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# A, B, and C each drove 100-mile legs of a 300-mile course at speeds of

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A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 07:20
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25% (medium)

Question Stats:

77% (02:20) correct 23% (02:22) wrong based on 387 sessions

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A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Project PS Butler : Question #07

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A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 07:33
3
A's Time = $$\frac{100}{40} = \frac{10}{4}$$
Total time = $$\frac{100}{40} + \frac{100}{50} + \frac{100}{60} = \frac{10}{4} + \frac{10}{5} + \frac{10}{6} =\frac{(150+120+100)}{60}=\frac{370}{60} = \frac{37}{6}$$
Hence A's fraction = $$\frac{10}{4} * \frac{6}{37} =\frac{15}{37}$$
Hence Ans C.
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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 07:43
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HKD1710 wrote:
A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Project PS Butler : Question #07

Time for A = 100/40 = 10/4

Time for B = 100/50 = 10/5

Time for C = 100/60 = 10/6

10/4 + 10/5 + 10/6 = 150/60 + 120/60 + 100/60

= 370/60

Time for A of total = 150/60 / 370/60 = 150/370 = 15/37

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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 07:52
1
HKD1710 wrote:
A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Project PS Butler : Question #07

Let the each runner runs 600 miles...

So, Time taken by A is 15 ; Time taken by B is 12 & Time taken by C is 10

Thus, Total time taken is 15 + 12 + 10 = 37

Fraction of the total time did A drove is 15/37 , Answer must be (C)
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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 07:57
1
HKD1710 wrote:
A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Project PS Butler : Question #07

Useful Formula:

Distance / speed = time.

At first we must find out the total time.

100/ 40 + 100/50 + 100/60

= 5/2 + 2 + 5/3

= 37/6.

A's time = 5/2.

required fraction: (5/2) / (37/6) = 15/37.

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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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09 Nov 2018, 09:23
HKD1710 wrote:
A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Project PS Butler : Question #07

A time 100/40 = 2.5 i.e 15/2

Total Time A+B+C= 2.5+6+5 = 18/5

15/2 / 18/5 = 15/37

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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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12 Nov 2018, 08:20
HKD1710 wrote:
A, B, and C each drove 100-mile legs of a 300-mile course at speeds of 40, 50, and 60 miles per hour, respectively. What fraction of the total time did A drive?

(A) 15/74
(B) 4/15
(C) 15/37
(D) 3/5
(E) 5/4

Using time = distance/rate, we can create the following expression, where the numerator is the total time that A drove and the denominator is the total time driven by A, B, and C combined:

(100/40)/(100/40 + 100/50 + 100/60)

(5/2)/(5/2 + 2 + 5/3)

Multiplying by 6/6, we have:

15/(15 + 12 + 10) = 15/37

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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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13 Nov 2018, 12:11
Hi All,

We're told that A, B and C each drive 100-mile legs of a 300-mile course at speeds of 40, 50 and 60 miles per hour, respectively. We're asked what FRACTION of the total drive time did A drive. This question can be solved in a couple of different ways; there's actually a great 'ratio shortcut' that can help you to avoid most of the math.

Since Person A drove his 1/3 of the distance SLOWEST, we know that he drove MORE than 1/3 of the DRIVE TIME. The three speeds (40 mph, 50 mph and 60 mph) are close enough to one another that we know Person A did NOT drive "most" of the time though. Knowing those deductions, let's consider the 5 answers...

Answer A: 15/74 - this is closer to 1/5; TOO SMALL
Answer B: 4/15 - this is a little more than 1/4; TOO SMALL
Answer C: 15/37 - this is a little more than 1/3; MATCHES what we're looking for
Answer D: 3/5 - this is 60% of the total; TOO BIG
Answer E: 5/4 - this is above 100%; TOO BIG (and not mathematically possible)

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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of  [#permalink]

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02 Dec 2019, 19:41
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Re: A, B, and C each drove 100-mile legs of a 300-mile course at speeds of   [#permalink] 02 Dec 2019, 19:41
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