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 5 of 21.
SURENDRA SINGH NEGI said:
6 years ago
Case 1:
Abhay's speed = Y.
Sameer has taken time = T,
Then, Y = 30km/T + 2.
Case 2:
Abhay's speed = 2Y.
Abhay's taken time = T-1.
Then, 2Y=30km/T-1.
We put the value of Y in case 2.
Then , 2*30km/T+2 = 30km/T+1.
60T+60=30T+60
30T=120.
T=4.
Put the value of T in case 1.
Y=30km/4hrs + 2hrs.
= 30km/6hrs.
= 5 km/hrs.
Abhay's speed = Y.
Sameer has taken time = T,
Then, Y = 30km/T + 2.
Case 2:
Abhay's speed = 2Y.
Abhay's taken time = T-1.
Then, 2Y=30km/T-1.
We put the value of Y in case 2.
Then , 2*30km/T+2 = 30km/T+1.
60T+60=30T+60
30T=120.
T=4.
Put the value of T in case 1.
Y=30km/4hrs + 2hrs.
= 30km/6hrs.
= 5 km/hrs.
Sabarish said:
7 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)
Anand said:
10 years ago
We know that:
D = ST.
Case 1 : Let's x be the time taken by Sameer.
So, 30 = s(x+2) ----- 1.
Case 2: If Abhay doubles the speed, 1 hr less than Sameer.
So 30 = 2s(x-1) ----- 2.
Equate 1 and 2.
We can get x+2 = 2x-1.
x = 4.
Put x = 4 in any of the equation.
So we will get s = 5 km/hr.
D = ST.
Case 1 : Let's x be the time taken by Sameer.
So, 30 = s(x+2) ----- 1.
Case 2: If Abhay doubles the speed, 1 hr less than Sameer.
So 30 = 2s(x-1) ----- 2.
Equate 1 and 2.
We can get x+2 = 2x-1.
x = 4.
Put x = 4 in any of the equation.
So we will get s = 5 km/hr.
Prakash said:
6 years ago
In simple words:
The time taken to cover the distance of 30 km is always the same in both conditions.
let the speed of Abhay be x.
Now the time taken for 1st condition =time taken for 2nd condition.
Therefore,(30/x)+2 = (30/2x)-1.
(30/x)-(30/2x) = 3.
x = 5.
The time taken to cover the distance of 30 km is always the same in both conditions.
let the speed of Abhay be x.
Now the time taken for 1st condition =time taken for 2nd condition.
Therefore,(30/x)+2 = (30/2x)-1.
(30/x)-(30/2x) = 3.
x = 5.
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)
Santhosh said:
7 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)
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.
(26)
Jithin said:
9 years ago
Let the time is taken by Sameer to cover 30 km be t. Abhay will take t+2 time to cover the same. If he doubles speed he takes t-1 time.
We know time = distance/speed.
So t+2 = 30/x. Where x is the speed of Abhay? t-1 = 30/2x. Now subtract both sides you can get the solution for x.
We know time = distance/speed.
So t+2 = 30/x. Where x is the speed of Abhay? t-1 = 30/2x. Now subtract both sides you can get the solution for 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.
Kishore said:
8 years ago
Speed = distance/time.
Let x be Abhay's speed and y be Sameer's time.
* first condition
speed s = 30/(y+2) ; ------- a.
speed 2s=30/(y-1); ---------b.
Put a ((s=30/(y+2)) in b.
Now you get y=4.
put y=4 in either a or b.
You will get speed as 5kmph.
Let x be Abhay's speed and y be Sameer's time.
* first condition
speed s = 30/(y+2) ; ------- a.
speed 2s=30/(y-1); ---------b.
Put a ((s=30/(y+2)) in b.
Now you get y=4.
put y=4 in either a or b.
You will get speed as 5kmph.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers