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M12-08

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Bunuel
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M12-08 [#permalink] New post  16 Sep 2014, 00:46
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If a speed-boat travels at 35 kmh, it will arrive at the port 2 hours late. If the boat travels at 50 kmh, it will arrive 1 hour early. How fast does the boat have to go to arrive at the port on time?

A. 42.00 kmh
B. 43.50 kmh
C. 43.75 kmh
D. 44.25 kmh
E. 44.50 kmh
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Bunuel
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Re M12-08 [#permalink] New post  16 Sep 2014, 00:46
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If a speed-boat travels at 35 kmh, it will arrive at the port 2 hours late. If the boat travels at 50 kmh, it will arrive 1 hour early. How fast does the boat have to go to arrive at the port on time?

A. 42.00 kmh
B. 43.50 kmh
C. 43.75 kmh
D. 44.25 kmh
E. 44.50 kmh

The equation is \((t + 2)*35 = (t - 1)*50\). From this equation, scheduled time \(t = \frac{120}{15} = 8\)hours. The distance the boat has to cover is \((8 - 1)*50 = 350\) km. The optimal speed is \(\frac{350}{8} = 43.75\) kmh.

Answer: C
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Re: M12-08 [#permalink] New post  20 Nov 2019, 00:00
Hi - I understand this solution, however, i’m confused how you got to 15?? can anyone help

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Re: M12-08 [#permalink] New post  06 Mar 2020, 07:50
strsfstr wrote:
Hi - I understand this solution, however, i’m confused how you got to 15?? can anyone help


I hope i understood your question right.
You should be getting 15 as the denominator because of the equation.

\((t + 2)*35 = (t - 1)*50\)
\(35t+70 = 50t-50\)
\(120 = 15t\)
\(\frac{120}{15} = t = 8\) hours
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Sri07
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Re: M12-08 [#permalink] New post  10 Jul 2020, 06:07
I used the LCM method.

I took the LCM of 35 and 50 i.e 350
Now, we have the (assumed) distance as 350

Now, T1=350/35=10 hrs (which is 2 hrs late, so actual time should be 8 Hrs)
Similarly T2 350/50=7 Hrs (Which is 1 hr early, so actual time should be 8 hrs)

We have distance we have time speed 350/8=42.75 m/hr

Although I think the approach by maths expert Bunuel is much better
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Re: M12-08 [#permalink] New post  29 Jul 2022, 03:37
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Re: M12-08 [#permalink]
29 Jul 2022, 03:37
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