Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing

# Aptitude - Time and Distance - Discussion

@ : Home > Aptitude > Time and Distance > General Questions - Discussion

"The secret to creativity is knowing how to hide your sources."
- Albert Einstein
12.

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

 [A]. 8 kmph [B]. 11 kmph [C]. 12 kmph [D]. 14 kmph

Explanation:

Let the distance travelled by x km.

 Then, x - x = 2 10 15

3x - 2x = 60

x = 60 km.

 Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs. 10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

 Required speed = 60 kmph. = 12 kmph. 5

 Arunkumar said: (Fri, Jul 2, 2010 10:22:15 AM) How to find that 2. We don't know the starting time. Then how that is calculated.

 Xyz said: (Fri, Jul 16, 2010 02:49:16 PM) Its 2 hours 12 noon to 2pm = 2 hours

 Das said: (Sun, Aug 22, 2010 09:34:43 AM) Hi, could u tell me why we have done 60/5 in last step ?

 Ashish said: (Sun, Aug 29, 2010 06:53:12 AM) 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: (Sun, Nov 28, 2010 02:41:19 AM) 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: (Wed, Jan 19, 2011 12:51:25 AM) How to solve the equation first?

 Aamir said: (Tue, Feb 22, 2011 02:12:52 AM) 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: (Fri, Apr 8, 2011 09:27:04 AM) 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: (Tue, May 3, 2011 01:31:14 AM) Thank you aamir. Is this applicable for all these type methods. ?

 Srinu said: (Thu, Jun 23, 2011 09:39:07 AM) 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: (Mon, Aug 1, 2011 09:41:49 PM) 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: (Fri, Aug 5, 2011 11:27:48 AM) Thank you Raja.

 Vishu said: (Mon, Sep 19, 2011 01:24:49 PM) 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: (Fri, Oct 28, 2011 08:06:08 PM) I agree to vishu comment. How that step came ?

 Anshu said: (Sat, Mar 17, 2012 02:40:59 PM) 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: (Mon, Mar 19, 2012 09:51:47 PM) Can any one give me the clear explanation? please.

 Ankit said: (Wed, May 23, 2012 09:22:56 AM) 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: (Sat, Jun 9, 2012 12:18:39 PM) 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: (Sat, Jan 19, 2013 01:31:03 PM) 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: (Sun, Feb 10, 2013 10:33:26 PM) 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: (Sun, Mar 17, 2013 02:36:16 PM) 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.