Time distance speed questions

Let the speed of travelling to office and back to home be x and y respectively.

So, his average speed is =\( \frac{2xy}{x+y} \)

\( \frac{2\;\times\;25\times4}{25+4} \)

= 200/29 km/hr

He covers the whole journey in 5 hours 48 minutes = \( 5\frac45 \) = 29/5 hrs

Therefore, total distance covered

= (\( \frac{200}{29} \) × \( \frac{29}5 \)) = 40 km

So, the distance from his home to office = \( \frac{40}2 \) = 20 km

A runs 1000 meters while B runs 900 meters and C runs 800 meters.

Therefore, B runs 900 meters while C runs 800 meters.

So, the number of meters that C runs when B runs 1000 meters = \(\frac{(1000 \times 800)}{900} =\frac{ 8000}{9} = 888.88 meters\)

Thus, B can give C (1000 - 888.88) = 111.12 meters start

Speed1 = 50 miles/hour
Time1 = 2.5 hour
Distance1 = Speed1 × Time1 = 50 x 2.5=125 miles
Speed2 = 70 miles/hour
Time2 = 1.5 hour
Distance2 = Speed2 × Time2 = 70×1.5=105 miles
Total Distance = Distance1 + Distance2 =125+105=230 miles

Let the distance be x km.
If travels at 10 kmph, Arun will reach point A at 2 pm.
If travels at 15 kmph, Arun will reach point 12 noon.
=> Time taken when traveling at 10 kmph = Time taken when traveling at 15 kmph + 2 hours ⇒\(\frac{x}{10}=\frac{x}{15}+2⇒\frac{x}{10}-\frac{x}{15}=2⇒3x−2x=2×30\)⇒x=60
⇒Distance = 60 km

If travels at 10 kmph, Arun will reach point A at 2 pm.

⇒Time taken = Distance/ Speed=\( \frac{60}{10} =6 \) hours
He must have started 6 hours back 2 pm, ie, at 8 am
Now he wants to reach at 1 pm. ie; time to be taken = 5 hours

Speed needed = Distance/Time=\( \frac{60}{5}=12 \) kmph.

Time taken by A to cover 400 m

=\(\frac{400}{16}\) = 25 s

B covers (400 — 16) = 384 m in (25 + 40) = 65 s

B's speed = \(\frac{384}{65}\) = 5.9 m/s

Let rate of stream be a km/h.

According to the question,

24 + a = 2(24 - a)

24 + a = 48 - 2a

3a = 48 - 24 = 24

a = 8kmph

Speed downstream = 48/20 = 2.4 kmph

Speed upstream = 48/24 = 2 kmph

Speed of boat in still water = \(\frac{1}{2}(2.4 + 2) = \frac{4.4}{2} = 2.2 kmph\)

Given, speed of boat = 10kmph

let speed of flow of river be x kmph

Upstream speed =(10 - x) kmph

Downstream speed = ( 10 + x) kmph

According to the question,

\(\frac{91}{10 - x}+\frac{91}{10 +x}\) = 20

On solving above equation,

\(x^2 = 9\)

x = 3kmph

Let the speed of Kavya in still water = x

Then, Kavya's speed downstream = (x + 6)

Kavya's speed upstream = (x - 6)

According to the question,

Distance travelled downstream = Distance traveled upstream

=> 12 (x + 6) = 18 (x - 6)

=> 2x + 12 = 3x - 18

=> 3x - 2x = 18 + 12

x = 30 km/h

Here, speed of man in still water = x =12 kmph

Speed of river = 2.4 kmph ; t = 1 hr

Required distance = \(\frac{t(x^2 - y^2)}{2x}\)

= \(\frac{1\times(12^2 - 2.4^2)}{2\times12}\)

= 5.76 km

Speed of Ist person = 4.5 km/h= 1.25 m/s

= Speed of 2nd person = 5.4 km/h = 1.5 m/s

Let the speed of the train be x m/s.

Then, (x —1.25) x 8.4 = (x —1.5) x 8.5

8.4x —10.5 = 8.5x —12.75

0,1x = 2.25

x= 22.5 m/s

Let Binu catches Arun after x h.

