John would have reduced the time it took him to drive from his home to

Question

John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

Source: GMAT Focus

(A) 25 
(B) 30 
(C) 40 
(D) 45 
(E) 50

Source: GMAT Focus

EMPOWERgmatRichC · Accepted Answer

Hi All,

This question is based on a math 'truism' that many Test Takers don't know: Increasing your speed by 50% decreases the amount of travel time by 1/3. Here's why:

Distance = (Rate)(Time)

D = (R)(T)

Since the distance remains the same (you're just changing the rate and time), any increase in rate or time is met with a change in the other term. Increasing the rate by 50% would give us...

D = (3/2)(R)(?)

The change in time has to 'offset' the (3/2) that is now part of the calculation. The fraction (2/3) creates that 'offset', which gives us....

D = (3/2)(R)(2/3)(T)

D = (6/6)(R)(T)

In this prompt, we're told that the time is decreased by 1/3 after the 15 miles/hour increase in speed. We're asked for the ORIGINAL speed, which means that the 15 miles/hour represents the 50% increase. Thus, the original speed must be 30 miles/hour.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

BrentGMATPrepNow · Answer

Let's start with a  word equation:  
 (John's travel time at faster speed)  = 2/3 (John's regular travel time)

Let  r = John's regular speed 
So,  r+15 = John's faster speed 
Let  d = the distance traveled at regular speed 
So  d = the distance traveled at regular speed

time = distance/speed 
So, we get:  d/(r + 15)  = (2/3) (d/r)  
Rewrite as: d/(r + 15)= 2d/3r
Cross multiply: (d)(3r) = 2d(r + 15)
Expand: 3rd = 2rd + 30d
Rearrange to get: rd = 30d
Divide both sides by d to get:  r = 30

Answer: B

RELATED VIDEO 
 https://www.youtube.com/watch?v=XwVu_TEXQqs

ENGRTOMBA2018 · Answer

Let v,t be the original speed and time taken by John

As the total distance (=speed*time) must be the same,

v*t = (2t/3)*(v+15) ---> v=30 miles per hour.

B is the correct answer.

pepo · Answer

I like the way you approached the question. I tried to solve the question in a similar way, but without success

the new time given is equal to x-1/3, while the distance is the same and the John's rate changed, which is equal to Disatnce/time - 1/3. In this way I got this equation: x-1/3=D/T-1/3... after that I was stucked. Where did I fail?

ENGRTOMBA2018 · Answer

The mistake you are doing is that when you are told that the time is reduced by 1/3 it's not x-1/3 but x-x/3 = 2x/3

EMPOWERgmatRichC · Answer

Hi pepo,

In this question, the changes in rate and time are based on 'ratios' (NOT on an absolute number).

For example, if you're traveling 60 miles/hour and you reduce THAT speed by 1/3, the calculation is NOT 60 - 1/3.... it's 60 - (1/3)(60) = 40.

Since most of your other math 'skills' seem fine, this issue is ultimately about your organization and how you take your notes. Instead of just writing down "- 1/3", you should think about what that difference represents (it represents a 1/3 decrease in speed) and add a bit more detail to your work.

GMAT assassins aren't born, they're made,
Rich

Mo2men · Answer

There a re different approaches for the question. Engr2012 & Rich demonstrated great algebraic and logic way.

Another way that we can mix and match between both.

distance is constant. Any decrease in time means increased speed (or vice versa)

So time decreased by 1/3..........>remaining time is 2/3.........> speed is 3/2= V2/V1= (V1+15)/V1 (where V2: new speed & V1:original speed)

TESTing The Answers, it is easily to find that

(30+15)/15=45/30=3/2

Answer: B

mvictor · Answer

suppose r=his rate, t his time, rt = distance traveled
new time 2t/3, new rate r+15, distance is the same = rt.

now, (r+15)* 2t/3 = rt
2rt/3 + 10t = rt - multiply by 3 to get rid of the fractions:
2rt +30t = 3rt
30t = rt | divide by t
30=r.

his rate was 30.

eakabuah · Answer

This is my approach. 
Time (t)= distance(d)/speed(s)
This means t is inversely proportional to speed
In this case the distance is the constant since it’s the same and can be ignored since it will cancel out in the equations.
So we can say that t=1/s ........(a)
Where t is the original time taken by John to travel home at a speed of s

We are told that he can reduce his speed by 1/3 had he increased his average speed by 15mph.

So new time would now be 2t/3 and will correspond to a speed of s+15
Hence 2t/3=1/(s+15)
So t=3/(2(s+15)) ........(b)

equating (a) to (b)

1/s=3/(2s+30)
3s=2s+30
Hence s=30

Answer is therefore B

Posted from my mobile device

DanTheGMATMan · Answer

https://www.youtube.com/watch?v=mFU4-NDA_O8

DanTheGMATMan · Answer

https://www.youtube.com/watch?v=mFU4-NDA_O8

AnishPassi · Answer

Speed x Time = Distance (S x T = D)

First, let me explain an underlying idea.

1. If I double my speed would I take as much time as before, more than before or less than before to cover the same distance?

If I go faster, I’d take lesser time.

2. Now, how would the time reduce?
If the speed is doubled, correspondingly the time would get halved to cover the same distance.

This can be understood using the equation too.

If the new speed is 2xS, to cover the same distance D, the time taken would be T/2.

2S x 1/2T = S x T = D

More generally, 
If S x T = D. And I  keep the distance the same , the speed and time would balance each other out.

If speed changes to 2S, time would change to T/2.
If speed changes to 3S, time would change to T/3.
If speed changes to 1/2S, time would change to 2T.
If speed changes to 3/2S, time would change to 2/3T.

