Aptitude - Time and Distance - Discussion

@All.

In the above question, only time and speed are varying so distance remains constant.


So, First convert time 40 min=40/60 hr.

Now let's begin.
we know v=d/t. therefore d=vt--> (1)

Now by the first condition.
v+3=d/(t-40/60) .
Therefore d=(v+3)(t-40/60).

By multiplying the brackets.

d=(vt)-(2v/3)+(3t)-2 --> (2).

Now we know distance is constant
Therefore equating equations (1) and (2).

vt = vt-2v/3+3t-2.
Solving this we get -2v/3+3t=2.
now multiply it by 3.
We get -2v+9t=6 ---> (3).

Now By the second condition.

v-2=d/(t+40/60),
therefore d=(v-2)(t+40/60),
therefore d=(vt) +(2v/3)-(2t)-(4/3)--> (4).

Now equating equations (1) and (4)
vt=vt+2v/3-2t-4/3
By solving we get 2v/3-2t=4/3
Now multiply by 3 on both sides
2v-6t=4 ---> (5).

Now adding equations (3) & (5)
-2v+9t=6
+2v-6t=4
We get 3t=10 therfore t=10/3 hr
Now put t=10/3 in equation (5).

2v-6(10/3)=4.
And 2v=4+20.

Therfore
v=12 kmph.

Now finally from equation (1)
d=vt=12*(10/3)
d=40 km..

Hope it helps.

For simplifying method, follow easy steps which are lengthy but clear understanding.

(x/y)-(x/y+3)=40/60 ---> (1)
(x/y-2)-(x/y)=40/60 ---> (2)

Let solve eq. (1)
(x/y)-(x/y+3) = 40/60,
(x(y+3)-y(x)) / y(y+3) = 2/3,
((xy+3x-xy) / (y^2+3y)) = 2/3,
3(3x)=2(y^2+3y),
9x=2y(y+3) ---------> (A).

Now let solve eqn. (2)
(x/y-2)-(x/y) = 40/60,
((xy-x(y-2)) /(y(y-2))) = 2/3,
((xy-xy+2x) / (y^2-2y)) = 2/3,
3(2x) = 2(y^2-2y)
6x = 2y(y-2)
now divide by 2 we get
3x = y(y-2) --------------> (B)
now we get two equations as per given in answer are
9x = 2y(y+3) -------------->(A).
3x = y(y-2) -------------->(B).

Now we divide eqn (A) by (B) we get;
(2y(y+3)) / y(y-2) = 9x/3x,
2(y+3) / (y-2) = 3,
y+3 / y-2 = 3/2,
2(y+3) = 3(y-2),
y = 12.

After putting y=12 in equation (A) we get x = 40.

Let Speed be 's' and Time be 't';
Original Distance= st--------------> (i)
If speed and time changes,
New Speed=s+3
New time= t-2/3 (40/60=2/3)
So again D= (s+3)(t-2/3) = st + 3t - 2/3s - 2
from (i), D=st
=> st= st + 3t - 2/3s - 2
=> 3t - 2/3s = 2--------------> (ii)
Similarly,
D= (s-2)(t+2/3) = st - 2t + 2/3s - 4/3
and D= st
putting value of D,
=> st = st - 2t + 2/3s - 4/3
=> - 2t + 2/3s = 4/3 --------------> (iii)
Now solving (ii) and (iii),
3t - 2/3s = 2
- 2t + 2/3s = 4/3
________________
t = 10/3
Putting value of t in (ii),
=> 3*10/3 - 2/3s = 2,
=> 10-2 = 2/3s,
=> 2/3s = 8,
=> s=12,
So, D=st,
=12*10/3,
= 40 km Ans..

I have easier solution to this:

Distance is same so first consider speed x and time t:

As per the question,
Increasing speed by 3 i.e. (x+3) will decrease time by 40 min i.e (t-40/60).
Therefore,
Distance = (x+3)(t-40/60).

Now Decreasing speed by 2 i.e. (x-2) will decrease time by 40 min i.e (t+40/60).
Therefore,

Distance = (x+3)(t-40/60).

