You can do it through algebraic way too, but a logical way that can get you to the answer fast is
(I) If the ratio of speed for same distance is 1:2/3, the time taken would be 2/3:1.
Difference = \(x-\frac{2x}{3} = \frac{x}{3} = 10....x=30\) minutes
Time taken if covered in original speed = \(30*\frac{2}{3}=20\) minutes
(II) If the ratio of speed for distance in second part is 1:2*2/3, the time taken would be 2*2/3:1.
Difference = \(\frac{4x}{3}-x= \frac{x}{3} = 10+5....x=45\) minutes { 15 comes from covering 10 minutes and overall 5 less minutes}
Time taken if covered in original speed = \(45*\frac{4}{3}=60\) minutes

Total time is 80 minutes through original speed, out of which 20 minutes route is done at 2/3rd of original speed => \(\frac{20}{80} = \frac{1}{4}\)


D
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assume the original speed is 60mph for 60 min meaning the total distance is 60 miles
if the answer is D
2/3 speed is 40mph for (45/2) min which is 15 miles (which is where we get the 15/60=1/4 of the distance)
4/3 speed is 80mph for (135/4) min which is the remaining 45 miles
but then the total time spent is greater than 55 min (even though question says he arrives 5 min earlier)

????

That is not correct. You cannot assume all values as some would be interconnected.

If you want to test whether the solution for D in the earlier post is correct, you will have to take time as 80 minutes as that is the total time as per the solution.


Hope it helps
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ALGEBRAIC way

Let 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: a
Time 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-a
Time 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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Hello Sir,
Can you please explain this step to me?

Hi Bunuel, please provide with a solution. Thanks.

Set D to total distance, D1 to intermediate distance and R to original speed.

The first part of the problem translates to

D1/(2R/3) = D1/R + 10

The second part translates to

D1/(2R/3) + (D-D1)/(4R/3) =
D/R - 5

Clearing R from the denominator of the first and solving for R yields

R = D1/20

Clearing R from the denominator of the second yields

3D1/2 +3(D-D1)/4 = D-5R

Substituting for R=D1/20 and expanding the above yields

3D1/2 - 3D1/4 + 3D/4 = D -
D1/4

Multiplying through by 4

6D1 - 3D1 + 3D = 4D - D1

So

4D1 = D and D1/D = 1/4

Posted from my mobile device

I could finally solve this question, all because of your solution Thank you so so so much!