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Rates and Inequalities

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Rates and Inequalities

by topspin360 » Thu Feb 06, 2014 10:17 pm
Can someone please show how to solve the following problem algebraically? I understand the concept of actual numbers vs. fractions but just don't know how to algebraically apply (2) to show sufficiency.

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x+60

(2) y/x > 4/3

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked

B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked

C) Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient

D) EACH statement ALONE is sufficient to answer the question asked

E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed
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by Bill@VeritasPrep » Fri Feb 07, 2014 12:10 am
Starting with the stem, the combined rate is equal to the total distance divided by the total time. Total distance is easy: x + y. Time takes more work, but Jane spends x/30 hours at 30 mph and y/40 hours at 40 mph. Her total time is then x/30 + y/40; using the LCM of 120 would make this 4x/120 + 3y/120 = (4x + 3y)/120.

Now we can use R = D/T to bring everything together and simplify:

R = (x + y)/( (4x + 3y)/120) (Invert the denominator and multiply)

R = (120(x + y)) / (4x + 3y)

At this point, we want to know if the expression on the right is greater than 35:

(120(x + y)) / (4x + 3y) > 35 (Multiply both sides by the denominator and distribute)

120x + 120y > 140x + 105y (Isolate x and y)

15y > 20x (Factor out a 5)

3y > 4x

So our question can be reframed as "Is 3y > 4x?", which fits very nicely with Statement 2. [/b]
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by sanju09 » Fri Feb 07, 2014 12:27 am
topspin360 wrote:Can someone please show how to solve the following problem algebraically? I understand the concept of actual numbers vs. fractions but just don't know how to algebraically apply (2) to show sufficiency.

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x+60

(2) y/x > 4/3

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked

B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked

C) Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient

D) EACH statement ALONE is sufficient to answer the question asked

E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed
Average Speed = Total Distance/Total Time

Total Distance = x + y

Total Time = x/30 + y/35 = (7x + 6y)/210

Average Speed = (x + y)/ [(7x + 6y)/210] = 210(x + y)/ (7x + 6y).

Simply ballpark that the fraction (x + y)/ (7x + 6y) is slightly less than 1/6, hence 210 (< 1/6) is always less than 35. The question never required any statement at all. I have profound doubt in the source of this question.
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by GMATGuruNY » Fri Feb 07, 2014 2:09 am
topspin360 wrote:On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x+60

(2) y/x > 4/3
This is a THRESHOLD problem.
Here, the threshold is 35mph.
35mph is HALFWAY between the two speeds (30mph and 40mph).
To exceed this threshold -- in other words, for the average speed to be GREATER than halfway between the two speeds -- the amount of time spent driving at the faster speed (y/40) must be GREATER than the amount of time spent driving at the slower speed (x/30):
y/40 > x/30
y/x > 4/3.

Question rephrased: Is y/x > 4/3?

Statement 1: y > x+60
If x=30 and y=1000, then y/x > 4/3.
If x=300 and y=400, then y/x = 4/3.
INSUFFICIENT.

Statement 2: y/x > 4/3
SUFFICIENT.

The correct answer is B.
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by GMATGuruNY » Fri Feb 07, 2014 9:34 am

Total Distance = x + y

Total Time = x/30 + y/35 = (7x + 6y)/210

Average Speed = (x + y)/ [(7x + 6y)/210] = 210(x + y)/ (7x + 6y).

Simply ballpark that the fraction (x + y)/ (7x + 6y) is slightly less than 1/6, hence 210 (< 1/6) is always less than 35. The question never required any statement at all. I have profound doubt in the source of this question.
The value in red should be 40 (the faster of the two speeds).
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by Bill@VeritasPrep » Fri Feb 07, 2014 11:21 am
GMATGuruNY wrote: To exceed this threshold -- in other words, for the average speed to be GREATER than halfway between the two speeds -- the amount of time spent driving at the faster speed (y/40) must be GREATER than the amount of time spent driving at the slower speed (x/30):
A key point, to be sure. A common misinterpretation would be that, to have an average speed of 35, Jane must travel an equal distance (instead of an equal time) at each speed.
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by topspin360 » Fri Feb 07, 2014 2:52 pm
Thank you all for such great explanations! Appreciate it.
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by sanju09 » Sat Feb 08, 2014 1:06 am
GMATGuruNY wrote:

Total Distance = x + y

Total Time = x/30 + y/35 = (7x + 6y)/210

Average Speed = (x + y)/ [(7x + 6y)/210] = 210(x + y)/ (7x + 6y).

Simply ballpark that the fraction (x + y)/ (7x + 6y) is slightly less than 1/6, hence 210 (< 1/6) is always less than 35. The question never required any statement at all. I have profound doubt in the source of this question.
The value in red should be 40 (the faster of the two speeds).
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by feedrom » Sat Feb 08, 2014 10:22 am
Brilliant, Mitch!
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