Due to construction, the speed limit along an 8-mile section of highwa

Let t1 is time taken @ 55 mph
t1 = 8/55
Let t2 is time taken @ 35 mph
t2 = 8/35
Difference in time = 8/35 - 8/55 = 8 * 20 / (35 * 55) hr = 160 * 60 / (35 * 55) mins = 4.7 = 5 (approx) mins

8/35 - 8/55 = 8/5 * ( 11 - 7)/77

= 8/5 * 4/77 * 60 min

= 8 * 12 * 4/77

= 384/77 ~ 4.9

Answer - A
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Old time in minutes to cross 8 miles stretch = 8*60/55 = 8*12/11 = 96/11= 8.72
New time in minutes to cross 8 miles stretch = 8*60/35 = 8*12/7 = 13.71

Time difference = 13.71-8.72= 5

Ans: "A"

I wondering if someone can use componendo / dividendo since t1/t2 is known. Hence average speed and average time can come into picture. Otherwise there is no takeaway from this problem.

8/35 - 8/55 ~= 5

Answer is A.

Since answer choices are fairly dispersed, we can use approximations here.

Difference is given by \(8*60/35 - 8*60/55\)

We can substitute 36 instead of 35 and 54 instead of 55 in expression above without sacrificing accuracy too much

We will get \(8*10/6-8*10/9 = 8*10*(1/6-1/9) = 8*10/18\) ~4.5 ~5

=>55 x t1 = 35 x t2
=>t2 = 11/7 x t1
=> t2- t1 = 4/7 t1 = 4 x 8 / (7 x 55)
=> 1/16 (approx) hours
1/10 hrs is 6 mins, and 1/20 hr is 3 mins so 1/16 is around 4 and 5
Ans = A
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Original time in minutes = \(\frac{d}{r} * 60 = \frac{8}{55} * 60 = \frac{8}{11} * 12 ≈ 9\) minutes.

\(\frac{new-speed}{original-speed} = \frac{35}{55} = \frac{7}{11}\)
Since the rate and time have a reciprocal relationship:
\(\frac{new-time}{original-time} = \frac{11}{7} ≈ \frac{3}{2}\)

The resulting time ratio indicates that the new time is about 50% greater than the original time.
Thus:
Time increase in minutes ≈ 50% of 9 = 4.5.


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Because we're trying to find the difference in approximate minutes it would take with the new vs old speed, we can use estimation
We are trying to find New Time - Old Time = X

Since we're dealing with Rates and Distance; Use the formula Rate * Time = Distance

Old speed limit: 55 mph * T1 = 8


T1 = \(\frac{8}{55} * 60\) minutes, we need to multiply by 60 to get it in mins

T1 \(\frac{480}{55}\) minutes


New speed limit: 35 mph * T1 = 8


T2 = \(\frac{8}{35} * 60\) minutes, we need to multiply by 60 to get it in mins

T2 \(\frac{480}{35}\) minutes

This is where estimation comes in handy:

T1 \(\frac{480}{55}\) minutes, we know that 55*10 = 550 mins, and 55*9 = 495. So T1 is approx 9 mins

T2 \(\frac{480}{35}\) minutes, we know that 35*10 = 350. You can subtract 480 - 350 = 130. 35 can go into 130 around 4 times (35*4 = 140). So 10 + 4 = approx 14 mins

~14 - ~9 = Approx 5

So the ANSWER IS A

distance = rate * time

distance/rate = time

\(\frac{8}{55} * 60 = \frac{96}{11} = 9\) minutes

\(\frac{8}{35} = \frac{96}{7} = 14\) minutes

\(14 - 9 = 5\) minutes

Answer is A.
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Given: Due to construction, the speed limit along an 8-mile section of highway is reduced from 55 miles per hour to 35 miles per hour.
Asked: Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ?

Change in time required to travel along the section of highway = (8/35 - 8/55)*60 = (8/7 - 8/11)*12 = (11-7)8*12/77 = 4*8*12/7*11 ~ 5 minutes

IMO A
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Quick approximation:

55 mph is almost 1 mile/minute. So at that speed, the trip takes just over 8 minutes. 35 mph is more than half the original speed, so a trip at that speed will take less than double the original time. That means the increase is less than 8 minutes, so A.
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Since the fractions are a bit tedious, I made one assumption.

- Let the distance be 80 miles instead of 8 miles.

Case 1 = @55 mph
Time = 80/55 gives 1.4 hours & for 8 miles we shift a decimal 0.14 hours

Case 2 = 35 mph
Time = 35/55 gives 2.2 hours & for 8 miles we shift a decimal 0.22 hours

Here the difference in 0.22 & 0.14 hours gives the time saved (i,e, 0.08 hours)
0.08 hours is nothing but 8% of 60 minutes and basically any number less than 10% i.e 6 minutes.

Hence answer is A = 5 minutes.