Distance, Time and Speed: CAT Previous Year Questions

Ans: D

Explanation:
Let the speeds in the four parts be $s_1, s_2, s_3, s_4$ and the time taken for each part be $t_1, t_2, t_3, t_4$.
The speeds form an arithmetic progression, so $s_i = s_1 + (i-1)d_s$ for $i=1, 2, 3, 4$.
The times form an arithmetic progression, so $t_i = t_1 + (i-1)d_t$ for $i=1, 2, 3, 4$.

The total distance traveled is 224 km = 224,000 meters.
The total travel time is 3 hours = 180 minutes.

For the first part:
Speed $s_1 = 960$ meters per minute.
Time $t_1 = 30$ minutes.
Distance of the first part $d_1 = s_1 \times t_1 = 960 \times 30 = 28800$ meters.

The total distance is the sum of the distances of the four parts:
$D = d_1 + d_2 + d_3 + d_4 = 224000$
$d_i = s_i \times t_i = (s_1 + (i-1)d_s) \times (t_1 + (i-1)d_t)$

The total time is the sum of the times of the four parts:
$T = t_1 + t_2 + t_3 + t_4 = 180$
$180 = 30 + (30 + d_t) + (30 + 2d_t) + (30 + 3d_t)$
$180 = 120 + 6d_t$
$60 = 6d_t$
$d_t = 10$ minutes.

So, the times are:
$t_1 = 30$ minutes
$t_2 = 30 + 10 = 40$ minutes
$t_3 = 30 + 2 \times 10 = 50$ minutes
$t_4 = 30 + 3 \times 10 = 60$ minutes

Check total time: $30 + 40 + 50 + 60 = 180$ minutes = 3 hours. This is correct.

Now let’s find the speeds.
The total distance is 224000 meters.
$d_1 = s_1 \times t_1 = 960 \times 30 = 28800$ meters.
$d_2 = s_2 \times t_2 = (960 + d_s) \times 40$
$d_3 = s_3 \times t_3 = (960 + 2d_s) \times 50$
$d_4 = s_4 \times t_4 = (960 + 3d_s) \times 60$

The sum of distances is 224000:
$28800 + (960 + d_s) \times 40 + (960 + 2d_s) \times 50 + (960 + 3d_s) \times 60 = 224000$
$28800 + 38400 + 40d_s + 48000 + 100d_s + 57600 + 180d_s = 224000$
$(28800 + 38400 + 48000 + 57600) + (40 + 100 + 180)d_s = 224000$
$172800 + 320d_s = 224000$
$320d_s = 224000 – 172800$
$320d_s = 51200$
$d_s = \frac{51200}{320} = \frac{5120}{32} = 160$ meters per minute.

So, the speeds are:
$s_1 = 960$ m/min
$s_2 = 960 + 160 = 1120$ m/min
$s_3 = 960 + 2 \times 160 = 960 + 320 = 1280$ m/min
$s_4 = 960 + 3 \times 160 = 960 + 480 = 1440$ m/min

The distance traveled in the fourth part is $d_4 = s_4 \times t_4$.
$d_4 = 1440 \times 60$
$d_4 = 86400$ meters.

Let’s check the distances:
$d_1 = 960 \times 30 = 28800$
$d_2 = 1120 \times 40 = 44800$
$d_3 = 1280 \times 50 = 64000$
$d_4 = 1440 \times 60 = 86400$
Total distance = $28800 + 44800 + 64000 + 86400 = 224000$ meters. This is correct.

The distance she travels in the fourth part is 86400 meters.

Correct_Option: D

Ans: 8

Explanation:Let $v_R$ be Rita’s speed in still water and $v_S$ be Sneha’s speed in still water.
We are given $v_R = 5$ km/h and $v_S = 6$ km/h.
Let $v$ be the velocity of the river.

For Sneha:
Speed downstream = $v_S + v = 6 + v$
Speed upstream = $v_S – v = 6 – v$

Time taken to row 14 km upstream = $\frac{14}{6-v}$ hours.
Time taken to row 14 km downstream = $\frac{14}{6+v}$ hours.

We are given that Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream.
48 minutes = $\frac{48}{60}$ hours = $\frac{4}{5}$ hours.

So, $\frac{14}{6-v} – \frac{14}{6+v} = \frac{4}{5}$
Multiply by $(6-v)(6+v)$ on both sides:
$14(6+v) – 14(6-v) = \frac{4}{5}(36 – v^2)$
$84 + 14v – 84 + 14v = \frac{4}{5}(36 – v^2)$
$28v = \frac{4}{5}(36 – v^2)$
Multiply by 5:
$140v = 4(36 – v^2)$
$140v = 144 – 4v^2$
$4v^2 + 140v – 144 = 0$
Divide by 4:
$v^2 + 35v – 36 = 0$
Factor the quadratic equation:
$(v+36)(v-1) = 0$
Since the speed of the river cannot be negative, $v = 1$ km/h.

Now for Rita:
Rita’s speed in still water is $v_R = 5$ km/h.
The river velocity is $v = 1$ km/h.

When Rita rows upstream, her speed is $v_R – v = 5 – 1 = 4$ km/h.
When Rita rows downstream, her speed is $v_R + v = 5 + 1 = 6$ km/h.

Let the distance Rita rows upstream be $d$ km.
Then the distance Rita rows downstream is also $d$ km.
Total distance covered by Rita is $2d$ km.

Time taken to row upstream = $\frac{d}{4}$ hours.
Time taken to row downstream = $\frac{d}{6}$ hours.

Total time taken by Rita = $\frac{d}{4} + \frac{d}{6}$ hours.
We are given that the total time taken is 100 minutes.
100 minutes = $\frac{100}{60}$ hours = $\frac{5}{3}$ hours.

So, $\frac{d}{4} + \frac{d}{6} = \frac{5}{3}$
Find a common denominator for the left side:
$\frac{3d + 2d}{12} = \frac{5}{3}$
$\frac{5d}{12} = \frac{5}{3}$
Multiply both sides by 12:
$5d = \frac{5}{3} \times 12$
$5d = 5 \times 4$
$5d = 20$
$d = 4$ km.

The total distance Rita will cover is $2d$.
Total distance = $2 \times 4 = 8$ km.

Final_Answer:8

Ans: 444

Let Ankita’s pace while walking be represented by $5x$. Consequently, her speed when running, which is $40\%$ faster than her walking pace, can be expressed as $1.4 \times 5x = 7x$.
The proportional relationship between her walking and running speeds is thus $5:7$. Consequently, the ratio of the durations Ankita requires to traverse a set distance while walking and running will be $7:5$.
Ankita’s journey from B to C on foot takes 3 hours and 30 minutes, equivalent to $3.5$ hours. In the alternate situation, when Ankita covers the distance from B to C at her running pace, her time will decrease proportionally to the inverse of the speed ratio. The duration of her run from B to C will be $\frac{3.5}{7} \times 5 = 2.5$ hours.
Her run from A to B requires 3 hours and 30 minutes, or $3.5$ hours. When Ankita undertakes the journey from A to B at her walking pace in the second scenario, her time will increase proportionally to the inverse of the speed ratio. The duration of her walk from A to B will be $\frac{3.5}{5} \times 7 = 4.9$ hours.
Therefore, the aggregate time for Ankita in the second scenario is $4.9 + 2.5 = 7.4$ hours. This translates to $7.4 \times 60 = 444$ minutes.

Ans: B

Let Rahul’s usual travel pace be represented by $x$ kilometers per hour. If his typical journey takes 6 hours (from 5 pm to 11 pm), the overall distance covered is $6 \times x = 6x$ kilometers.

Let $y$ kilometers be the distance covered before Rahul takes a break in both situations.

In the first instance, a 20-minute break, equivalent to $\frac{1}{3}$ of an hour, means his actual travel time is $6 – \frac{1}{3} = \frac{17}{3}$ hours. The equation representing this scenario is:
$\dfrac{y}{x} + \dfrac{6x-y}{x+3} = \dfrac{17}{3}$
$\Rightarrow \dfrac{6x^2+3y}{x^2+3x} = \dfrac{17}{3}$
$\Rightarrow 18x^2 + 9y = 17x^2 + 51x$
$\Rightarrow x^2 = 51x – 9y$ …..(1)

In the second instance, a break of 20 minutes plus an additional 10 minutes, totaling 30 minutes or $\frac{1}{2}$ of an hour, means his actual travel time is $6 – \frac{1}{2} = \frac{11}{2}$ hours. The equation for this situation is:
$\dfrac{y}{x} + \dfrac{6x-y}{x+5} = \dfrac{11}{2}$
$\Rightarrow \dfrac{6x^2+5y}{x^2+5x} = \dfrac{11}{2}$
$\Rightarrow 12x^2 + 10y = 11x^2 + 55x$
$\Rightarrow x^2 = 55x – 10y$ …..(2)

Equating expressions for $x^2$ from equations (1) and (2):
$55x – 10y = 51x – 9y$
$4x = y$

Substituting the value of $y$ into equation (1):
$x^2 = 51x – 36x$
Since $x$ represents speed and must be positive, $x = 15$. Thus, the correct option is B.

Ans: D

Explanation:Let the distance between A and B be D = 45 km.
Let the speed of Anil be $S_A$ km/hr and the speed of Sunil be $S_S$ km/hr.
Anil goes from A to B, and Sunil goes from B to A. They start at the same time.
They cross each other in 1 hour 30 minutes, which is 1.5 hours.
When they cross each other, the sum of the distances they have traveled is equal to the total distance between A and B.
Distance traveled by Anil = $S_A \times 1.5$
Distance traveled by Sunil = $S_S \times 1.5$
So, $1.5 S_A + 1.5 S_S = 45$
Dividing by 1.5, we get $S_A + S_S = 45 / 1.5 = 30$ (Equation 1)

Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A.
1 hour 15 minutes = 1.25 hours.
Time taken by Anil to travel from A to B = $D / S_A = 45 / S_A$
Time taken by Sunil to travel from B to A = $D / S_S = 45 / S_S$

According to the problem statement:
Time taken by Anil = Time taken by Sunil + 1.25 hours
$45 / S_A = 45 / S_S + 1.25$ (Equation 2)

From Equation 1, we have $S_S = 30 – S_A$. Substitute this into Equation 2:
$45 / S_A = 45 / (30 – S_A) + 1.25$
$45 / S_A – 45 / (30 – S_A) = 1.25$
$45 [(30 – S_A) – S_A] / [S_A (30 – S_A)] = 1.25$
$45 (30 – 2 S_A) = 1.25 S_A (30 – S_A)$
$1350 – 90 S_A = 37.5 S_A – 1.25 S_A^2$
Rearrange the terms to form a quadratic equation:
$1.25 S_A^2 – 90 S_A – 37.5 S_A + 1350 = 0$
$1.25 S_A^2 – 127.5 S_A + 1350 = 0$
Multiply by 4 to get rid of the decimal:
$5 S_A^2 – 510 S_A + 5400 = 0$
Divide by 5:
$S_A^2 – 102 S_A + 1080 = 0$

We can solve this quadratic equation using the quadratic formula $S_A = [-b \pm \sqrt{b^2 – 4ac}] / 2a$, where a=1, b=-102, c=1080.
$S_A = [102 \pm \sqrt{(-102)^2 – 4 \times 1 \times 1080}] / (2 \times 1)$
$S_A = [102 \pm \sqrt{10404 – 4320}] / 2$
$S_A = [102 \pm \sqrt{6084}] / 2$
$S_A = [102 \pm 78] / 2$

Two possible values for $S_A$:
$S_A1 = (102 + 78) / 2 = 180 / 2 = 90$
$S_A2 = (102 – 78) / 2 = 24 / 2 = 12$

If $S_A = 90$ km/hr, then from Equation 1, $S_S = 30 – 90 = -60$ km/hr, which is not possible as speed cannot be negative.
Therefore, $S_A = 12$ km/hr.

Let’s verify this.
If $S_A = 12$ km/hr, then $S_S = 30 – 12 = 18$ km/hr.
Time for Anil to reach B = $45 / 12 = 3.75$ hours.
Time for Sunil to reach A = $45 / 18 = 2.5$ hours.
The difference in their arrival times = $3.75 – 2.5 = 1.25$ hours, which is 1 hour 15 minutes. This matches the condition.

The speed of Anil is 12 km per hour.

Correct_Option:D

Ans: A

Let’s denote the typical travel duration of the bus as $t$.

Based on the first scenario (bus traveling at 60 km/h), the total distance covered can be expressed as $60(t + 3.5)$.

The second scenario indicates that the bus covered two-thirds of its usual journey distance in one-third of its usual travel time. This implies that the remaining one-third of the distance was covered in the remaining two-thirds of the usual travel time.

Specifically, if $\frac{1}{3}$ of the distance is covered in $\frac{2}{3}t$ time, then the speed during this segment is $\frac{\frac{1}{3} \text{distance}}{\frac{2}{3}t} = \frac{1}{2} \times \frac{\text{distance}}{t}$. Since the usual speed is $\frac{\text{distance}}{t}$, this speed is half of the usual speed.

We are given that this speed is 40 km/h. Therefore, half of the usual speed is 40 km/h, which means the usual speed of the bus is $2 \times 40 = 80$ km/hr.

Now, using the relationship from the first scenario, we have $60(t + 3.5) = 80t$.
Solving for $t$:
$60t + 210 = 80t$
$210 = 20t$
$t = \frac{210}{20} = 10.5$ hours.

Thus, the bus typically completes its journey in 10.5 hours.

If the journey begins at 9:00 AM, it will conclude at 7:30 PM.
Consequently, Option A is the correct choice.

Ans: A

Explanation:Let the distance travelled by the train be $D$ km and the original uniform speed be $S$ km per hour. Let the original time taken be $T$ hours.
We know that distance = speed × time, so $D = S \times T$.

According to the first condition:
If the speed had been 6 km per hour more, i.e., $S+6$, it would have needed 4 hours less, i.e., $T-4$.
So, $D = (S+6)(T-4)$
$ST = (S+6)(T-4)$
$ST = ST – 4S + 6T – 24$
$0 = -4S + 6T – 24$
$4S – 6T = -24$
Dividing by 2, we get:
$2S – 3T = -12$ (Equation 1)

According to the second condition:
If the speed had been 6 km per hour less, i.e., $S-6$, it would have needed 6 hours more, i.e., $T+6$.
So, $D = (S-6)(T+6)$
$ST = (S-6)(T+6)$
$ST = ST + 6S – 6T – 36$
$0 = 6S – 6T – 36$
$6S – 6T = 36$
Dividing by 6, we get:
$S – T = 6$ (Equation 2)

Now we have a system of two linear equations with two variables:
1) $2S – 3T = -12$
2) $S – T = 6$

From Equation 2, we can express $S$ in terms of $T$:
$S = T + 6$

Substitute this expression for $S$ into Equation 1:
$2(T+6) – 3T = -12$
$2T + 12 – 3T = -12$
$-T + 12 = -12$
$-T = -12 – 12$
$-T = -24$
$T = 24$ hours

Now substitute the value of $T$ back into the equation for $S$:
$S = T + 6$
$S = 24 + 6$
$S = 30$ km per hour

The distance travelled by the train is $D = S \times T$.
$D = 30 \times 24$
$D = 720$ km

Let’s check the conditions with the calculated values:
Original speed $S = 30$ km/h, original time $T = 24$ h, distance $D = 720$ km.

Condition 1: Speed $S+6 = 30+6 = 36$ km/h. Time $T-4 = 24-4 = 20$ h.
Distance $= 36 \times 20 = 720$ km. This matches.

Condition 2: Speed $S-6 = 30-6 = 24$ km/h. Time $T+6 = 24+6 = 30$ h.
Distance $= 24 \times 30 = 720$ km. This matches.

Therefore, the distance travelled by the train is 720 km.

Correct_Option:A

Ans: C

The odometer reading after the journey is 26862 km, and the trip itself lasted for 8 hours.
Let ‘s’ represent the average speed of the car during the trip in km/hr.
The distance covered before the trip began can be calculated as the total distance minus the distance covered during the trip: 26862 – 8s. This prior distance must also be a palindromic number.

Considering the provided options:
If s = 110, the distance before the trip is 26862 – (110 * 8) = 26862 – 880 = 25982. This is not a palindrome.
If s = 100, the distance before the trip is 26862 – (100 * 8) = 26862 – 800 = 26062. This is a palindrome.

Therefore, s = 100 is the correct value for the average speed.

Ans: 972

Let the pace of Arvind be represented by the variable ‘a’, and the pace of Surbhi be represented by the variable ‘s’.
Assume they encounter each other after a duration of ‘t’ hours.

According to the problem statement:
Arvind covered a distance of ‘s*t’ in 6 hours.
Surbhi covered a distance of ‘a*t’ in 24 hours.

This can be translated into the following equations:
Equation 1: s*t = a*6
Equation 2: a*t = s*24

By manipulating these equations, we can deduce the value of $t^2$:
From Equation 1, we can express ‘s’ as: s = (a*6)/t
Substitute this expression for ‘s’ into Equation 2:
a*t = ((a*6)/t) * 24
a*t = (a*144)/t
Multiply both sides by ‘t’:
a*$t^2$ = a*144
Since ‘a’ is a speed and is not zero, we can divide both sides by ‘a’:
$t^2$ = 144
Taking the square root of both sides, we find:
t = 12 hours

We are given that Arvind’s speed (a) is 54. Now we can find Surbhi’s speed (s) using Equation 1:
s*12 = 54*6
s*12 = 324
Divide both sides by 12:
s = 324 / 12
s = 27

The total separation between the starting points of Arvind and Surbhi is the sum of the distances they covered until they met. This distance is equal to (Arvind’s speed + Surbhi’s speed) * time to meet.
Total distance = (s + a) * t
Total distance = (27 + 54) * 12
Total distance = 81 * 12
Total distance = 972 Kms.

Ans: C

It is stated that Ravi’s velocity is 40 km/h, equivalent to $\frac{100}{9}$ m/s. Ashok’s velocity is given as 50 km/h, which converts to $\frac{125}{9}$ m/s.

The initial separation between Ravi and Ashok is 225 meters. Their combined velocity (relative speed when approaching each other) is calculated as $\frac{125}{9}+\frac{100}{9}=\frac{225}{9}=25$ m/s.

Therefore, the time until they converge is $\frac{225 \text{ meters}}{25 \text{ m/s}} = 9$ seconds.

In these 9 seconds, Ravi covers a distance of $\frac{100}{9} \text{ m/s} \times 9 \text{ s} = 100$ meters.

Given that Vijay began 54 meters behind Ravi, Vijay needs to cover the distance Ravi traveled plus this initial deficit. Thus, Vijay must travel (100 + 54) = 154 meters in the same 9 seconds.

Consequently, Vijay’s velocity is $\frac{154 \text{ meters}}{9 \text{ seconds}} = \frac{154}{9}$ m/s. Converting this to km/h: $\frac{154}{9} \times \frac{18}{5} = \frac{154 \times 2}{5} = \frac{308}{5} = 61.6$ km/h.

The corresponding answer choice is C.

Ans: D

Let’s denote the speed of the first vessel as $v_1$, the speed of the second vessel as $v_2$, and the speed of the current as $v_c$.
Let the distance between points A and B be represented by $D$.

According to the problem statement, we have two expressions for the distance $D$:
$D = 2(v_1 + v_c)$
$D = 3(v_1 – v_c)$

From these, we can derive:
$v_1 + v_c = D/2$
$v_1 – v_c = D/3$

Subtracting the second equation from the first gives us:
$2v_c = D/2 – D/3$
$2v_c = (3D – 2D)/6$
$2v_c = D/6$
$v_c = D/12$

The problem also provides the following equation:
$\dfrac{D}{v_2 + v_c} + \dfrac{D}{v_2 – v_c} = 6$

Substitute the value of $v_c$ we found:
$\dfrac{D}{v_2 + \dfrac{D}{12}} + \dfrac{D}{v_2 – \dfrac{D}{12}} = 6$

This simplifies to:
$D \left( \dfrac{1}{v_2 + \dfrac{D}{12}} + \dfrac{1}{v_2 – \dfrac{D}{12}} \right) = 6$
$D \left( \dfrac{(v_2 – \dfrac{D}{12}) + (v_2 + \dfrac{D}{12})}{(v_2 + \dfrac{D}{12})(v_2 – \dfrac{D}{12})} \right) = 6$
$D \left( \dfrac{2v_2}{v_2^2 – \dfrac{D^2}{144}} \right) = 6$
$\dfrac{2Dv_2}{v_2^2 – \dfrac{D^2}{144}} = 6$

Rearranging and multiplying by $v_2^2 – \dfrac{D^2}{144}$:
$2Dv_2 = 6 \left( v_2^2 – \dfrac{D^2}{144} \right)$
$2Dv_2 = 6v_2^2 – \dfrac{6D^2}{144}$
$2Dv_2 = 6v_2^2 – \dfrac{D^2}{24}$

Multiplying the entire equation by 24 to clear the fraction:
$48Dv_2 = 144v_2^2 – D^2$

Rearranging this into a quadratic equation in terms of $v_2$:
$144v_2^2 – 48Dv_2 – D^2 = 0$

Using the quadratic formula to solve for $v_2$:
$v_2 = \dfrac{-(-48D) \pm \sqrt{(-48D)^2 – 4(144)(-D^2)}}{2(144)}$
$v_2 = \dfrac{48D \pm \sqrt{2304D^2 + 576D^2}}{288}$
$v_2 = \dfrac{48D \pm \sqrt{2880D^2}}{288}$
$v_2 = \dfrac{48D \pm D\sqrt{2880}}{288}$

We can simplify $\sqrt{2880}$:
$\sqrt{2880} = \sqrt{576 \times 5} = 24\sqrt{5}$

So,
$v_2 = \dfrac{48D \pm 24D\sqrt{5}}{288}$
$v_2 = D \left( \dfrac{48 \pm 24\sqrt{5}}{288} \right)$
$v_2 = D \left( \dfrac{2 \pm \sqrt{5}}{12} \right)$

Since speed must be positive, we take the positive root:
$v_2 = D \left( \dfrac{2 + \sqrt{5}}{12} \right)$

We need to find the value of $\dfrac{D}{v_2 + v_c}$.
Substitute the expressions for $v_2$ and $v_c$:
$v_2 + v_c = D \left( \dfrac{2 + \sqrt{5}}{12} \right) + \dfrac{D}{12}$
$v_2 + v_c = D \left( \dfrac{2 + \sqrt{5} + 1}{12} \right)$
$v_2 + v_c = D \left( \dfrac{3 + \sqrt{5}}{12} \right)$

Now, calculate $\dfrac{D}{v_2 + v_c}$:
$\dfrac{D}{v_2 + v_c} = \dfrac{D}{D \left( \dfrac{3 + \sqrt{5}}{12} \right)}$
$\dfrac{D}{v_2 + v_c} = \dfrac{1}{\dfrac{3 + \sqrt{5}}{12}}$
$\dfrac{D}{v_2 + v_c} = \dfrac{12}{3 + \sqrt{5}}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate $(3 – \sqrt{5})$:
$\dfrac{12}{3 + \sqrt{5}} \times \dfrac{3 – \sqrt{5}}{3 – \sqrt{5}} = \dfrac{12(3 – \sqrt{5})}{3^2 – (\sqrt{5})^2}$
$= \dfrac{12(3 – \sqrt{5})}{9 – 5}$
$= \dfrac{12(3 – \sqrt{5})}{4}$
$= 3(3 – \sqrt{5})$

Ans: C

Explanation:Let the speed of the slower ship be $s$ km per hour.
Then the speed of the faster ship is $s+6$ km per hour.
After 2 hours, the distance travelled by the slower ship is $2s$ km (south).
After 2 hours, the distance travelled by the faster ship is $2(s+6)$ km (west).
Since one ship goes south and the other goes west, their paths form the two perpendicular sides of a right-angled triangle. The distance between them after 2 hours is the hypotenuse of this triangle.
According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, $(2s)^2 + (2(s+6))^2 = 60^2$.
$4s^2 + 4(s+6)^2 = 3600$.
Divide by 4:
$s^2 + (s+6)^2 = 900$.
$s^2 + (s^2 + 12s + 36) = 900$.
$2s^2 + 12s + 36 – 900 = 0$.
$2s^2 + 12s – 864 = 0$.
Divide by 2:
$s^2 + 6s – 432 = 0$.
We can solve this quadratic equation for $s$ using the quadratic formula $s = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$, where $a=1$, $b=6$, $c=-432$.
$s = \frac{-6 \pm \sqrt{6^2 – 4(1)(-432)}}{2(1)}$.
$s = \frac{-6 \pm \sqrt{36 + 1728}}{2}$.
$s = \frac{-6 \pm \sqrt{1764}}{2}$.
To find the square root of 1764:
$40^2 = 1600$
$42^2 = (40+2)^2 = 1600 + 2(40)(2) + 4 = 1600 + 160 + 4 = 1764$.
So, $\sqrt{1764} = 42$.
$s = \frac{-6 \pm 42}{2}$.
We have two possible values for $s$:
$s_1 = \frac{-6 + 42}{2} = \frac{36}{2} = 18$.
$s_2 = \frac{-6 – 42}{2} = \frac{-48}{2} = -24$.
Since speed cannot be negative, we take the positive value.
The speed of the slower ship is $s = 18$ km per hour.
The speed of the faster ship is $s+6 = 18+6 = 24$ km per hour.
Let’s check:
Distance travelled by slower ship in 2 hours = $2 \times 18 = 36$ km.
Distance travelled by faster ship in 2 hours = $2 \times 24 = 48$ km.
Distance between them = $\sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60$ km.
This matches the given information.

Correct_Option:C

Ans: B

Explanation:Let the speeds of the two cars be $s_1$ and $s_2$ km/hr, where $s_1$ is the speed of the slower car and $s_2$ is the speed of the faster car. Let the distance between the two locations be D km.

When the cars travel towards each other, their relative speed is the sum of their speeds, $s_1 + s_2$. They meet after 1.5 hours.
So, $D = (s_1 + s_2) \times 1.5$ (Equation 1)

When the cars travel in the same direction, the faster car catches up with the slower car. Their relative speed is the difference between their speeds, $s_2 – s_1$. They meet after 10.5 hours.
So, $D = (s_2 – s_1) \times 10.5$ (Equation 2)

We are given that the speed of the slower car is 60 km/hr, so $s_1 = 60$.

Now, we can substitute $s_1 = 60$ into Equations 1 and 2:
From Equation 1: $D = (60 + s_2) \times 1.5$
From Equation 2: $D = (s_2 – 60) \times 10.5$

Since both expressions equal D, we can set them equal to each other:
$(60 + s_2) \times 1.5 = (s_2 – 60) \times 10.5$

Divide both sides by 1.5:
$60 + s_2 = \frac{10.5}{1.5} (s_2 – 60)$
$60 + s_2 = 7 (s_2 – 60)$
$60 + s_2 = 7s_2 – 420$

Now, solve for $s_2$:
$60 + 420 = 7s_2 – s_2$
$480 = 6s_2$
$s_2 = \frac{480}{6}$
$s_2 = 80$ km/hr

So, the speed of the faster car is 80 km/hr.

The question asks for the distance traveled by the slower car when it meets the other car while traveling towards each other.
The slower car travels at a speed of $s_1 = 60$ km/hr.
They meet after 1.5 hours when traveling towards each other.
Distance traveled by the slower car = speed of slower car $\times$ time
Distance = $60 \times 1.5 = 90$ km.

Correct_Option:B

Ans: 60

Explanation:Let $S$ be the length of the escalator and $v_m$ be Moody’s normal walking speed. Let $v_e$ be the speed of the escalator.

When Moody walks at his normal speed, the relative speed of Moody with respect to the ground is $v_m + v_e$. The time taken is 30 seconds.
So, the length of the escalator $S$ can be expressed as:
$S = (v_m + v_e) \times 30$ (Equation 1)

When Moody walks at twice his normal speed, his speed is $2v_m$. The relative speed of Moody with respect to the ground is $2v_m + v_e$. The time taken is 20 seconds.
So, the length of the escalator $S$ can be expressed as:
$S = (2v_m + v_e) \times 20$ (Equation 2)

We have two equations for $S$:
1) $30v_m + 30v_e = S$
2) $40v_m + 20v_e = S$

Equating the two expressions for $S$:
$30v_m + 30v_e = 40v_m + 20v_e$

Now, we solve for the relationship between $v_m$ and $v_e$:
$30v_e – 20v_e = 40v_m – 30v_m$
$10v_e = 10v_m$
$v_e = v_m$

This means Moody’s walking speed is equal to the speed of the escalator.

Now, we want to find the time it takes for Moody to finish riding the escalator if he stands still. In this case, his speed with respect to the ground is just the speed of the escalator, $v_e$. Let this time be $T$.
$S = v_e \times T$

We can use Equation 1 and substitute $v_m = v_e$:
$S = (v_e + v_e) \times 30$
$S = (2v_e) \times 30$
$S = 60v_e$

Now, we equate this expression for $S$ with the expression when Moody stands still:
$60v_e = v_e \times T$

Since $v_e$ is the speed of the escalator and must be greater than 0, we can divide both sides by $v_e$:
$T = 60$

Therefore, Moody takes 60 seconds to finish riding the escalator if he stands still.

Final_Answer:60

Ans: C

Explanation:Let the length of the faster train be $L_f$ and its speed be $S_f$.
Let the length of the other train be $L_s$ and its speed be $S_s$.

We are given that the faster train crosses a lamp post in 12 seconds. When a train crosses a lamp post, the distance covered is equal to the length of the train.
So, $L_f = S_f \times 12$.
We are given $L_f = 160$ m.
Therefore, $160 = S_f \times 12$.
$S_f = \frac{160}{12} = \frac{40}{3}$ m/s.

We are given that the speed of the other train is 6 km/hr less than the faster one.
First, convert $S_f$ to km/hr.
$S_f = \frac{40}{3} \times \frac{18}{5} = \frac{40 \times 6}{5} = 8 \times 6 = 48$ km/hr.

Now, $S_s = S_f – 6$ km/hr.
$S_s = 48 – 6 = 42$ km/hr.

Convert $S_s$ back to m/s.
$S_s = 42 \times \frac{5}{18} = \frac{42 \times 5}{18} = \frac{7 \times 5}{3} = \frac{35}{3}$ m/s.

We are given that the two trains cross each other in 14 seconds when running in opposite directions.
When two trains cross each other, the relative speed is the sum of their speeds, and the distance covered is the sum of their lengths.
Relative speed = $S_f + S_s$.
Distance = $L_f + L_s$.
Time = 14 seconds.

So, $L_f + L_s = (S_f + S_s) \times 14$.
We know $L_f = 160$ m.
$S_f = \frac{40}{3}$ m/s.
$S_s = \frac{35}{3}$ m/s.

$160 + L_s = \left(\frac{40}{3} + \frac{35}{3}\right) \times 14$.
$160 + L_s = \left(\frac{40 + 35}{3}\right) \times 14$.
$160 + L_s = \left(\frac{75}{3}\right) \times 14$.
$160 + L_s = 25 \times 14$.
$160 + L_s = 350$.
$L_s = 350 – 160$.
$L_s = 190$ m.

Correct_Option: C

Ans: A

Explanation:Let the speed of train A be $5x$ and the speed of train B be $3x$. Since they are moving in opposite directions, their relative speed is $5x + 3x = 8x$.
Let the length of train A be $L_A$ and the length of train B be $L_B$.

When the front ends of the trains crossed each other, let’s consider this as time $t=0$.
The front end of A crossed the rear end of B 46 seconds after the front ends crossed. This means that in 46 seconds, the relative distance covered by the front end of A to clear the rear end of B is equal to the length of train B.
So, $8x \times 46 = L_B$.
$L_B = 368x$ (Equation 1)

It took another 69 seconds for the rear ends of the trains to cross each other, after the front ends had crossed. This means the total time from the moment the front ends crossed until the rear ends crossed is $46 + 69 = 115$ seconds.
In this total time, the relative distance covered to clear both trains is the sum of their lengths.
So, $8x \times 115 = L_A + L_B$.
$L_A + L_B = 920x$ (Equation 2)

Now we can substitute the value of $L_B$ from Equation 1 into Equation 2:
$L_A + 368x = 920x$
$L_A = 920x – 368x$
$L_A = 552x$

We need to find the ratio of the length of train A to that of train B, which is $L_A : L_B$.
$\frac{L_A}{L_B} = \frac{552x}{368x} = \frac{552}{368}$

To simplify the ratio, we can find the greatest common divisor (GCD) of 552 and 368.
Divide both by 8: $552/8 = 69$, $368/8 = 46$.
So the ratio is $69 : 46$.
Now divide both by 23 (since $69 = 3 \times 23$ and $46 = 2 \times 23$).
$69/23 = 3$, $46/23 = 2$.
The ratio of the length of train A to that of train B is $3 : 2$.

Correct_Option:A

Ans: 8

Explanation:Let $v_M$ be Mira’s speed and $v_A$ be Amal’s speed. Let the length of the circular track be $L$.
When they walk in the same direction, the relative speed is $|v_A – v_M|$.
In 45 minutes, Amal completes 3 more rounds than Mira. This means the distance covered by Amal is 3 times the length of the track more than the distance covered by Mira.
Distance covered by Amal in 45 minutes = $v_A \times 45$
Distance covered by Mira in 45 minutes = $v_M \times 45$
So, $v_A \times 45 – v_M \times 45 = 3L$
$(v_A – v_M) \times 45 = 3L$
$v_A – v_M = \frac{3L}{45} = \frac{L}{15}$

When they walk in opposite directions, their relative speed is $v_A + v_M$.
They meet for the first time after 3 minutes. This means the sum of the distances they covered in 3 minutes is equal to the length of the track.
Distance covered by Amal in 3 minutes = $v_A \times 3$
Distance covered by Mira in 3 minutes = $v_M \times 3$
So, $v_A \times 3 + v_M \times 3 = L$
$(v_A + v_M) \times 3 = L$
$v_A + v_M = \frac{L}{3}$

We have a system of two linear equations:
1) $v_A – v_M = \frac{L}{15}$
2) $v_A + v_M = \frac{L}{3}$

Add equation (1) and (2):
$(v_A – v_M) + (v_A + v_M) = \frac{L}{15} + \frac{L}{3}$
$2v_A = \frac{L}{15} + \frac{5L}{15}$
$2v_A = \frac{6L}{15} = \frac{2L}{5}$
$v_A = \frac{L}{5}$

Subtract equation (1) from equation (2):
$(v_A + v_M) – (v_A – v_M) = \frac{L}{3} – \frac{L}{15}$
$2v_M = \frac{5L}{15} – \frac{L}{15}$
$2v_M = \frac{4L}{15}$
$v_M = \frac{2L}{15}$

We need to find the number of rounds Mira walks in one hour.
Speed of Mira is $v_M = \frac{2L}{15}$ units per minute.
In one hour (60 minutes), the distance Mira walks is $v_M \times 60$.
Distance = $\frac{2L}{15} \times 60 = 2L \times 4 = 8L$.
The number of rounds Mira walks in one hour is the total distance covered divided by the length of the track:
Number of rounds = $\frac{8L}{L} = 8$.

Final_Answer:8

Ans: B

Explanation:Let the distance between point A and point B be $D$ km.
Let the speed of Car 1 be $S_1$ km/hr and the speed of Car 2 be $S_2$ km/hr.
We are given that $S_1 = 60$ km/hr.
Let the time when the two cars meet be $t$ hours after they start.
When the cars meet, the distance covered by Car 1 from A is $S_1 \times t$ and the distance covered by Car 2 from B is $S_2 \times t$.
The sum of the distances covered by both cars is equal to the total distance between A and B.
So, $S_1 \times t + S_2 \times t = D$.
This can be written as $(S_1 + S_2) \times t = D$.

After meeting, Car 1 takes 45 minutes to complete its journey. This means Car 1 travels the remaining distance from the meeting point to B in 45 minutes.
45 minutes = 45/60 hours = 3/4 hours.
The distance covered by Car 1 after meeting is $S_1 \times (3/4)$. This distance is equal to the distance covered by Car 2 before meeting, which is $S_2 \times t$.
So, $S_1 \times (3/4) = S_2 \times t$. (Equation 1)

After meeting, Car 2 takes 20 minutes to complete its journey. This means Car 2 travels the remaining distance from the meeting point to A in 20 minutes.
20 minutes = 20/60 hours = 1/3 hours.
The distance covered by Car 2 after meeting is $S_2 \times (1/3)$. This distance is equal to the distance covered by Car 1 before meeting, which is $S_1 \times t$.
So, $S_2 \times (1/3) = S_1 \times t$. (Equation 2)

From Equation 1, we have $t = \frac{3S_1}{4S_2}$.
Substitute this value of $t$ into Equation 2:
$S_2 \times (1/3) = S_1 \times \left(\frac{3S_1}{4S_2}\right)$
$\frac{S_2}{3} = \frac{3S_1^2}{4S_2}$
Multiply both sides by $12S_2$:
$4S_2^2 = 9S_1^2$
Take the square root of both sides:
$2S_2 = 3S_1$
$S_2 = \frac{3}{2} S_1$

We are given $S_1 = 60$ km/hr.
$S_2 = \frac{3}{2} \times 60$
$S_2 = 3 \times 30$
$S_2 = 90$ km/hr.

Alternatively, using the formula derived from such problems:
If two cars start from A and B towards each other and meet after some time, and then take $t_1$ and $t_2$ respectively to complete their journeys, then the ratio of their speeds is given by:
$\frac{S_1}{S_2} = \sqrt{\frac{t_2}{t_1}}$
Here, $t_1 = 45$ minutes (time taken by Car 1 after meeting) and $t_2 = 20$ minutes (time taken by Car 2 after meeting).
$\frac{S_1}{S_2} = \sqrt{\frac{20}{45}}$
$\frac{S_1}{S_2} = \sqrt{\frac{4}{9}}$
$\frac{S_1}{S_2} = \frac{2}{3}$
We are given $S_1 = 60$ km/hr.
$\frac{60}{S_2} = \frac{2}{3}$
$2S_2 = 60 \times 3$
$2S_2 = 180$
$S_2 = \frac{180}{2}$
$S_2 = 90$ km/hr.

Correct_Option:B

Ans: B

Explanation:Let the distance between Amal’s home and office be D km.
Let the time taken to reach the office when traveling at 8 km/hr be T1 hours.
Let the time taken to reach the office when traveling at 15 km/hr be T2 hours.

When Amal travels at 8 km/hr, he reaches the office at 10:15 am.
When Amal travels at 15 km/hr, he reaches the office at 9:40 am.

The difference in arrival times is 10:15 am – 9:40 am = 35 minutes.
35 minutes = 35/60 hours = 7/12 hours.

We know that Time = Distance / Speed.
So, T1 = D/8
And T2 = D/15

The difference in time is T1 – T2 = 7/12 hours.
D/8 – D/15 = 7/12

To solve for D, find a common denominator for the left side, which is 120.
(15D – 8D) / 120 = 7/12
7D / 120 = 7/12

Multiply both sides by 120:
7D = (7/12) * 120
7D = 7 * 10
7D = 70
D = 10 km.

Now, Amal leaves home at 9:10 am and wants to reach the office exactly at 10:00 am.
The total time available for travel is 10:00 am – 9:10 am = 50 minutes.
50 minutes = 50/60 hours = 5/6 hours.

Let the required speed be S km/hr.
We know that Speed = Distance / Time.
S = D / (5/6)
S = 10 / (5/6)
S = 10 * (6/5)
S = (10 * 6) / 5
S = 60 / 5
S = 12 km/hr.

Correct_Option:B

Ans: C

Explanation:Let the usual speed of the train be $S$ and the usual time taken to reach the destination be $T$. The distance to the destination is $D = S \times T$.

In the first journey, the train travelled at one-third of its usual speed, so its speed was $\frac{S}{3}$.
The time taken for this journey was $\frac{D}{\frac{S}{3}} = \frac{3D}{S}$.
Since $D = S \times T$, the time taken is $\frac{3(S \times T)}{S} = 3T$.
The train reached the destination 30 minutes after the scheduled time, so $3T = T + 30$ minutes.
This implies $2T = 30$ minutes, so $T = 15$ minutes.
Thus, the usual time taken for the journey is 15 minutes.

Now consider the return journey. The distance is the same, $D$.
The train initially travelled at its usual speed $S$ for 5 minutes.
Distance covered in the first 5 minutes = $S \times 5$ minutes.
The remaining distance is $D – S \times 5$ minutes.
The train then stopped for 4 minutes.

The total time elapsed so far is 5 minutes (travel) + 4 minutes (stop) = 9 minutes.
The scheduled time for the return journey is also $T = 15$ minutes.
So, the remaining time to reach the destination at the scheduled time is $15 – 9 = 6$ minutes.

The remaining distance is $D – 5S$. Since $D = ST$, the remaining distance is $ST – 5S = S(T-5)$.
Given $T=15$ minutes, the remaining distance is $S(15-5) = 10S$.

Let the new speed of the train for the remaining journey be $S_{new}$.
This new speed must cover the remaining distance $10S$ in the remaining time of 6 minutes.
So, $S_{new} \times 6 \text{ minutes} = 10S \times 1 \text{ minute}$.
$S_{new} = \frac{10S}{6} = \frac{5}{3}S$.

We need to find the percentage increase in the usual speed.
The usual speed is $S$. The new speed is $\frac{5}{3}S$.
The increase in speed is $S_{new} – S = \frac{5}{3}S – S = \frac{2}{3}S$.
The percentage increase is $\frac{\text{Increase in speed}}{\text{Usual speed}} \times 100 = \frac{\frac{2}{3}S}{S} \times 100 = \frac{2}{3} \times 100 = \frac{200}{3} \%$.
$\frac{200}{3} \% \approx 66.67 \%$.

The nearest percentage is 67%.

Correct_Option:C

Ans: B

Explanation:Let the length of the train be L meters and the speed of the train be S km/hr.
Let the speeds of the two persons be v1 = 2 km/hr and v2 = 4 km/hr.
The speeds are given in km/hr, but the time is in seconds. It’s best to convert speeds to meters per second (m/s).
1 km/hr = 1000 meters / 3600 seconds = 5/18 m/s.
So, v1 = 2 * (5/18) = 10/18 m/s = 5/9 m/s.
And, v2 = 4 * (5/18) = 20/18 m/s = 10/9 m/s.
The speed of the train S in m/s can be written as S_m/s.

When the train crosses a person walking in the same direction, the relative speed of the train with respect to the person is the difference between their speeds.
Relative speed of the train with respect to the first person = S_m/s – v1 = S_m/s – 5/9 m/s.
The time taken to cross the first person is t1 = 90 seconds.
The distance covered by the train in this time is its own length L.
So, L = (S_m/s – 5/9) * 90 (Equation 1)

Relative speed of the train with respect to the second person = S_m/s – v2 = S_m/s – 10/9 m/s.
The time taken to cross the second person is t2 = 100 seconds.
So, L = (S_m/s – 10/9) * 100 (Equation 2)

Now we have two equations for L. Let’s equate them:
(S_m/s – 5/9) * 90 = (S_m/s – 10/9) * 100
90 * S_m/s – 90 * (5/9) = 100 * S_m/s – 100 * (10/9)
90 * S_m/s – 50 = 100 * S_m/s – 1000/9
1000/9 – 50 = 100 * S_m/s – 90 * S_m/s
1000/9 – 450/9 = 10 * S_m/s
550/9 = 10 * S_m/s
S_m/s = (550/9) / 10 = 55/9 m/s.

Now we can find the length of the train L using either Equation 1 or Equation 2. Using Equation 1:
L = (55/9 – 5/9) * 90
L = (50/9) * 90
L = 50 * 10
L = 500 meters.

Now, we need to find the time taken by the train to cross an electric post. When a train crosses an electric post, it means the train covers its own length. The speed of the electric post is considered zero.
So, the time taken to cross an electric post = Length of train / Speed of train
Time = L / S_m/s
Time = 500 meters / (55/9 m/s)
Time = 500 * (9/55) seconds
Time = (500/55) * 9 seconds
Time = (100/11) * 9 seconds
Time = 900 / 11 seconds.

Let’s calculate the value of 900/11:
900 / 11 = 81.8181… seconds.

The question asks for the nearest time.
Option A: 87
Option B: 82
Option C: 78
Option D: 75

The calculated time is approximately 81.82 seconds. This is nearest to 82 seconds.

Let’s double check the calculations.

Equation 1: L = (S – 5/9) * 90
Equation 2: L = (S – 10/9) * 100

90S – 50 = 100S – 1000/9
1000/9 – 50 = 10S
(1000 – 450)/9 = 10S
550/9 = 10S
S = 55/9 m/s

L = (55/9 – 5/9) * 90 = (50/9) * 90 = 500 m

Time to cross post = L/S = 500 / (55/9) = 500 * 9 / 55 = 100 * 9 / 11 = 900/11 seconds.

900/11 = 81.818…

The nearest integer value to 81.818… is 82.

The closest option is B.

Final check of options:
A: 87 – Difference = |87 – 81.82| = 5.18
B: 82 – Difference = |82 – 81.82| = 0.18
C: 78 – Difference = |78 – 81.82| = 3.82
D: 75 – Difference = |75 – 81.82| = 6.82

Option B is clearly the nearest.

Correct_Option:B

Ans: A

Explanation:Let the distance from A to B be $d$.
Then the distance from B to C is $3d$.
Let the speed of train 1 from A to B be $v$.
Then the speed of train 2 from A to B is $2v$.

The time taken by train 1 to travel from A to B is $t_{1AB} = \frac{d}{v}$.
The time taken by train 2 to travel from A to B is $t_{2AB} = \frac{d}{2v}$.

While traveling from B to C, the speeds are interchanged.
So, the speed of train 1 from B to C is $2v$.
The speed of train 2 from B to C is $v$.

The time taken by train 1 to travel from B to C is $t_{1BC} = \frac{3d}{2v}$.
The time taken by train 2 to travel from B to C is $t_{2BC} = \frac{3d}{v}$.

The total time taken by train 1 to travel from A to C is $T_1 = t_{1AB} + t_{1BC} = \frac{d}{v} + \frac{3d}{2v} = \frac{2d + 3d}{2v} = \frac{5d}{2v}$.
The total time taken by train 2 to travel from A to C is $T_2 = t_{2AB} + t_{2BC} = \frac{d}{2v} + \frac{3d}{v} = \frac{d + 6d}{2v} = \frac{7d}{2v}$.

The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is:
$\frac{T_1}{T_2} = \frac{\frac{5d}{2v}}{\frac{7d}{2v}} = \frac{5}{7}$

Thus, the ratio is 5:7.

Correct_Option:A

Ans: B

Explanation:Let the distance of the race be D km.
Let the speeds of car A, B, and C be $v_A$, $v_B$, and $v_C$ respectively.
Let the time taken by car A to complete the race be $t_A$.
Let the time taken by car B to complete the race be $t_B$.
Let the time taken by car C to complete the race be $t_C$.

When car A beats car B by 45 km, it means that when car A completes the distance D, car B has covered D – 45 km.
So, $D/v_A = t_A$ and $(D-45)/v_B = t_A$.
Therefore, $D/v_A = (D-45)/v_B$, which implies $v_B/v_A = (D-45)/D$. (Equation 1)

When car B beats car C by 50 km, it means that when car B completes the distance D, car C has covered D – 50 km.
So, $D/v_B = t_B$ and $(D-50)/v_C = t_B$.
Therefore, $D/v_B = (D-50)/v_C$, which implies $v_C/v_B = (D-50)/D$. (Equation 2)

When car A beats car C by 90 km, it means that when car A completes the distance D, car C has covered D – 90 km.
So, $D/v_A = t_A$ and $(D-90)/v_C = t_A$.
Therefore, $D/v_A = (D-90)/v_C$, which implies $v_C/v_A = (D-90)/D$. (Equation 3)

From Equation 1, we have $v_B = v_A * (D-45)/D$.
From Equation 2, we have $v_C = v_B * (D-50)/D$.
Substitute the expression for $v_B$ from Equation 1 into this equation:
$v_C = [v_A * (D-45)/D] * (D-50)/D = v_A * (D-45)*(D-50)/D^2$.

Now, substitute this expression for $v_C$ into Equation 3:
$v_C/v_A = [(D-45)*(D-50)/D^2]$.
From Equation 3, $v_C/v_A = (D-90)/D$.

Equating the two expressions for $v_C/v_A$:
$(D-45)*(D-50)/D^2 = (D-90)/D$.
Multiply both sides by $D^2$ (assuming D is not 0):
$(D-45)*(D-50) = D*(D-90)$.
Expand both sides:
$D^2 – 50D – 45D + 2250 = D^2 – 90D$.
$D^2 – 95D + 2250 = D^2 – 90D$.
Subtract $D^2$ from both sides:
$-95D + 2250 = -90D$.
Add 95D to both sides:
$2250 = -90D + 95D$.
$2250 = 5D$.
Divide by 5:
$D = 2250 / 5$.
$D = 450$.

The distance over which the race has been conducted is 450 km.

Correct_Option:B

Ans: C

Let the velocity of Ram be denoted by $v_R$ and the velocity of Rahim be denoted by $v_H$. Suppose they encounter each other after a duration of $t$ from their start.

In this time $t$, Ram will traverse a distance of $v_R t$, and Rahim will traverse a distance of $v_H t$.

Following their meeting, Ram arrives at his destination in 1 minute. This means Ram covered the distance Rahim had traveled ($v_H t$) in 1 minute. Therefore, $v_R(1) = v_H t$.

Similarly, Rahim arrives at his destination in 4 minutes. This implies Rahim covered the distance Ram had traveled ($v_R t$) in 4 minutes. Hence, $v_H(4) = v_R t$.

Dividing the first derived equation by the second equation yields:
$ \frac{v_R(1)}{v_H(4)} = \frac{v_H t}{v_R t} $
$ \frac{v_R}{4v_H} = \frac{v_H}{v_R} $
Rearranging this equation, we get:
$ \frac{v_R}{v_H} = 2 $
Thus, the ratio of their velocities is 2.

Ans: C

Let the distance be $D$.
The problem states that the difference in time taken at two different speeds for the same distance $D$ is $\frac{6}{60}$ hours.
This can be represented as:
$ \frac{D}{35} – \frac{D}{40} = \frac{6}{60} $
Solving this equation for $D$:
$ D = 28 \text{ km} $
The normal time taken to cover this distance of 28 km at a speed of 40 km/hr is:
$ \text{Normal time} = \frac{28 \text{ km}}{40 \text{ km/hr}} = \frac{7}{10} \text{ hours} = 42 \text{ minutes} $
The time spent on the first segment of the journey (58/3 km) is $\frac{1}{3}$ of the normal time:
$ \text{Time for first segment} = \frac{1}{3} \times 42 \text{ minutes} = 14 \text{ minutes} $
Following this, a break of 8 minutes was taken.
To arrive at the destination on schedule, the remaining distance of $\frac{28}{3}$ km needs to be covered in the remaining time.
Total scheduled time = 42 minutes.
Time already spent = 14 minutes (travel) + 8 minutes (break) = 22 minutes.
Remaining time = 42 minutes – 22 minutes = 20 minutes.
Therefore, the required speed to cover the remaining $\frac{28}{3}$ km in 20 minutes (which is $\frac{20}{60}$ hours) is:
$ \text{Required speed} = \frac{\frac{28}{3} \text{ km}}{\frac{20}{60} \text{ hours}} = \frac{\frac{28}{3}}{\frac{1}{3}} \text{ km/hr} = 28 \text{ km/hr} $

Ans: C

Explanation:
Let the distance between stations A and B be $D = 90$ km.
Train 1 leaves station A at 9:00 am with a speed $v_A = 40$ km/hr towards B.
Train 2 leaves station B at 10:30 am with a speed $v_B = 20$ km/hr towards A.

First, let’s determine the position of Train 1 at 10:30 am, which is when Train 2 starts its journey.
The time elapsed for Train 1 from 9:00 am to 10:30 am is $10:30 – 9:00 = 1.5$ hours.
The distance covered by Train 1 in this time is $d_A = v_A \times \text{time} = 40 \text{ km/hr} \times 1.5 \text{ hours} = 60$ km.

At 10:30 am, Train 1 is 60 km away from A.
The remaining distance between the two trains at 10:30 am is $D’ = D – d_A = 90 \text{ km} – 60 \text{ km} = 30$ km.

Now, both trains are moving towards each other from a distance of 30 km apart.
Train 1 is moving at 40 km/hr and Train 2 is moving at 20 km/hr.
Their relative speed when moving towards each other is the sum of their speeds:
$v_{rel} = v_A + v_B = 40 \text{ km/hr} + 20 \text{ km/hr} = 60$ km/hr.

The time it will take for them to meet from 10:30 am is:
$\text{time to meet} = \frac{\text{Remaining distance}}{\text{Relative speed}} = \frac{D’}{v_{rel}} = \frac{30 \text{ km}}{60 \text{ km/hr}} = 0.5$ hours.

0.5 hours is equal to 30 minutes.
So, the trains will meet 30 minutes after 10:30 am.
Meeting time = 10:30 am + 30 minutes = 11:00 am.

Let’s check the options:
Option A: 11:45 am
Option B: 11:20 am
Option C: 11:00 am
Option D: 10:45 am

The calculated meeting time is 11:00 am, which corresponds to Option C.

Correct_Option:C

Ans: A

Explanation:The length of the circular path is 3 km.
Anil’s speed = 15 km/hr
Sunil’s speed = 10 km/hr
Ravi’s speed = 8 km/hr

Anil and Sunil will meet again at the starting point when both have completed an integer number of laps and have returned to the starting point simultaneously. This means their relative speed is not what matters for meeting at the starting point, but rather the time it takes for each of them to complete a full lap.

Time taken by Anil to complete one lap = Distance / Speed = 3 km / 15 km/hr = 1/5 hr
Time taken by Sunil to complete one lap = Distance / Speed = 3 km / 10 km/hr = 3/10 hr

They will meet again at the starting point for the first time after a time that is the Least Common Multiple (LCM) of their individual lap times.
LCM of (1/5, 3/10).
To find the LCM of fractions, we use the formula: LCM(a/b, c/d) = LCM(a, c) / GCD(b, d).
Here, a = 1, b = 5, c = 3, d = 10.
LCM(1, 3) = 3
GCD(5, 10) = 5
So, LCM of (1/5, 3/10) = 3 / 5 hr.

This is the time when Anil and Sunil will meet again for the first time at the starting point.

Now, we need to find the distance Ravi will have run during this time.
Ravi’s speed = 8 km/hr
Time = 3/5 hr
Distance run by Ravi = Speed × Time = 8 km/hr × (3/5) hr = 24/5 km = 4.8 km.

Correct_Option:A

Ans: A

Explanation:Let $D$ be the distance traveled by both cars.
Let $t_1$ be the time taken by the first car and $t_2$ be the time taken by the second car.
Let $s_1$ be the speed of the first car and $s_2$ be the speed of the second car.

We know that distance = speed × time. So, $D = s_1 t_1$ and $D = s_2 t_2$.
Therefore, $s_1 t_1 = s_2 t_2$.

The first car starts at 10:00 am and the second car starts at 11:00 am.
They reach their common destination at the same point of time.
This means the second car travels for 1 hour less than the first car.
So, $t_2 = t_1 – 1$.

From $s_1 t_1 = s_2 t_2$, we get:
$s_1 t_1 = s_2 (t_1 – 1)$
$\frac{s_2}{s_1} = \frac{t_1}{t_1 – 1}$

We are given that the first car travelled for at least 6 hours, which means $t_1 \ge 6$.

We want to find the highest possible value of the percentage by which the speed of the second car could exceed that of the first car.
The percentage by which the speed of the second car exceeds that of the first car is given by:
Percentage increase = $\left(\frac{s_2 – s_1}{s_1}\right) \times 100 = \left(\frac{s_2}{s_1} – 1\right) \times 100$

Substitute $\frac{s_2}{s_1} = \frac{t_1}{t_1 – 1}$:
Percentage increase = $\left(\frac{t_1}{t_1 – 1} – 1\right) \times 100$
Percentage increase = $\left(\frac{t_1 – (t_1 – 1)}{t_1 – 1}\right) \times 100$
Percentage increase = $\left(\frac{1}{t_1 – 1}\right) \times 100$

To maximize this percentage increase, we need to minimize the denominator $t_1 – 1$.
Since $t_1 \ge 6$, the minimum value of $t_1$ is 6.
When $t_1 = 6$, the minimum value of $t_1 – 1$ is $6 – 1 = 5$.

The highest possible value of the percentage increase is:
Percentage increase = $\left(\frac{1}{5}\right) \times 100 = 20\%$

The second car’s speed exceeds the first car’s speed by 20%.

Correct_Option:A

Ans: C

Explanation:Let $r_A$ be the radius of wheel A and $r_B$ be the radius of wheel B.
Given $r_A = 30$ cm and $r_B = 40$ cm.
Let $n_A$ be the number of revolutions for wheel A and $n_B$ be the number of revolutions for wheel B to cover a certain distance $d$.
We are given that $n_A = n_B + 5000$.

The distance covered by a wheel in one revolution is its circumference.
Circumference of wheel A, $C_A = 2 \pi r_A = 2 \pi (30) = 60 \pi$ cm.
Circumference of wheel B, $C_B = 2 \pi r_B = 2 \pi (40) = 80 \pi$ cm.

The total distance $d$ can be expressed as:
$d = n_A \times C_A$
$d = n_B \times C_B$

Substituting the expressions for $C_A$ and $C_B$:
$d = n_A \times 60 \pi$
$d = n_B \times 80 \pi$

Equating the two expressions for $d$:
$n_A \times 60 \pi = n_B \times 80 \pi$
$60 n_A = 80 n_B$
$3 n_A = 4 n_B$

Now, substitute $n_A = n_B + 5000$ into this equation:
$3 (n_B + 5000) = 4 n_B$
$3 n_B + 15000 = 4 n_B$
$n_B = 15000$ revolutions.

Now we can find the distance $d$ using $n_B$:
$d = n_B \times C_B = 15000 \times 80 \pi$ cm.
$d = 1200000 \pi$ cm.

We need to convert this distance to kilometers.
1 km = 1000 m = 1000 * 100 cm = 100000 cm.
$d = \frac{1200000 \pi}{100000}$ km = $12 \pi$ km.

Bicycle B traveled this distance in 45 minutes.
Time $t = 45$ minutes.
We need to convert this time to hours.
$t = \frac{45}{60}$ hours = $\frac{3}{4}$ hours.

Speed of bicycle B = $\frac{\text{Distance}}{\text{Time}}$
Speed = $\frac{12 \pi \text{ km}}{\frac{3}{4} \text{ hours}}$
Speed = $12 \pi \times \frac{4}{3}$ km/hour
Speed = $4 \pi \times 4$ km/hour
Speed = $16 \pi$ km/hour.

Comparing this with the given options:
Option A: $18 \pi$
Option B: $14 \pi$
Option C: $16 \pi$
Option D: $12 \pi$

The calculated speed matches Option C.

Correct_Option:C

Ans: 880

Let the distance of the race track be denoted by ‘D’ and the speeds of the three horses be ‘S1’, ‘S2’, and ‘S3’ respectively.

The time taken by the first horse to complete the race is D/S1.
The time taken by the second horse to complete the race is D/S2.
The time taken by the third horse to complete the race is D/S3.

According to the problem statement:

When the first horse finishes the race, the second horse is 11 units behind. This means the second horse has covered D-11 units in the same time the first horse covered D units.
Therefore, we can write the equality of time:
$ \frac{D}{S1} = \frac{D-11}{S2} $ ……(Equation 1)

When the first horse finishes the race, the third horse is 90 units behind. This means the third horse has covered D-90 units in the same time the first horse covered D units.
Therefore, we can write the equality of time:
$ \frac{D}{S1} = \frac{D-90}{S3} $ ……(Equation 2)

When the second horse finishes the race, the third horse is 80 units behind. This means the third horse has covered D-80 units in the same time the second horse covered D units.
Therefore, we can write the equality of time:
$ \frac{D}{S2} = \frac{D-80}{S3} $ ……(Equation 3)

From Equation 1 and Equation 2, since both are equal to D/S1, we can equate them:
$ \frac{D-11}{S2} = \frac{D-90}{S3} $ …..(Equation 4)

Now, let’s divide Equation 3 by Equation 4. We are dividing the left sides and the right sides:
$ \frac{\frac{D}{S2}}{\frac{D-11}{S2}} = \frac{\frac{D-80}{S3}}{\frac{D-90}{S3}} $

This simplifies to:
$ \frac{D}{D-11} = \frac{D-80}{D-90} $

Now, we cross-multiply:
$ D(D-90) = (D-11)(D-80) $

Expanding both sides:
$ D^2 – 90D = D^2 – 80D – 11D + 880 $
$ D^2 – 90D = D^2 – 91D + 880 $

Subtracting D^2 from both sides:
$ -90D = -91D + 880 $

Adding 91D to both sides:
$ 91D – 90D = 880 $
$ D = 880 $

Thus, the length of the race course is 880 units.

Ans: A

Explanation:Let the total distance from A to B be D and the total time taken by Amal be T.
Amal’s journey:
Amal took each mode of transport for 1/3rd of his total journey time (T).
Time for each mode = T/3
Speed of the three modes are S1 = 10 kmph, S2 = 20 kmph, S3 = 30 kmph.
Distance covered by each mode for Amal:
D1 = S1 * (T/3) = 10 * (T/3)
D2 = S2 * (T/3) = 20 * (T/3)
D3 = S3 * (T/3) = 30 * (T/3)
Total distance D = D1 + D2 + D3 = (10T/3) + (20T/3) + (30T/3) = (60T/3) = 20T.
So, T = D/20. This means Amal’s average speed is 20 kmph.

Bimal’s journey:
Bimal took each mode of transport for 1/3rd of the total distance (D).
Distance for each mode = D/3
Time taken for each mode by Bimal:
Time1 = (D/3) / S1 = (D/3) / 10 = D/30
Time2 = (D/3) / S2 = (D/3) / 20 = D/60
Time3 = (D/3) / S3 = (D/3) / 30 = D/90
Total time taken by Bimal (T_Bimal) = Time1 + Time2 + Time3 = D/30 + D/60 + D/90
To sum these fractions, find a common denominator, which is 180.
T_Bimal = (6D/180) + (3D/180) + (2D/180) = 11D/180.

Now we need to compare Bimal’s travel time with Amal’s travel time.
Amal’s total time T = D/20.
We need to find the percentage by which Bimal’s travel time exceeds Amal’s travel time.
Percentage Increase = ((T_Bimal – T) / T) * 100
Percentage Increase = ((11D/180 – D/20) / (D/20)) * 100
First, calculate the difference in time:
11D/180 – D/20 = 11D/180 – 9D/180 = 2D/180 = D/90
Now, divide the difference by Amal’s time and multiply by 100:
Percentage Increase = ((D/90) / (D/20)) * 100
Percentage Increase = (D/90) * (20/D) * 100
Percentage Increase = (20/90) * 100
Percentage Increase = (2/9) * 100
Percentage Increase = 200/9
Percentage Increase ≈ 22.22%

The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to 22%.

Correct_Option:A

Ans: D

Upon their initial encounter at 10:00 AM, individual A had traversed 60% of the course.
Consequently, individual B would have covered 40% of the course.
We are informed that A completes a return to point P at 10:12 AM, indicating that A covers 40% of the course in 12 minutes.
A covers 60% of the course in 18 minutes.
Individual B covers 40% of the course during the same period that A covers 60% of the course.
Therefore, B covers 40% of the course in 18 minutes.
B will then cover the remaining 60% in 27 minutes, leading to their arrival back at B at 10:27 AM.

Ans: C

Starting at 10:01 am, a motorcycle departs from point A towards point B every minute.
By 11:00 am, a total of forty-five motorcycles have arrived at B.
This indicates that the forty-fifth motorcycle departed from A at 10:45 am and arrived at B at 11:00 am, completing the journey in 15 minutes.
As all motorcycles maintain the same velocity, each motorcycle’s travel time from A to B is consistently 15 minutes.
If the velocity of a motorcycle were doubled, its travel time would be halved, resulting in an arrival at B by 10:30 am.
Given that each motorcycle requires 15 minutes for the trip, by the time a motorcycle completes its journey, fifteen motorcycles would have already reached B.

Ans: 48

Velocity of John = 6 kilometers per hour
Velocity of Mary = 7.5 kilometers per hour
Combined length of paths A and B = 325 meters
Assume the dimension of path A is denoted by ‘a’, then the dimension of path B is 325-a meters.
Completing 9 circuits on path A by John is equivalent to 5 circuits on path B by Mary.
$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$
Upon resolution, we find that 13a equals 1300.
Thus, a = 100.
The dimension of path A = 100 meters, path B = 225 meters.
Time taken by Mary to complete one circuit of path A = $\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$
= 48 seconds.

