Aptitude - Time and Distance - Discussion

Case 1: Abhays speed=x
Abhays time =t+2
velocity(x)=30/t+2

Case 2: abhays speed=2x
Abhays time =t-1
velocity doubled 2x=30/t-1

now substitute value of x in case 2
t=4 substitute this value in any of the equations above
x=5 is the ans

Let the abhay speed be x then,

30=x*(t+2).........I
30 2x(t-1)..........II

t-1=30/2x
t=30/2x+1 ....Put this value in equation I , we get
30=x*(15/x+3)
30=15+3x
x=15/3.

x=5 km/hrs.

Let, abhay takes X km/hr to cover 30km.
Sameer takes Y km/h to cover 30km.

So first case,
(30/X)-(30/Y) = 2;[time taken by abhay - time taken by sammer = 2hr]

Now,
Abhay doubles his speed
(30/Y)-(30/2X) = 1;

[time taken by sammer - time taken by abhay after doubling the speed = 1 ]

Solving both equation
x=5km/hr.

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 Time Taken by Sameer is T.
Let abhay speed be x Km/hrs.

Distance=30km.
Acc to Question.

When speed is x km.

Then,
Time taken by abhay is 30/x = T+2.

So T=30/x - 2 ------------eq 1.

When speed is x km.

Then,
Time taken by abhay is 30/x = T-1.
So T=30/x + 1 ------------eq 2.

Equating value of the 1 And 2.
We get,

30/x - 2 = 30/2x +1.
=> 30/x - 30/2x =2+1.
=> (60-30)/2x = 3 (LCM).
=>15/x = 3.
=> x=5 Km/Hrs.

When Abhay doubled the speed he gained a total of 3 hours (he was 2 was behind but when doubled he gained 1 hour. ).

This means when he was in his actual speed he was taking 6 hours.

So 30/6 = 5 kmph.

Let Abhay's speed is x km/hr.

Than time he taken to travel 30 km is = 30/x.

But when he doubles his speed, he takes the time 1 hr less than sameer,

Which create a total of 3 hr difference in time he taken,

So, [30/x-30/(2x)] = 3.

x = 15/3 = 5 km/hr.

@Abhi.

I still don't get it. The "3" from the original solution. He was supposed to be 2 hours behind. How does doubling his speed Which would means he takes 1 hour less than sameer end up being "3".

Lets time taken by abhay be x and sameer be y.

Now acc to first line abhay is taking 2hrs more.
Hence , 30/x - 30/y = 2..... eqn 1 // time taken by abhay-sameer = 2.

Acc to second line speed of abhay got increased double.
So now sameer is taking 1 hour more.

30/y - 0/2x = 1.... eqn 2,

Adding eqns 1 and 2.

30/x - 30/2x = 3.

(60-30)/2x = 3 .

6x = 30.

x = 5.

@Devi, why do you subtract 30/x and 30/2x and t+2 and t-1?