"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
Answer: Option B
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.
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.
3x - 2x = 60
hrs
Required speed =