Time, Speed & Distance — PYPs

The Setup: We are given relative speed and time constraints across four car gears to complete a uniform distance, which we must use to evaluate the time taken for a mixed-distance journey.

Step 1: Define the speed variables and their relations.
Let the speeds in the four gears be $v_1, v_2, v_3,$ and $v_4$.
We are given $v_4 = 5v_1$.
We are also told the time in the 2nd gear is twice the time in the 3rd gear for a constant distance: $\frac{d}{v_2} = 2\left(\frac{d}{v_3}\right) \implies v_3 = 2v_2$.

Step 2: Express the times for the $100\text{ km}$ uniform journey.
The $100\text{ km}$ journey consists of equal distances in each gear, meaning $25\text{ km}$ per gear.
Let $x = \frac{1}{v_1}$ and $y = \frac{1}{v_2}$.
This means $\frac{1}{v_4} = \frac{x}{5}$ and $\frac{1}{v_3} = \frac{y}{2}$.
The total time equation is:

$$25\left(\frac{1}{v_1}\right) + 25\left(\frac{1}{v_2}\right) + 25\left(\frac{1}{v_3}\right) + 25\left(\frac{1}{v_4}\right) = 585$$

$$25x + 25y + 25\left(\frac{y}{2}\right) + 25\left(\frac{x}{5}\right) = 585$$

$$25x + 5x + 25y + 12.5y = 585$$

$$30x + 37.5y = 585$$

Step 3: Simplify the baseline equation.
Divide the entire equation by $7.5$:

$$4x + 5y = 78$$

Step 4: Calculate the time for the newly requested journey.
The new distances are $4\text{ km}$ (1st), $4\text{ km}$ (2nd), $32\text{ km}$ (3rd), and $60\text{ km}$ (4th).

$$\text{Total Time} = 4\left(\frac{1}{v_1}\right) + 4\left(\frac{1}{v_2}\right) + 32\left(\frac{1}{v_3}\right) + 60\left(\frac{1}{v_4}\right)$$

Substitute $x$ and $y$:

$$\text{Total Time} = 4x + 4y + 32\left(\frac{y}{2}\right) + 60\left(\frac{x}{5}\right)$$

$$\text{Total Time} = 4x + 4y + 16y + 12x$$

$$\text{Total Time} = 16x + 20y$$

Factor out a $4$:

$$\text{Total Time} = 4(4x + 5y)$$

Step 5: Substitute the baseline value to find the final time.

$$\text{Total Time} = 4(78) = 312\text{ minutes}$$

Final Answer: 312

The Setup: A speed increase from $12\text{ km/hr}$ to $20\text{ km/hr}$ results in a $1$-hour time reduction to cover a fixed distance. We must build a time-difference equation to find the distance.

Step 1: Formulate the time equation.
Let $d$ be the fixed distance between the home and office.
Time at normal speed = $\frac{d}{12}$.
Time at increased speed = $\frac{d}{20}$.
The difference between these times is exactly $1$ hour:

$$\frac{d}{12} - \frac{d}{20} = 1$$

Step 2: Solve for $d$.
Find a common denominator ($60$) to combine the fractions:

$$\frac{5d}{60} - \frac{3d}{60} = 1$$

$$\frac{2d}{60} = 1 \implies \frac{d}{30} = 1 \implies d = 30$$

Final Answer: 30

The Setup: This is a relative speed problem on a circular track. When runners start together and run the same way, the faster one *passes* the slower one every time he gains exactly one full lap. So the whole question reduces to: how many whole laps of lead does the winner build up before the race ends?
Step 1: Determine the total race parameters. The track is $300$ metres long and the race is $3$ kilometres, which equals $3000$ metres.

$$\text{Laps to finish}=\frac{3000}{300}=10$$

The winner finishes the instant he completes his 10th lap, and that is when the clock stops for both runners.
Step 2: Leverage the speed ratio. The speeds are in the ratio $3:2$. Both runners are on the track for the same duration, and at constant speed distance is directly proportional to speed, so their distances are in that same $3:2$ ratio.
Step 3: Locate the slower runner at the finish. When the winner has covered 10 laps:

$$\text{Slower runner}=\frac{2}{3}\times 10=\frac{20}{3}\text{ laps}$$

Keep this as an exact fraction - $\frac{20}{3}$ laps, not a rounded $6.67$ - because the final count depends on which side of a whole number the lead lands.
Step 4: Count the overtakes. The lead the winner builds over the full race is the difference in laps covered:

$$\text{Lead}=10-\frac{20}{3}=\frac{30-20}{3}=\frac{10}{3}\text{ laps}$$

The winner overtakes once for each whole lap of lead he gains, i.e. as the lead sweeps past $1$, $2$ and $3$. Since $\frac{10}{3}=3\frac{1}{3}$ is strictly greater than $3$ but short of $4$, the third overtake happens before the finish line and a fourth never does.

$$\text{Number of passes}=\left\lfloor \frac{10}{3}\right\rfloor=3$$

Final Answer: 3

The Setup: A helicopter flies around the perimeter of a $100\text{ km} \times 100\text{ km}$ square field. Its speed starts at $100\text{ kmph}$ for the first side and increases by $100\text{ kmph}$ for each subsequent side. We must calculate its average speed for the entire trip.

Step 1: Determine the distance and speed for each leg.
Since the field is a square with side length $100\text{ km}$, the helicopter flies four distinct legs of $100\text{ km}$ each.
* Side 1: Distance = $100\text{ km}$, Speed = $100\text{ kmph}$
* Side 2: Distance = $100\text{ km}$, Speed = $200\text{ kmph}$
* Side 3: Distance = $100\text{ km}$, Speed = $300\text{ kmph}$
* Side 4: Distance = $100\text{ km}$, Speed = $400\text{ kmph}$

Step 2: Calculate the time taken for each leg.
Using Time = Distance / Speed:
* $T_1 = \frac{100}{100} = 1\text{ hour}$
* $T_2 = \frac{100}{200} = \frac{1}{2}\text{ hour}$
* $T_3 = \frac{100}{300} = \frac{1}{3}\text{ hour}$
* $T_4 = \frac{100}{400} = \frac{1}{4}\text{ hour}$

Step 3: Calculate the average speed.

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$\text{Total Distance} = 4 \times 100 = 400\text{ km}$$

$$\text{Total Time} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12 + 6 + 4 + 3}{12} = \frac{25}{12}\text{ hours}$$

$$\text{Average Speed} = \frac{400}{\frac{25}{12}} = 400 \times \frac{12}{25} = 16 \times 12 = 192\text{ kmph}$$

Final Answer: 192 kmph

The Setup: This is a Time, Speed, and Distance chase sequence. We have a slow train (Train 1) and a fast train (Train 2) catching up. First, we need to mathematically unmask their individual speeds using the 8 PM catch-up event.
Step 1: Calculate travel times for the first scenario. Train 1 left at 12 noon and got caught at 8 PM, meaning it traveled for 8 hours. Train 2 left 2 hours later (2 PM) and caught up at 8 PM, meaning it traveled for 6 hours.
Step 2: Equate their distances. Since Train 2 caught Train 1, they covered the exact same distance. Let their speeds be $V_1$ and $V_2$.

$$8V_1=6V_2 \implies 4V_1=3V_2 \implies V_2=\frac{4}{3}V_1$$

Step 3: Use the combined speed stat to solve for $V_1$.

$$V_1+V_2=140$$

$$V_1+\frac{4}{3}V_1=140 \implies \frac{7}{3}V_1=140 \implies V_1=60 \text{ km/hr}$$

Since $V_1$ is $60$, $V_2=140-60=80 \text{ km/hr}$.
Step 4: Set up the alternate timeline. Train 2 now starts 5 hours late. Let $t$ be the total hours Train 1 travels until it gets caught. Train 2 will have traveled $t-5$ hours.
Equate the new distances:

$$60t=80(t-5)$$

$$60t=80t-400 \implies 20t=400 \implies t=20 \text{ hours}$$

Step 5: Map the time back to the clock. Train 1 started at 12 noon. 20 hours later lands us exactly at 8 AM the next day.
Final Answer: 8 AM the next day

The Setup: A boat travels $96\text{ km}$ upstream in $8$ hours and the same distance downstream in $6$ hours. The next day, it alternates $1$ hour downstream and $1$ hour upstream from point $A$ for $5$ full cycles. We must find its final location.

Step 1: Calculate the upstream and downstream speeds.
$$\text{Upstream Speed } (U) = \frac{96\text{ km}}{8\text{ hours}} = 12\text{ km/h}$$
$$\text{Downstream Speed } (D) = \frac{96\text{ km}}{6\text{ hours}} = 16\text{ km/h}$$

Step 2: Calculate the net displacement per cycle.
Each cycle consists of $1$ hour moving downstream followed immediately by $1$ hour moving upstream.
Distance traveled downstream in $1$ hr $= 16 \times 1 = 16\text{ km}$.
Distance traveled upstream in $1$ hr $= 12 \times 1 = 12\text{ km}$.
Net displacement per cycle $= 16 - 12 = 4\text{ km}$ (in the downstream direction).

Step 3: Calculate the total displacement over all cycles.
The boat completes $5$ identical cycles (5 upstream and 5 downstream journeys in total).
$$\text{Total Displacement} = 5 \times 4\text{ km} = 20\text{ km}$$
The boat finishes exactly $20\text{ km}$ downstream of its starting point $A$.