Since Distance would be same:

(x+3)(t-40/60)=(x+3)(t-40/60) -----> (1)

solving this, we get

-2x/3 + 3t -2 = 2x/3 + -2t - 4/3.

3t -2x/3 =2 -->(2)
2x/3 + -2t = 4/3 --->(3)

Solving two-equation, we get t=10/3 and x=12. Putting this value in eq. 1, we get,

Distance = 40.

If 3km/hr faster than actual speed, it takes 40 minutes less time to complete.
If 2km/hr slower, it takes 40 minutes more.

So,

If we take condition 1, it would be 2km more than the actual distance at the actual time.(3km/hr, so 2km in 40 minutes)

If we take condition 2, it would be 4/3 km less than the actual distance at the actual time.(2km/hr, so 4/3 km in 40 minutes)

Now the difference between the two condition is 80 minutes and the distance is 2 + 4/3.
Distance = speed * time.
So, we equate, 10/3 = x/80 * 60(multiplying 60 is for hour conversion).
So, x = 40.

I am not understanding as not able to calculate this lines from the given explanation:
2y(y + 3) = 9x ....(i)
y(y - 2) = 3x ....(ii)

Somebody tell me step by step calculation of how to get above answer.

And also @ Amit :
not understanding this line
How you got Y=12?
and as per your equation:
why didn u consider x/y in the whole sum?
As in the question it is given a certain distance at some speed.
So we should take x/y also.
Your method looks bit simple but I got all this doubts so am asking here .
Please kindly someone explain me explicitly.
thank you :)

Let the actual speed be xkmph and time be t hours.
Then distance (d) = x*t.
And so t = d/x.......(i).

Then from the question, d=(x+3)(t-40/60).......(ii).

Putting (i) in (ii),
d=(x+3)(d/x-40/60).
Then 40/60 =d/x-d/(x+3).....(iii).

Then taking the second condition from the given question,
d=(x-2)(t+40/60).......(iv).

Putting (i) in (iv),
d=(x-2)(d/x+40/60).

Then on solving 40/60 =d/(x-2)-d/x......(v).

Now on dividing (iv) and (v) , and on solving v get,
x=12kmph.

And putting the value of 'x' in (iii) and solving we get 'd=40km'.

Let distance = D, speed = S, time taken = T (formula, we know S*T=D).

Let's assume the man covered the distance twice(2D), one D with S+3 kmph and another D with S-2 kmph.

So, D + D = (S + 3) (T - 40) + (S - 2) (T + 40).
2D = ST + 3T - 40S - 120 + ST - 2T + 40S - 80.
=> 2D = 2ST + T - 200.
But ST = D,
=> T = 200 min.

Now with S + 3 speed, time taken = 200 - 40 = 160.
and with S - 2 speed, time was taken = 200 + 40.
= 240.
But we have a problem in our earlier format, solve it. You will get D, distance = 40 km.

3 km/hr Increase: 40 min less
2 km/hr Decrease: 40 min more

First, we calculate the original speed by the following trick:
=2 X increase in speed X decrease in speed / Difference of increase and decrease in speeds.
=2 x 3 x 2/(3-2).
=12 km/hr.

Now Scenario is as under:
3 km/hr Increase = 12+3 = 15 km/hr: 40 min less.
2 km/hr Decrease =12- 2 = 10 km/hr: 40 min more.

Now the distance
= products of speed / difference of speed X total time difference.
= 80/60 ( in hours ) X 15 X 10 /(15-10 ).
= 40 Km.

Let distance be the x km and y be the usual speed and t be the time,
=> t = x/y......(1).

Since, speed is inversely prop. to time.
i.e. faster speed -> slower time.

Now, original time(t)- slow time(t1).

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

And,

fast time(t2) - original time(t)

=> x/y-2 - x/y = 40/60 => 6x = 2y(y-2)........(3)

So dividing 2 by 3 we get y = 12.

Now put y= 12 in (2) we get,

9x = 24(12+3).

=> x = 24*15/9 = 5*8 = 40.

Hence distance = 40.