Aptitude - Time and Distance - Discussion

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 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.

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, 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 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.

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

Thanks devi. You made it simply.

If sameer take 5hr than abhay take 7 hr
when he increases his speed double than he take 4hr hence different is 7-4=3
mean (30/x)-(30/2x)=3(solve)
30/x speed when he take more time,30/2x when he double his speed
hence diff. between them=3

Let us look at ths prb this way:
Let abhays's speed be x and sameers time be y
Case 1: Abhays speed=x
Abhays time =t+2

Case 2: abhays speed=2x
Abhays time =t-1
We knw : Speed is inversely proportional to time
So, S2/S1=t1/t2
2x/x=t+2/t-1

2=t+2/t-1
solving we get t=4
hence we can calculate
Speed of abhay=Distance/time
=30/(4+2) ..Since t=4
30/6=5

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