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.
Pem said:
4 years ago
Let's assume Abhay's speed as y and time taken as x.
So,
y= 30/x+2 -----> equ.1
2y= 30/x-1 -----> equ.2
Substitute equ.1 in 2.
2(30/x+2)= 30/x-1.
We get x=4hr ( which is time).
Now, substitute x value in equ.1
Y = 30/(2+4),
Y = 5 km/hr.
So,
y= 30/x+2 -----> equ.1
2y= 30/x-1 -----> equ.2
Substitute equ.1 in 2.
2(30/x+2)= 30/x-1.
We get x=4hr ( which is time).
Now, substitute x value in equ.1
Y = 30/(2+4),
Y = 5 km/hr.
(5)
Yamuna said:
8 months ago
ta = x+2
ts = x
d = 30 km
Sa = 2Sa , ta = x-1.
Sa = 30/(x+2) --> xSa + 2Sa = 30.
2Sa = 30/(x-1) --> 2xSa - 2Sa = 30.
Since the distance is the same
xSa + 2Sa = 2xSa - 2Sa,
xSa = 4Sa.
x = 4,
ta = x+2,
ta = 6.
s = d/t = 30/6 = 5.
ts = x
d = 30 km
Sa = 2Sa , ta = x-1.
Sa = 30/(x+2) --> xSa + 2Sa = 30.
2Sa = 30/(x-1) --> 2xSa - 2Sa = 30.
Since the distance is the same
xSa + 2Sa = 2xSa - 2Sa,
xSa = 4Sa.
x = 4,
ta = x+2,
ta = 6.
s = d/t = 30/6 = 5.
(5)
Himanshu said:
5 years ago
@All.
Simply, the solution is;
Abhay's doubled speed is = to the speed required to cover 30km in 3 hrs.
2x=30/3,
x= 10/2,
x= 5 km/hr,
Another way to solve is;
(30/x)-(30/2x) = 3.
30((1/x)-(1/2x)) = 3.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.
Simply, the solution is;
Abhay's doubled speed is = to the speed required to cover 30km in 3 hrs.
2x=30/3,
x= 10/2,
x= 5 km/hr,
Another way to solve is;
(30/x)-(30/2x) = 3.
30((1/x)-(1/2x)) = 3.
1/2x = 1/10.
2x = 10.
x = 5 km/hr.
Amit Singh Rajput said:
1 decade ago
If sameer take 5hr than abhay take 7 hr
when he increases his speed double than he take 4hr hence different is 7-4=3
mean (30/x)-(30/2x)=3(solve)
30/x speed when he take more time,30/2x when he double his speed
hence diff. between them=3
when he increases his speed double than he take 4hr hence different is 7-4=3
mean (30/x)-(30/2x)=3(solve)
30/x speed when he take more time,30/2x when he double his speed
hence diff. between them=3
Anon said:
9 years ago
Why can the two equations not be just divided, you'll be able to cancel all unwanted constants and find time directly, but it gives a different answer.
30/x = t + 2
30/2x = t - 1
Dividing both (30/x) / (30/2x) = (t + 2)/(t - 1).
t = 4.
30/x = t + 2
30/2x = t - 1
Dividing both (30/x) / (30/2x) = (t + 2)/(t - 1).
t = 4.
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.
Siddhu said:
1 decade ago
Case 1: Abhays 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 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 value of x in case 2
t=4 substitute this value in any of the equations above
x=5 is the ans
Ayush Kumar Yadav said:
1 year ago
@All.
Here is the explanation:
Let speed of Abhay = t1.
and speed of Sameer = t2.
t1 = t +2 (t is actual Speed).
t2 = t -1 (t is actual speed).
t1 - t2 = t+2 - (t - 1) = 3.
30/x - 30/2x = 3.
30 - 15 = 3x.
15 = 3x.
x = 5 kmh^-1.
Here is the explanation:
Let speed of Abhay = t1.
and speed of Sameer = t2.
t1 = t +2 (t is actual Speed).
t2 = t -1 (t is actual speed).
t1 - t2 = t+2 - (t - 1) = 3.
30/x - 30/2x = 3.
30 - 15 = 3x.
15 = 3x.
x = 5 kmh^-1.
(15)
Mariyada Ramesh said:
10 months ago
Let Abhay's speed=x, Sameer's time=y
Case 1: 30/x = 2+y.
Case 2: 30/2x = y-1.
30/x-30/2x = (2+y)-(y-1),
= 2+y-y+1,
= 3.
We get the equation:
30/x-30/2x = 3.
By solving we will get the answer.
Case 1: 30/x = 2+y.
Case 2: 30/2x = y-1.
30/x-30/2x = (2+y)-(y-1),
= 2+y-y+1,
= 3.
We get the equation:
30/x-30/2x = 3.
By solving we will get the answer.
(27)
Anoo said:
1 decade ago
Time in first case= 30/x(30 is distance and x is speed)
Time in second case=30/2x(30 is distance and speed is doubled)
Total difference in time is 3 hrs(2 hrs slow and one hour fast)
(30/x)-(30/2x)=3
Solve and you get 5km/hr.
Time in second case=30/2x(30 is distance and speed is doubled)
Total difference in time is 3 hrs(2 hrs slow and one hour fast)
(30/x)-(30/2x)=3
Solve and you get 5km/hr.
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