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

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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink]
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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 [ 3.6 KiB | Viewed 9363 times ]

2T + 2 = 18

So, 2T = 16

Or, t = 8
Attachment:

Table 2.PNG [ 3.76 KiB | Viewed 9354 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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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink]
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

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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink]
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?

(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]
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Re: Bill spends two days driving from Point A to Point B. On the first day [#permalink]
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