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Carlos runs a lap around the track in x seconds. His second
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19 Aug 2018, 03:49
19 Aug 2018, 03:49
Expert Reply
00:00
Question Stats:
34% (01:30) correct
65% (01:07) wrong based on 55 sessions
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Carlos runs a lap around the track in x seconds. His second lap is five seconds slower than the first lap, but the third lap is two seconds faster than the first lap. What is Carlos’s average (arithmetic mean) number of minutes per lap, in terms of x?
(A) \(x - 1\)
(B) \(x + 1\)
(C) \(\frac{x-1}{60}\)
(D) \(\frac{x+1}{60}\)
(E) \(\frac{x+3}{60}\)
(A) \(x - 1\)
(B) \(x + 1\)
(C) \(\frac{x-1}{60}\)
(D) \(\frac{x+1}{60}\)
(E) \(\frac{x+3}{60}\)
Re: Carlos runs a lap around the track in x seconds. His second
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16 Sep 2018, 00:05
16 Sep 2018, 00:05
answer must be c , if not pls. explain??
Re: Carlos runs a lap around the track in x seconds. His second
[#permalink]
16 Sep 2018, 14:06
16 Sep 2018, 14:06
1
Expert Reply
Explanation
Carlos’s lap times can be expressed as x, x + 5, and x – 2. (Remember, slower race times are greater numbers, so “five seconds slower” means plus 5, not minus 5!) Average the lap times:
\(\frac{x+(x+5)+(x-2)}{3}=\frac{3x+3}{3}=x+1\)
His average time is x + 1 seconds. But the question requires minutes. Since there are 60 seconds in a minute, divide by 60 to get \(\frac{x+1}{60}\), or choice (D).
Carlos’s lap times can be expressed as x, x + 5, and x – 2. (Remember, slower race times are greater numbers, so “five seconds slower” means plus 5, not minus 5!) Average the lap times:
\(\frac{x+(x+5)+(x-2)}{3}=\frac{3x+3}{3}=x+1\)
His average time is x + 1 seconds. But the question requires minutes. Since there are 60 seconds in a minute, divide by 60 to get \(\frac{x+1}{60}\), or choice (D).
