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Bill spends two days driving from Point A to Point B. On the first day

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grantcke
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Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  24 Oct 2016, 09:50
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Bill spends two days driving from Point A to Point B. On the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. If during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

A) 20
B) 25
C) 28
D) 30
E) 35
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BrentGMATPrepNow
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Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  Updated on: 20 Dec 2019, 08:41
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grantcke wrote:
Bill spends two days driving from Point A to Point B . On the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. If during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

A) 20
B) 25
C) 28
D) 30
E) 35


First recognize that we have TWO pieces of information regarding the time Bill spent driving each day.
On day 1, Bill drove 2 hours longer than he drove on day 2.
So, let x = # of driving hours on day 2
Then x + 2 = # of driving hours on day 1

Bill drove a TOTAL of 18 hours
So, x + (x + 2) = 18
Simplify: 2x + 2 = 18
Solve, x = 8
So, Bill drove 10 hours on day 1 and he drove 8 hours on day 2

Now let's solve the question by starting with a word equation.
Let x = speed driven on day 2
So, x + 5 = speed driven on day 1

(Distance traveled on day 1) + (Distance traveled on day 2) = 680
Distance = (rate)(time)
We get: (x+ 5)(10) + (x)(8) = 680
Expand: 10x + 50 + 8x = 680
Simplify: 18x + 50 = 680
18x = 630
x = 35 (mph)

Answer:
Show Spoiler
E


Cheers,
Brent

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Originally posted by BrentGMATPrepNow on 24 Oct 2016, 10:09.
Last edited by BrentGMATPrepNow on 20 Dec 2019, 08:41, edited 1 time in total.
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gauravk
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  24 Oct 2016, 10:02
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10 (x+5) + 8x = 680

Solve for X = 35

Explanation:

Lets say he drove at X mi/hr on second day and he drove for Y hours

Question says he drove 2 hrs extra on day 1 so day 1 hours = Y + 2

Speed on day 1 is 5 mi/hr faster so, speed on day 1 = x + 5

Total time he drove (Day 1 + Day 2) = 18

Y + (Y+2) = 18

Y = 8 (Second day's time)

Y + 2 = 10 (First day's time)

Total distance covered = 680 = Speed of day 1* time for day + Speed of day2* time of day2

Putting values we get the very first equation.

HTH
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Abhishek009
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  25 Oct 2016, 08:56
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grantcke wrote:
Bill spends two days driving from Point A to Point B . On the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. If during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

A) 20
B) 25
C) 28
D) 30
E) 35

Hi does anyone have an easier way to try solve this?


Forma table -
Attachment:
TABLE.PNG
TABLE.PNG [ 3.6 KiB | Viewed 8222 times ]


2T + 2 = 18

So, 2T = 16

Or, t = 8
Attachment:
Table 2.PNG
Table 2.PNG [ 3.76 KiB | Viewed 8219 times ]


Now, 10s + 50 + 8s = 680

Or, 18s = 630

Or, s = 35

Hence, average speed on day 2 was 35, answer will be (E)...
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ScottTargetTestPrep
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  13 Dec 2017, 17:26
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grantcke wrote:
Bill spends two days driving from Point A to Point B . On the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. If during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

A) 20
B) 25
C) 28
D) 30
E) 35


We are given that on the first day, Bill drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. We can let the rate on the second day = r and the rate on the first day = r + 5. Also, we can let the time on the second day = t and the time on the first day = t + 2.

Since the total time is 18, we can create the following equation to determine t:

t + t + 2 = 18

2t = 16

t = 8

Thus, the distance on day 2 is 8r and the distance on day 1 is (r + 5)(10) = 10r + 50

We can create the following equation to determine r:

8r + 10r + 50 = 680

18r = 630

r = 35

Answer: E
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avocado310
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  09 Dec 2018, 19:28
Abhishek009 wrote:
grantcke wrote:
Bill spends two days driving from Point A to Point B . On the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. If during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

A) 20
B) 25
C) 28
D) 30
E) 35

Hi does anyone have an easier way to try solve this?


Forma table -
Attachment:
TABLE.PNG


2T + 2 = 18

So, 2T = 16

Or, t = 8
Attachment:
Table 2.PNG


Now, 10s + 50 + 8s = 680

Or, 18s = 630

Or, s = 35

Hence, average speed on day 2 was 35, answer will be (E)...


I'm probably confusing formulas, but aren't you allowed to "add rates" together?

I had created the formula

(2s+5) x 18 = 680
Clearly got the wrong answer but hoping someone can refresh me onto a strategy that shouldn't be used for this!
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink] New post  05 Feb 2023, 14:27
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