XAT Time, Speed and Distance Questions

Let's assume that all the dustbins are at equal distances and the distance between two dustbin's be l.

We are told that a minimum 800 steps to reach every second dustbin. 

800$$\le$$2l $$\Rightarrow$$ 400 $$\le$$ l

Similarly, we are also told that a maximum 1260 steps to reach every third dustbin.

So, 3l$$\le$$1260

400$$\le$$l$$\le$$420

The circumference will be 5l.

2000$$\le$$ 5l $$\le$$ 2100

2000 and 2100 are both in steps of Ramesh.  

We are told that each of his step is 0.77m.

So, 2000 steps = 2000*0.77= 1540m

2100 steps = 2100*0.77 = 1617

So, the circumference will lie between 1540 and 1617.

1540 $$\le 2\pi r \le$$ 1617.

$$\frac{1540}{2\pi}\le r\le\frac{1617}{2\pi}$$

$$245.22\le r\le\ 257.48$$

So, 250m is the correct option.

Let's assume the initial speed of the flight is F. And the time taken is t.

Given the distance to be travelled, 11200.

We know distance = speed *time.

So, 11200=Ft-->1

Due to bad weather, the flight started with a three-hour delay. Given that the flight speed increased by 100kmph, the delay would be reduced to one hour. So, it is nothing but the travel time is reduced by 2 hours.

In that case, the equation will become 11200=(F+100)*(t-2)-->2

Equation 1 can be re-written as t=$$\frac{11200}{F}$$-->3

Now Equation 2: $$\frac{11200}{F+100}$$=t-2

$$\Rightarrow$$ = $$\frac{11200}{F+100}$$+2=t

$$\frac{11200+2\left(F+100\right)}{F+100}=t$$

$$t=\frac{11400+2F}{F+100}$$-->4

Divide equation 3 by equation 4:'

$$\frac{t}{t}=\frac{\frac{11200}{F}}{\frac{11400+2F}{F+100}}$$

$$\Rightarrow$$ $$\frac{\left(11400+2F\right)}{F+100}=\frac{11200}{F}$$

$$\left(11400+2F\right)\cdot F=\left(F+100\right)\cdot11200$$

$$\left(11400F+2F^2\right)=\left(11200F+1120000\right)$$

$$\ 2F^2+200F-1120000=0\ \Rightarrow\ F^2+100F-560000=0$$

So, F=700 or -800. Speed can't be negative. So, the speed is 700km/hr.

Now, if the speed is increased by 350, the new speed will be 1050.

Hence, the time taken will be $$\frac{11200}{1050}=\frac{32}{3}=10\ \frac{2}{3\ }hrs=10hrs\ 40\min$$

The flight has to start at 6:30 am, but it is delayed by 3 hrs. So, the actual time is 9:30 am.

10hrs: 40mins from that will be 8:10 pm

The pace of Jim's watch is slower by 3 minutes in comparison with Kim's watch for every one hour. The difference increases as the hours pass by.

The time and "am" and "pm" of the watch coincide when the difference between the two clocks reduces to 24 hours. The two clocks display the same time including am and pm.

For a difference of 24 hours, the clock needs to lag by 1440 minutes.

For every one hour, the clock lags by 3 minutes. Hence in order to have a difference of 1440 minutes, it takes $$\frac{1440}{3}=\ 480\ hours$$. This is equivalent to 20 days of time.

Hence for every twenty days, they display the same time.

In the period of Jan 1 and June 1, there are 150 days which includes 7, 20-day intervals. Along with the 7 times once on Jan 1st and once on June 1st, they display the same time.

A total of 9 times.

Let speed of deer = 5steps/second and speed of tiger = 6 steps/sec

Let deer cover 1 m in a step => tiger covers 2/3 m in a step

Hence speed of deer = 5m/s and spped of tiger = 6 x 2/3 m/s = 4m/s

Hence time taken by a deer to catch tiger = 40 seconds

Distance travelled by deer in 40 seconds = 5 x 40 =200 steps

All the four friends will meet at the starting point after LCM(60,90,40,80) = 720 seconds.

