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 3 of 21.
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)
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)
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)
Ketan said:
4 years ago
Thanks @Devi.
Meenakshi said:
4 years ago
Actually use simple logic.
Initially, he took 2 hrs more than Sameer, then he took one hr less.
which means the difference in time that has occurred in 3hrs.
So, Intial time - final time = 3hrs.
Initially, he took 2 hrs more than Sameer, then he took one hr less.
which means the difference in time that has occurred in 3hrs.
So, Intial time - final time = 3hrs.
Hasan said:
4 years ago
Distance = 30km/hr.
The time taken by Sameer is t.
Abhay traveling with a Speed X.
Case 1:
Time taken by Abhay to cover 30km is 30/X = t+2.
Case 2(doubles speed):
Time taken by Abhay to cover 30km is 30/2X = t-1.
(30/X)- (30/2X) = (t+2)-(t-1),
(30/X) - (30/2X) = t+2-t+1,
(30 * 2X) - (30 * X) = 3 (2X^2) //LCM.
60X - 30X = 6 X^2,
30X = 6 X^2,
30 = 6 X,
X = 5.
The time taken by Sameer is t.
Abhay traveling with a Speed X.
Case 1:
Time taken by Abhay to cover 30km is 30/X = t+2.
Case 2(doubles speed):
Time taken by Abhay to cover 30km is 30/2X = t-1.
(30/X)- (30/2X) = (t+2)-(t-1),
(30/X) - (30/2X) = t+2-t+1,
(30 * 2X) - (30 * X) = 3 (2X^2) //LCM.
60X - 30X = 6 X^2,
30X = 6 X^2,
30 = 6 X,
X = 5.
(2)
Rohit said:
5 years ago
By applying direct formulation speed - distance /time.
30/3 * 2,
= 5.
30/3 * 2,
= 5.
(1)
Paawan said:
5 years ago
Thanks for the solution @Devi.
Harsh said:
5 years ago
Let's make it simple.
Let Abhay speed be x and Sameer be y
Equation 1be like this:
30/x - 30/y = 2.
Equation 2 be like this;
30/2x - 30/y = - 1(its because the double speed of Abhay than Sameer make a difference of 1 hour, i.e, Abhay take less time than Sameer that's why I put - 1).
On Solving both the equation you get something like this;
-30y= - 6xy.
6x=30 or x=5.
Let Abhay speed be x and Sameer be y
Equation 1be like this:
30/x - 30/y = 2.
Equation 2 be like this;
30/2x - 30/y = - 1(its because the double speed of Abhay than Sameer make a difference of 1 hour, i.e, Abhay take less time than Sameer that's why I put - 1).
On Solving both the equation you get something like this;
-30y= - 6xy.
6x=30 or x=5.
Palash said:
5 years ago
Abhay speed is getting double it means it will become Sameer = 2 Abhay and let us take Abhay as x then sameer will become 2x then add speed x+2x = 3x.
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