BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

The number of straight line miles traveled downriver...

Expert replies
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.
Join the discussion
Source: — Problem Solving |

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
Join the discussion

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]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Join the discussion