Number of laps by A in 720 seconds = 12

Number of laps by B in 720 seconds = 8

Number of laps by C in 720 seconds = 18

Number of laps by D in 720 seconds = 9

Together they complete = 47 laps

The difference between the hour and minute hand of a clock is given by $$\left|30H-5.5m\right|$$. Here H is the current hour and m represents the number of completed minutes in the current hour.

In the given time frame of 2: 30 to 3: 00 pm.

At 2 : 30 pm the angle = $$\left|30\cdot2-5.5\cdot30\right|\ =\ 105\ \deg rees$$

At 3: 00 pm the angle = $$\left|30\cdot3-5.5\cdot0\right|\ =\ 90\ \deg rees$$

The function of $$\left|30\cdot H-5.5\cdot m\right|\ =\ $$ constantly increases as the value of m increases from 31, 32................ 59.

Because of the modulus function, the net value of the function remains positive

Between 2: 30 to 2: 59 the angle is constantly increasing. The minimum value is 2: 30 which is equal to 105 degrees which is greater than the 90 degrees when the time is 3: 00.

Hence 90 degrees is the minimum angle.

Let the speed of the stream be x

$$\frac{12}{5-x}=\frac{12}{5+x}+1$$

The value of x satisfying the above equation is 1

Now,

$$\frac{N}{5+1}+\frac{N}{5-1}\ge2$$

$$\frac{2N+3N}{12}\ge2$$

=> $$N\ge4.8$$

2500=$$\left(5+10t\right)^2+\left(10+20t\right)^2$$ => $$\left(1+2t\right)^2=20$$

$$AB^2=25+100\left(1+2t\right)^2$$ => AB= 45km.

The total distance to be travelled is 6+6 = 12km

We know the speeds of the hare and the tortoise.

Time taken by the Tortoise to finish one round = $$\frac{12}{1}=12hr$$. 

We know the speed of the Hare is 12 km/h. Let's find the distance travelled by the hare in 12 hrs. = $$12\cdot12=144km$$

This is nothing but $$\frac{144}{12}=12\ \text{rounds}\ $$. So, Hare will make 12 rounds of OP.

In the first round, both have started from point O. After some time, the distance between them will be 1 km. 

After some more time, when the hare is returning from P to O, before and after crossing the tortoise, the hare will be two more times 1km apart from the tortoise. So, in the first round, there are three such occurrences.

In the second round, when the hare starts from point O, while going and returning, there will be four occurrences when, before and after crossing the tortoise, the hare will be exactly 1 km apart. But the first occurrence of round 2 is already counted in round 1. So, in second round as well, there will be total 3 occurrences.

In the third, fourth and fifth rounds, there will be 4 such occurrences.

In the sixth round, because the tortoise will be at point P, there will be only 2 cases.

Now, till around 6, there are 20 such occurrences. And from round 7 to 12, it will be exactly the same but in reverse order of 2, 4, 4, 4, 3, 3. Hence, total such occurrences = 20 * 2 = 40.

The man walks with a speed of V1 for 30 minutes. 
=> Distance covered = 30*V1.
On a particular day, he walks for 10 minutes at V1, takes a rest of 5 minutes, and then covers the distance by walking at V2 for 30 minutes. 
=> 10*V1 + 30*V2 = 30*V1
20*V1 = 30*V2
V1 = 1.5V2---------(1)
=> Total distance = 30*1.5*V2 = 45V2.
Now, we know that the person took 10 + 5 + 30 = 45 minutes to cover the entire distance. 
=> Average speed = 45V2/45 = V2.
Ratio of V2 and the average speed = 1:1.
Therefore, option A is the right answer. 

One boat is stationed to the North of the light house and the other boat is stationed to the East of the light house. 