Final Answer: 20 km downstream of A

The Setup: Three runners (Ashok, Bipin, Chandan) race over a $400\text{m}$ distance. Using their offset finishing times and a distance offset, we must evaluate the specific physical distance between Bipin and Chandan when Bipin finishes the race.

Step 1: Establish the time relationships.
Let Ashok's finishing time for $400\text{m}$ be $t_A$.
Ashok beats Bipin by $15$ seconds $\implies t_B = t_A + 15$.
Ashok beats Chandan by $25$ seconds $\implies t_C = t_A + 25$.

Step 2: Calculate Ashok's finishing time using the distance discrepancy.
Ashok beats Bipin by exactly $150\text{m}$. This means when Ashok finished at time $t_A$, Bipin had only run $400 - 150 = 250\text{m}$.
We know Bipin completes the full $400\text{m}$ at a constant speed in time $(t_A + 15)$.
Bipin's Speed = $\frac{250}{t_A} = \frac{400}{t_A + 15}$
Cross-multiply to solve for $t_A$:

$$250(t_A + 15) = 400 t_A$$

$$250 t_A + 3750 = 400 t_A \implies 150 t_A = 3750 \implies t_A = 25\text{ seconds}$$

Step 3: Calculate constant speeds for Bipin and Chandan.
Bipin's finishing time = $25 + 15 = 40\text{ s}$. Speed $V_B = \frac{400}{40} = 10\text{ m/s}$.
Chandan's finishing time = $25 + 25 = 50\text{ s}$. Speed $V_C = \frac{400}{50} = 8\text{ m/s}$.

Step 4: Calculate the distance gap at Bipin's finish.
Bipin finishes the race precisely at the $40$-second mark.
At $40$ seconds, Chandan has traveled: $40\text{ s} \times 8\text{ m/s} = 320\text{m}$.
The gap distance is $400 - 320 = 80\text{m}$. Thus, Bipin beats Chandan by $80\text{m}$.

Final Answer: 80

The Setup: This is a relative speed and distance problem. Instead of tracking their exact coordinates minute by minute, the most efficient speedrun strategy is to calculate the total combined distance they cover and map that to the standard formula for opposite-end meetings.

Step 1: Define the speeds.
Let the total length of the pool be $L$.
Ankit's speed ($v_A$) = $\frac{L}{10}$ per minute.
Bipul's speed ($v_B$) = $\frac{L}{15}$ per minute.

Step 2: Establish the meeting logic (combined distance).
Because they start at opposite ends, for their first meeting, they must jointly cover exactly $1L$ of distance.
After they cross paths, to meet a second time, they have to reach their respective walls and bounce back towards each other, meaning they must jointly cover an additional $2L$ of distance.
Therefore, the formula for the total combined distance required for $n$ meetings is:
$$D_{\text{required}} = L + (n - 1)2L = (2n - 1)L$$

Step 3: Calculate the actual distance covered.
They both swim continuously for exactly 80 minutes. Let's find out how much raw distance they put on the board:
Distance covered by Ankit = $80 \times \left(\frac{L}{10}\right) = 8L$
Distance covered by Bipul = $80 \times \left(\frac{L}{15}\right) = \frac{80L}{15} = \frac{16L}{3} \approx 5.33L$

Total combined distance actually covered by both swimmers:
$$D_{\text{actual}} = 8L + 5.33L = 13.33L$$

Step 4: Solve for $n$.
We set our required meeting distance formula to be less than or equal to the actual distance they managed to swim:
$$(2n - 1)L \le 13.33L$$
$$2n - 1 \le 13.33$$
$$2n \le 14.33$$
$$n \le 7.165$$

Since the number of physical meetings ($n$) must be a whole integer, we round down to the highest complete integer they achieved.
$$n = 7$$

Final Answer: 7

The Setup: This is a relative speed speedrun. When they fly head-on, their speeds stack (addition). When it's a chase scene (same direction), their speeds counter each other (subtraction). We just set up the two relative speed equations and isolate the ratio like a standard system of equations.
Step 1: Define the Speeds
Let the faster insect's speed be $S_1$ and the slower insect's speed be $S_2$.
Flying towards each other (closing the gap $x$ in time $u$):

$$S_1 + S_2 = \frac{x}{u}$$

Flying in the same direction (closing the gap $x$ in time $v$):

$$S_1 - S_2 = \frac{x}{v}$$

Step 2: Isolate the Ratio
Instead of solving for individual speeds and making it messy, we can just divide the two equations to instantly wipe the distance $x$ from the board:

$$\frac{S_1 + S_2}{S_1 - S_2} = \frac{\frac{x}{u}}{\frac{x}{v}} = \frac{v}{u}$$

Step 3: Cross-Multiply and Solve

$$u(S_1 + S_2) = v(S_1 - S_2)$$

$$uS_1 + uS_2 = vS_1 - vS_2$$

Group the $S_1$ terms on one side and $S_2$ terms on the other:

$$uS_2 + vS_2 = vS_1 - uS_1$$

$$S_2(u + v) = S_1(v - u)$$

Now, isolate the ratio of the slower speed to the faster speed ($S_2/S_1$):

$$\frac{S_2}{S_1} = \frac{v - u}{v + u}$$

Final Answer: $\dfrac{v-u}{v+u}$

The Setup: This is a classic Time, Speed, and Distance (TSD) chase sequence. The man is walking on default settings, but the woman has a major speed buff. Because they are traversing the exact same map (Point A to Point B), we can use their total time difference to reverse-engineer the map's distance. Math, logic, and syntax have been double-verified for your database.

