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27 Dec 2012, 10:25
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
53% (03:24) correct
47%
(02:58)
wrong
based on 221
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The number of straight line miles traveled downriver in one hour by Lucy's boat is approximated within a linear range by 3n + 4, where n represents her fuel consumption in units/hr. Suppose that traveling x miles requires k hours at a fuel rate of 7 units/hr, but that increasing her fuel consumption by 5 units/hr would allow her to travel 40% further in 1 fewer hour. How far would she travel in k hours at a fuel rate of 10 units/hr?
A. 8
B. 200
C. 225
D. 236
E. 272
A. 8
B. 200
C. 225
D. 236
E. 272
28 Dec 2012, 02:15
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danzig
The wording of the question is a little complicated so take it one step at a time.
"The number of straight line miles traveled downriver in one hour by Lucy's boat "
miles driven in one hour gives you speed
"is approximated within a linear range by 3n + 4, where n represents her fuel consumption in units/hr."
So speed = 3n+4 where n is the fuel consumption
"Suppose that traveling x miles requires k hours at a fuel rate of 7 units/hr,"
Speed = x/k = 3*7 + 4 = 25
Distance traveled = 25k
"but that increasing her fuel consumption by 5 units/hr would allow her to travel 40% further in 1 fewer hour. "
Speed = 3*12 + 4 = 40
Distance traveled = 40(k-1) = (7/5)*25k (You multiply by 7/5 because 40% extra distance is traveled in this case)
k = 8
"How far would she travel in k hours at a fuel rate of 10 units/hr?"
Speed = 3*10 + 4 = 34
Distance traveled = 34*k = 34*8 = 272 miles
ConnectTheDots
GMAT 2: 770 Q50 V47
Posts: 239
27 Dec 2012, 10:56
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We need to find: traveled miles k hours and 10 units of fuel.
From the linear equation: Answer = \(k*(3*10+4)\) = 34k
Data:
x miles in k hours at 7 units/hr: equation: \(x =k*(3*7+4)\) = 25k
=> x=25k or k = x/25 (equation 1)
40% extra miles = 1.4 x
one hour less = k-1 hours
5 units/hr extra fuel = 7+5 = 12
Equation \(1.4x=(k-1)*(3*12+4)\)
\(=> 14x = (k-1)*(40)*10\)
\(=> 14x= 400k-400\)
\(=> 14x = 16x - 400\) (substitute k= x/25)
\(=> 2x = 400\)
\(=> x = 200\)
\(=> k = 200/25 = 8\) (substitute back from equation 1: k=x/25)
Required Answer = 34k = 34*8 = 272
While solving the equation, you can either substitute x or k, depending on the numbers in hand and involved calculation.
I substituted k = x/25, because my numbers were becoming large and 400 was easily divisible by 25.
It was quicker to solve x first than k.
From the linear equation: Answer = \(k*(3*10+4)\) = 34k
Data:
x miles in k hours at 7 units/hr: equation: \(x =k*(3*7+4)\) = 25k
=> x=25k or k = x/25 (equation 1)
40% extra miles = 1.4 x
one hour less = k-1 hours
5 units/hr extra fuel = 7+5 = 12
Equation \(1.4x=(k-1)*(3*12+4)\)
\(=> 14x = (k-1)*(40)*10\)
\(=> 14x= 400k-400\)
\(=> 14x = 16x - 400\) (substitute k= x/25)
\(=> 2x = 400\)
\(=> x = 200\)
\(=> k = 200/25 = 8\) (substitute back from equation 1: k=x/25)
Required Answer = 34k = 34*8 = 272
While solving the equation, you can either substitute x or k, depending on the numbers in hand and involved calculation.
I substituted k = x/25, because my numbers were becoming large and 400 was easily divisible by 25.
It was quicker to solve x first than k.
