When a certain stretch of highway was rebuilt and straightened, the di

Question

When a certain stretch of highway was rebuilt and straightened, the distance along the stretch was decreased by 20 percent and the speed limit was increased by 25 percent. By what percent was the driving time along this stretch reduced for a person who always drives at the speed limit?

A. 16%
B. 36%
C. 37 1/2%
D. 45%
E. 56 1/4%

PS12502.01

A. 16%
B. 36%
C. 37 1/2%
D. 45%
E. 56 1/4%

PS12502.01

CareerGeek · Accepted Answer

Let the initial distance and speed be 100
—> Initial time = distance/speed = 100/100 = 1

Final distance = 80
Final speed = 125
—> Final time = 80/125 = 0.64

So, decreased by 36%

IMO Option B

Mayukhca2018 · Answer

Let D be distance of the original and S be original speed. 
Therefore time before rebuilding is D/S

Post Rebuilding:
Distance reduced by 20% thus Distance becomes 0.8D
Speed increased by 25%, thus Speed becomes 1.25S
Therefore new time= 0.8D/1.25S=0.64D/S

Therefore decrease in time = (D/S-0.64D/S)/D/S = 36%

Posted from my mobile device

Archit3110 · Answer

earlier distance = 100
new distance = 80
earlier speed = 10
new speed = 12.5
time earlier ; 10
time new ; 80/12.5 ; 6.4
∆ change in time % ; 10-6.4/10 ; 36%
IMO B

TheNightKing · Answer

Let the original distance be = 200miles
Let the original speed be= 20miles per hour.
Original time taken: 10 hours.

New Distance=.8*200=160
New Speed=1.25*20=25
New time=160/25=6.4 hours

Reduction 3.6 hours over 10 hours. So 36%.

Hence B.

Thank you!

ScottTargetTestPrep · Answer

We can let the original distance = 100 miles and the original speed limit = 100 mph, so the original time is 100/100 = 1 hour.

With the decreases and increases, the new distance is 80 miles, and the new speed limit is 125 mph, so the new time is 80/125 = 16/25 = 0.64 hour.

Thus, the percent decrease uses the formula:

(New - Old)/Old x 100 = (0.64 - 1)/1 = -0.36 = -36%

We see that the driving time is reduced by 36%.

Answer: B

akadiyan · Answer

Lets consider initial distance as 100 Miles and initial speed as 40 Miles/Hour

Current distance and speed = 100 Miles = 40 Miles/Hour = 2.5 Hours = 150 Minutes
New distance and speed = 80 Miles = 50 Miles/Hour = 1 Hour and 36 Minutes = 96 Minutes

Time decrease in percentage = ((150-96)/150) *100) = 36%

Ans:   B

Pritishd · Answer

Let's assume values for speed and distance.

Prior to reconstruction  
Speed  =   100 
Distance  =   100 
Time  =   \frac{Distance}{Speed}   =   \frac{100}{100}   =   1

After reconstruction  
Speed  =   125  (We are given that speed increases by  25%  i.e.  100 + 100 * 25%   =   100 + 100 * \frac{25}{100}   =   125 )

Distance  =   80  (We are given that distance decreases by  20%  i.e.  100 - 100 * 20%   =   100 - 100 * \frac{20}{100}  =  80 )

Time  =   \frac{Distance}{Speed}   =   \frac{80}{125}   =   \frac{16}{25}

% Change in time taken

\frac{New - Original}{Original}   * 100

\frac{16 - 25}{25}   * 100

(16 - 25) * 4

9 * 4 = 36%

Ans.  B

hemantbafna · Answer

Distance decreased by 20%= 0.80
Speed increased by 25%=1.25
Time=Distance/Speed= 0.80/1.25= 0.64

1.00-0.64(is equivalent to decrease of 36%)=0.36 
The decrease is of 36%.
 B

Basshead · Answer

let total distance = 100
speed limit = 100

distance was decreased by 20 percent = 80
speed limit was increased by 25 percent = 126

80/126 = 64%

Speed was reduced by 36% -- answer is B

BharatTj · Answer

Bunuel   VeritasKarishma  , Could you please explain why this method is not fetching the correct answer (why this is not an appropriate approach)

Distance Decreased by 20% = 4/5

Speed Increased by 25% = 5/4

Time is inversely proportional to Speed = 4/5

Old time = 1

Change in Time = 1-4/5
 = 20%

KarishmaB · Answer

The relation that time has with distance (time and distance are in the same ratio) holds when speed is constant. So when distance becomes 4/5th, time taken becomes 4/5th too assuming same speed is maintained. So if initial time taken was t, now 4t/5 time is taken. Next, the speed limit is changed. Speed is increased to 5/4 initial speed. Ratio of time taken is inverse of ratio of speed, assuming distance covered is the same. Now we keep distance constant. For the reduced distance at initial speed time taken was 4t/5. For the same reduced distance, if speed becomes 5/4, time taken will become 4/5th of previous time taken i.e. (4/5)*(4t/5) = 16t/25 So time taken reduces by 9/25 which is 36%. Check for TSD using ratios: https://youtu.be/7ASEIvxYPCM

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

When a certain stretch of highway was rebuilt and straightened, the di

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

When a certain stretch of highway was rebuilt and straightened, the di

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards