Marcel is taking a trip, driving at a constant speed of X miles per ho

Question

Data Insights (DI) Butler 2023-24       Click here for Details ]

Marcel is taking a trip, driving at a constant speed of X miles per hour for the first 2 hours of his trip, and at a constant speed of Y miles per hour after the first 2 hours.

In terms of X and Y, select the expression that represents Marcel’s average speed if he drives for a total of 5 hours, and select the expression that represents Marcel’s average speed if he drives for a total of 5X miles. Make only one selection in each column.

Sajjad1994 · Accepted Answer

Official Explanation

The first question is easier to answer. If Marcel drives for 5 hours, then he goes at a constant speed of X for 2 hours and at a constant speed of Y for the remaining 3 hours. Thus, his average speed is  \frac{2X+3Y}{5}

Now, if Marcel travels a total of 5X miles, then over the first 2 hours he covers 2X miles, so he has 3X miles left to travel at a speed of Y miles per hour:

\frac{Y miles}{1 hour}  =  \frac{3X miles}{?miles}

?= \frac{3X}{Y}

So he covers the remaining 3X miles in  \frac{3X}{Y}  hours. Therefore, in total his trip lasts  2 + \frac{3X}{Y}  hours. You can now find his average speed for the trip:

S =  \frac{Total Distance}{Total Time}

S =  \frac{5X}{2+(3X/Y)}

S =  \frac{5X}{(2y+3X)/Y}

S =  \frac{5XY}{2Y+3X}

karanbadalia · Answer

Can someone please tell me how did we get the answer to 'Average speed over 5 hours' problem?

Gemmie · Answer

Average speed over 5 hours 
 
 Total travel: 2x + 3y 
 Total time: 5 
 Avg speed:  \frac{2x+3y}{5}

Average speed over 5X miles 
 
 Total travel: 5x 
 Total time:  2 + \frac{5x-2x}{y} = 2 + \frac{3x}{y}  
 Avg speed:  
 
 5x \div (2 + \frac{3x}{y}) 
=  5\div (\frac{2y+3x}{y}) 
=  5x * \frac{y}{(2y+3x)} 
=  \frac{5xy}{2y+3x}

gabgol1998 · Answer

Let's plug in numbers:

X = 3 km/h
Y = 2 Km/h

Average speed = total distance / total time.

Marcel drove the first two hours at X, and the last 3 at Y, then in the two first hours Marcel drove 6 km, and for the last 3 hours, 6 km.
(You can do it with mental math, or just apply the formula D = V x T)

Average speed = 12/5 km/h

Option 1 - 3Y/5

3x2/5 is diferrent from 12/5

Option 2 - (2X + 3Y)/5

(2x3 + 3x2)/5 = 12/5.

So we have our first option conclued, let's move on to second.

The distance Marcel traveled was 5X, hence 5x3 = 15 km.

Consider he always drive the first 2 hours at 3km/h and the remaining at 2km/h

Now it's just logic and convertion.

In two hours he drove 6 km, there's 9 left to run at 2 km/h, hence 8 km in 4 hours + 1 km in 0.5 hours.
(Or just apply the formula T = D/V for the second part of the trip: T = 9/2, Total time = 4,5 + 2 from the first parte of the trip.)

So he drove 15 miles in 6.5 hours, then his average speed is 15/6.5 or 30/13.

Let's check the options, we already know the correct answer for the second part can't be the two first ones.

Option 3 - (3X - 2Y)/5

Don't even calculate, we know it yields a smaller number than we need.

Look at option 5, it also yields a smaller number, hence we only have the fourth option as the correct one.

Want to check it out? Ok.

Option 4 - 5YX/(2Y + 3X)

5x2x3/(2x2 + 3x3) = 30/13.

Average speed over 5 hours	Average speed over 5X miles
		\(\frac{3Y}{5}\)
		\(\frac{2X+3Y}{5}\)
		\(\frac{3X-2Y}{5}\)
		\(\frac{5XY}{2Y+3X}\)
		\(\frac{5}{3Y}\)

Marcel is taking a trip, driving at a constant speed of X miles per ho

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GMAT Two-part Analysis (TPA) Questions

Marcel is taking a trip, driving at a constant speed of X miles per ho

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards