Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 12)
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.?
Answer: Option
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 |
Discussion:
93 comments Page 2 of 10.
Shahul said:
9 years ago
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.
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.
Ammad Arshad said:
1 decade ago
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.
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.
Mohamed said:
1 decade ago
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.
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.
Raman Sharma said:
7 years ago
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.
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.
Ishwar said:
5 years ago
Let calculate distance firs say d.
D/10-d/15=2 (he reach 12noon 2hr early).
By calculating we get d = 60km.
Let's speed is S to calculate S difference between travel at first speed rate and S rate is.
60/10-60/S=1 (because he reaches by 1 am, 1hr early).
By calculating this we get.
5S=60.
S = 12kmph.
D/10-d/15=2 (he reach 12noon 2hr early).
By calculating we get d = 60km.
Let's speed is S to calculate S difference between travel at first speed rate and S rate is.
60/10-60/S=1 (because he reaches by 1 am, 1hr early).
By calculating this we get.
5S=60.
S = 12kmph.
Prerna23 said:
5 years ago
Time reached -- time taken -- Speed
2 p.m. -- t -- 10 km/hr
12 noon -- t-2 -- 15 km/hr
1 p.m. -- t-1 -- ?.
Distance is constant.
Writing in terms of time and speed, t x 10 = (t-2) x 15 which gives t = 6.
Distance is 60 km, so speed required to reach in t-1= 5 hrs i.e. at 1 p.m. is 60/5 = 12 km/hr.
2 p.m. -- t -- 10 km/hr
12 noon -- t-2 -- 15 km/hr
1 p.m. -- t-1 -- ?.
Distance is constant.
Writing in terms of time and speed, t x 10 = (t-2) x 15 which gives t = 6.
Distance is 60 km, so speed required to reach in t-1= 5 hrs i.e. at 1 p.m. is 60/5 = 12 km/hr.
(7)
Harsha said:
1 decade ago
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.
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.
Sree said:
1 decade ago
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.
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.
Tarun said:
1 decade ago
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.
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.
Srinu said:
1 decade ago
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
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
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