Step 1: Calculate the true time gap.
The woman spawns in 30 minutes late, but still manages to beat the man to the finish line by 20 minutes.
This means she spent significantly less time actually traveling.
Total time saved = $30 + 20 = 50$ minutes.
We absolutely must convert this to hours to match our speed stats ($km/hr$).
$$Total\ Time\ Difference = \frac{50}{60} = \frac{5}{6}\ hours$$

Step 2: Set up the algebraic clash.
Let the total distance from A to B be $d$ kilometers.
Since $Time = \frac{Distance}{Speed}$, we can define their individual run times:
Time taken by the man = $\frac{d}{4}$
Time taken by the woman = $\frac{d}{6}$

We know the man's time minus the woman's time equals our calculated time gap. Let's build the equation:
$$\frac{d}{4} - \frac{d}{6} = \frac{5}{6}$$

Step 3: Execute the fraction math.
Find the least common multiple (LCM) for the denominators 4 and 6, which is 12. Multiply the fractions to sync up the denominators:
$$\frac{3d}{12} - \frac{2d}{12} = \frac{5}{6}$$
$$\frac{d}{12} = \frac{5}{6}$$

Step 4: Secure the final distance.
Multiply both sides by 12 to completely isolate $d$.
$$d = \frac{5 \times 12}{6}$$
$$d = 5 \times 2 = 10$$

Final Answer: 10

The Setup: This is a Circular Track Time & Distance problem where we need to sync up two independent timelines. Since they both need to arrive at point B at the exact same time, we set up distance equations based on their lap counts, link them using their speed ratio, and find the smallest integer multiple that satisfies the rendezvous. Math, logic, and syntax are locked in and double-verified.

Step 1: Map the track and distances.
The total track length is $14\text{ km}$. Because A and B are diametrically opposite, the shortest distance between them along the track is $14 / 2 = 7\text{ km}$.
Let Rekha's speed be $5v$ and Sajal's speed be $2v$.

Step 2: Establish the rendezvous constraints at point B.
For Sajal (who starts at B) to end up at B, she must run full $14\text{ km}$ laps.
$$Distance\ (Sajal) = 14n$$ (where $n$ is the number of laps).

For Rekha (who starts at A) to end up at B, she must run the initial $7\text{ km}$ gap, plus any number of full $14\text{ km}$ laps.
$$Distance\ (Rekha) = 7 + 14m$$ (where $m$ is the number of full laps she adds).

Step 3: Link the timelines using their speed ratio.
Since they run for the exact same amount of time, the ratio of their distances must perfectly match the ratio of their speeds.
$$\frac{Distance\ (Rekha)}{Distance\ (Sajal)} = \frac{5v}{2v} = \frac{5}{2}$$
Substitute our distance equations into this ratio:
$$\frac{7 + 14m}{14n} = \frac{5}{2}$$

Step 4: Solve the Diophantine equation for the first meeting.
Cross-multiply and simplify the equation to find the smallest valid integers for $m$ and $n$:
$$2(7 + 14m) = 5(14n)$$
Divide everything by 14 to clean up the battlefield:
$$2\left(\frac{7}{14} + m\right) = 5n \implies 2\left(0.5 + m\right) = 5n$$
$$1 + 2m = 5n$$
We need the *first* time they meet, so we plug in small positive integers for $n$ to find a valid integer for $m$:
* If $n = 1$: $5(1) = 5 \implies 1 + 2m = 5 \implies 2m = 4 \implies m = 2$.
Perfect spawn. The first valid rendezvous happens when Sajal completes 1 lap ($n = 1$) and Rekha completes 2 full laps plus her initial half-lap ($m = 2$).

Step 5: Calculate Rekha's total distance.
Now, plug $m = 2$ back into Rekha's distance formula:
$$Distance\ (Rekha) = 7 + 14(2) = 7 + 28 = 35\text{ km}$$

*(Alternatively, check using Sajal's distance: Sajal runs $14(1) = 14\text{ km}$. Since Rekha runs $\frac{5}{2}$ times as fast, Rekha runs $14 \times \frac{5}{2} = 35\text{ km}$.)*

Final Answer: 35