Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

23 Dec 2015, 14:17

1

8

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

73% (02:12) correct 27% (01:56) wrong based on 172 sessions

HideShow timer Statistics

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?

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

27 Dec 2015, 08:48

2

6

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.

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

23 Dec 2015, 14:36

1

1

RSOHAL wrote:

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

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

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

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

23 Jan 2016, 09:26

Engr2012 wrote:

RSOHAL wrote:

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

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.

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.

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

23 Jan 2016, 10:01

1

pepo wrote:

Engr2012 wrote:

RSOHAL wrote:

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

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.

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?

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

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

23 Jan 2016, 12:32

1

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.

Re: John would have reduced the time it took him to drive from his home to
[#permalink]

Show Tags

02 Mar 2018, 07:49

Top Contributor

RSOHAL wrote:

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

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