Aptitude - Time and Distance - Discussion

Let's assume that Sameer's speed is S km/h and Abhay's speed is A km/h.
We know that Sameer takes time T to cover 30 km, so:
T = 30/S

We also know that Abhay takes 2 hours longer than Sameer, so:
T + 2 = 30/A

If Abhay doubles his speed, he will cover 30 km in half the time, which is:
T/2

We also know that this time is 1 hour less than Sameer's time, so:
T/2 = T - 1

Solving for T in terms of S:
30/S/2 = 30/S - 1
15/S = 30/S - 1

Multiplying both sides by S(S-15):
15(S-15) = 30(S-15) - S(S-15)
15S - 225 = 30S - 450 - S^2 + 15S
S^2 - 60S + 225 = 0
(S-15)^2 = 0
S = 15 km/h
Now we can use the equation T + 2 = 30/A to solve for Abhay's speed:
T + 2 = 30/A
30/15 + 2 = 30/A
4 = 30/A
A = 7.5 km/h
Therefore, Abhay's speed is 7.5 km/h.

If suppose x speet abay cover in y+2 time .

then 2x speed he covered in y-1.(here y is sameer to cover time).
then diirent between time = (y+2)-(y-1)=3.so
first covered time - second covered time = 3.

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.

Here's a ChatGpt answer method to solve the problem:

Let's denote Abhay's speed as "a" and Sameer's speed as "s".
We know that distance = speed × time.

According to the problem, Abhay takes 2 hours more than Sameer to cover 30 km. So, we can write:
30 = s × (t + 2) (Equation 1)
where t is Sameer's time taken to cover the distance.

If Abhay doubles his speed, he would take 1 hour less than Sameer to cover 30 km. So, we can write:
30 = 2a × (t - 1) (Equation 2)
where t - 1 is Abhay's time taken to cover the distance at double the speed.

Now, we can solve these two equations for a and s:
From Equation 1, we get:
t = (30/s) - 2

Substituting this value of t in Equation 2, we get:
30 = 2a × ((30/s) - 3).
Simplifying this equation, we get:
a = (45s)/(4s - 60).

Now, substituting this value of a in Equation 1, we get:
30 = s × ((30/s) + 2).

Simplifying this equation, we get:
s^2 = 450.

Therefore, s = 15 km/h.
Substituting this value of s in the expression for a, we get:

a = (45 × 15)/(4 × 15 - 60) = 45 km/h
Therefore, Abhay's speed is 45 km/h.

Speed => 1 : 2.
Time => 2 : 1 = 6 : 3.
So, Required Distance = 30/6 = 5 km/h (Ans.)

Hello I'm not getin from were do that "3" comes in second line.

Please explain in detaail!

Take L.C.M arun ji.... u can understand from whr do 6x came...!!!!

30/x-30/2x=3
=> 30/x-15/x=3
=> 15/x=3
=> 15=3x
=> 15/3=x
=> 5=x

A bit confusing.

Distance=30km/hr
time taken by Sameer is t
Abhay travelling 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