Ans: 12

Explanation:Let the length of AP be $x$. Then the length of BP is $3x$. The total length of AB is $AP + BP = x + 3x = 4x$.
Let the speed of car 1 be $v_1$ and the speed of car 2 be $v_2$.
We are given that the speed of car 2 is half that of car 1, so $v_2 = \frac{1}{2} v_1$.

Car 1 starts from A and moves towards B. The time taken by car 1 to reach P is $t_1 = \frac{\text{Distance AP}}{\text{Speed of car 1}} = \frac{x}{v_1}$.

Car 2 starts from B and moves towards A. The time taken by car 2 to reach P is $t_2 = \frac{\text{Distance BP}}{\text{Speed of car 2}} = \frac{3x}{v_2}$.

We are given that car 2 reaches P one hour after car 1 reaches P. So, $t_2 = t_1 + 1$ hour.

Substitute the expressions for $t_1$ and $t_2$:
$\frac{3x}{v_2} = \frac{x}{v_1} + 1$

Now substitute $v_2 = \frac{1}{2} v_1$:
$\frac{3x}{\frac{1}{2} v_1} = \frac{x}{v_1} + 1$
$\frac{6x}{v_1} = \frac{x}{v_1} + 1$

Subtract $\frac{x}{v_1}$ from both sides:
$\frac{6x}{v_1} – \frac{x}{v_1} = 1$
$\frac{5x}{v_1} = 1$

We are asked to find the time, in minutes, taken by car 1 in reaching P from A, which is $t_1 = \frac{x}{v_1}$.
From the equation $\frac{5x}{v_1} = 1$, we can find $\frac{x}{v_1}$:
$\frac{x}{v_1} = \frac{1}{5}$ hours.

To convert this time to minutes, multiply by 60:
$t_1 (\text{in minutes}) = \frac{1}{5} \times 60 = 12$ minutes.

Final_Answer:12

Ans: 15

Explanation:Let the distance between station X and station Y be D.
Let the speed of train T be $v_T$.
Let the speed of train S be $v_S$.
We are given that $v_S = \frac{3}{4} v_T$.

Train T leaves X at 3 pm.
Train S leaves Y at 4 pm.

The trains pass each other at station Z.
The distance between X and Z is $\frac{3}{5}$ of the distance between X and Y.
So, the distance XZ = $\frac{3}{5} D$.

Since the trains meet at Z, the distance traveled by train T from X to Z is $\frac{3}{5} D$.
The distance traveled by train S from Y to Z is $D – \frac{3}{5} D = \frac{2}{5} D$.

Let $t$ be the time (in hours) that train T travels until it meets train S at station Z.
Train T starts at 3 pm and meets train S at Z. So, train T travels for $t$ hours.
The distance traveled by train T is $distance_T = v_T \times t$.
We know this distance is $\frac{3}{5} D$.
So, $\frac{3}{5} D = v_T \times t$ (Equation 1).

Train S leaves station Y at 4 pm. Train T leaves at 3 pm.
This means train S starts 1 hour after train T.
So, train S travels for $t-1$ hours when it meets train T at Z.
The distance traveled by train S is $distance_S = v_S \times (t-1)$.
We know this distance is $\frac{2}{5} D$.
So, $\frac{2}{5} D = v_S \times (t-1)$ (Equation 2).

Now we have a system of two equations:
1. $\frac{3}{5} D = v_T \times t$
2. $\frac{2}{5} D = v_S \times (t-1)$

Substitute $v_S = \frac{3}{4} v_T$ into Equation 2:
$\frac{2}{5} D = \frac{3}{4} v_T \times (t-1)$ (Equation 3).

Now, divide Equation 1 by Equation 3 to eliminate D and $v_T$:
$\frac{\frac{3}{5} D}{\frac{2}{5} D} = \frac{v_T \times t}{\frac{3}{4} v_T \times (t-1)}$

$\frac{3/5}{2/5} = \frac{t}{\frac{3}{4} (t-1)}$
$\frac{3}{2} = \frac{t}{\frac{3}{4} (t-1)}$

Multiply both sides by $\frac{3}{4} (t-1)$:
$\frac{3}{2} \times \frac{3}{4} (t-1) = t$
$\frac{9}{8} (t-1) = t$

Distribute the $\frac{9}{8}$:
$\frac{9}{8} t – \frac{9}{8} = t$

Subtract $t$ from both sides:
$\frac{9}{8} t – t = \frac{9}{8}$
$\frac{9}{8} t – \frac{8}{8} t = \frac{9}{8}$
$\frac{1}{8} t = \frac{9}{8}$

Multiply both sides by 8:
$t = 9$ hours.

The question asks for the total time train T takes for its journey from X to Y.
Train T travels $\frac{3}{5}$ of the distance in 9 hours.
Let the total time for train T’s journey from X to Y be $T_{total}$.
The distance for the whole journey is D.
Distance = Speed × Time
D = $v_T \times T_{total}$

From Equation 1: $\frac{3}{5} D = v_T \times 9$
This means $D = \frac{5}{3} v_T \times 9 = v_T \times (5 \times 3) = v_T \times 15$.
So, $T_{total} = 15$ hours.

Alternatively, if train T travels $\frac{3}{5}$ of the distance in 9 hours, then to travel the full distance (which is 1 unit or $\frac{5}{5}$ of the distance), it will take:
$\frac{1}{3/5}$ times the time it took for $\frac{3}{5}$ of the distance.
$T_{total} = \frac{5}{3} \times 9$ hours
$T_{total} = 5 \times 3$ hours
$T_{total} = 15$ hours.

Final_Answer:15

Ans: D

Explanation:Let the speed of Partha be $S_P$ km/hr and the speed of Narayan be $S_N$ km/hr.
The distance from A to B is 60 km.

Partha takes four hours more than Narayan to reach B.
Time taken by Partha to reach B = $T_P = \frac{60}{S_P}$
Time taken by Narayan to reach B = $T_N = \frac{60}{S_N}$
According to the problem: $T_P = T_N + 4$
$\frac{60}{S_P} = \frac{60}{S_N} + 4$ (Equation 1)

Partha reaches the mid-point of A and B (30 km) two hours before Narayan reaches B.
Time taken by Partha to reach the mid-point = $\frac{30}{S_P}$
According to the problem: $\frac{30}{S_P} = T_N – 2$
$\frac{30}{S_P} = \frac{60}{S_N} – 2$ (Equation 2)

From Equation 2, we can express $\frac{60}{S_N}$ in terms of $S_P$:
$\frac{60}{S_N} = \frac{30}{S_P} + 2$

Substitute this expression for $\frac{60}{S_N}$ into Equation 1:
$\frac{60}{S_P} = (\frac{30}{S_P} + 2) + 4$
$\frac{60}{S_P} = \frac{30}{S_P} + 6$

Now, solve for $S_P$:
$\frac{60}{S_P} – \frac{30}{S_P} = 6$
$\frac{30}{S_P} = 6$
$S_P = \frac{30}{6}$
$S_P = 5$ km/hr

Let’s check if this speed is consistent with the given options. Yes, 5 km/hr is option D.

We can also find $S_N$ to verify the conditions.
From Equation 1: $T_P = \frac{60}{5} = 12$ hours.
$12 = T_N + 4 \implies T_N = 8$ hours.
$S_N = \frac{60}{8} = 7.5$ km/hr.

Check the second condition:
Time for Partha to reach mid-point = $\frac{30}{5} = 6$ hours.
Narayan reaches B in 8 hours.
Is 6 hours equal to 8 hours – 2 hours? Yes, $6 = 8 – 2$.
The speeds are consistent with the conditions.

The speed of Partha is 5 km per hour.

Correct_Option:D

Ans: D

Explanation:Let the speed of car 1 be $v_1$, the speed of car 2 be $v_2$, and the speed of car 3 be $v_3$.
The distance between A and B is 300 km.
Points P and Q are at distances of 100 km and 200 km from A, respectively.
Car 1 and car 2 start from A towards B. Car 3 starts from B towards A.
Car 3 meets car 1 at Q.
The distance covered by car 1 when they meet is the distance from A to Q, which is 200 km.
The distance covered by car 3 when they meet is the distance from B to Q, which is 300 km – 200 km = 100 km.
Since they travel for the same amount of time until they meet, let this time be $t_1$.
So, $200 = v_1 * t_1$ and $100 = v_3 * t_1$.
From these two equations, we get $v_1 / v_3 = 200 / 100 = 2$. So, $v_1 = 2 * v_3$.

Car 3 meets car 2 at P.
The distance covered by car 2 when they meet is the distance from A to P, which is 100 km.
The distance covered by car 3 when they meet is the distance from B to P, which is 300 km – 100 km = 200 km.
Since they travel for the same amount of time until they meet, let this time be $t_2$.
So, $100 = v_2 * t_2$ and $200 = v_3 * t_2$.
From these two equations, we get $v_2 / v_3 = 100 / 200 = 1/2$. So, $v_2 = (1/2) * v_3$.

We need to find the ratio of the speed of car 2 to that of car 1, which is $v_2 / v_1$.
We have $v_2 = (1/2) * v_3$ and $v_1 = 2 * v_3$.
So, $v_2 / v_1 = ((1/2) * v_3) / (2 * v_3) = (1/2) / 2 = 1/4$.
The ratio of the speed of car 2 to that of car 1 is 1:4.

Correct_Option:D

Ans: 5

Explanation:
The total distance between points A and B is 150 km.
The speed of each car is divided into three segments:
Segment 1: First 50 km at 100 kmph
Segment 2: Next 50 km at 50 kmph
Segment 3: Last 50 km at 25 kmph

Let’s calculate the time taken by car 1 to reach point B.
Time for the first 50 km = Distance / Speed = 50 km / 100 kmph = 0.5 hours
Time for the next 50 km = Distance / Speed = 50 km / 50 kmph = 1 hour
Time for the last 50 km = Distance / Speed = 50 km / 25 kmph = 2 hours

Total time taken by car 1 to reach B = 0.5 hours + 1 hour + 2 hours = 3.5 hours.

Car 2 starts from A when car 1 is already 20 km away from A.
Let’s calculate the time taken by car 1 to travel the first 20 km.
Time = Distance / Speed = 20 km / 100 kmph = 0.2 hours.
So, car 2 starts 0.2 hours after car 1.

Car 1 reaches B at time T_1 = 3.5 hours from its start.
Car 2 starts at time T_start_2 = 0.2 hours from car 1’s start.
Car 2 will travel for a duration of (T_1 – T_start_2) when car 1 reaches B.
Duration car 2 travels = 3.5 hours – 0.2 hours = 3.3 hours.

Now, let’s calculate the distance covered by car 2 in 3.3 hours.
We need to see which segment of speed car 2 is in.
For the first 50 km, car 2 takes 50 km / 100 kmph = 0.5 hours.
After 0.5 hours, car 2 has covered 50 km. Remaining time for car 2 is 3.3 – 0.5 = 2.8 hours.

For the next 50 km (from 50 km to 100 km), car 2 takes 50 km / 50 kmph = 1 hour.
After another 1 hour, car 2 has covered 50 + 50 = 100 km. Remaining time for car 2 is 2.8 – 1 = 1.8 hours.

For the last 50 km (from 100 km to 150 km), car 2 travels at 25 kmph.
The time taken to cover this last 50 km is 50 km / 25 kmph = 2 hours.
Car 2 has 1.8 hours remaining. In this time, car 2 will cover:
Distance = Speed × Time = 25 kmph × 1.8 hours = 45 km.

So, when car 1 reaches B, car 2 has covered 100 km (first two segments) + 45 km (part of the third segment) = 145 km from A.
The total distance to B is 150 km.
The distance between car 2 and B is 150 km – 145 km = 5 km.

Final_Answer:5

Ans: 50

Explanation:Let the speed of the car starting from A be $v_A$ km/hr and the speed of the car starting from B be $v_B$ km/hr.
The distance between A and B is 350 km.
Case 1: They move towards each other.
When they move towards each other, their relative speed is the sum of their speeds, i.e., $v_A + v_B$.
They meet after 1 hour.
Distance = Speed × Time
350 km = $(v_A + v_B)$ km/hr × 1 hr
$v_A + v_B = 350$ (Equation 1)

Case 2: They both move towards east.
B is 350 km west of A. This means A is 350 km east of B.
If both cars move towards east, the car starting from B will be behind the car starting from A. For them to meet, the car starting from B must be faster than the car starting from A, so $v_B > v_A$.
When they move in the same direction, their relative speed is the difference between their speeds, i.e., $v_B – v_A$.
They meet after 7 hours.
The distance that the faster car (from B) needs to cover to meet the slower car (from A) is the initial distance between them, which is 350 km.
Distance = Speed × Time
350 km = $(v_B – v_A)$ km/hr × 7 hrs
$v_B – v_A = \frac{350}{7}$
$v_B – v_A = 50$ (Equation 2)

We are asked to find the difference between their speeds, which is $|v_A – v_B|$. From Equation 2, we have $v_B – v_A = 50$. This implies that the speed of the car from B is 50 km/hr more than the speed of the car from A. Therefore, the difference between their speeds is 50 km/hr.

Let’s verify by finding $v_A$ and $v_B$.
Add Equation 1 and Equation 2:
$(v_A + v_B) + (v_B – v_A) = 350 + 50$
$2v_B = 400$
$v_B = 200$ km/hr

Substitute $v_B = 200$ into Equation 1:
$v_A + 200 = 350$
$v_A = 350 – 200$
$v_A = 150$ km/hr

The difference between their speeds is $v_B – v_A = 200 – 150 = 50$ km/hr.
Or $|v_A – v_B| = |150 – 200| = |-50| = 50$ km/hr.

Final_Answer:50

Ans: 20

Explanation:Let $d$ be the distance from the man’s home to the railway station in km.
Let $t$ be the scheduled departure time of the train in hours.

Case 1: The man walks at a speed of 12 km/h.
The time taken to reach the station is $\frac{d}{12}$ hours.
He reaches the station 10 minutes after the train had departed.
10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours.
So, the time he reached the station is $t + \frac{1}{6}$ hours.
Therefore, $\frac{d}{12} = t + \frac{1}{6}$ (Equation 1)

Case 2: The man walks at a speed of 15 km/h.
The time taken to reach the station is $\frac{d}{15}$ hours.
He reaches the station 10 minutes before the train’s departure.
So, the time he reached the station is $t – \frac{1}{6}$ hours.
Therefore, $\frac{d}{15} = t – \frac{1}{6}$ (Equation 2)

We have a system of two linear equations with two variables $d$ and $t$:
1) $\frac{d}{12} = t + \frac{1}{6}$
2) $\frac{d}{15} = t – \frac{1}{6}$

We can eliminate $t$ by subtracting Equation 2 from Equation 1:
$(\frac{d}{12}) – (\frac{d}{15}) = (t + \frac{1}{6}) – (t – \frac{1}{6})$
$\frac{d}{12} – \frac{d}{15} = t + \frac{1}{6} – t + \frac{1}{6}$
$\frac{d}{12} – \frac{d}{15} = \frac{1}{6} + \frac{1}{6}$
$\frac{d}{12} – \frac{d}{15} = \frac{2}{6}$
$\frac{d}{12} – \frac{d}{15} = \frac{1}{3}$

To solve for $d$, find a common denominator for the fractions on the left side, which is 60:
$\frac{5d}{60} – \frac{4d}{60} = \frac{1}{3}$
$\frac{5d – 4d}{60} = \frac{1}{3}$
$\frac{d}{60} = \frac{1}{3}$

Multiply both sides by 60:
$d = 60 \times \frac{1}{3}$
$d = 20$

Alternatively, we can find the value of $t$ first. From Equation 1, $t = \frac{d}{12} – \frac{1}{6}$. Substitute this into Equation 2:
$\frac{d}{15} = (\frac{d}{12} – \frac{1}{6}) – \frac{1}{6}$
$\frac{d}{15} = \frac{d}{12} – \frac{2}{6}$
$\frac{d}{15} = \frac{d}{12} – \frac{1}{3}$
$\frac{1}{3} = \frac{d}{12} – \frac{d}{15}$
This leads to the same equation as before.

Let’s calculate the time difference. The difference in arrival times at the station is 10 minutes + 10 minutes = 20 minutes = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours.
This difference in time is due to the difference in speeds.
Let $t_1$ be the time taken at 12 km/h and $t_2$ be the time taken at 15 km/h.
$t_1 = \frac{d}{12}$ and $t_2 = \frac{d}{15}$.
We know that $t_1 – t_2 = \frac{1}{3}$ hours.
$\frac{d}{12} – \frac{d}{15} = \frac{1}{3}$
$\frac{5d – 4d}{60} = \frac{1}{3}$
$\frac{d}{60} = \frac{1}{3}$
$d = \frac{60}{3} = 20$ km.

Final_Answer:20

Ans: B

Let the river’s current speed be denoted by $x$ and the boat’s speed in still water be denoted by $u$. Let $d$ represent the one-way distance covered, and let $t$ be the initial time taken for the round trip.

The problem statement provides the following initial condition:
$t = \frac{d}{u – x} + \frac{d}{u + x}$ … (Equation 1)

It is also given that under new conditions, the time taken is $t/4$. The new boat speed is $2u$ and the river speed remains $x$. Thus, the second condition is:
$\frac{t}{4} = \frac{d}{2u – x} + \frac{d}{2u + x}$
Multiplying by 4, we get:
$t = \frac{4d}{2u – x} + \frac{4d}{2u + x}$ … (Equation 2)

Now, we equate the expressions for $t$ from Equation 1 and Equation 2:
$\dfrac{d}{u – x} + \dfrac{d}{u + x} = \dfrac{4d}{2u – x} + \dfrac{4d}{2u + x}$

We can simplify the left side of the equation:
$\frac{d(u+x) + d(u-x)}{(u-x)(u+x)} = \frac{2du}{u^2 – x^2}$

And simplify the right side of the equation:
$\frac{4d(2u+x) + 4d(2u-x)}{(2u-x)(2u+x)} = \frac{8du + 4dx + 8du – 4dx}{4u^2 – x^2} = \frac{16du}{4u^2 – x^2}$

Now, equating the simplified expressions:
$\frac{2du}{u^2 – x^2} = \frac{16du}{4u^2 – x^2}$

Assuming $d$ and $u$ are non-zero, we can divide both sides by $2du$:
$\frac{1}{u^2 – x^2} = \frac{8}{4u^2 – x^2}$

Cross-multiplying:
$4u^2 – x^2 = 8(u^2 – x^2)$
$4u^2 – x^2 = 8u^2 – 8x^2$

Rearranging the terms to group $u^2$ and $x^2$:
$8x^2 – x^2 = 8u^2 – 4u^2$
$7x^2 = 4u^2$

Now, we want to find the ratio $u/x$. Divide both sides by $4x^2$:
$\frac{7x^2}{4x^2} = \frac{4u^2}{4x^2}$
$\frac{7}{4} = \frac{u^2}{x^2}$

Taking the square root of both sides:
$\sqrt{\frac{7}{4}} = \sqrt{\frac{u^2}{x^2}}$
$\frac{\sqrt{7}}{2} = \frac{u}{x}$

Ans: 1900

When A covered a distance of 10 KM, B covered 9 KM.
This implies the ratio of their speeds is:
$Speed_A : Speed_B = 10:9$
In a similar fashion, when B covered 10 KM, C covered 9 KM.
This gives us the ratio of their speeds as:
$Speed_B : Speed_C = 10:9$
To find the combined ratio of all three, we synchronize the common term ($Speed_B$):
$Speed_A : Speed_B = 10 \times 10 : 9 \times 10 = 100:90$
$Speed_B : Speed_C = 9 \times 10 : 9 \times 9 = 90:81$
Thus, the combined ratio of their speeds is:
$Speed_A : Speed_B : Speed_C = 100:90:81$
This means that when A covers 100 units of distance, C covers 81 units of distance.
If A covers 10 KM (which is 100 units if we scale down), C would cover 8.1 KM (81 units scaled down).
The difference in distance covered by A and C is:
10 KM – 8.1 KM = 1.9 KM
Converting this difference to meters:
1.9 KM $\times$ 1000 M/KM = 1900 M
Therefore, A wins against C by 1900 meters.

Ans: B

Explanation:Let the speed of the motorbike be $v$ km/hr.
The car’s speed is $2v$ km/hr.

The motorbike leaves point A at 1 pm and meets the car at 3:40 pm.
The total time the motorbike travels is from 1 pm to 3:40 pm, which is 2 hours and 40 minutes.
2 hours and 40 minutes = $2 + \frac{40}{60}$ hours = $2 + \frac{2}{3}$ hours = $\frac{8}{3}$ hours.

The distance covered by the motorbike when they meet is 168 km.
Distance = Speed × Time
168 km = $v \times \frac{8}{3}$
$v = \frac{168 \times 3}{8}$
$v = 21 \times 3$
$v = 63$ km/hr.

The speed of the motorbike is 63 km/hr.
The speed of the car is $2v = 2 \times 63 = 126$ km/hr.

The car leaves point B at 2 pm and meets the motorbike at 3:40 pm.
The total time the car travels is from 2 pm to 3:40 pm, which is 1 hour and 40 minutes.
1 hour and 40 minutes = $1 + \frac{40}{60}$ hours = $1 + \frac{2}{3}$ hours = $\frac{5}{3}$ hours.

The distance covered by the car is:
Distance = Speed × Time
Distance covered by car = $126 \times \frac{5}{3}$
Distance covered by car = $42 \times 5$
Distance covered by car = 210 km.

The meeting point is 168 km from A and 210 km from B.
The distance between A and B is the sum of the distances covered by the motorbike and the car.
Distance AB = Distance covered by motorbike + Distance covered by car
Distance AB = 168 km + 210 km
Distance AB = 378 km.

The distance between A and B is 378 km.

Correct_Option:B