### Discussion :: Time and Distance - General Questions (Q.No.12)

Arunkumar said: (Jul 2, 2010) | |

How to find that 2. We don't know the starting time. Then how that is calculated. |

Xyz said: (Jul 16, 2010) | |

Its 2 hours 12 noon to 2pm = 2 hours |

Das said: (Aug 22, 2010) | |

Hi, could u tell me why we have done 60/5 in last step ? |

Ashish said: (Aug 29, 2010) | |

Actually 5 comes as: Robert start travelling at 8 am. Now difference between 8 am to 1 pm is 5 hrs. So that 60/5. |

Mohamed said: (Nov 28, 2010) | |

This is another way of doing it: In the last step what you can do instead is (60 / S) - (60 / 15) = 1 following the same logic as the predecessor example while S is the speed that will be arriving @ 1pm. This will bring us to (900 - 60S) / 15S = 1 -> multiply this two sides to 15S, this will reap 15S = 900 - 60S. then 75S = 900, S = 12 km/hr. |

Santhosh said: (Jan 19, 2011) | |

How to solve the equation first? |

Aamir said: (Feb 22, 2011) | |

We can also solve through avearge speed its easy .x=10,y=15 2xy/x+y 2*10*15/10+15= 300/25=12 |

Vyshu said: (Apr 8, 2011) | |

Why we are subtracting here i.e., x/10-x/15=2;. And in the next problem why we are adding. Can anyone explain when should add and when should subtract. |

Pardhu said: (May 3, 2011) | |

Thank you aamir. Is this applicable for all these type methods. ? |

Srinu said: (Jun 23, 2011) | |

Hi vyshu 2 is time between 12noon and 1pm why subtracting 12noon - 2 pm = 2 it is time x/15-x/10=2 x=60km 12noon time =60/15=4hrs(before) mean 8am find speed 1pm----------8am to 1pm time is 5 hrs ,distance 60km,speed =? speed=60/5= 12kmph |

Raja said: (Aug 1, 2011) | |

Simple one 2pm-12=2 hour Similarly 15km-10km=5km 5km/2hour=2.5km/h initial 10+2.5=12.5 nearer to 12 |

Venkat said: (Aug 5, 2011) | |

Thank you Raja. |

Vishu said: (Sep 19, 2011) | |

x/10-x/15=2 =>3x-2x=60 I didn't get the proper method. How it comes anyone could tell me briefly it will be helpful for me. |

Mano said: (Oct 28, 2011) | |

I agree to vishu comment. How that step came ? |

Anshu said: (Mar 17, 2012) | |

By, speed 1 (10 kmph) he would reach at point A at 2 P.M. speed 2 (15 kmph) he would reach at point A at 12 noon. we know that speed = distance time ^ -1 let the distanc be x then x/10 - x/15 = 2 3x - 2x /30 = 2 x/30 = 2 x = 60 the distance is 60 km @the speed of 10 kmph he would require 10 hours to reach point A. therefore he left his point of starting 6 hours ago. time was 14:00 hours - 6 hours = 8 hours. or 8 A.M. total between 6 a.m. and 08:00 hours - 13:00 hours = 05:00 hours time = 05:00 hours distance = 60 km [ we have already found out] speed = ? speed = distance time^-1 60 / 5 - 12 therefore, the speed = 12 kmph. |

Teju said: (Mar 19, 2012) | |

Can any one give me the clear explanation? please. |

Ankit said: (May 23, 2012) | |

Hi, Suppose Robert has started traveling on his cycle = x hr Suppose distance to be covered is = y km case 1:- speed = 10 kmph time to reach point A = 2 pm or 14 (no am or pm : just standard time) hence distance = speed * time y = 10 (14 - x) -- eq 1 case 2 :- speed = 15 kmph time to reach point A = 12 noon or 12 (no am or pm : just standard time) hence distance = speed * time y = 15 (12 - x) -- eq 2 since distance traveled is the same hence equating eq1 and eq2. 10(14 - x) = 15(12 -x) hence solving 2(14 - x) = 3(12 -x) 28 - 2x = 36 -3x x = 8 hence Robert started at 8 hrs i.e 8 AM. Now equating the value of x in eq1 or eq2 to get the value of distance travelled. y = 10(14 -8) hence y = 60 and hence distance = 60 km. Now, time to reach at A : - 1 pm or 13 hrs speed = z kmph (??) distance = 60 km 60 = z(13 -8) 60 = 5z z = 60/5 hence z = 12 or speed = 12 kmph. Hope this helps. Thanks |

Saba Azmat said: (Jun 9, 2012) | |

We can also solve dis que by this method, (2*10*15)/(10+15)=average speed in covering same distance =12 kmph |

Rahul Mathur said: (Jan 19, 2013) | |

Take 2 pm as a reference time say--- t. Let the distance to be covered be--- d. Case I d/t=10. Case II d/t-2=15. Solving both equations we have t=10 and d=60. To find d/t-1=? Replacing values of d and t will give speed as 12kmph. |

Aarti said: (Feb 10, 2013) | |

Nice method @Rahul Mathur. The only thing is you have calculated value of t=10 wrongly. It comes out to be 6. Check out. |

Harsha said: (Mar 17, 2013) | |

Speed1=10km/hr. Reach at=2 p.m. Speed2=15km/hr. Reach at = 12 a.m. Let dist. be="x"km. Now, By formula dist/speed= time. x/10-x/15=2-12 3x-2x = 60 x = 60km. Now time taken when travelled with speed 15 km/hr. 15=60/t t=4 hr. i.e started at 8am Then 60/5=x x=12 kmph Answer. |

Tbabe said: (Jul 2, 2013) | |

Hi all, it was not specified that he traveled at the same distance? Am still confused with the solutions. |

Abhi said: (Aug 11, 2013) | |

It travelled the same distance. Let total time taken in first case is t. Therefore, Distance = 10*t..........{1st case}. = 15*(t-2) .....{2nd case}. = x*(t-1).......{3rd case} let x be the speed. Now,equating 1st case and 2nd case as both are the same distance. 10*t=15(t-2). => 10t=15t-30. => 5t=30. => t=6. So distance = 10*6=60. Therefore 60=x*(6-1)......{3rd case}. =>60=5x. =>x=12 kmph. |

Hyndavi said: (Aug 15, 2013) | |

Easy method: Better take an average speed. avg speed = 2xy/x+y. x is 1st speed and y is 2nd speed. So. here if we c...1pm is between 12pm and 2pm. Calculating avg speed: 2xy/x+y = (2*10*15)/(10+15). = 300/25. = 12km/hr. |

Divya said: (Aug 16, 2013) | |

Lets try a hit and trial method : By increasing his speed by 5 km/hr (15-10) , he is saving 2 hours (2pm-12pm) , so when he reaches at 1 pm he increased speed by 2.5 (half of 5) = 12.5, from the options 12 is closest so that must be the answer :). |

Ajit said: (Oct 13, 2013) | |

We need to calculate the distance, then we can calculate the time and finally speed . Lets solve this, Let the distance travelled by x km. Time = Distance/Speed. x10-15 = 2[because, 2 pm - 12 noon = 2 hours] 3x-2x=60 =>x=60. Time = Distance*Speed, Time@10km/hr = 60/10 = 6 hours So 2 P.M. - 6 = 8 A.M i.e. Robert starts at 8 A.M. He have to reach at 1 P.M. i.e,diff between 8 A.M to 1P.M in 5 hours So, Speed = distance/time => 60/5 = 12 km/hr |

Leelakrishna said: (Oct 17, 2013) | |

Since the distance traveled to be same equate d = v*t. i.e. 10*(14-t) = 15*(12-t). From above eq find t i.e. t = 8. Substitute in 10*(14-8) = 60km. So velocity = 60/5 =12kmph. |

