Bunuel wrote:
Every morning Jim walks to his office at a certain constant speed which enables him to arrive exactly on time. One fine morning he started walking at two-thirds his usual speed but after some time he realized that he was already 10 minutes behind at that point so he doubled his speed and reached his office 5 minutes early. What fraction of the total distance had Jim covered when he doubled his speed?
(A) 5/12
(B) 3/8
(C) 1/3
(D) 1/4
(E) 7/24
ALGEBRAIC wayLet the distance traveled be d at a constant speed of x mph, and a out of total d be traveled at 2x/3 mph.
DISTANCE: aTime taken on a normal day = \(\frac{a}{x}\)
Time taken on the specific day = \(\frac{a}{\frac{2x}{3}}=\frac{3a}{2x}\)
Thus, \(\frac{3a}{2x}-\frac{a}{x}=\frac{10}{60}.........\frac{a}{x}=\frac{1}{3}\)
DISTANCE: d-aTime taken on a normal day = \(\frac{d-a}{x}\)
Time taken on the specific day = \(\frac{d-a}{\frac{4x}{3}}=\frac{3a}{4x}\)
Time saved is 5 minutes, so the time taken at increased speed would have saved 10+5 or 15 minutes.Thus, \(\frac{d-a}{x}-\frac{3(d-a)}{4x}=\frac{(d-a)}{4x}=\frac{15}{60}.........\frac{d}{x}-\frac{a}{x}=1\)
\(\frac{d}{x}-\frac{1}{3}=1.........\frac{d}{x}=\frac{4}{3}\)
We are looking for the value of \(\frac{a}{d}\)
Now, \(\frac{\frac{a}{x}}{\frac{d}{x}}=\frac{a}{d}=\frac{\frac{1}{3}}{\frac{4}{3}}=\frac{1}{4}\)
D
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