Edwin is planning to drive from Boston to New Orleans. By what percent would his travel time be reduced if Edwin decides to split the driving time equally with his friend George, instead of making the trip alone?

(1) The driving distance from Boston to New Orleans is 1500 miles.
(2) George’s driving speed is 1.5 times Edwin’s driving speed.


Statement 1 : Does not tell us either Edwin'speed or George's. Hence we cannot calculate time taken in any scenario. INSUFFICIENT

Statement 2 : Let distance be d

Time taken originally = d/s

Now in second scenario Edwin drives k miles and George drives d-k miles

Therefore time taken = k/s + d-k/(1.5s)

k/s = (d-k) / (1.5)s ... Given

d=2.5k

Therefore original time taken = 2.5k /s

New time taken = k/s + (1.5k/1.5s)

= (2k)/s

Reduction is 0.5 k/s

Therefore percent reduction is 20%

SUFFICIENT

B.

1) only distance is given, speed/time is not given so insufficient.

2) let edwin's speed be x. --> george's speed becomes 1.5x
so when he travels alone, he must have taken d/x amount of time
now when he splits the driving time equally with george, then lets assume he drove for k miles
=> k/x = (d-k)/1.5x
=> 2.5k = d

earlier he travelled for 2.5k/x hours, now he travels for k/x hours. so time reduction rate can be calculated. hence, sufficient.
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I got this question on Mgmat too.
Time was less so i just did it very fast.
A- We have to see the time but only distance given. Cant take any thing out of it. So A and D gone.
B- Driving speed relation given and we know from question stem time is split equally. Bingo Time and speed relation can be established and there by the reduced time percent.
So answer is B

A is not sufficient as the distance alone does not tell us the respective speeds of the two.

B alone is sufficient because ratio os speed also gives a ratio of time, as the distance is constant

When distance is constant, Speed is inversely proportional to time
When speed is constant, Distance is proportional to time
When time is constant, distance is proportional to spee.
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rishi02 - You are taking both Edwin and George's time into consideration while calculating the reduction, whereas only Edwin's Driving time reduction is asked. Earlier Edwin was taking 2.5k/s time alone for driving and after taking help from George, his driving time has reduced to k/s. So, percentage of Edwin's driving time reduction can be calculated.
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The question asks what is the reduction in travel time as a result of the drive being split with George who from statement 2 we know is a faster driver. But the reduction should be set up as comparing it between him driving alone (2.5k/x) vs his total time by splitting the drive with George, not just his share of it; in the 2nd scenario, his travel time would include the period where George drives too, wouldn't it?
I'm not sure if I'm right, please let me know Bunuel ?

Statement 1 : Does not tell us either Edwin'speed or George's. Hence we cannot calculate time taken in any scenario. INSUFFICIENT
Statement 2: Tells us the speed rate of edwin and george.
It's already given that both of them drove for an equal amount of time , so u can find the % amount of time reduced irrespective of the distances because of the reasons provided.

So, statement 2 alone is sufficient

Wanted (T_old - T_new)/T_old = ?

1) D = 1500
Not sufficient

2) Use formula: R1T1 + R2T2 = W
T2 = T1 (Given: "split the driving time equally")
R1 = S
R2 = 1.5S
ST1 + 1.5ST1 = D
T1 (2.5S) = D
T1 = D/(2.5S)

T_new = T1 + T2 = 2*T1 = 2D/(2.5S)

Now we have both T_new and T_old so we can stop here => Sufficient

Just to show the calculations:
(D/S - 2D/2.5S)/(D/S) = ((2.5D - 2D)/(2.5S))/(D/S) = (0.5D/2.5S)/(D/S) = 0.5/2.5 = 0.2 = 20% => Edwin's travel time decreases by 20%

ANSWER: B

we start by giving the time when split driving and algebraic equation: T2 = D/2*1.5*Speed George + Distance/2*Speed George

After simplifying this equation we get T2=D/Speed George * 1/5 => T2=T1*1/5

which means that the difference between T1 which is time if George drive alone and T2 time when they split driving is T1 (1-1/5) a 20% decrease

I am stuck at one point please help:

original time of edwin : D/x

New time : Edwin + George
: D/2x + D/2*1.5x (x being the speed of edwin and also since the distance travelled got halved)

On solving : D/x - (D/2x+ D/2*1.5x) = D/x- 2.5D/2*1.5x = 0.5D/3x .....1

then,

0.5D/3x / D/x ..... being the original speed

But I am not getting 20% decrease. Can someone help ????

They decided to split the travel time not the travel distance. Might be clearer with some sample numbers:

Consider Edwin's speed as 1 unit/hr and George's speed as 1.5 unit/hr and assume distance is 15 units. Average speed if each drove for same amount of time would be 1.25 unit/hr.

Time taken to cover distance by 1) Edwin alone = 15 hours i.e. d/s = 15/1
2) Combined with each driving equal amount of time = 12 hours i.e. d/s = 15/1.25

Reduction = 3 hours = 3/15 = 20%