And similarly, 
If the time taken changes to 2T, the speed would have been S/2.
If the time taken changes to 3T, the speed would have been S/3.
If the time taken changes to 1/2T, the speed would have been 2S.
If the time taken changes to 3/2T, the speed would have been 2/3S.

For a constant Distance, out of T and S, whatever  multiple  one quantity changes by, the other would change by the reciprocal of that multiple.

Notice: The change is in terms of multiples and not addition or subtraction.

- - 
This understanding gives a way to solve this question.

It is important to notice that John would have reduced the time it took him    by    1/3, and not    to    1/3. So, the time would have reduced to 2/3 of the actual time taken. (If you consider the hypothetical time to be 1/3T and correspondingly the hypothetical speed to be 3S, that would a mistake.)

John would have reduced the time to 2/3T (if T is the actual time taken).
So, correspondingly, the speed would have been 3/2S (if S is the actual average speed).

Now comes the next bit to be careful about.

The question mentions: “if he had increased his average speed  by 15 miles per hour ”.

How do we reconcile this increase of 15mph with the fraction I mentioned above (3/2S)?

If the speed changed from S to 3/2S, the speed increased by 1/2S.

This 1/2S is the 15 mph mentioned in the question.

1/2S = 15 mph
-> S = 30 mph

Answer: B.

An alternate approach:

(For problem solving questions, I already know that the answer has to be one of the answer choices. So, I also like to start with the options to see if I can figure out the correct answer that way - by either eliminating all the wrong ones and/ or by finding the one that fits.)

(A) If the speed changes from 25mph to 25+15=40 mph, the speed has changed by a multiple of 40/25=8/5. So, the time would change by a multiple of 5/8. But according to the information, the time would have reduced to 2/3 of the original time. 2/3, not 5/8. So, I can reject (A).

(B) If the speed changes from 30mph to 30+15=45 mph, the speed has changed by a multiple of 45/30=3/2. So, the time would change by a multiple of 2/3. This one fits! (Time reducing by 1/3 is the same as time becoming 2/3 of the original time. e.g. x - 1/3x = 2/3x)

I could very well stop here, mark B and move on. There will be only one multiple answer choices that fit all the given information. For some additional practice for you, I’ll go through the rest.

(C) If the speed changes from 40mph to 40+15=55 mph, the speed has changed by a multiple of 55/40=11/8. So, the time would change by a multiple of 8/11. But according to the information, the time would have reduced to 2/3 of the original time. So, I can reject (C).

(D) If the speed changes from 45mph to 45+15=60 mph, the speed has changed by a multiple of 60/45=4/3. So, the time would change by a multiple of 3/4. But according to the information, the time would have reduced to 2/3 of the original time. So, I can reject (D).

(E) If the speed changes from 50mph to 50+15=65 mph, the speed has changed by a multiple of 65/50=13/10. So, the time would change by a multiple of 10/13. But according to the information, the time would have reduced to 2/3 of the original time. So, I can reject (E).

A modification to the alternate approach:

If this whole fraction and multiple business is still feeling overwhelming, here is another approach.

I’ll assume that the original time taken was 3 hours. (I wanted to choose a multiple of 3 because I’m going to multiple this figure by 1/3, and I wanted to keep my life simple.)
Also, I can assume the time to any figure since the question anyway doesn’t go into those specifics of time and distance.

If the original time taken was 3 hours, after the 1/3 reduction, the time taken would become 2 hours.

Now I’m going to check if these fit with the answers choices.

(A) Original speed: 25mph. Hypothetical speed: 40mph. 
If at 25mph, one took 3 hours, the total distance would be 75 miles. Would they take 2 hours to cover 75 miles at 40mph? No. 75/40 is less than 2 (80/40 would have been 2). Rejected.

(B) Original speed: 30mph. Hypothetical speed: 45mph. 
If at 30mph, one took 3 hours, the total distance would be 90 miles. Would they take 2 hours to cover 90 miles at 45mph? Yes. 90/45 = 2.

(Again, we could just mark B at this point and move on. I’ll go through the rest just so you can get some more practice.)

(C) Original speed: 40mph. Hypothetical speed: 55mph. 
If at 40mph, one took 3 hours, the total distance would be 120 miles. Would they take 2 hours to cover 120 miles at 55mph? No. 120/60 would have been 2. 120/55 would be greater. Rejected.

(D) Original speed: 45mph. Hypothetical speed: 60mph. 
If at 45mph, one took 3 hours, the total distance would be 135 miles. Would they take 2 hours to cover 135 miles at 60mph? No. 120/60 would have been 2. 135/60 would be greater. Rejected.

(E) Original speed: 50mph. Hypothetical speed: 65mph. 
If at 50mph, one took 3 hours, the total distance would be 150 miles. Would they take 2 hours to cover 150 miles at 65mph? No. 130/65 would have been 2. 150/65 would be greater. Rejected.

Something we could learn after answering the question this way:

Notice how for answer choices C, D and E - the ones greater than 30mph - the time taken was coming to more than 2 hours for each. Would this continue for higher values of original speed as well? Yes. And this can be understood why that would be happening.

As the original speed increases, the 15mph would become less and less significant. So the corresponding time reduction would also become less and less significant.

Think about it:
Say you’re jogging at 5mph v/s you drive a scooter at 20mph, how would your time taken to go to your office change? This one a pretty significant change, right? The speed’s become 4x. The time would become 1/4 the original time.

Say a rocket is traveling at 10,000mph v/s the rocket travels at 10,015mph. Assuming that the machinery can work fine at both these speeds, would the time taken to go to Mars change drastically? Nah. The speed hasn’t even increased by 1%, the corresponding time would not even reduce by 1%.

John would have reduced the time it took him to drive from his home to

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