Sidhu said: (Oct 19, 2013) | |

It can be solve by using 2xy/x+y. |

Pinky said: (Oct 24, 2013) | |

When came to know that on where we have to use this avg formula ? Please anyone answer. |

Sree said: (Jan 8, 2014) | |

Average formula can be used when a person travels two equal distances at different speeds. |

Sree said: (Jan 8, 2014) | |

For example : A man travels 20 km at a speed of 15 km/hr and returns back at a speed of 10 km/hr. What is the average speed of the man in his journey ? Sol: Speed = 2*15*10/15+10. = 300/25. =12 km/hr. I Think now you came to know where we can use the average formula. |

Shilpa said: (Jan 22, 2014) | |

Why we are subtracting here i.e., x/10-x/15=2;? And in the next problem why we are adding. Can anyone explain when should add and when should subtract. |

Kishore said: (Feb 5, 2014) | |

@Shilpa. We have to go with subtraction in case the objects moving in same direction and go with addition in case the objects moving in opposite direction. |

Shashank said: (May 6, 2014) | |

We can use net speed formula = 2xy/x+y. Because the distance travelled is same, So => 2*10*15/(10+15) = 12. |

Mohit said: (Sep 29, 2014) | |

Let time at 12noon be x hours. Time at 2pm will be x+2. Speed = distance*time. So, 15*x = 10*(x+2). x = 20/5 = 4km. Distance = 15*4 = 60km. Speed at 1pm = 60/5 = 12kmph. |

Karthika said: (Oct 30, 2014) | |

Same distance but with different speeds. The difference in time is 2 hours. Let the distance be x. X/10-x/15 = 2. => x = 60 kms. Calculate time in first case i.e. 60/10 = 6 hours. => he took 6 hours to reach the destination. The question is at 1 p.m => speed to reach destination in 5 hours. We know that speed is inversely proportional to time. s1/s2 = t2/t1. 10/x = 5/6. => x = 12 kmph. |

Nancy said: (Nov 1, 2014) | |

Yes, but why we are considering only 10 km/hr speed? Why not 15 km/hr? |

Ammad Arshad said: (Nov 29, 2014) | |

10km/h = d/x; -- (1); 15km /h = d/(x-2);--- (2); By comparing both equation 1 and 2 we get total time. 15(x-2) = 10x; 15x -30 = 10x; 5x = 30; x = 6h; Put the value of x in equation 1 we got distance. d = 60km. As from the state we know that. y = d/(x-1); -- (3); Put the value of d and x in equation 3. y = 60/(6-1). y = 60/5. y = 12 kmph Answer. |

Geeta said: (Jan 16, 2015) | |

Simple. 10 = d/t. d = 10t. 15 = d/t-2. 15 = 10t/t-2. 15t-30 = 10t. 15t-10t = 30. 5t = 30. t = 6. S = d/t-1. S = 10t/t-1 substitute from above. S = 10*6/6-1. S = 60/5. S = 12 km/h. |

Harish said: (Mar 31, 2015) | |

At 15 kmph ---> 2 p.m. _?_kmph ---> 1 p.m. 10 kmph ---> 12 a.m. So in between 12 am and 2 pm, 1 pm lies in between. (10+15)/2 = 12.5 which is nearer to 12. |

Param said: (May 4, 2015) | |

Distance is same in both the cases: So :: let's the time taken to reach at 2 pm be t. And then time taken to reach at 12 pm will be t-2(2-12 = 2 hours). As distance is same: t*10 = t-2*15. Giving t = 6 hours (to reach at 2, time taken is 6 hours). Total distance = 6*10 = 60 KM. Now to reach at 1 PM speed = d/t. s = 60/t-1 = 60/6-1 = 12. Why 6-1? Because (2 pm - 1 pm = 1 hour and we calculated t as 6). Hope this helps. |

