ab}{a+b}$ (harmonic mean, not arithmetic mean).
Equal Times Shortcut: When two equal time intervals are spent at speeds $a$ and $b$, the average speed IS $\frac{a+b}{2}$ (simple average — this is the rare case where averaging works).
Relative Speed and Trains
Relative speed is the effective speed between two moving objects. When objects move in opposite directions, their speeds add. When they move in the same direction, subtract the slower from the faster.
$$\begin{cases} s_{\text{rel}} = s_1 + s_2 & \text{(opposite directions)} \\ s_{\text{rel}} = |s_1 - s_2| & \text{(same direction)} \end{cases}$$
Relative speed depends on direction — objects approaching each other close the gap faster.
$s_{\text{rel}}$=Relative speed between two objects(km/h, m/s)
$s_1, s_2$=Individual speeds of the two objects(km/h, m/s)
Opposite Directions: Two cars at 50 km/h and 40 km/h heading toward each other close the gap at 90 km/h.
Same Direction: A faster car at 70 km/h chasing a slower car at 50 km/h closes the gap at only 20 km/h.
Meeting Time: Time to meet = initial gap $\div$ relative speed. Time to overtake = gap $\div$ relative speed (same direction).
Train problems add one extra layer: the train's own length counts as distance. When a train crosses a pole, it travels its own length. When it crosses another train, it travels the sum of both lengths.
$$t = \frac{L_1 + L_2}{s_{\text{rel}}}$$
Time to cross = total length to cover divided by relative speed.
$L_1$=Length of the first train(metres)
$L_2$=Length of the second train (or 0 for a pole/person)(metres)
$s_{\text{rel}}$=Relative speed (convert to m/s first)(m/s)
Crossing a pole or a person
→$L_2 = 0$, so time = $\frac{L_1}{s}$.
Crossing a platform
→Distance = train length + platform length.
Always Convert to m/s: Train lengths are in metres but speeds are usually given in km/h. Multiply by $\frac{5}{18}$ to convert.
Crossing vs Overtaking: Two trains moving in the same direction overtake each other; in opposite directions, they cross each other.
Platform Length: When a train crosses a platform, the total distance = train length + platform length, because the train's last carriage must clear the platform's far end.
Boats and Streams
A boat's effective speed depends on whether it moves downstream (with the current) or upstream (against the current). The stream adds to the boat's speed going downstream and subtracts going upstream.
$$s_{\text{down}} = b + c \qquad s_{\text{up}} = b - c$$
Downstream speed is boat speed plus stream speed; upstream is boat speed minus stream speed.
$s_{\text{down}}$=Speed of the boat going downstream(km/h)
$s_{\text{up}}$=Speed of the boat going upstream(km/h)
$b$=Speed of the boat in still water(km/h)
$c$=Speed of the stream/current(km/h)
Finding Boat and Stream Speeds: If you know $s_{\text{down}}$ and $s_{\text{up}}$: boat speed $b = \frac{s_{\text{down}} + s_{\text{up}}}{2}$ and stream speed $c = \frac{s_{\text{down}} - s_{\text{up}}}{2}$.
Time for Round Trip: Total time = $\frac{d}{b+c} + \frac{d}{b-c}$ where $d$ is the one-way distance.
Stream Speed Must Be Less Than Boat Speed: If $c \geq b$, the boat cannot move upstream ($s_{\text{up}} \leq 0$).
Circular Track Problems
On a circular track, two runners starting from the same point meet whenever the faster runner gains exactly one full lap on the slower one. The time to meet depends on the relative speed and the track length.
$$t_{\text{meet}} = \frac{L}{|s_1 - s_2|}$$
First meeting time equals track length divided by the relative speed (same direction).
$t_{\text{meet}}$=Time until they meet for the first time(seconds)
$L$=Length of the circular track(metres)
$s_1, s_2$=Speeds of the two runners(m/s)
Opposite directions
→Use $s_1 + s_2$ instead of $|s_1 - s_2|$: $t_{\text{meet}} = \frac{L}{s_1 + s_2}$.
$n$-th meeting
→Time = $n \times t_{\text{meet}}$ (each meeting takes the same time interval).
