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 6 of 21.
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).
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.
Manish Pillai said:
9 years ago
Only consider Abhay the ratio of his speeds is 1 : 2.
So, the ratio of time he would take would be 2 : 1.
2x for 1 case and x for 2.
Given 2x - x = 3
so, x = 3hrs.
So, for 1 case he will take 6 hrs to complete the journey.
His speed will be 30/6 = 5km/hr.
So, the ratio of time he would take would be 2 : 1.
2x for 1 case and x for 2.
Given 2x - x = 3
so, x = 3hrs.
So, for 1 case he will take 6 hrs to complete the journey.
His speed will be 30/6 = 5km/hr.
Govindraj said:
1 decade ago
Let Abhay's speed is x km/hr.
Than time he taken to travel 30 km is = 30/x.
But when he doubles his speed, he takes the time 1 hr less than sameer,
Which create a total of 3 hr difference in time he taken,
So, [30/x-30/(2x)] = 3.
x = 15/3 = 5 km/hr.
Than time he taken to travel 30 km is = 30/x.
But when he doubles his speed, he takes the time 1 hr less than sameer,
Which create a total of 3 hr difference in time he taken,
So, [30/x-30/(2x)] = 3.
x = 15/3 = 5 km/hr.
SAPTARSHI DUTTA said:
1 decade ago
Let the speed be x, we will move with Abhay only at the outset.
A/Q D=S*T or, 30 = (t+2)*x or, 30/x = t+2.
If he doubles his speed then,
30 = (t-1)*2x or, 30/2x = t-1.
REMEMBER t is the time taken by Sameer.
D/S = T => 30/x-30/2x = 3 => x = 5.
A/Q D=S*T or, 30 = (t+2)*x or, 30/x = t+2.
If he doubles his speed then,
30 = (t-1)*2x or, 30/2x = t-1.
REMEMBER t is the time taken by Sameer.
D/S = T => 30/x-30/2x = 3 => x = 5.
DIPS said:
8 years ago
Case 1: Abhay's speed=x.
Abhays time =t+2,
velocity(x)=30/t+2,
Case 2: abhays speed=2x.
Abhays time =t-1,
velocity doubled 2x=30/t-1.
now substitute the value of x in case 2.
t=4 substitute this value in any of the equations above.
x=5 is the ans.
Abhays time =t+2,
velocity(x)=30/t+2,
Case 2: abhays speed=2x.
Abhays time =t-1,
velocity doubled 2x=30/t-1.
now substitute the value of x in case 2.
t=4 substitute this value in any of the equations above.
x=5 is the ans.
Ameer said:
7 years ago
Abay = 2hour + Sameer, [you can analyze this equation by taking sameer=5, which gives Abay=7]
And; 2 * Abay = Sameer - 1, [similarly in this equation, suppose sameer=5, which results in Abay=4]
Both the equation satisfies the above statement.
And; 2 * Abay = Sameer - 1, [similarly in this equation, suppose sameer=5, which results in Abay=4]
Both the equation satisfies the above statement.
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.
Yamini Dhanasekar said:
7 years ago
We can also solve it by finding a time,
Let, Abhay's speed-x, the time taken by Sameer-t.
X=30/t+2 --> (1)
When speed doubles
2X= 30/t-1..,
X=15/t-1 --> (2).
Equating (1) &(2)
30/t+2 = 15/t-1.
t=4.
Sub in eqn (1),
X = 30/4+2,
X = 5 kmph.
Let, Abhay's speed-x, the time taken by Sameer-t.
X=30/t+2 --> (1)
When speed doubles
2X= 30/t-1..,
X=15/t-1 --> (2).
Equating (1) &(2)
30/t+2 = 15/t-1.
t=4.
Sub in eqn (1),
X = 30/4+2,
X = 5 kmph.
Apoorv said:
7 years ago
(30/x)-(30/2x)=3.
In 1st case LCM is x and in 2nd case LCM is 2x.now make both LCM equal.
(2*(30/x))-30/2x=3,
(60/2x)-(30/2x)=3,
(60-30)/2x=3,
(30/2x)=3,
30= 3*2x,
30=6x,
30/6=x,
5=x.
Note- multiply by whole fraction (2*(30/x) and by only x.
In 1st case LCM is x and in 2nd case LCM is 2x.now make both LCM equal.
(2*(30/x))-30/2x=3,
(60/2x)-(30/2x)=3,
(60-30)/2x=3,
(30/2x)=3,
30= 3*2x,
30=6x,
30/6=x,
5=x.
Note- multiply by whole fraction (2*(30/x) and by only x.
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