Aptitude - Time and Distance - Discussion

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.

@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.

d=(S1S2/S1-S2)*t.
(S(S+3) / 3)*(40/60)=(S(S-2)/2)*(40/60).
2S+6= 3S-6.
S=12Kmph.
d=( (12*15)/3)*40/60.
d=40km.

d = st--->1.
d = (s+3)(t-2/3)------>2.
d = (s-2)(t+2/3)------>3.
Solving 1and 2 we get 2s/3 = 3t-2----->4.
2s/3 = 2t-4/3--------->5.
By solving 4and 5 we get t = 10/3 and s = 12 so d = st = 40km.

Thank you @Sachin bhagat.

@Pauli.
For Jake sum.
S*t=90.....1.
(S+15)(t-30/60)=90.......2.

Solving 2 eqn, answer: 45km/hr.

Thanks @Sachin Bhagat.

Thanks @Sachin Bhagat.

(40*3)-(40*2),
120-80 = 40.

The solution is;

2y(y + 3) = 9x --> (i)
y(y - 2) = 3x --> (ii)

So, 2y(y + 3) = 3*3x = 2y(y + 3) = 3* y(y - 2).

After Solving y=12 put y value any above 1 and 2 equation, and get x=40.