Then, distance travelled by Arun in (x + 2) h = Distance travelled by Binu in x h

According to the question,

2 (x + 2) = 5x

2x + 4 = 5x

3x = 4

So distance travelled by binu in x h = \(\frac{4}{3}\) x 5 = \(6\frac{2}{3}\) km

Explanation:

Let the trains meet after x hours, then 60x – 50x =120

=>10x =120

=> x =12hrs

Distance AB =(Distance covered by slow train) + (Distance covered by fast train)

=[(50 x 12)+ (60x12)]km

=(600+720)km

= 1320 km

Explanation:

A's 1 day's work = \( \frac1{15} \)

B's 1 day's work =\( \frac1{20} \)

(A + B)'s 1 day's work =(1/15+1/20)=7/60

(A + B)'s 4 day's work = (7/60) X 4= $\frac{7}{15}$

Therefore, Remaining work = ( 1 -\( \frac7{15} \))

= \( \frac8{15} \)

Relative speed = (\(5.5\) - \(5\)) = \(0.5\) kmph

(because they walk in the same direction)

Distance = \(8.5\) km

time = \(\frac{distance}{speed}\) = \(\frac{8.5}{0.5}\) = \(17\) hr

The distance covered by the train to cross the platform and bridge = Length of the platform + Length of bridge + Distance between platform and bridge + length of the train

= \(165 + 135 + 30+110 = 440\)m

Speed of train = \(\frac{ 110}{3} \)m/s

.•. Required time taken = \(\frac{440}{\frac{110}{3}} = 12 s\)

We know speed= distance/ time

here Distance= Speed x time

= 1100x22/5= 4840 feet

Distance = (AB + BC) = 2x

= 16 x \(\frac{1}{2} = 8 km\)

x= 8/2 = 4 km

Area of the field = 4 x 4 = 16 sq km

To reach the winning post, L will have to cover a distance of (1000 — 280) = 720 m

When L covers 3 m, then M covers 4 m.

When L covers 720 m, then M covers (4/3 x 720) = 960 m

Thus, when L reaches the winning post, M covers 960 m and therefore, remains 40 m behind.

So L wins by 40 m.

Consider actual speed as 100 % , so if he decrease the speed by 40%, then remaining is 60% of original speed . Then he increases the speed by 50 %, ie, the speed increases to 150%.

\(Required \ average \ speed = 75 \times\frac{60}{100}\times\frac{150}{100} \)

= 67.5 km/h

Here we have b = 20 km/hr and w = 5 km/hr

So the downstream speed = b + w = 20 + 5 = 25km/hr

The upstream speed = b – w = 20 - 5 = 15 km/hr

Let they meet after x h.

Then, according to the question,

3x + 4x =17.5

7x = 17.5

\(x = \frac{17.5}{7}=2.5 h\)

So, they meet 2.5 h after 8 : 00 am.

It means they meet at 10 : 30 am.

Required average speed =\(\frac{Total distance covered}{Total time taken}\)

= \(\frac{3\times39}{\frac{39}{26} + \frac{39}{39}+\frac{39}{52}}\)

=\(\frac{3\times13\times12}{13}\) = 36kmph

Here, S1 = 8 km/h

and S2 = 16 km/h

d1 = S1 x t1 = 8t1 ...(i)

and d2 = S2 X t2 = 16t2

We know that, t1 + t2 = 8 ...(iii)

and d1 + d2 = 80 (given) ...(iv)

From Eqs. (i) and (ii) put the value of d1 and d2 in Eq. (iv), we get

d1+ d2 = 80

8t1 + 16t2 = 80

8t1 + 8t2 + 8t2 = 80

8 (t1 + t2) + 8t2 = 80

8 x 8 + 8t2 = 80

From Eq (iii), 8t2 = 80 — 64 = 16

t2 = 2 h

t1 = 8-2 = 6 hrs

d1 = 8 x 6 = 48 km

Let the speed of asok be a kmph and speed of stream be b kmph

Let AB =BC = x km

first condition from the question is,

\(\frac{x}{a + b} + \frac{x}{a - b} = 10\) ...........(i)

From second statement,

\(\frac{2x}{a + b} = 4\)

\(\frac{x}{a + b} = 2\).............(ii)

substituting (ii) in (i),

\(2 + \frac{x}{a - b} = 10\)

\(\frac{x}{a - b} = 8\) ...............(iii)

On dividing eqn. (iii) by eqn. (ii), we get

\(\frac{a + b}{a - b} = \frac{8}{2} = 4\)

a + b = 4a - 4b

3a = 5b

a : b = 5 : 3

The 2nd train leaving Goa starts its journey earlier and it travels = 60 x (3:45 pm - 2:35)

= 60 x \(1\frac{1}{6}\) h = 70 km, when the 1st train (that leaves Manglore ) starts its journey.

Now, both the trains cover (510 - 70) km i.e., 440 km with relative speed (50 + 60) km/h = 110 km/h.

Therefore, the trains meet after \(\frac{440}{110} = 4 h\)

after the 1st train starts at 3:45 pm.

Now, the 1st train covers 4 x 50 km = 200 km to meet the 2nd train.

Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec

Distance = (35 x 4) m = 140 m.

Time taken = 140 x (2/5) sec= 56 sec

4 km downstream is covered in 15 min.

∴ speed of downstream = 16 km/hr.

So speed of upstream = 8km/hr.

Total time taken for downstream journey = 15 min (given).

Now total time taken for upstream journey = 8/8 = 1 hr = 60 min.

Hence total time taken from A to B = 15 + 60 = 75 min.

As average speed = Total distance /total time,

so average speed from A to B = (12/75)*60 = 48/5 = 9.6kmph.

Let x be the speed of man in still water and y be the speed of stream.

∴ Speed of man (x) = 60 km/hr and speed of downstream = 75 km/hr.

∴ Speed of stream = 15 km/hr.

Hence upstream speed = 60 – 15 = 45 km/hr.

So time taken to cover 20 km = 20/45*60 = 26.67min.

speed of the train = \(75 mph = 75\times 1.6 = 120 kmph\)

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)

This is an A.P with first term, a=35, number of terms,

n= 12,d=2.

Required distance

= 12/2[2 x 35+{12-1) x 2]

=6(70+23)

= 552 kms.

Let distance he drove be x. Time = \(\frac{Distance}{Speed}\)

From given condition,

\(\frac{x}{40} -\frac{x}{45} = 1\)

On solving this, \(5x = 40 \times 45\)

x = \(360\) km

Man's rate in still water = (15 - 3) km/hr = 12km/hr.

Man's rate against the current = (12- 3) km/hr = 9 km/hr.

Speed of boat = \(8\) km/hr
Speed of River = \(3\) km/hr
So speed of Downstream = \(11\) km/hr
Speed of Upstream =\(5\) km/hr
Distance traveled = \(x\)
\(\frac{x}{5}=\frac{x}{11}+3\)
\(\frac{6x}{55}=3\)
\(x=27.5\) km

Number of gaps between 41 poles = 40

So the total distance between 41 poles = 40*50

= 2000 meter = 2 km

In 1 minute train is moving 2 km/minute.

Speed in hour = 2*60 = 120 km/hour

Let the two trains meet at X km distance.

Time taken by the first train to cover X km = X/75 hours

Time taken by 2nd train to cover X km = X/60 hours

Now, X/60 – X/75 = 1/2

thus, X =150 km

Here, a= 125 km/h, b = 75 km/h and d= 50 km

According to the formula, Distance between the stations M and N = \((\frac{a +b}{a - b})\times 50\)

=\(\frac{125+75}{125-75}\times50\)

=\(\frac{200}{50}\times50\) = 200 km

Let x is the length of the train in meter and v is its speed in kmph.
\( \frac {x}{9} = ( v-2) \times (\frac{10}{36}) \) ---(1)
\( \frac{x}{10} = ( v-4) \times (\frac{10}{36} )\) --- (2)
Dividing equation 1 with equation 2,
\( \frac {10}{9} =\frac{ (v-2)} {(v-4)} \)
\( \implies 10v - 40 = 9v - 18 \)
\( \implies v = 22 \)
Substituting in equation 1, \( \frac {x}{9} = \frac{200}{36} \) \( \implies x = \frac{9×200}{36} = 50 \) m

After servicing speed of car = 60 km/h.

.'. Distance covered in 6 h = (60 X 6) km = 360 km.

When not serviced, then time taken to cover 360 km= \(\frac{360}{50}\) = 7.2 hrs

Distance travelled in first 2 hours =10 × 2 = 20 km.

Distance travelled in next 1 hour =13 × 1 = 13 km.

Total Distance travelled = 20 + 13 = 33 km . Total time taken = 2 + 1 = 3 hrs.

Average Speed =Total Distance Travelled/Total Time Taken =33/3 = 11 km/hr.

Speed of downstream =\( (22 + 4) = 26 \) km/h
Time = \( 24 \) minutes = \(\frac{24}{60}\) hour = \(\frac{2}{5}\) hour
Distance travelled = Time × speed
= \(\frac{2}{5}×26 = 10.4\)km

37.5 kmph is incorrect as the time traveled is different in both the cases and only the distances are same. Let the distance between the office and Pimpri class be x km.

∴Time taken on my onward journey

= x/30 hours and time taken on my return journey = x/45 .

∴The total time taken for my onward and return journey = x/30 + x/45 = 5x/90 hours.

The total distance traveled both ways = 2x km

∴ Average speed = 2x/(5x/90) = 36 kmph

40 x 15/60 = 10 km.

The important point to note is that time given was in minutes, whereas the speed was in kmph.