Same Direction: The faster runner must gain one full lap to meet. Relative speed = $|s_1 - s_2|$.
Opposite Direction: They approach each other, so relative speed = $s_1 + s_2$. They meet more quickly.
Starting at Different Points: If they start $k$ metres apart, first meeting time = $\frac{k}{s_{\text{rel}}}$ (same direction) or $\frac{L - k}{s_{\text{rel}}}$ (opposite direction, depending on which gap you measure).
Application Strategy
Worked Examples
Problem Statement
A typist can type a 200-page manuscript in 10 hours. How many pages does she type in 3.5 hours?
work rate
Strategic Toolkit
$$W = R \times T$$
Step 1
Find the rate
Divide total pages by total time to get the typing rate.
$R = \frac{200}{10} = 20 \text{ pages/hour}$
Step 2
Apply the formula
Multiply the rate by 3.5 hours to find pages typed.
$W = 20 \times 3.5 = 70 \text{ pages}$
Problem Statement
A can complete a project in 12 days and B can complete it in 18 days. If they work together, how many days will the project take?
combined work
Strategic Toolkit
$$R_{\text{total}} = R_1 + R_2$$
Step 1
Find individual rates
Express each person's rate as fraction of the job per day.
$R_A = \frac{1}{12}, \quad R_B = \frac{1}{18}$
Step 2
Add the rates
The combined rate is the sum of individual rates, using a common denominator of 36.
$R_{\text{total}} = \frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \text{ job/day}$
Step 3
Find the combined time
Invert the combined rate to get the number of days to finish the project.
$T = \frac{1}{R_{\text{total}}} = \frac{36}{5} = 7.2 \text{ days}$
Problem Statement
A is three times as efficient as B. If B alone can complete a task in 24 days, how long will it take if both work together?
efficiency
Strategic Toolkit
$$R_A : R_B = 3 : 1$$
Step 1
Find B's rate
B completes the job in 24 days, so B's rate is the reciprocal.
$R_B = \frac{1}{24}$
Step 2
Find A's rate from efficiency ratio
A is three times as efficient, so A's rate is three times B's rate. This means A alone would finish in 8 days.
$R_A = 3 \times \frac{1}{24} = \frac{1}{8}$
Step 3
Compute combined rate and time
Add both rates and invert to find the time working together.
$R_{\text{total}} = \frac{1}{8} + \frac{1}{24} = \frac{3}{24} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6} \Rightarrow T = 6 \text{ days}$
Problem Statement
Pipe A fills a tank in 12 minutes, Pipe B fills it in 15 minutes, and Pipe C drains it in 20 minutes. If all three pipes are opened together, how long does it take to fill the tank?
pipes and cisterns
Strategic Toolkit
$$R_{\text{net}} = R_A + R_B - R_C$$
Step 1
Identify each pipe's rate
Assign positive rates to fill pipes and a negative rate to the drain pipe.
$R_A = \frac{1}{12}, \quad R_B = \frac{1}{15}, \quad R_C = -\frac{1}{20}$
Step 2
Calculate net fill rate
Add the two fill rates and subtract the drain rate, using a common denominator of 60.
$R_{\text{net}} = \frac{1}{12} + \frac{1}{15} - \frac{1}{20} = \frac{5}{60} + \frac{4}{60} - \frac{3}{60} = \frac{6}{60} = \frac{1}{10}$
Step 3
Find fill time
The net rate is positive, so invert it to get the time to completely fill the tank.
$T = \frac{1}{\frac{1}{10}} = 10 \text{ minutes}$
Problem Statement
A driver travels from City X to City Y at 40 km/h and returns from Y to X at 60 km/h. What is the average speed for the entire round trip?
average speed
Strategic Toolkit
$$\text{Avg Speed} = \frac{2ab}{a+b}$$
Step 1
Set a convenient distance
Choose a one-way distance that divides evenly by both speeds to keep arithmetic clean.
$d = 120 \text{ km each way}$
Step 2
Calculate time for each leg
Find the time taken for the onward and return journeys separately.