Tarun said: (May 9, 2015) | |

Let the traveled distance X. Then X/10-X/15 = 2. X = 60 kmph. Time taken to travel 60 km at 10 kmph is 6 hours. So the Mr stylish Yankee Robert Peterson a billionaire person started 6 hours before 2 pm. i.e 8 pm. Speed of Mr equals to 60/(8 pm-1 pm) = 12 kmph. |

Bharat said: (Jul 3, 2015) | |

x/10-x/15=2. How did 60 comes? |

Kuru said: (Aug 10, 2015) | |

@Amir that was nice, and easy step to solve. Thanks. |

Marco said: (Apr 11, 2016) | |

The average speed : 2*10*15/10 + 15. = 300/25. = 12 km/h. |

K.Naveena said: (Jul 28, 2016) | |

2xy/(x + y) = [(2 * 10 * 15)/(10 + 15)] = 12kmph. |

Naveen said: (Jul 31, 2016) | |

Can anyone explain in easy way? It is hard to understand. |

Vinaykumar said: (Aug 25, 2016) | |

Simple :. 2 speeds calculate average speed 2xy/(x + y). So, 2 * 10 * 15/(10 + 15). 12km/hr. |

Suma said: (Aug 31, 2016) | |

If two distance is equal, here two distance is equal. Then speed = 2s1s2/s1 + s2. 2 * 10 * 15/(10 + 15) = 12. |

Shahul said: (Sep 2, 2016) | |

1st case-. Speed = 10. Time = t. Distance = 10 * t -----> (1). 2nd case. Speed = 15. Time = t - 2 (he reach at 2pm at t time so take t-2 for 12pm). D = 15 * (t - 2) -----> (2). From 1 and 2. 10 * t = 15 * (t - 2). 15t - 10t = 30. t = 6. So he need 5hrs to reach at 1pm (take 6hrs to reach 2pm and will take 5hrs to reach at 1). D = 10 * t = 10 * 6 = 60. Speed = 60/5 = 12. |

Ashish said: (Sep 5, 2016) | |

Let the distance be covered 'd'. Case I: d/t = 10. Case II: d/t - 2 = 15. Solving both equations we have t = 10 and d = 60. To find d/(t-1) = ? Replacing values of 'd' and 't' will give speed as 12 kmph. |

Neha said: (Dec 3, 2016) | |

We can solve this problem by using avg.speed 2xy/x+y. 2 * (10 + 15)/10 + 15, =12 km/h. |

Prawesh Pradhan said: (Mar 18, 2017) | |

After finding out the distance. 60/10 - 60/x = 1, 5/x = 1, x = 5. |

Prawesh Pradhan said: (Mar 18, 2017) | |

After finding the distance. 60/10-60/x = 1, 6x-60 = x, 5x = 60, x = 12. |

Aarusi said: (Apr 18, 2017) | |

It can be solved easily by applying the average speed formula: x = 10(given)(speed) y = 15(given)(speed) For average speed: 2xy/x + y 2 * 10 * 15/10 + 15 = 12. |

Sagar said: (May 19, 2017) | |

8 am to 1 pm= 5 hours. Distance:-60 and time:-5. Therefore: 60/5 is taken at last step. i.e: speed:-distance/ time. |

Ranjith said: (Jul 3, 2017) | |

How you are saying starting 8am @Ashish. |

Bikash Mahato said: (Aug 10, 2017) | |

Let the distance has been travelled be x km. At the speed of 10 kmph, it will reach 2pm.(equation 1.) At the speed of 15, it will reach at 12 noon.(equation 2) And, At the speed of ?, it will reach at 1pm. (equation 3) Let's balance the equation.1 and 2, x/10-x/15=2 hours, i.e. x=60km. Now again balance the equation 2 and 3, 60/?-60/15=1 hours, ?=12, Thus, 12 kmph is the answer. |

Tirtha said: (Aug 26, 2017) | |

If speed is 10 then he will reach at .2. If speed 15 then he will reach at 12. so for 1 pm then we can take avg (10+15)/2=12.5 aprox 12. Is this correct? |

