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 1 of 21.
Mostafa kamal Abir said:
4 weeks ago
Speed => 1 : 2.
Time => 2 : 1 = 6 : 3.
So, Required Distance = 30/6 = 5 km/h (Ans.)
Time => 2 : 1 = 6 : 3.
So, Required Distance = 30/6 = 5 km/h (Ans.)
(2)
Asit Kumar Chand said:
1 month ago
30/A-30/B = 2 ---> (1)
30/B-30/2A = 1 ---> (2)
By adding two equations
30/A - 30/2A = 3,
60 - 30/2A = 3,
30 = 6A.
A = 30/6.
A = 5km/h.
30/B-30/2A = 1 ---> (2)
By adding two equations
30/A - 30/2A = 3,
60 - 30/2A = 3,
30 = 6A.
A = 30/6.
A = 5km/h.
(6)
Aditya said:
5 months ago
@Arun Kumar
Use 2 variables x and y, then it will be super easy.
Speed = distance/time.
Time taken by sameer = x,
Time taken by abhay = x + 2,
Speed of Abhay = y.
y = 30/x+2.
2y = 30/x -1.
as 30 i.e distance is constant,
xy + 2y = 2xy - 2y.
x = 4,
putting x = 4 in xy + 2y = 30,
y = 5.
Use 2 variables x and y, then it will be super easy.
Speed = distance/time.
Time taken by sameer = x,
Time taken by abhay = x + 2,
Speed of Abhay = y.
y = 30/x+2.
2y = 30/x -1.
as 30 i.e distance is constant,
xy + 2y = 2xy - 2y.
x = 4,
putting x = 4 in xy + 2y = 30,
y = 5.
(25)
Arun Kumar r said:
5 months ago
I am not getting this. Anyone, please help me.
(13)
Sabarish said:
6 months ago
t is actual time (Let's take Sameer's speed as actual).
When Abhay travels normally;
30/x = t+2 , t = 30/x - 2 ----(> t)
when Abhay travels 2x speed.
30/2x = t-1.
30/2x +1 = t.
30/2x +1 = 30/x - 2.
30+2x/2x = 30 -2x/x.
30x+2x^2 = 60x - 4x^2.
6x^2 = 30x.
x = 30/6.
x = 5.
When Abhay travels normally;
30/x = t+2 , t = 30/x - 2 ----(> t)
when Abhay travels 2x speed.
30/2x = t-1.
30/2x +1 = t.
30/2x +1 = 30/x - 2.
30+2x/2x = 30 -2x/x.
30x+2x^2 = 60x - 4x^2.
6x^2 = 30x.
x = 30/6.
x = 5.
(4)
Santhosh said:
6 months ago
Abhay takes 2 hrs more than Sameer to cover 30 km.
Abhay's speed (s) = 30/t+2 ---> (1)
If Abhay doubles his speed he takes 1 hr less.
Abhay's double speed (2s) = 30/t-1 ---> (2)
Equate (1) and (2) we get,
2[30/t+2] = 30/t-1.
t = 4 hrs.
T = t+2 = 4+2 = 6 hrs.
S = 30/6 = 5 km/hr.
Abhay's speed (s) = 30/t+2 ---> (1)
If Abhay doubles his speed he takes 1 hr less.
Abhay's double speed (2s) = 30/t-1 ---> (2)
Equate (1) and (2) we get,
2[30/t+2] = 30/t-1.
t = 4 hrs.
T = t+2 = 4+2 = 6 hrs.
S = 30/6 = 5 km/hr.
(13)
Yamuna said:
7 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)
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.
(26)
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)
JaSHaN SaGaR said:
2 years ago
Take the ratio of the speed of Abhay:
1:2
Then the inverse time we get Time ratio,
So:
2:1.
The Gap is 1 unit and this one is given as the sum of hours which is 3 hours.
So 2=6 hours and 1= 3 hours.
So we get the ans by applying a formula that is S= D/T
S= 30/6, which is equal to 5.
1:2
Then the inverse time we get Time ratio,
So:
2:1.
The Gap is 1 unit and this one is given as the sum of hours which is 3 hours.
So 2=6 hours and 1= 3 hours.
So we get the ans by applying a formula that is S= D/T
S= 30/6, which is equal to 5.
(65)
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