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M22-05

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M22-05 [#permalink]

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The rental charge for a car is 34 cents for the first \(\frac{1}{4}\) mile driven and 6 cents for every \(\frac{1}{5}\) mile driven over the initial \(\frac{1}{4}\) mile. If a man paid $1.24 in rental charges, how many miles did he drive?

A. 2.50
B. 3.00
C. 3.25
D. 3.75
E. 4.00
[Reveal] Spoiler: OA

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Official Solution:

The rental charge for a car is 34 cents for the first \(\frac{1}{4}\) mile driven and 6 cents for every \(\frac{1}{5}\) mile driven over the initial \(\frac{1}{4}\) mile. If a man paid $1.24 in rental charges, how many miles did he drive?

A. 2.50
B. 3.00
C. 3.25
D. 3.75
E. 4.00

The driver went \((\frac{1.24 - 0.34}{0.06})/5 = 3\) miles in excess of the first fourth of a mile. The overall distance is 3.25 miles.

Answer: C
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Collection of Questions:
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Re: M22-05 [#permalink]

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New post 14 Oct 2015, 19:08
Unable to come at this equation ....total fare= No. of km * Fare / Km----->1.24 = 0.34 + 0.06*(1/5) X
Here X is the No of 1/5 interval.So X= 75 n total distance = 1/5 X = 15+0.25=15.25

The above equation differs from my equation. Pls let me know where i am wrong

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Re: M22-05 [#permalink]

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New post 12 Jul 2016, 05:21
suyashtcs wrote:
Unable to come at this equation ....total fare= No. of km * Fare / Km----->1.24 = 0.34 + 0.06*(1/5) X
Here X is the No of 1/5 interval.So X= 75 n total distance = 1/5 X = 15+0.25=15.25

The above equation differs from my equation. Pls let me know where i am wrong



The equation is, where x = total miles

-> 124 = 34 + (6)(5)(x-0.25)

solve for x

Since the charge is 6 cents for 1/5 mile, it will be 30 cents for the entire mile. I think thats where you're making the error. Also, i didn't see where you are subtracting the 0.25 miles.

Cheers

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Re: M22-05 [#permalink]

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New post 16 Apr 2017, 20:46
Elaborating further on what kaaaaku said:
1.24 - 0.34 = 90 cents left.
Given:
(1/5)th mile cost = 6 cents
1 mile cost = ? => 6/(1/5) = 6*5 = 30 cents.
Therefore,
1 mile = 30 cents
?------= 90 cents
90/30 = 3 miles.
Total miles = 3.25 miles. | C

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Re: M22-05 [#permalink]

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New post 14 Aug 2017, 08:35
The only mistake you are making after obtaining 15 intervals of 1/5 - is that you are considering it to be 15 miles whereas its 3 miles (15 intervals of 1/5 mile each equals 3 miles ). 3 miles + 0.25 mile is 3.25 mile.


Hope that helps
PS: I made the same mistake :P

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Re: M22-05   [#permalink] 14 Aug 2017, 08:35
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