Aptitude - Time and Distance - Discussion

> LET THE TIME TAKEN BY SAMEER = T.
> LET THE SPEED OF ABHAY = X.

TIME = DIST/SPEED.

>Abhay takes 2 hours more than,
Sameer TO COVER 30 KM = T+2 = 30/X.

>Abhay doubles his speed, then he Would take 1 hour less,
Than Sameer TO COVER 30 KM = T-1 = 30/2X.

NOW CROSS MULTIPLY THE BELO EQUATIONS:
T+2 = 30/X.
T-1 = 30/2X.
(T+2)30/2X = (T-1)30/X.
(2TX+4X+30)/2X = (TX-X+30)/X.
(2TX+4X+30)/2 = (TX-X+30).

2TX+4X+30 = 2TX-2X+60.
2TX+4X+30-2TX+2X-60=0.
6X-30=0 => 6X=30 .

Ans) X = 5KMPH.

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

Then 2x speed he covered in y-1. (here why is sameer to cover time).

Then different between time = (y+2) - (y-1) =3. So.

First covered time - second covered time = 3.

Distance = 30.

Abhay's normal speed = x.
Abhay doubles speed = 2x.
Sameer's time = t.

t+2 = 2 hrs + sameer's time.

t-1 = 1 hr less than sameer's time.

[30/x] - [30/2x] = [t+2] - [t-1].
[30/x] - [30/2x] = 3.

[1/x] - [1/2x] = 1/10.
(2-1) /2x = 1/10.

1/2x = 1/10.
2x = 10.

x = 5 km/hr.

Let time taken by Sameer be "t".

First case: when the time taken by Sameer is "t", time taken by Abhay is "t + 2".

Let the speed of Abhay be "s".

Second case: when the speed of Abhay is "2s" time taken by him is "t - 1".

Now from case one and case two, it can be said as,

S = 30/(x + 2) -----> equation (1).

2s = 30/(x - 1) -----> equation (2).

Now substitute "s" value in equation (2).

Thus we get the value of t = 4 hours.

Now Speed of Abhay = 30/(x + 2) = 30/(4 + 2) = 30/6 = 5 hours.

Let the speed of Abhay be X kmph.
Let the Time taken by Sameer to cover a distance of 30 km be T hr.
According to problem;

Abhay takes 2hr more than Sameer i.e, (T+2)
Speed =Distance/Time .
Abhay's speed = total distance/ time taken by Abhay.

X=30/(T+2) --------> eqn1.

Now,
If Abhay doubles his speed i.e, 2X he would take 1 hr less than Sameer i.e (T-1)
2X=30/(T-1)--------> eqn1.

Now divide eqn1/ eqn2.
You will get T = 4hr.
Now put T=4 in eqn1. You will get X = 5kmph.

Let total time saved by Abhay is ( 2+1=3).

Since, in first part of the question, Abhay takes 2 hours more
In the second part of the question Sameer takes 1 hour more, it means Abhay takes 1 hour less than Sameer.
When coming to remaining part of the question,
total distance= 30km
Let Abhay speed= x km/hr
In the second part, he doubles his speed= 2x km/hr.
by using formula t= D/S.
30/x - 30/2x = 3.
Here why we subtracting means the relative speed of the Abhay?

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

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.

In this problem we need to calculate two different 'Time values' of Abhay only(consider it like two different journeys of a single person)& Sameer is to get clue or value comparison.

Let's consider the time of Abhay and Sameer be x and y.
So,In Time(1) x = y+2hr
In Time(2) x= y-1hr (after speed doubles).

Now simply calculate the problem.
T1 - T2 = x
D/S - D/S = x

(y+2) - ( y-1) = 3

30/x - 30/2x = 3,
30 = 6x,
X=5.

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.

Lets take distance = 30.

Speed=S;.
Time=T;.

D = S*T;.
30 = S*T;.

HERE,

30 = S*(T+2).
30 = ST+2S------------------1.

Next equation,

30 = 2S*(T-1).
30 = 2ST-2S-------------------2.

TO SOLVE 1 AND 2 EQUATION.

60 = 2ST+4S-------------------1.
30 = 2ST-2S-------------------2 (-).
----------------------------------.
30 = 0+6S.
-----------------------------------.

THEREFORE,

6S = 30.
S = 5.

Ans = 5.