$t_1 = \frac{120}{40} = 3 \text{ h}, \quad t_2 = \frac{120}{60} = 2 \text{ h}$
Step 3
Apply the average speed formula
Divide total distance by total time. This equals the harmonic mean of the two speeds.
$\text{Avg Speed} = \frac{240}{5} = 48 \text{ km/h} \quad \left(\text{or } \frac{2 \times 40 \times 60}{40+60} = 48\right)$
Problem Statement
Train A is 200 metres long and travels at 72 km/h. Train B is 300 metres long and travels at 54 km/h in the opposite direction. How many seconds does it take for the two trains to completely pass each other?
relative speedtrains
Strategic Toolkit
$$t = \frac{L_1 + L_2}{s_{\text{rel}}}$$
Step 1
Convert speeds to m/s
Multiply each speed by $\frac{5}{18}$ to convert from km/h to m/s for consistency with metre lengths.
$s_A = 72 \times \frac{5}{18} = 20 \text{ m/s}, \quad s_B = 54 \times \frac{5}{18} = 15 \text{ m/s}$
Step 2
Find relative speed
Since the trains move in opposite directions, their speeds add up.
$s_{\text{rel}} = 20 + 15 = 35 \text{ m/s}$
Step 3
Calculate crossing time
The total distance to cover is the sum of both train lengths, divided by the relative speed.
$t = \frac{200 + 300}{35} = \frac{500}{35} = \frac{100}{7} \approx 14.3 \text{ seconds}$
Problem Statement
A boat travels 30 km downstream in 2 hours and the same 30 km upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.
boats and streams
Strategic Toolkit
$$b = \frac{s_{\text{down}} + s_{\text{up}}}{2}, \quad c = \frac{s_{\text{down}} - s_{\text{up}}}{2}$$
Step 1
Calculate downstream and upstream speeds
Divide the distance by time for each leg of the journey.
$s_{\text{down}} = \frac{30}{2} = 15 \text{ km/h}, \quad s_{\text{up}} = \frac{30}{3} = 10 \text{ km/h}$
Step 2
Solve for boat and stream speeds
Average the downstream and upstream speeds for the boat speed; half the difference gives the stream speed.
$b = \frac{15+10}{2} = 12.5 \text{ km/h}, \quad c = \frac{15-10}{2} = 2.5 \text{ km/h}$
Problem Statement
Two runners start at the same point on a 400-metre circular track. Runner A runs at 5 m/s and Runner B runs at 3 m/s, both in the same direction. After how many seconds will they meet for the first time?
circular trackrelative speed
Strategic Toolkit
$$t_{\text{meet}} = \frac{L}{|s_1 - s_2|}$$
Step 1
Find the relative speed
Both runners move in the same direction, so subtract the slower speed from the faster.
$s_{\text{rel}} = 5 - 3 = 2 \text{ m/s}$
Step 2
Calculate meeting time
The faster runner must gain one full lap (400 m) on the slower runner to meet.
$t_{\text{meet}} = \frac{400}{2} = 200 \text{ seconds}$
Memorize formulas
Work = Rate × Time
Fundamental work equation — set total job = 1 and rate = 1/T.
Formula
$W = R \times T$
Combined Work (Two Workers)
Time for two workers together when individual times are known.
Formula
$T_{\text{together}} = \frac{t_1 \cdot t_2}{t_1 + t_2}$
Speed–Distance–Time
Rearrange to find speed, distance, or time from the other two.
Formula
$s = \frac{d}{t}, \quad d = s \times t, \quad t = \frac{d}{s}$
Average Speed (Equal Distances)
Harmonic mean of two speeds when equal distances are covered at each speed.
Formula
$\text{Avg Speed} = \frac{2ab}{a+b}$
Relative Speed
Add speeds for opposite directions, subtract for same direction.
Formula
$s_{\text{rel}} = s_1 + s_2 \text{ (opposite)}, \quad s_{\text{rel}} = |s_1 - s_2| \text{ (same)}$
Boat and Stream Speeds
Find boat and stream speeds from downstream/upstream measurements.
