Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 11)
11.
In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
Answer: Option
Explanation:
Let Abhay's speed be x km/hr.
Then, | 30 | - | 30 | = 3 |
x | 2x |
6x = 30
x = 5 km/hr.
Discussion:
209 comments Page 4 of 21.
Bhavana said:
1 decade ago
Good explanation by rahul and amith.
Srikara said:
1 decade ago
Good Explanation Devi.. Thanx.. :)
Harish said:
1 decade ago
thanks
Saranya said:
1 decade ago
Please explain how 3 comes shortly?
Suresh said:
1 decade ago
Superb explanation devi, anoo, amit sakshi.
VIMAL said:
1 decade ago
Lets abhays speed x km/hr and sameer takes T hour to complete 30 km journey
X=30/T+2
2X= 30/T-1
THEN 2(30/T+2)=30/T-1
2T-2=T+2
T=4 hour
X= 30/T+2
30/4+2= 5 HOURS
X=30/T+2
2X= 30/T-1
THEN 2(30/T+2)=30/T-1
2T-2=T+2
T=4 hour
X= 30/T+2
30/4+2= 5 HOURS
Shro said:
1 decade ago
Thanks Abdus :).
Praveen said:
1 decade ago
Thanks Devi
Ankit said:
1 decade ago
Good explanation Devi.
Another Method.
Let Abhay's speed :- x kmph
Let Sameer's speed:- y kmph
According to the question
30/x - 30/y = 2 -- 1st equation
&
30/y - 30/2x = 1 -- 2nd equation
Solving above two we get:-
(y-x)/xy = 1/15 --3rd equation
&
(2x -y)/xy = 1/15 -- 4th equation
since the rhs of both the equations is equal equating the lhs of the both equations is also correct: so
y -x = 2x - y
i.e
2y = 3x or y = 3x/2
now equate the value of y in 3rd or the 4th equation as follows:-
(2x - (3/2)x)/(x*(3/2)x) = 1/15
we get
x/3x^2 = 1/15
so
1/x = 1/15
so
x = 5 km/hr
Hope this helps.
Another Method.
Let Abhay's speed :- x kmph
Let Sameer's speed:- y kmph
According to the question
30/x - 30/y = 2 -- 1st equation
&
30/y - 30/2x = 1 -- 2nd equation
Solving above two we get:-
(y-x)/xy = 1/15 --3rd equation
&
(2x -y)/xy = 1/15 -- 4th equation
since the rhs of both the equations is equal equating the lhs of the both equations is also correct: so
y -x = 2x - y
i.e
2y = 3x or y = 3x/2
now equate the value of y in 3rd or the 4th equation as follows:-
(2x - (3/2)x)/(x*(3/2)x) = 1/15
we get
x/3x^2 = 1/15
so
1/x = 1/15
so
x = 5 km/hr
Hope this helps.
K Powar said:
1 decade ago
The solution may be found with 2 variables, explained nicely above with x and y.
However, the above solution is done with 1 variable only.
The figure of 3 comes from the total savings in hrs if the speed is doubled.
Assuming Abhay's speed is = x ; time taken is 2 hrs more
If Abhay's speed is doubled = 2x ; time saved is 1 hrs more
Therefore difference in speed results in total time savings = 3 hrs (2 hrs + 1 hr)
Now, to calculate the two speeds:
Given distance = 30 km
Assuming Abhay's speed = x
Therefore Time1 = 30/x
Abhay's speed is doubled = 2x
Therefore Time2 = 30/2x
Since difference in time is the total time saved :
Hence,
Time1 - Time2 = 3 hrs
30/x - 30/2x = 30
Solving :
6x = 30
x = 5 kmph
However, the above solution is done with 1 variable only.
The figure of 3 comes from the total savings in hrs if the speed is doubled.
Assuming Abhay's speed is = x ; time taken is 2 hrs more
If Abhay's speed is doubled = 2x ; time saved is 1 hrs more
Therefore difference in speed results in total time savings = 3 hrs (2 hrs + 1 hr)
Now, to calculate the two speeds:
Given distance = 30 km
Assuming Abhay's speed = x
Therefore Time1 = 30/x
Abhay's speed is doubled = 2x
Therefore Time2 = 30/2x
Since difference in time is the total time saved :
Hence,
Time1 - Time2 = 3 hrs
30/x - 30/2x = 30
Solving :
6x = 30
x = 5 kmph
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