The boat stationed to the East subtends an angle of 45 degrees and the boat stationed to the North subtends an angle of 30 degrees. 
Now, distance between the boat stationed to the East and the light house,d1 = tan 45
$$300/d1 = 1$$
=> $$d1 = 300$$ feet

Distance between the boat stationed to the North and the light house,d2 = tan 30
$$300/d2 = 1/\sqrt{3}$$
=> $$d2=300\sqrt{3}$$

Shortest distance between the 2 boats = $$\sqrt{300^2+(300\sqrt{3})^2}$$
                                                           = $$300*\sqrt{4}$$
                                                           =  $$600$$ feet.
Therefore, option C is the right answer.

We draw BD perpendiculat to AC.
In right angled triangle BDC,  BD / BC = sin 30°
or, BD = (V2 * T)/2 ......(i)
In right angled triangle BDA, BD / BA = sin 45°
Or, BD = (V1 * T)/$$\sqrt{2}$$ ......(ii)
From (i) and (ii), we get
V2 / V1 = $$\sqrt{2}$$
Total distance to be travelled from C to A = CD + DA  = $$\sqrt{3}$$BD + BD
= BD($$1 + \sqrt{3}$$)
Replacing BD = (V2 * T)/2 in the avove equation,
CA = $$\dfrac{\text{(V2 * T)}}{2} (1 + \sqrt{3}$$)
Time taken at speed V2 = $$0.5(\sqrt{3}+1)$$T
Hence, option C is the correct answer.

Both of them row the boat for the same time. Therefore, we can ignore the time taken to row the boat and add it to the final answer. 
Arup will walk 2 km in 2/4 = 30 minutes.
Swarup will walk 2 km in 2/5 = 24 minutes.

Therefore, Swarup will start driving 30-24 = 6 minutes earlier than Arup. 
In 6 minutes, Swarup will gain a lead of 6*40/60 = 4 km. 
Arup drives at 50 kmph (i.e, 10 kmph faster than Swarup). Therefore, Arup will cover the 4 km advantage that Swarup has in 4/10*60 = 24 minutes. 
Therefore, the time after which both of them will meet is 30 (to walk) + 30 (to row the boat) + 24 = 1 hour 24 minutes after Arup starts. 
Since both of them start at 8 AM, they will meet again at 9:24 AM. 
Therefore, option D is the right answer. 

AB + BC + CD = 70 ------------(i)

AB + BD = 45 ------------------(ii)

AC + CD = 30 ------------------(iii)

AC + CB + BD = 65 ------------(iv)

Adding (i) & (iv)

=> AB + BC + CD + AC + CB + BD = 70 + 65

(AB + BD) + 2 BC + (AC + CD) = 135

From (ii) & (iii)

=> 45 + 2 BC + 30 = 135

=> 2 BC = 135 - 75 = 60

=> BC = $$\frac{60}{2} = 30$$ min

10 hour analog clock is used. 

Time in mirror = 3 hours 42 minutes and 20 seconds

=> Actual time = 10 - (3 hrs 42 min 20  sec)

= 6 hrs 17 min 20 sec

$$\therefore$$ Time after 5 minutes = 6 hrs 22 min 20 sec

Let walking speed = $$a$$

=> Driving speed = $$8a$$ 

Let the distance between temple and office = x

Acc. to ques,

=> $$\frac{x}{a} = \frac{1}{a} + \frac{x + 1}{8 a}$$

=> $$8x = 8 + x + 1$$

=> $$x = \frac{9}{7}$$

Scenario I : Devanand's position after $$t$$ hours is $$(50 - 3t)$$ km west  of Pradeep's house, while Pradeep's position is $$4t$$ km south of his own house.

If $$d$$ is the distance between them, then

=> $$d^2 = (50 - 3t)^2 + (4t)^2$$

=> $$d^2 = 2500 - 300t + 25t^2$$

=> $$d^2 = 25 (t^2 - 12t + 36) + 1600$$

=> $$d^2 = 25 (t - 6)^2 + 1600$$

Thus, minimum distance is 40 km after 6 hours.

