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 7 of 21.
Prakash said:
1 decade ago
Its given that Abhay takes 2 hours more than Sameer. Then it should be x+2 instead of 2x. Am I right?
Hrisheek said:
1 decade ago
Where does 30/2x come from?
Arghyadeep Debnath said:
1 decade ago
Let, @Sameer takes x hour. So, Abhay will take (x+2)hour.
Speed = 30.2/(x+2). Now according to the Q. 30.2/(x+2) = 30/x-1
Here we get x = 4. So abhay takes (x+2). is equal to (4+2) = 6.
So, speed of Abhay is 30/6 = 5km/hr.
Speed = 30.2/(x+2). Now according to the Q. 30.2/(x+2) = 30/x-1
Here we get x = 4. So abhay takes (x+2). is equal to (4+2) = 6.
So, speed of Abhay is 30/6 = 5km/hr.
Bizu said:
1 decade ago
A person covers a particular distance with a speed x in time t1 and in his second trip if he covers the same distance with speed y at time t2. Shortcut formula:
Distance covered= time gap X product of speed/difference in speed.
In this case: (t1+t2)X(xy)/(y-x).
Distance covered= time gap X product of speed/difference in speed.
In this case: (t1+t2)X(xy)/(y-x).
GIS said:
1 decade ago
Distance = 30.
Abhays normal speed = x.
Abhay doubles speed = 2x.
Sameers time = t.
t + 2 = 2 hrs + sameers time.
t - 1 = 1 hr less than sameers time.
[30/x] - [30/2x] = [t+2] - [t-1].
[30/x] - [30/2x] = 3.
[1/x] - [1/2x] = 1/10.
(2-1)/2x = 1/10.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.
Abhays normal speed = x.
Abhay doubles speed = 2x.
Sameers time = t.
t + 2 = 2 hrs + sameers time.
t - 1 = 1 hr less than sameers time.
[30/x] - [30/2x] = [t+2] - [t-1].
[30/x] - [30/2x] = 3.
[1/x] - [1/2x] = 1/10.
(2-1)/2x = 1/10.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.
Lipu said:
1 decade ago
Let speed of Abhay's be X kmph.
And time taken by Sameer be Y.
So time taken by Abhay's is Y+2.
So 30/X = Y+2.
Again speed doubles means 2X.
So 30/2X = Y-1.
By subtraction of two equation we get.
30/X - 30/2X = (Y+2) - (Y-1) = 3.
=> X=5.
And time taken by Sameer be Y.
So time taken by Abhay's is Y+2.
So 30/X = Y+2.
Again speed doubles means 2X.
So 30/2X = Y-1.
By subtraction of two equation we get.
30/X - 30/2X = (Y+2) - (Y-1) = 3.
=> X=5.
SATISH said:
1 decade ago
> LET THE TIME TAKEN BY SAMEER = T.
> LET THE SPEED OF ABHAY = X.
TIME = DIST/SPEED.
>Abhay takes 2 hours more than,
Sameer TO COVER 30 KM = T+2 = 30/X.
>Abhay doubles his speed, then he Would take 1 hour less,
Than Sameer TO COVER 30 KM = T-1 = 30/2X.
NOW CROSS MULTIPLY THE BELO EQUATIONS:
T+2 = 30/X.
T-1 = 30/2X.
(T+2)30/2X = (T-1)30/X.
(2TX+4X+30)/2X = (TX-X+30)/X.
(2TX+4X+30)/2 = (TX-X+30).
2TX+4X+30 = 2TX-2X+60.
2TX+4X+30-2TX+2X-60=0.
6X-30=0 => 6X=30 .
Ans) X = 5KMPH.
> LET THE SPEED OF ABHAY = X.
TIME = DIST/SPEED.
>Abhay takes 2 hours more than,
Sameer TO COVER 30 KM = T+2 = 30/X.
>Abhay doubles his speed, then he Would take 1 hour less,
Than Sameer TO COVER 30 KM = T-1 = 30/2X.
NOW CROSS MULTIPLY THE BELO EQUATIONS:
T+2 = 30/X.
T-1 = 30/2X.
(T+2)30/2X = (T-1)30/X.
(2TX+4X+30)/2X = (TX-X+30)/X.
(2TX+4X+30)/2 = (TX-X+30).
2TX+4X+30 = 2TX-2X+60.
2TX+4X+30-2TX+2X-60=0.
6X-30=0 => 6X=30 .
Ans) X = 5KMPH.
Azam said:
1 decade ago
How 3 comes is the confusion to many of us.
Say, in 1st case- Sameer reaches at 10:00 Hrs, so Abhay must reach at 12:00 Hrs.
In 2nd case-Sameer reaches at 10:00 Hrs but Abhay reaches at 09:00 Hrs.
So, Abhay's time difference is 3 hrs due to change of speed.
Say, in 1st case- Sameer reaches at 10:00 Hrs, so Abhay must reach at 12:00 Hrs.
In 2nd case-Sameer reaches at 10:00 Hrs but Abhay reaches at 09:00 Hrs.
So, Abhay's time difference is 3 hrs due to change of speed.
Jay said:
1 decade ago
Lets take distance = 30.
Speed=S;.
Time=T;.
D = S*T;.
30 = S*T;.
HERE,
30 = S*(T+2).
30 = ST+2S------------------1.
Next equation,
30 = 2S*(T-1).
30 = 2ST-2S-------------------2.
TO SOLVE 1 AND 2 EQUATION.
60 = 2ST+4S-------------------1.
30 = 2ST-2S-------------------2 (-).
----------------------------------.
30 = 0+6S.
-----------------------------------.
THEREFORE,
6S = 30.
S = 5.
Ans = 5.
Speed=S;.
Time=T;.
D = S*T;.
30 = S*T;.
HERE,
30 = S*(T+2).
30 = ST+2S------------------1.
Next equation,
30 = 2S*(T-1).
30 = 2ST-2S-------------------2.
TO SOLVE 1 AND 2 EQUATION.
60 = 2ST+4S-------------------1.
30 = 2ST-2S-------------------2 (-).
----------------------------------.
30 = 0+6S.
-----------------------------------.
THEREFORE,
6S = 30.
S = 5.
Ans = 5.
Geeta said:
1 decade ago
Best one is T+2 T-1.
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