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SravnaTestPrep
A car travelled the first quarter of a certain distance at twice the speed it traveled the remaining distance. What proportion of the total time travelled, was the time taken to travel the first quarter of the distance?

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8


Solution:

1.\(S1, T1, D1\) be the speed, time of travel and the distance in the first quarter. \(S2, T2\) be the speed and time of travel in the remaining distance and\(D2\) be the remaining distance.

2. We need to find \(T1/ (T1 + T2)\) = \((D1/S1) / (D1/S1 + D2/S2)\)

3. \(S1= 2S2\), \(D2=3D1\)

4. Substituting (3) in RHS of (2), the RHS becomes: \((D1/2S2) / (D1/2S2 + 3D1/S2)\)

\(= (D1/2S2) / (7D1/ 2S2)\)
\(=1/7\)

Therefore the answer is choice D.
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An excellent rates problem. Thanks SravnaTestPrep.

Alright, let's solve it :-D

The first notion you've got to keep in mind is : distance (d) = speed (S) x time (t)

The total distance can be represented as follows :
Attachment:
PS1.jpg
PS1.jpg [ 26.06 KiB | Viewed 7558 times ]

So taking into account the fact that the car travelled d1 at twice the speed it travelled d2, ergo S1 = 2 x S2 we get :

The speed used to travel d1 is : \(S1 = \frac{d1}{t1} = \frac{d}{(4*t1)}\)
The speed used to travel d2 is : \(S2 = \frac{d2}{t2}= \frac{(3*d)}{(4*t2)}\)

Therefore, by using S1 = 2 x S2 we get :
\(\frac{d}{(4*t1)} = 2*\frac{(3*d)}{(4*t2)}\)

Simplifying the "d"s and dividing 4 by 2 yields :
\(\frac{1}{(4*t1)} = \frac{3}{(2*t2)}\)

Which gives us :
\(2*t2 = 12*t1\) which is equivalent to \(t2 = 6*t1\)

And since we're interested in the proportion of the total time travelled necessary to travel the first quarter of the distance then :

Knowing that the total time is : \(t = t1 + t2\)

We substitute t2 in the expression which gives us : \(t = t1 + 6*t1\)

Meaning that : \(t = 7*t1\) or \(t1 =\frac{t}{7}\) which is answer choice D.

Hope that helped :)
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Lets say 1st quarter distance = \(\frac{1}{4}\) & speed = 2s

So, remaining distance = \(\frac{3}{4}\) & speed = s

Time required in First Quarter\(= \frac{\frac{1}{4}}{2s} = \frac{1}{8s}\)

Total Time required for travelling complete distance\(= \frac{1}{8s} + \frac{3}{4s} = \frac{7}{8s}\)

\(Fraction = \frac{\frac{1}{8s}}{\frac{7}{8s}} = \frac{1}{7}\)

Answer = D
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Substitution method-

let distance be 120 . So, quarter of 120= 30 km.

Let speed at which 120-30 = 90 km traveled be 10 kmph.. So, speed at which quarter portion traveled be twice of 10 = 20 kmph .

Now,
time taken to travel quarter distance = 30/20 = 3/2 hours.

time taken to travel rest of distance = 90/10 = 9 hours.

Hence, total time = 9 + 3/2 = 21/2

therefore, time required to travel quarter distance / total time

= 3/2 / 21/2

= 1/7

Answer option D.
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distance first quarter = 0.25d
speed first quarter = 2x.

distance remaining part = 0.75d
speed remaining part = x.

time first quarter is d/8x. time for the remaining part 3d/4x

now, total time: 7d/8x.

since we are asked the ratio of the first portion to the total time: d/8x * 8x/7d => 1/7
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let d=first quarter distance
2s=first quarter speed
(d/2s)+(3d/s)=7d/2s total time
(d/2s)/(7d/2s)=1/7 ratio of first quarter time to total time
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