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John would have reduced the time it took him to drive

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by BTGmoderatorDC » Fri Mar 02, 2018 2:32 am
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

Can some experts show me the formula on how to solve it?

OA B
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by GMATGuruNY » Fri Mar 02, 2018 3:06 am
lheiannie07 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
To reduce the time by 1/3 = to take 2/3 his actual time.
Time and rate have a RECIPROCAL RELATIONSHIP.
To take 2/3 his actual time, John must travel at 3/2 his actual speed.
Since 3/2 his actual speed increases the speed by 15 miles per hour, we get:
(3/2)s = s + 15
3s = 2s + 30
s = 30.

The correct answer is B.
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by Brent@GMATPrepNow » Fri Mar 02, 2018 6:47 am
lheiannie07 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
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

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by mbawisdom » Fri Mar 02, 2018 6:51 am
lheiannie07 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

Can some experts show me the formula on how to solve it?

OA B
S = Speed
D = Distance
T = Time

We know that S = D/T
From the question we know that S + 15 = D/(2/3)T (if the speed goes up by 15, the time is cut by a third)

S + 15 = D/(2/3)T
S + 15 = (3/2)(D/T)
S + 15 = (3/2)(S) <--- S = D/T
15 = (1/2)S
S = 30
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by Scott@TargetTestPrep » Tue Mar 06, 2018 9:10 am
lheiannie07 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
We can let John's speed from his home to the store = r and his time = t. Thus, we have:

rt = (r + 15)(2/3)t

r = (r + 15)(2/3)

3r = 2r + 30

r = 30

Answer: B

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by [email protected] » Tue Mar 06, 2018 5:45 pm
Hi lheiannie07,

We're told that 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. We're asked for John's actual average speed, in miles per hour, when he drove from his home to the store. This question can be solved by TESTing THE ANSWERS (with a little TESTing VALUES thrown in).

We're never told what the distance traveled is, so we can TEST any Value we choose. In addition, it's interesting that increasing his speed by 15 miles/hour would lead to an exact 1/3 decrease in time. Usually, numbers don't 'interact' so nicely (and you end up with weird fractions); here though, the 1/3 decrease heavily implies that John's current speed is some multiple of 15 (so that when we add 15 to that number, we're increasing by a 'nice' percentage).

Let's TEST Answer B: 30 miles/hour
IF.... John travels 30 miles at 30 miles/hour, then the travel time will be 1 hour
increasing that speed by 15 miles/hour would have John traveling 30 miles at 45 miles/hour, so the travel time will be 2/3 of an hour.
This is a decrease of exactly 1/3, which matches what we were told - so this MUST be the answer.

Final Answer: B

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