The number of straight line miles traveled downriver...

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

The number of straight line miles traveled downriver...

by BTGmoderatorLU » Tue Feb 06, 2018 7:32 am

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

The OA is E.

I'm really confused with this PS question. Experts, any suggestion please? Thanks in advance.

Quote

Source: Beat The GMAT — Problem Solving | Tue Feb 06, 2018 7:32 am

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

The number of straight line miles traveled downriver...

by regor60 » Thu Feb 08, 2018 12:25 pm

LUANDATO wrote: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

The OA is E.

I'm really confused with this PS question. Experts, any suggestion please? Thanks in advance.

Using the formula, X/K = 3(7) + 4 for the 7 units/hour rate = 25 miles/hour

Similarly, for the 7+5 = 12 units/hour rate, 3(12) + 4 = 1.4X/(K-1) > to reflect 40% farther and 1 hour less compared to 7 units/hr

So now you have two equations: X/K=25 miles/hour and 1.4X/(K-1)=40 miles/hour

Solve for K from the first equation: K=X/25. Substitute this into the second equation:

1.4X/((X/25)-1) = 40. Simplify: 1.4X/((X-25)/25 = 35X/(X-25) = 40

Solve for X: 40X-1000 = 35X, therefore X = 200 (miles) at the 7 units/hour fuel rate.

That means that since X/K = 25 miles/hour , K= X/25 = 200/25 =8 hours

Using the formula to find distance traveled/hour at the new fuel rate of 10 units/hour:

3(10)+4 = 34 miles/hour = X/K = X/8 miles/hour

Therefore, X = 8 hours * 34 miles/hour = 272 miles, E

Quote

Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Thu Feb 08, 2018 3:55 pm

LUANDATO wrote: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

The first sentence of the problem really is saying that the speed of the boat is approximated by 3n + 4, where n is the fuel consumption in units/hr. Thus we are saying that at a fuel rate of 7 units/hr,

x/k = 3(7) + 4

x/k = 25

x = 25k

At a fuel rate of 7 + 5 = 12 units/hr,

1.4x/(k - 1) = 3(12) + 4

1.4x/(k - 1) = 40

Since x = 25k, we have:

1.4(25k)/(k - 1) = 40

35k = 40(k - 1)

35k = 40k - 40

40 = 5k

8 = k

Since we know now k = 8, let's determine m, the number of miles the boat travels in 8 hours at a fuel rate of 10 units/hr:

m/8 = 3(10) + 4

m/8 = 34

m = 272

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]