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 2 of 21.
Arun Kumar r said:
5 months ago
I am not getting this. Anyone, please help me.
(13)
Riyaz said:
4 years ago
Speed = Distance/Time.
Let;
Time taken by Sameer = x.
therefore abay,s time = x+2,
let speed of Abhay = y.
Now d = 30 , time = x+2.
The speed of Abhay
y = 30/x+2 ---(1)
Now Abhay speed = 2y , time = x-1.
Speed of Abhay 2y = 30/(x-1)----(2).
sub y from eq(1) to eq(2).
We get x =4 , which is scammers time , Abhays tie is( x+2) => 6.
sub x+2 value in eq 1.
speed of Abhay => y = 30/6==> 5km/hr.
Let;
Time taken by Sameer = x.
therefore abay,s time = x+2,
let speed of Abhay = y.
Now d = 30 , time = x+2.
The speed of Abhay
y = 30/x+2 ---(1)
Now Abhay speed = 2y , time = x-1.
Speed of Abhay 2y = 30/(x-1)----(2).
sub y from eq(1) to eq(2).
We get x =4 , which is scammers time , Abhays tie is( x+2) => 6.
sub x+2 value in eq 1.
speed of Abhay => y = 30/6==> 5km/hr.
(12)
Sarang said:
3 years ago
d = (s1*s2*time diff)/( s1-s2).
30 = x * 2x * 3/x * 2x.
x = 5.
30 = x * 2x * 3/x * 2x.
x = 5.
(7)
Asit Kumar Chand said:
2 months 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)
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)
Shivam said:
4 years ago
This is how I understand the problem.
Let Abhay's speed be 'x' kmph.
Let time taken by sameer be 't' hr.
Distance = 30km.
According to first condition,
Sameer time (t) = 30/x - 2 hr.
According to second condition,
Sameer time (t) = 30/2x + 1 hr.
Therefore,
30/x - 2 = 30/2x + 1.
After soving this,
X = 5kmph.
Therefore Abhay's speed is 5kmph.
Let Abhay's speed be 'x' kmph.
Let time taken by sameer be 't' hr.
Distance = 30km.
According to first condition,
Sameer time (t) = 30/x - 2 hr.
According to second condition,
Sameer time (t) = 30/2x + 1 hr.
Therefore,
30/x - 2 = 30/2x + 1.
After soving this,
X = 5kmph.
Therefore Abhay's speed is 5kmph.
(5)
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)
Anomie said:
3 years ago
Good explanation. Thanks all.
(4)
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)
Navin said:
4 years ago
s = d/t.
x=30/60,
= 2.
Therefore : 2 + 2 {dbl spd} + 1 {1 hour less} = 5.
x=30/60,
= 2.
Therefore : 2 + 2 {dbl spd} + 1 {1 hour less} = 5.
(3)
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