Thus, scenario I is possible

Scenario II & III are not possible as minimum distance in that case would be 50 km as after that distance will keep on increasing between the two.

As it took Prof. Mandal 30 minutes for round trip to the market by auto, we can infer that the auto takes 15 minutes for one way trip.
When he walks one way and return by auto, it took him 90 minutes.
We know the auto would have taken 15 minutes while returning.
So, the professor takes (90 - 15) = 75 minutes one way while walking to the market.
So, for a round trip by walking, he will take 2 * 75 minutes = 150 minutes.

Hence, option D is the correct.

For the first 2 km, fare = Rs. 20
For the next 8 km, fare = Rs. 5 * 8 = Rs. 40
For the next 4 km, fare = Rs. 8 * 4 = Rs. 32
Total fare up to 14 km = Rs. 20 + Rs. 40 + Rs. 32 = Rs. 92
Total fare paid by Sardar Khan = Rs. 102
Money spent on bullock cart = Rs. 102 - Rs. 92 = Rs. 10
Charge for bullock cart = Rs. 2/km
Distance travelled by bullock cart = 5 km
Total distance travelled by Sardar Khan = (14 + 5) km = 19 km
Hence, option C is the correct answer.

Let 'A' be the point from where all started and D is the destination. At 'x' hours both Ram and Hari reached point C, Hari got off and Ram turned back to pick up Shyam. At the same instant, Shyam was at point B.

Let 'T' is the total amount of time taken by all three to reach point D.

For Hari,

T = $$x+\dfrac{100-25x}{5}$$  ... (1)

For Ram,

T = $$x+\dfrac{20x*5/6}{25}$$ + $$\dfrac{20x*5/6+100-25x}{25}$$  ... (2)

By equating (1) and (2),

$$x+\dfrac{100-25x}{5}$$ = $$x+\dfrac{20x*5/6}{25}$$ + $$\dfrac{20x*5/6+100-25x}{25}$$

$$\Rightarrow$$ $$x = 3$$ hours.

Therefore, time taken by all to reach destination  = $$x+\dfrac{100-25x}{5}$$

$$\Rightarrow$$ 3 + $$\dfrac{100-25*3}{5}$$

$$\Rightarrow$$ 3 + 5 

$$\Rightarrow$$ 8 hours

Hence, we can say that option A is the correct answer. 

Let the distance between two termini = $$P$$ km
Let the speed of the bus started from Terminus A be x and that of the bus started from terminus B be y.
The two buses met at a distance of 7km from Terminus A
Since the time of travel for both buses is the same.
$$\frac{7}{x}$$=$$\frac{P-7}{y}$$
$$\frac{x}{y}$$=$$\frac{7}{P-7}$$  ----------- Eq (1)
They met again at a distance of 4 km from terminus B.
Distance travelled by bus which started from Terminus A = P+4
Distance travelled by bus which started from Terminus A = 2P-4
So $$\frac{x}{y}$$ = $$\frac{P+4}{2P-4}$$   --Eg (2)
On solving Eq 1& 2, we get P=17 km
Each bus covers a distance of 17*2=34 km on a round trip.
Each bus makes 5 round trips in a day =34*5=170 km
Cost of running one bus = $$170 \times 20 = 3400$$
$$\therefore$$ Cost of running both buses = $$3400 \times 2 = Rs. 6,800$$

Shortest distance between A and B = $d = \sqrt{(AM)^2 + (BM)^2}$ ----------Eqn(1)

In $$\triangle$$ OPN

=> $$sin 45 = \frac{PN}{PO}$$

=> $$\frac{1}{\sqrt{2}} = \frac{PN}{50}$$

=> $$PN = \frac{50}{\sqrt{2}} = 25 \sqrt{2}$$

=> $$BM = PN = 25 \sqrt{2}$$

Again, $$tan 45 = \frac{PN}{ON}$$

=> $$ON = 25 \sqrt{2}$$

=> $$AM=AO\ +\ ON\ +\ NM\ =\ 50\ +\ 25\sqrt{2}\ +\ 50\ =\ 100\ +\ 25\sqrt{2}$$

Using eqn(I), we get :

