A car travelling from city A to B takes an hour lesser to cover the di

given
d=s*t
(t-1)*(s+10) = (t+1)*(s-6)
LHS = 10t-s=10
and RHS = s-6t=6
add both ; we get t= 4 hrs and s= 30 kmph
d= 120 km
IMO A

We can create the equations:

rt = d (Eq. 1)

and

(r + 10)(t - 1) = d (Eq. 2)


and

(r - 6)(t + 1) = d (Eq. 3)

Equating the left hand sides of Eq. 1 and Eq. 2, we have:

rt = (r + 10)(t - 1)

rt = rt - r + 10t - 10

r = 10t - 10 ⟶ Eq. 4

Equating the left hand sides of Eq. 2 and Eq. 3, we have:

(r + 10)(t - 1) = (r - 6)(t + 1)

rt - r + 10t - 10 = rt + r - 6t - 6

16t - 4 = 2r

8t - 2 = r ⟶ Eq. 5

Equating the right hand side of Eq. 4 and the left hand side of Eq. 5, we have:

10t - 10 = 8t - 2

2t = 8

t = 4

Since r = 10t - 10, r = 4(10) - 10 = 30. Finally, since d = rt, d = 30(4) = 120.

Answer: A

Hi Sarkar93,

Distance = Speed*Time
Let's say Original Speed = 's' (small s) and Original Time taken = 't' (small t)
It gives Distance = s*t or D = st --> Eq(1)

Case 1 : If Speed increases by 10 kmph, Time taken is reduced by 1 hr.
So, New speed (S) will be = (s+10) and New Time (T) = (t-1)
--> Distance = S*T = (s+10)*(t-1) = (st - s +10t - 10) ---> Eq(2)
--> Notice we have the 'st' component here which is equal to D (from eq.1)

Rewriting the equations we get, D = ST or D = st (where S<>s and T <>t; but ST = st)
(That is, when the Distance remains the same, the Speed and Time can take different sets of values)

Therefore, from Eq.2 --> ST = st - s + 10t -10
Since ST = st, we can cancel them off on either side of the above equation.

That gives us ST - st = (-s+10t-10) or (-s+10t-10) = 0

Hope this clarifies.

navaneethmmb , ScottTargetTestPrep

Thanks for the detailed explanation. I get it now. Previously, I was ignoring the equation (i), D = S * t in my calculation.

My pleasure!!

ScottTargetTestPrep

Hello!

Maybe im a bit tired but I have been reading the problem por a whole hour and I dont understand why cant we equate equations 2 and 3.

Kind regards!

Using ST=D equation
according to question
ST = D ------(i)
(S+10)(T-1) = D ------(ii)
(S-6)(T+1) = D -------(iii)

Equating (i) and (ii)
(S+10)(T-1) = ST
S = 10T - 10 -------(iv)

Equating (ii) and (iii)
(S+10)(T-1) = (S-6)(T+1)
S = 8T - 2 -------(v)

from (iv) and (v)
T = 4
S = 30
D = 120

A is correct.

We actually did equate equations 2 and 3; we obtained equation 5 as a result of that.

can this be done under 2 mins?

It's a tough problem. I'd say closer to 3 minutes.

This was not very clear

Might want to break down the steps as little




Posted from my mobile device

This is the clearest explanation by far

Everyone upvote this one

Posted from my mobile device

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