Formula
$b = \frac{s_d + s_u}{2}, \quad c = \frac{s_d - s_u}{2}$
Circular Track Meeting Time
Time for first meeting on a circular track (same direction).
Formula
$t_{\text{meet}} = \frac{L}{|s_1 - s_2|}$
km/h to m/s Conversion
Convert between km/h and m/s for train and distance problems.
Formula
$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$
Review
1
A factory machine produces 120 units of a product in 6 hours. At the same constant rate, how many units will it produce in 15 hours?
Ans
300 units
Review
Tap to reveal
Find the work rate
Divide the total output by the time taken to get the rate of production per hour.
$$R = \frac{120}{6} = 20 \text{ units/hour}$$
Calculate output for 15 hours
Multiply the rate by the new time to find the total work done.
$$W = 20 \times 15 = 300 \text{ units}$$
2
Pipe A can fill a tank in 12 hours and Pipe B can fill it in 8 hours. If both pipes are opened simultaneously, how long will it take to fill the tank?
Ans
4.8 hours
Review
Tap to reveal
Find the combined filling rate
Add the individual rates of the two pipes since they work together.
$$R_{\text{combined}} = \frac{1}{12} + \frac{1}{8} = \frac{5}{24} \text{ tank/hour}$$
Compute the filling time
Take the reciprocal of the combined rate to get the total time.
$$T = \frac{24}{5} = 4.8 \text{ hours}$$
3
Worker A can complete a task in 15 days and Worker B can complete it in 10 days. They work together for 4 days, after which B leaves. How many additional days will A need to finish the remaining work?
Ans
5 days
Review
Tap to reveal
Set up using the LCM method
Take LCM of 15 and 10 as 30 units of total work, then find each worker's daily rate.
$$W = 30, \quad R_A = 2 \text{ units/day}, \quad R_B = 3 \text{ units/day}$$
Compute work done together and remaining work
In 4 days at 5 units/day they complete 20 units, leaving 10 units.
$$\text{Remaining} = 30 - 5 \times 4 = 10 \text{ units}$$
Find A's time to finish alone
Divide the remaining work by A's individual rate.
$$T = \frac{10}{2} = 5 \text{ days}$$
4
Worker A is twice as efficient as Worker B. If they work together and earn a total of Rs. 1800 for completing a job, what is B's share of the earnings?
Ans
Rs. 600
Review
Tap to reveal
Determine the efficiency ratio
Since A is twice as efficient as B, their work rate ratio is $2:1$.
$$A : B = 2 : 1$$
Split the earnings in the efficiency ratio
The total of 3 parts equals Rs. 1800, so B receives 1 part.
$$\text{B's share} = \frac{1}{3} \times 1800 = \text{Rs. } 600$$
5
An inlet pipe can fill a tank in 20 hours, while a drain pipe can empty the full tank in 30 hours. If both pipes are opened together, how many hours will it take to fill the tank?
Ans
60 hours
Review
Tap to reveal
Calculate the net filling rate
Subtract the drain rate from the inlet rate to get the net rate of filling.
$$R_{\text{net}} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \text{ tank/hour}$$
Find the total filling time
Take the reciprocal of the net rate to determine how long it takes to fill one full tank.
$$T = \frac{1}{1/60} = 60 \text{ hours}$$
6
Two inlet pipes A and B can fill a tank in 10 hours and 15 hours respectively. A drain pipe C can empty the full tank in 12 hours. If all three pipes are opened together, in how many hours will the tank be completely filled?
Ans
12 hours
Review
Tap to reveal
Combine the three pipe rates
Add the inlet rates and subtract the drain rate using a common denominator of 60.
$$R = \frac{1}{10} + \frac{1}{15} - \frac{1}{12} = \frac{6+4-5}{60} = \frac{5}{60} = \frac{1}{12}$$
Compute the filling time
The reciprocal of the net rate gives the total time to fill the tank.
$$T = 12 \text{ hours}$$
7
A bus travels at a constant speed of 60 km/h for 3 hours. What total distance does it cover?
Ans
180 km
Review
Tap to reveal
Apply the speed-distance-time formula
Multiply the speed by the time to find the distance traveled.
$$d = s \times t = 60 \times 3 = 180 \text{ km}$$
8
A train 200 metres long passes a platform 300 metres long in 25 seconds. What is the speed of the train in km/h?
Ans
72 km/h
Review
Tap to reveal
Find the total distance the train must cover
Add the train's length and the platform's length since the train must fully clear the platform.
$$d = 200 + 300 = 500 \text{ m}$$
Calculate speed in m/s and convert to km/h
Divide distance by time, then multiply by $\frac{18}{5}$ to convert from m/s to km/h.
$$s = \frac{500}{25} = 20 \text{ m/s} = 20 \times \frac{18}{5} = 72 \text{ km/h}$$
9
A delivery driver travels from Warehouse P to Warehouse Q at 80 km/h and returns along the same route at 120 km/h. What is the average speed for the entire round trip?
Ans
96 km/h
Review
Tap to reveal
Identify the correct average speed formula
For equal distances traveled at two different speeds, use the harmonic mean formula.
$$\text{Avg Speed} = \frac{2 v_1 v_2}{v_1 + v_2}$$
Substitute and calculate
Plug in $v_1 = 80$ and $v_2 = 120$ to find the average speed for the round trip.
$$\text{Avg Speed} = \frac{2 \times 80 \times 120}{80 + 120} = \frac{19200}{200} = 96 \text{ km/h}$$
10
A motorboat can travel 30 km downstream in 2 hours and 30 km upstream in 3 hours. What is the speed of the boat in still water?
Ans
12.5 km/h
Review
Tap to reveal
Find downstream and upstream speeds
Divide each distance by its respective travel time to get $v_d$ and $v_u$.
$$v_d = 15 \text{ km/h}, \quad v_u = 10 \text{ km/h}$$
Calculate the still water speed
Average the downstream and upstream speeds to isolate the boat's own speed.
$$v_b = \frac{v_d + v_u}{2} = \frac{15 + 10}{2} = 12.5 \text{ km/h}$$
11
Two trains, 150 m and 200 m long, run in the same direction on parallel tracks at 72 km/h and 54 km/h respectively. How many seconds will it take for the faster train to completely overtake the slower train?
Ans
70 seconds
Review
Tap to reveal
Find the relative speed
Subtract the slower speed from the faster speed since both trains move in the same direction, then convert to m/s.
$$v_{\text{rel}} = 72 - 54 = 18 \text{ km/h} = 5 \text{ m/s}$$
Determine the distance to cover and compute time
The overtaking distance equals the sum of both train lengths.
$$d = 150 + 200 = 350 \text{ m}, \quad T = \frac{350}{5} = 70 \text{ s}$$
12
Two trains 120 m and 180 m long approach each other on parallel tracks at 50 km/h and 70 km/h respectively. In how many seconds will they completely clear each other?
Ans
9 seconds
Review
Tap to reveal
Calculate the relative speed
Add the two speeds since the trains approach from opposite directions, then convert to m/s.
$$v_{\text{rel}} = 50 + 70 = 120 \text{ km/h} = \frac{100}{3} \text{ m/s}$$
Compute the crossing time
Divide the combined length of both trains by the relative speed.
$$T = \frac{120 + 180}{100/3} = \frac{300 \times 3}{100} = 9 \text{ seconds}$$
13
Two runners start from the same point on a circular track of 800 m length and run in the same direction with speeds 5 m/s and 3 m/s respectively. After how many seconds will they meet for the first time?
Ans
400 seconds
Review
Tap to reveal
Apply the circular track formula
For same-direction runners, the first meeting occurs when the faster runner gains exactly one lap on the slower one.
$$T = \frac{L}{|v_1 - v_2|} = \frac{800}{5 - 3} = 400 \text{ s}$$
14
Two cyclists start from the same point on a circular track of circumference 1200 m and ride in opposite directions at 12 m/s and 8 m/s respectively. When will they first meet?
Ans
60 seconds
Review
Tap to reveal
Use the opposite-direction meeting formula
The combined closing speed is the sum of both speeds, and they meet after covering the full track length.
$$T = \frac{L}{v_1 + v_2} = \frac{1200}{12 + 8} = 60 \text{ s}$$
15
If 5 painters can paint a house in 8 days, how many days will 10 painters take to paint the same house, assuming each painter works at the same rate?
