Aptitude - Time and Distance - Discussion

Hi, Simran & Venish,
I have a soln of this problm :-
Let Distns = X, & Spd = Y
Now fm 1st half of quesn
Spd = Y+3
Then Time = X/Y+3
2nd half of qesn
Spd = y-2
Then Time = X/Y-2
Now from quesn
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
Then, 45x+300 = 30x-300
15x = - 600
X = - 40 Km
"After neglect the minus value of X"
X = 40 Km Answer

x/y - x/y+3 = 40/60 => xy+3x-xy = 2/3y(y+3) => 9x = 2y(y+3)......(2).


x/y-2 -x/y=40/60 => xy-xy+2x = 2/3y(y-2) => 3x = y(y-2)........(3).


Multiply eq(3)in 2 and solve (2)&(3).

We got y = 3x/10.


Then substitute y = 3x/10 in eq(3).

3x = 3x/10(3x/10 -2).

3x = 3x/10(3x-20/10).

3x = 3x(3x)-60x/100.

300x = 3x(3x)-60x.

3x(3x)-360x = 0.

x(x)-40x = 0.

x(x-40) = 0.

x = 40.

First we find the speed using this concept.

2s1*s2/s1-s2.
2*3*2/3-2=12km/hr.

Now, if we increase speed by 3km/hr, it will become 12+3=15km/hr and 40min less
same as,if we decrease the speed by 2km/hr then it become 12-2=10km/hr and 40min more
we have s1=15kmhr and s2=10kmhr and the time difference is 80/60.

Let distance be x.
x/10 - x/15 =80/60,
3x-2x/30=4/3,
x=4/3*30=40km/hr.

LET'S SOLVE IN VERY EASY MANNER.

Easiest formula :- DISTANCE = (s1*s2)/(speed difference) * time difference
So, now
D= {s*(s+3)/3} * 40/60 = {s*(s-2)/2} * 40/60.
=> (s+3)/3 = (s-2)/2 [ s and 40/60 will cancel out each other at both side],
=> 2s+6 = 3s-6,
=> s = 12 km/hr.
And now,
D= (12*15)/3 * 40/60 = 40 km Ans. [ according to formula].

Short trick :

3 * (+40) : 2 *(-40)
120: 80 ==> 3 : 2

Then take Avg for the both speed of the Ratio
2(2*3)/1 ==> same direction so 2(XY) / X-Y.

Then we get original speed 12kmhr.
then ,12+3 = 15kmhr
12-2= 10kmhr

Diff for both time 40 - (-40) = 80mins.
The product of speed/ diff of speed *time
15*10/5 * 80/60.
Finally, we get 40 km distance.

See take distance =x and rate=y and t= time.
Now overall x/y=t.

We have one condition,
x/y+3 at t-40.

We have second condition,
x/y-y at t+40.

Now equate,
x/y=t and x/y+3=t-40 sub them,
You get x/y- x/y+3=40/60.

Now equate,
x/y=t and x/y-2=t+40 sub them,
You get x/y-2- x/y=40/60.

You will see t getting canceled.

40/60 because convert min into hr.

@JAHNAVI.

Initially, distance is the same so to find time and speed, we take speed x and time and now implement the distance formula.

First, speed is greater than 3 we take x+3 and if the speed is increased time is decreased so t-40/60 we convert 40 min to an hour divided by 60. Second x-2 and t+40/60 solve these equations you get the why and then x.

Let distance be x speed be y and time taken by him in actual speed is p, ie., x/y=p (distance/speed=time) (km/kmphr=hr).

Then go for 1st condition 3 km faster hence 40 min less, x/(y+3)=p-40/60 (converting 40 min to hr).

Already we know p=x/y sub it we get x/(y+3)= x/y-40/60, that's how we get x/y -x/(y+3)= 40/60.

Hi, I'm giving a practical example on it.

Suppose initial time is 5minute and after increasing speed is 3 minute so 5-3 = 2minute that 2 minute is time taken to cover the distance at 3kmph faster speed.

Now make an equation for 5 i.e. as per question and subtract equation 3 i.e as per question and 2 minute is 40/60.

What if decreasing speed and increasing speed would be same?

When a person is moving with speed x km/hr is reaching office in t time. When he increases speed by 20km/hr, he reaches 40 min early. But when he decreases speed by 20 km/hr, he reaches 40 min late. What is the distance?

Can anyone tell me?