Therefore, either speed will have to be expressed as km/min or time will have to be expressed in hours to apply the relationship.

In this case we converted time into hours to get the answer.

Conversely, converting speed into km/min, we get 40 kmph = 40/60 km/min = 2/3 km/min.

Therefore, distance traveled = 15 x 2/3 = 10 km.

Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours = Distance covered by express train in 4 hours

∴ 10x = 4×90 or x = 36.

So, speed of goods train = 36 kmph.

Speed of train1 = \(\frac{120}{10} = 12 m/s\)

Speed of train2 = \(\frac{120}{15} = 8 m/s\)

If they travel in opposite direction, relative speed = \(12+8 = 20 m/s\)

Distance covered = \(120+120 = 240 m\)

Time = \(\frac{distance}{speed} = \frac{240}{20} = 12 s\)

Let in x hours,he move 4 km downstream.
Speed downstream =\(\frac{4}{x}\) km/hr
Speed upstream = \(\frac{3}{x}\) km/hr

\(\frac{48}{\frac{4}{x}}+\frac{48}{\frac{3}{x}}\) =14
X=1/2
Speed downstream = 8 km/hr,
Speed upstream = 6 km/hr.

Rate of the stream =\(\frac{8-6}{2}\)=1 km/hr

Time taken to cover first 75km =\(\frac{distance\ travelled }{ Speed}\) = 75/25 = 3h

Time taken to cover next 25 km =\(\frac{distance\ travelled }{ Speed}\)= 25/ 5= 5 h

Time taken to cover last 50 km of its journey =\(\frac{distance\ travelled }{ Speed}\)= 50/25 = 2 h

Total distance travelled = 75 + 25 + 50 = 150 km

Total time taken = 3 + 5 + 2 = 10 h

Required average speed =\(\frac{Total distance}{Total time taken}\) = 150/10 = 15 kmph

Average speed of train = 240/12 = 20 km/h

Average speed of bus = \(\frac{3}{4}\times 20 = 3 \times 5 = 15 km/h\)

Required distance = 15 x 7 = 105 km

Distance = Speed x Time

=\( 96 \times \frac{100}{60}\) = 160 km

New time = \(\frac{80}{60}\) h = \(\frac{4}{3}\) h

As speed is inversely proportional to time,

New speed = 160 x \(\frac{3}{4}\) = 40 x 3 = 120kmph

We can safely assume that the policeman is running in the same direction as the thief.

Speed of policeman w.r.t thief = (10 – 8) = 2 km/hr.

Time taken by policeman to cover the 100m distance between him and the thief =

(100/1000) / 2 = \( \frac1{20} \) hr.

Therefore, the distance covered by thief in \( \frac1{20} \) hr.= 8 × \( \frac1{20} \) hr.

= \( \frac25 \)km = 400 meters.

• If A runs 600 m, then B runs 570 m. If A runs 400 m, B then runs

• \((\frac{570}{600}\times400)\) m = 380 m

• When B runs 500 m, then C runs 475 m

• When B runs 380 m, then C runs \((\frac{475}{500})\times380\) m = 361 m

• A beats C by (400 — 361) = 39 m

• Hence, option C is the correct answer.

• Let the speed of the two trains be 7x and 8x.

• Then, 8x = 400 / 4

⇒ 8x = 100 ⇒ x = 12.5 km/h.

• Hence, speed of the first train = 7x

= 7 × 12.5 = 87.5 km/h.

Speed of the truck = \(\frac{550}{1000}\times60=33 kmph\)
Speed of the train = \(\frac{33}{45}\times60=44 kmph \)
Ratio of their speeds \(= 33 : 44 = 3 : 4\)

Usual time taken = 50 minutes = 5/6 hours.

The distance = 60 km.

Usual Speed = Distance ÷ Usual Time → 60/(5/6) = 72 kmph.

Speed on this occasion = Distance ÷ Time on this occasion

= 60/1 = 60 kmph.

The ratio between the usual speed to the speed on this occasion = 72/60 = 6/5

The ratio of the usual time taken to the time taken on this occasion = 50min/60min = 5/6 .

• Here, a = 15 km/h, b = 5 km/ h,

  t1 = 20 min and t2 = 5 min

• Since the answer is given in km , we have to change the time in to hours

• Required distance = (t1 — t2) (a + b) \(\frac{a}{b}\)

  = \(\frac{20 - 5 }{60}(15 + 5)\frac{15}{5}\)

  = \(\frac{15\times20\times3}{60}\)

  = 15 km

In 1 hour Peter covers 5 km and Beckon covers 6 km.

So, they are 1 km apart after 1 hour.

Therefore, they are 17 km apart after 17 hours