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# If a car had traveled 20 kmh faster than it actually did, the trip

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Math Expert
Joined: 02 Sep 2009
Posts: 44644
If a car had traveled 20 kmh faster than it actually did, the trip [#permalink]

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26 Jun 2016, 03:20
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25% (medium)

Question Stats:

78% (02:27) correct 22% (02:59) wrong based on 121 sessions

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If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh
[Reveal] Spoiler: OA

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Posts: 1619
Location: India
Concentration: Strategy, General Management
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Re: If a car had traveled 20 kmh faster than it actually did, the trip [#permalink]

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26 Jun 2016, 03:38
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Time = Distance/Speed

Difference in time = 1/2 hrs

60/x - 60/(x + 20) = 1/2

Substitute the value of x from the options. --> x = 40 --> 60/40 - 60/60 = 3/2 - 1 = 1/2

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Joined: 22 May 2016
Posts: 1564
If a car had traveled 20 kmh faster than it actually did, the trip [#permalink]

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30 Dec 2017, 11:03
Bunuel wrote:
If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

Using answer choices, it is much easier if you use 30 minutes = $$\frac{1}{2}$$ hour

If original speed was 50, then time, t = D/r:
Time at 50 mph: $$\frac{60}{50}=\frac{6}{5}$$ hours
20 mph faster? (50 + 20) = 70 mph. t = D/r
Time at 70 mph: $$\frac{60}{70}=\frac{6}{7}$$ hours

That hour fraction in minutes will not yield an integer (30 minutes), but find the difference in times to get a benchmark
$$(\frac{6}{5} - \frac{6}{7})=\frac{(42-30)}{35}=\frac{12}{35}$$ hours
$$\frac{12}{35}\approx{\frac{1}{3}}$$ hour
$$\frac{1}{3} < \frac{1}{2}$$ hr

We need time to be longer. That means speed must be slower (decreased speed = increased travel time). Eliminate answers D and E.

Try B) 40
Time at 40 mph: $$\frac{60}{40} = \frac{3}{2}$$ hour
20 mph faster: (40 + 20) = 60 mph
Time at 60 mph: $$\frac{60}{60}=1$$ hour
Time difference? $$\frac{3}{2} - 1 = \frac{1}{2}$$ hour = 30 minutes

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Re: If a car had traveled 20 kmh faster than it actually did, the trip [#permalink]

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31 Dec 2017, 10:06
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Bunuel wrote:
If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

Let the actual speed be "S" kmh

The increased speed will be "S + 20"

The difference in timing has been given as 30 minutes or 1/2 hour

Actual distance travelled = 60 km

Thus, we can write :

$$\frac{60}{S} - \frac{60}{(S+20)} = 0.5$$

Though we can solve the equation, a quicker way would be to look at the options, which either gives us an integer when $$\frac{60}{S}$$ or when $$\frac{60}{(S+20)}$$.

35, 50 and 65 would not help get integer values of time.

A good option would be 40, because $$\frac{60}{40} = 1.5$$ and $$\frac{60}{(40+20)} = 1$$ and the difference is 1.

Thus, the correct answer would be Option B.
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Joined: 07 Dec 2014
Posts: 965
Re: If a car had traveled 20 kmh faster than it actually did, the trip [#permalink]

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31 Dec 2017, 14:00
Bunuel wrote:
If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

let r=actual kmh
r+20=60/[(60/r)-1/2]➡
r^2+20r-2400=0
r=40 kmh
B
Re: If a car had traveled 20 kmh faster than it actually did, the trip   [#permalink] 31 Dec 2017, 14:00
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