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Bob’s pickup truck averages 20 miles/gallon. If Bob fills th
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03 Oct 2017, 20:42
03 Oct 2017, 20:42
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Bob’s pickup truck averages 20 miles/gallon. If Bob fills the 12-gallon gas tank to capacity, how many miles can he travel once the ratio of unused gas to used gas is 3:1?
(A) 60 miles
(B) 140 miles
(C) 170 miles
(D) 180 miles
(E) 240 miles
Kudos for correct solution.
(A) 60 miles
(B) 140 miles
(C) 170 miles
(D) 180 miles
(E) 240 miles
Kudos for correct solution.
Re: Bob’s pickup truck averages 20 miles/gallon. If Bob fills th
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03 Oct 2017, 22:47
03 Oct 2017, 22:47
1
Given he has filled his truck with 12 gallons, the unused gas is \(\frac{12}{4}*3 = 9\), where 4 is the sum of the parts in the ratio and 3 the part unused.
Then, he can drive for \(9*20 = 180\) miles more.
Answer D
Then, he can drive for \(9*20 = 180\) miles more.
Answer D
Re: Bob’s pickup truck averages 20 miles/gallon. If Bob fills th
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03 Oct 2017, 23:17
03 Oct 2017, 23:17
1
Bunuel wrote:
Bob’s pickup truck averages 20 miles/gallon. If Bob fills the 12-gallon gas tank to capacity, how many miles can he travel once the ratio of unused gas to used gas is 3:1?
Let us consider Bob uses x gallons
Therefore we can set the ratio as -
\(\frac{(12-x)}{x} =\frac{3}{1}\) (since 12 gallons is full tank and (12-x) is the unused gas)
or \(x= 3\).
Now we have to consider in the ratio of \(\frac{3}{1}\)
i.e \(\frac{3x}{x} = \frac{9}{3}\) (putting the value of x = 3, because we have assumed he uses x gallons and so in every 3parts used gas we have 9 parts unsed gas)
Therefore Bob can travel with 9 gallons remaining fuel = 9*20 = 180 miles