Ans
4 days
Review
Tap to reveal
Express total work in worker-time units
Multiply the number of workers by the number of days to get the total work constant.
$$W = 5 \times 8 = 40 \text{ painter-days}$$
Find the new time with more workers
Divide the total work by the new number of painters.
$$T = \frac{40}{10} = 4 \text{ days}$$
16
A boat takes 4 hours to travel 48 km downstream and 6 hours to travel the same 48 km upstream. What is the speed of the stream?
Ans
2 km/h
Review
Tap to reveal
Find downstream and upstream speeds
Compute the effective speed in each direction by dividing distance by time.
$$v_d = \frac{48}{4} = 12 \text{ km/h}, \quad v_u = \frac{48}{6} = 8 \text{ km/h}$$
Solve for the stream speed
The stream speed is half the difference between downstream and upstream speeds.
$$v_s = \frac{v_d - v_u}{2} = \frac{12 - 8}{2} = 2 \text{ km/h}$$
17
Two friends start from the same location at the same time. One walks north at 4 km/h and the other walks east at 3 km/h. After how many hours will they be 25 km apart?
Ans
5 hours
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Express distances in terms of time
After $t$ hours, the two friends have traveled $4t$ km north and $3t$ km east respectively.
$$d_{\text{north}} = 4t, \quad d_{\text{east}} = 3t$$
Apply the Pythagorean theorem and solve
The straight-line distance between them forms the hypotenuse of a right triangle.
$$\sqrt{(4t)^2 + (3t)^2} = 25 \implies 5t = 25 \implies t = 5 \text{ hours}$$
18
Two workers A and B together can complete a job in 6 days. Working alone, A can do it in 10 days. How many days will B need to complete the job alone?
Ans
15 days
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Set up rate equations
Express the combined rate as the sum of individual rates.
$$\frac{1}{R_A} + \frac{1}{R_B} = 6, \quad \frac{1}{R_A} = 10$$
Isolate B's rate
Subtract A's rate from the combined rate to find B's contribution.
$$R_B = \frac{1}{6} - \frac{1}{10} = \frac{1}{15}$$
Find B's time
Take the reciprocal of B's rate to get the number of days B needs alone.
$$T_B = \frac{1}{1/15} = 15 \text{ days}$$
19
Pipe A can fill a tank in 24 hours and Pipe B can fill it in 36 hours. Pipe A is opened first for 6 hours, and then Pipe B is also opened. How much longer will it take to completely fill the tank?
Ans
10.8 hours
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Find work done by A alone
Pipe A fills at $\frac{1}{24}$ tank per hour, so in 6 hours it fills a quarter of the tank.
$$\text{Work by A} = \frac{6}{24} = \frac{1}{4} \text{ of the tank}$$
Calculate the combined rate
Both pipes together fill at the sum of their individual rates.
$$R_{\text{combined}} = \frac{1}{24} + \frac{1}{36} = \frac{5}{72} \text{ tank/hour}$$
Find time to fill the remaining portion
Divide the remaining $\frac{3}{4}$ of the tank by the combined rate.
$$T = \frac{3/4}{5/72} = \frac{3 \times 72}{4 \times 5} = 10.8 \text{ hours}$$
20
On a circular track of 1500 m, Runner A and Runner B start together from the same point in the same direction with speeds 10 m/s and 4 m/s respectively. How many times will A overtake B (including the first time they meet) in 45 minutes?
Ans
10
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Find the time between consecutive meetings
A gains one full lap on B every $\frac{L}{v_1 - v_2}$ seconds.
$$T = \frac{1500}{10 - 4} = 250 \text{ seconds}$$
Count full meetings in 45 minutes
Convert 45 minutes to seconds and divide by the meeting interval, taking only complete meetings.
$$\frac{2700}{250} = 10.8 \implies 10 \text{ full overtakes}$$
Question 1 of 20
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1
A factory machine produces 120 units of a product in 6 hours. At the same constant rate, how many units will it produce in 15 hours?