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

Question

A car travelling from city A to B takes an hour lesser to cover the distance if it moved 10 kilometres per hour faster. On the other hand it will take an hour more to cover the distance if it moved 6 kilometres per hour slower. What is the distance between the two cities?

A. 120 kilometres
B. 150 kilometres
C. 180 kilometres
D. 360 kilometres
E. 380 kilometres

Are You Up For the Challenge: 700 Level Questions

A. 120 kilometres
B. 150 kilometres
C. 180 kilometres
D. 360 kilometres
E. 380 kilometres

Are You Up For the Challenge: 700 Level Questions

navaneethmmb · Accepted Answer

Distance = Speed (s) x time (t) 
Case 1: If speed is (s+10), then time taken is (t-1)
Case 2: If speed is (s-6), then time taken is (t+1)

From Case 1,
Distance = st = (s+10)*(t-1) = st - s + 10t -10 Or,  s-10t+10 = 0  
From Case 2, 
Distance = st = (s-6)*(t+1) = st + s - 6t - 6 Or,  s - 6t - 6 = 0

Solving the above two equations, we get t = 4 h
Plugging in value of t in the above equations gives s = 30 kmph

Therefore Distance = 30*4 = 120 km

Ans (A)

ScottTargetTestPrep · Answer

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

Sarkar93 · Answer

Can you kindly explain why you equated the two equations in case 1 and case 2 to 0? 
Thanks

Archit3110 · Answer

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

navaneethmmb · Answer

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.

Sarkar93 · Answer

navaneethmmb  ,  ScottTargetTestPrep

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

ScottTargetTestPrep · Answer

My pleasure!!

jfranciscocuencag · Answer

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!

QuantMadeEasy · Answer

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.

ScottTargetTestPrep · Answer

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

hellothere88 · Answer

can this be done under 2 mins?

ScottTargetTestPrep · Answer

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

Andrewcoleman · Answer

This was not very clear

Might want to break down the steps as little

Posted from my mobile device

Andrewcoleman · Answer

This is the clearest explanation by far

Everyone upvote this one

Posted from my mobile device

bumpbot · Answer

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