Aptitude - Time and Distance - Discussion

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.

Let the time taken by Abby be Ta and that of Sam be Tb.

So, according to question,

[making time equation]

Ta-Tb = 2 ---------(i)

and Tb-Ta = 1 ---------(ii),

Since time of Abby is less than Sam for the second case.

Let the speed of Abby be x and that of Sam be y.

Now from eqn(1) we have, (by using speed formula:

time = distance/speed) we have,

30/x - 30/y = 2 -------(iii).

and from eqn (ii) we have,

30/y - 30/2x = 1 .

=> 30/y = 1+30/2x -------(iv).

Now putting this value 30/y of eqn(iv) in eqn(iii) we get.

30/x - (1+30/2x) = 2.

=> (60-30-2x)/2x = 2 [taking LCM].

=> 30/2x - 1 = 2.

=> 30/2x = 3.

=> 15/x = 3.

=> 15/3 = x.

=> x = 5 [ remember x ? it's speed].

Ans: 5 km/hr.

And that's the required answer!

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.

One more method.

Let Sameer's speed is y kmph and time taken by Sameer is t.
So Abhay takes t+2 time more than summer's time at y kmph.

Then,
Abhay doubles his speed so the speed is 2y so the time taken by Sameer is t-1hr less than Sameer.

Speed = dis/time formula when abhay has normal speed.
y = 30km / t+2 ............... (1).
Speed = dis / time formula when abhay doubles his speed.
2y = 30km / t-1 .................(2).

Put the value of y from 1st equation into 2nd equaton.
2(30/t+2) = 30/t-1.
2/t+2 = 1/t-1.
2t-2 = t+2.
2t-t = 2+2.
t = 4 ................... (3).

Put the value of t in eq. 2.
2y = 30/ 4-1.
2y = 30/3.
2y = 10.
y = 5 kmph.

You can put the value of t in eq.1 also you will get the same answer 5kmph.

Good explanation Devi.

Another Method.

Let Abhay's speed :- x kmph
Let Sameer's speed:- y kmph

According to the question

30/x - 30/y = 2 -- 1st equation

&

30/y - 30/2x = 1 -- 2nd equation

Solving above two we get:-

(y-x)/xy = 1/15 --3rd equation

&

(2x -y)/xy = 1/15 -- 4th equation

since the rhs of both the equations is equal equating the lhs of the both equations is also correct: so

y -x = 2x - y

i.e

2y = 3x or y = 3x/2

now equate the value of y in 3rd or the 4th equation as follows:-

(2x - (3/2)x)/(x*(3/2)x) = 1/15

we get

x/3x^2 = 1/15

so

1/x = 1/15

so

x = 5 km/hr

Hope this helps.

The solution may be found with 2 variables, explained nicely above with x and y.

However, the above solution is done with 1 variable only.

The figure of 3 comes from the total savings in hrs if the speed is doubled.

Assuming Abhay's speed is = x ; time taken is 2 hrs more
If Abhay's speed is doubled = 2x ; time saved is 1 hrs more

Therefore difference in speed results in total time savings = 3 hrs (2 hrs + 1 hr)

Now, to calculate the two speeds:

Given distance = 30 km
Assuming Abhay's speed = x
Therefore Time1 = 30/x

Abhay's speed is doubled = 2x
Therefore Time2 = 30/2x

Since difference in time is the total time saved :
Hence,

Time1 - Time2 = 3 hrs

30/x - 30/2x = 30

Solving :
6x = 30
x = 5 kmph

@All.


Here, is my explanation;

When Abhay moves at normal speed;

Let the time taken by Sameer, time = X.
Then, Abhay needs 2 more hours time than Sameer. ie...time=X+2.
Consider the speed of Abhay, speed =Y

Now, Abhay moves at double speed, so he reaches 1 hour before Sameer.

So speed becomes speed=2Y and time =X-1

Now you can make the equation for distance
Distance is already given d=30

Case 1 (during the normal speed of Abhay)
Distance =speed*time
ie...30=Y*(X+2)----> equation 1

Similarly,
Case 2(during the double speed of Abhay)
30=2Y*(X-1)--->equation 2

Now solve equations 1 and 2 with your logic;
You, will get the answer 5.

Let Sammeer original time = x.

Abhay speed: y = 30/x+2 (let Abhay speed be y, x+2 is written because Abhay takes 2 hrs more than Sammer). This equation can be derived when you analyses the first statement alone.

So, let y = 30/x+2....> (1).

Case 2:

Abhay doubles his speed.

2y = 30/x-1(x-1 is because Abhay takes 1 hr less than Sammer).

So, 2y = 30/x-1....> (2).

From (2) , get the value of x and substitute in (1) , you will get the value of why which is Abhay speed.

2y = 30/x-1.
2y(x-1) = 30.
x-1 = 30/2y.

So, x = (30/2y)+1.

Substitute the x value in (1).

y = 30/x+2.
y =30/((30/2y)+1)+2.
y = 5.

Let us assume first that the velocity of Abhay is V km/hr and say Sameer takes x hours to cover the distance 30km.

Now, we all know S = V*t
=> 30 = V*(x + 2) ; where (x + 2) is the time taken by Abhay to cover the distance 30 km with speed Vkm/hr.

For the second case, i.e, for V' = 2*V, the double speed of Abhay

The above relation changes to:
30 = V'*t = (2*V)*(x-1); where (x-1) is the time taken by Abhay to cover 30km.

Now equating both equations :
=> V*(x+2) = 2*V*(x - 1).
=> x + 2 = 2*x - 2
=> x = 4
Putting x=4 in any of the above two equations we get the velocity V=5 km/hr.

Given:

d=30;
Sameer speed=s1;
Abay speed=s2;

Time taken for covering 30 km is =t;
Speed*time=distance
s1*t=30
s2*(t+2)=30 -----> 1
s2*(t-1)=30 -----> .2

Abay have a different speed.
when he travelled in speed 1 he needs two hours more to get the distance
when he travelled in speed 2 he needs 1 hour less then Sameer's speed to get the distance
So we need to find the difference of time for both speeds,
equation 1-2 provides the time different.
(30/s2)-(30/2s2) = (t+2)-(t-1),
(30/s2)-(30/2s2) = 3,
(15/s2) = 3,
s2=5.