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Math Expert

Joined: 02 Sep 2009

Posts: 61484

Car X traveled from city A to city B in 30 minutes. The first half of [#permalink]

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16 Nov 2019, 14:07

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Difficulty:

35% (medium)

Question Stats:

78% (02:00) correct

22% (02:19) wrong

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Car X traveled from city A to city B in 30 minutes. The first half of the distance was covered at 50 miles per hour, and the second half of the distance was covered at 60 miles per hour. What was the average speed of car X?

A. 200/11
B. 400/11
C. 500/11
D. 600/11
E. 700/11

Spoiler: OA

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Re: Car X traveled from city A to city B in 30 minutes. The first half of [#permalink]

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16 Nov 2019, 17:39

Bunuel wrote:

Here equal distance is covered in two different speeds, so the overall average =2ab/(a+b)=2*50*60/(50+60)=6000/110=600/11

D
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Car X traveled from city A to city B in 30 minutes. The first half of [#permalink]

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Updated on: 01 Dec 2019, 04:06

\(Ave.Speed= \frac{Distance}{ (t1+ t2)}\)

\(—> \frac{D}{2}= 50* t1–> t1= \frac{D}{100}\)

—> \(\frac{D}{2}= 60* t2–> t2= \frac{D}{120}\)

\(Ave.speed= \frac{D}{(D/100 + D/120)}= \frac{D}{(11D/600)}= \frac{600}{11} \)

The answer is D

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Originally posted by lacktutor on 17 Nov 2019, 01:29.
Last edited by lacktutor on 01 Dec 2019, 04:06, edited 3 times in total.

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Re: Car X traveled from city A to city B in 30 minutes. The first half of [#permalink]

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18 Nov 2019, 20:11

Bunuel wrote:

Answer is D

Explanation: total Distance (d) = d1 + d2 = 50 mph * t1 + 60mph * t2;
We are also told that d1 = d2;
So: 50 t1 = 60t2;

t = 30 minutes = 0.5 hour;
t1 + t2 = 0.5 hour;
So: t1 = 0.5 - t2;

Solve 2 equations, 2 unknowns:
(1) 50t1 = 60t2 and (2) t1 = 0.5 - t2

t2 = 25/110

t1 = 0.5 - t2 = 55/110 - 25/110 = 30 /110 = 3/11;

Ave speed = d / t = (50t1 + 60t2) / 0.5
= (50 * 3/11 + 60*25/110 ) /0.5
= (150/11 + 150/11) / 0.5
= (300/11)/ 0.5
= 600/11

Please give kudos if you find my Explanation, Really need it

Re: Car X traveled from city A to city B in 30 minutes. The first half of [#permalink] 18 Nov 2019, 20:11

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