=> $$d^2 = (100 + 25 \sqrt{2})^2 + (25 \sqrt{2})^2$$

=> $$d^2 = 10000 + 5000 \sqrt{2} + 1250 + 1250 $$ = $$12500 + 5000 \sqrt{2}$$

=> $$d^2 = 2500 (5 + 2 \sqrt{2}) = 2500 (5 + \sqrt{8})$$ -----------Eqn(II)

Also, it is given that : $$d = a\sqrt{b+\sqrt{c}}$$

=> $$d^2 = a^2 (b + \sqrt{c})$$ -----------Eqn(III)

Comparing, eqn(II) & (III), we get :

=> $$a^2 (b + \sqrt{c}) = 2500 (5 + \sqrt{8})$$

=> $$a = 50 , b = 5 , c = 8$$

$$\therefore a + b + c = 50 + 5 + 8 = 63$$

At the start, Ramya has 10 litres petrol. She purchased 20 litres, 15 litres and 10 litres.

At the end, she has left 5 litres of petrol.

Petrol used = $$(10 + 20 + 15 + 10 - 5) = 50$$ litres

Distance travelled = $$800 - 400 = 400$$ km

$$\therefore$$ Mileage of Ramya's car = $$\frac{400}{50} = 8$$ km/litre

Ramya's car has 5 litres in the tank. She can fill a maximum of 30 litres more as tank capacity is 35 litres.

The cost of petrol in Rampur is Rs. 45/litre. As the cost of petrol is lower at all the succeeding petrol pumps and hence to minimise the cost, she will fill enough petrol to reach first petrol pump i.e. 150 km.

Initially she can travel = $$8 \times 5 = 40$$ km using 5 litres of petrol

Hence to travel 110 km, she will need = $$\frac{110}{8} = 13.75$$ litre at the rate of Rs. 45/litre

On reaching the first petrol pump in the reverse journey, she will fill up enough petrol to reach the second petrol pump as the cost of petrol in the second pump is less than the cost of first pump and the distance is 50 km.

=> She needs = $$\frac{50}{8} = 6.25$$ litres at the rate of Rs. 40/litre

For the rest of the journey (200 km), she will need = $$\frac{200}{8} = 25$$ litres at the rate of Rs. 35/litre

$$\therefore$$ Total cost = $$(13.75 \times 45) + (6.25 \times 40) + (25 \times 35)$$

= $$618.75 + 250 + 875 = 1743.75 \approx Rs. 1744$$

Amarendra took 25 minutes to cover 15 km. In the same time Dharmendra travel 2500m less, i,e, 12.5 km

=> Speed of Amarendra = $$\frac{15}{25} = 0.6$$ km/min

Speed of Dharmendra = $$\frac{12.5}{25} = 0.5$$ km/min

=> Time taken by Dharmendra to reach the station = $$\frac{15}{0.5} = 30$$ minutes

Next day, Dharmendra started 7 minutes early, so he will reach the station = 30 - 25 - 7 = -2 minutes

=> 2 minutes or 120 seconds before Amarendra.

Angle covered by the hour hand in 12 hours = 360°

In 1 hour = $$\frac{360}{12} = 30^{\circ}$$

and in 1 minute = $$\frac{30}{60} = \frac{1}{2}^{\circ}$$

Similarly, angle covered by minute hand in 1 hour = 360°

In 1 minute = $$\frac{360}{60} = 6^{\circ}$$

=> Every minute, the angle between the two hands changes by = $$6 - \frac{1}{2} = \frac{11}{2}^{\circ}$$

$$\therefore$$ From 7:45 A.M. to 7:47 A.M.,i.e. in 2 minutes the angle between the two hands will change by

= $$2 \times \frac{11}{2} = 11^{\circ}$$