Elanagai said: (Dec 22, 2017) | |

How did 60/10 came in the last? Can anyone explain after that? |

Rinn Sailo said: (Feb 8, 2018) | |

Just take average speed. Required speed. = 2xy/(x+y), =(2 x 10 x 15)/(10+15), = 300/25. = 12Km/hr. That's it. |

Ishwarya said: (May 15, 2018) | |

What is the logical and easiest method for solving this problem? |

Abishek Jesuraj said: (Jul 28, 2018) | |

Well I would make it simple !! first, let us find the time (in hrs) let t be the time distance/speed = time, distance we don't know so take it as x. here there are two references given with respect to time. he once reached at 2pm which can be taken as t . he again reached at 12pm so that can be taken as t - 2. So now there are two times , (t) and (t-2). substitute this in formula (above distance\speed = time). x/10 - x/15 = (t) - (t-2) (difference of two times, two distances etc), now, (15x - 10x) / 150 = 2, 5x = 300, x=60km. Now we have found the distance travelled Again substitute this in formula speed = d/t. 1st case 10= 60 / t = 6hrs. 2nd case 15 = 60 / t = 4hrs so if in the frst case he has travelled for 6 hours and reached at 2pm. so he would have started at 8am. Now the main problem, at what speed he must travel so that he can reach at 1pm? he starts at 8 am and he should reach at 1 pm so his travel must be 5 hrs. Now substitute 5hrs in the formula speed =d/t. speed = 60/5, speed = 12 kmph. |

Raman Sharma said: (Aug 26, 2018) | |

Distance = speed * time. For time ,let say time taken to travel to point A = t. d = 15*t ------- eqn.1 d = 10(t+2) ------eqn. 2 d = s* (t+1)--------- eqn. 3. From eqn 1 and eqn 2 15* t = 10(t+2). after solving, t = 4 means time to travel is 4 hours. now, we put value of t = 4 in eqn 3 and eqn. 2. 60 = s(4+1). s = 12 kmph. |

Bhushan Suriya said: (Oct 29, 2018) | |

Consider starting time as x. We know that time = (distance/speed), When he reach at 2 P.M i.e.,14.00 (railway time) (14-x) = (distance/ 10) When he reach at 12 P.M (12-x) = (distance/15) As we know that., distance is the same. So, from above equations we get x=8. so he started cycling at 8 A.M. So from this, we can calculate distance = 60 km. To reach at 1 P.M, the time taken is 5 hrs. Speed = (60/5) = 12 kmph. |

Subash said: (Jan 3, 2019) | |

Thanks for the answer @Raja. |

Pavan Sai said: (Jan 10, 2019) | |

It's very easy by using relative speed concept there are two things up there. So, now by using the formula 2xy/x+y. where x=10 and y=15, => 2*10*15/10+15, =>300/25, =>12, So the answer is 12kmph. |

Shivasai said: (Mar 5, 2019) | |

Then, x/10 - x/15 = 2 why we have taken this? What is the logic behind it |

Abhi said: (Mar 21, 2019) | |

How 60/5 comes, does any one please tell me? |

Muhammad Ali said: (Apr 5, 2019) | |

Why don't you just find the mean value? With 10kmph reach at 2 p.m. Witn 15kmph reach at 12noon. Find mean of both for 1p.m required time. So, 10+15 = 25/2 = 12.5kmph should be the correct answer. |

Ram Sai said: (May 24, 2019) | |

Thanks for the answer @Ashish. |

Noorandeous said: (Oct 11, 2019) | |

6 hours before 2 PM was not given how you added it? |

Erigela Vani said: (Jan 3, 2020) | |

@Shivasai. I think here the difference between the two times I. E12noon and 2pm is 2, and also time =d÷s, so we should take distance as "X" and speeds which they are given so x÷10-x÷15=2. |

Vani said: (Jan 3, 2020) | |

@Bhushan suriya. How 5 hours will get at the last at 1:00pm? |

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