An excellent rates problem. Thanks SravnaTestPrep.

Alright, let's solve it

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 :

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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