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A tricky PS here

by dddanny2006 » Fri Mar 14, 2014 9:43 am
The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph

[spoiler]OA : D[/spoiler]

Here's my method to it

First 20 miles R=50mph D=20miles T=(D/R)=(20/50)

Next 20 miles R=Unknown D=20miles

Total trip Average must be 60

So we can solve using the weighted average method

50(2/5)+x(3/5)= 60

x=70

But the answer should be 75.Please explain this case and why weighted averages dont work here.

Thanks

Dan
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by theCodeToGMAT » Fri Mar 14, 2014 10:25 am
First 20 miles
D = 20
S = 50
T = 2/5

Avg Speed = (40)/(2/5 + t)

60 ( 2/5 + t) = 40

24 + 60t = 40

t = 16/60 = 4/15

Speed = 20/4/15 = 75

[spoiler]{D}[/spoiler]
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by dddanny2006 » Fri Mar 14, 2014 11:05 am
Thanks for this,but the bigger question is:why my method doesnt get me the answer?
theCodeToGMAT wrote:First 20 miles
D = 20
S = 50
T = 2/5

Avg Speed = (40)/(2/5 + t)

60 ( 2/5 + t) = 40

24 + 60t = 40

t = 16/60 = 4/15

Speed = 20/4/15 = 75

[spoiler]{D}[/spoiler]
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by GMATGuruNY » Fri Mar 14, 2014 11:06 am
dddanny2006 wrote:The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph
Half the total distance is traveled at 50 miles per hour.
We can plug in ANY VALUE for the total distance, as long as half the total distance is traveled at 50 miles per hour.

Let the total distance = 300 miles.
Time to travel 300 miles at an average speed of 60 miles per hour = 300/60 = 5 hours.
Time to travel the first 150 miles at an average speed of 50 miles per hour = 150/50 = 3 hours.
Time remaining for the next 150 miles = total time - time for the first 150 miles = 5-3 = 2 hours.

To travel the next 150 miles in 2 hours, the required rate = 150/2 = 75 miles per hour.

The correct answer is D.
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by dddanny2006 » Fri Mar 14, 2014 11:12 am
Mitch,thanks for that.Can you tell me why weighted averages dont work here?Im very confused.I cant distinguish between straight and weighted average problems.
GMATGuruNY wrote:
dddanny2006 wrote:The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph
Half the total distance is traveled at 50 miles per hour.
We can plug in ANY VALUE for the total distance, as long as half the total distance is traveled at 50 miles per hour.

Let the total distance = 300 miles.
Time to travel 300 miles at an average speed of 60 miles per hour = 300/60 = 5 hours.
Time to travel the first 150 miles at an average speed of 50 miles per hour = 150/50 = 3 hours.
Time remaining for the next 150 miles = total time - time for the first 150 miles = 5-3 = 2 hours.

To travel the next 150 miles in 2 hours, the required rate = 150/2 = 75 miles per hour.

The correct answer is D.
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by Patrick_GMATFix » Fri Mar 14, 2014 11:34 am
ddanny wrote:50(2/5)+x(3/5)= 60

x=70

But the answer should be 75.Please explain this case and why weighted averages dont work here.
Hi Daniel,

The guys above explained good ways to solve; I just wanted to address your question about why your method did not work. In fact, weighted average does work. It is good that you recognized that it should; you just did not apply it correctly.

Remember that the weights in weighted averages are the % distribution of the different groups. Consider your equation
50(2/5)+x(3/5)= 60
You used 2/5 and 3/5 as the weights. The problem is that 2/5 is the amount of time it took to cover the first leg, not the percentage of total time spent on the first leg. If you wish to solve using weighted average, you will need to compute the % distribution of times. Thankfully that's easy.

Leg 1 takes 20miles/50mph = 2/5 hour = 6/15 hour
Total trip takes 40miles/60mph = 2/3 hour = 10/15 hour
Leg 2 must take the difference: 10/15 - 6/15 = 4/15 hour

Now we have the amounts of time for each leg: 6/15 hour and 4/15 hour. This is a ratio of 6:4, or 3:2. This ratio has 5 total parts, so the correct weights distribution for is 3/5 for Leg1 and 2/5 for Leg 2.

Rebuild your formula for weighted average with the correct weight distribution:
50*(3/5) + x*(2/5) = 60. If you solve you will find that x=75

To recap: do not mistake actual amounts for percentage distribution if you are using the weighted average formula Ax + By = Avg where x and y are weights.
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by dddanny2006 » Fri Mar 14, 2014 11:54 am
Thanks Patrick,that was brilliant.I find it hard when it comes to identifying weights.Any tips for me?
Patrick_GMATFix wrote:
ddanny wrote:50(2/5)+x(3/5)= 60

x=70

But the answer should be 75.Please explain this case and why weighted averages dont work here.
Hi Daniel,

The guys above explained good ways to solve; I just wanted to address your question about why your method did not work. In fact, weighted average does work. It is good that you recognized that it should; you just did not apply it correctly.

Remember that the weights in weighted averages are the % distribution of the different groups. Consider your equation
50(2/5)+x(3/5)= 60
You used 2/5 and 3/5 as the weights. The problem is that 2/5 is the amount of time it took to cover the first leg, not the percentage of total time spent on the first leg. If you wish to solve using weighted average, you will need to compute the % distribution of times. Thankfully that's easy.

Leg 1 takes 20miles/50mph = 2/5 hour = 6/15 hour
Total trip takes 40miles/60mph = 2/3 hour = 10/15 hour
Leg 2 must take the difference: 10/15 - 6/15 = 4/15 hour

Now we have the amounts of time for each leg: 6/15 hour and 4/15 hour. This is a ratio of 6:4, or 3:2. This ratio has 5 total parts, so the correct weights distribution for is 3/5 for Leg1 and 2/5 for Leg 2.

Rebuild your formula for weighted average with the correct weight distribution:
50*(3/5) + x*(2/5) = 60. If you solve you will find that x=75

To recap: do not mistake actual amounts for percentage distribution if you are using the weighted average formula Ax + By = Avg where x and y are weights.
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by Patrick_GMATFix » Fri Mar 14, 2014 12:29 pm
dddanny2006 wrote:You have reversed my ratio.The ratio I considered was 2:3,but you say its 3:2.Why?Also,I find it hard to identify and fill weights in the formulas.Leg 1 takes (20/50)hrs.So this leg 1 takes the 2/5.Can you please explain.Thanks
Daniel,

It is true that the correct ratio is the reverse of your original ratio, but that's just a coincidence. I didn't actually reverse your ratio. My post explains how I came up with the right ratio.

As for your 2nd question, the weights should not be the actual amounts of time, but rather the percentage distribution of those amounts.

I think my posts answers your questions. Maybe you read through it too quickly?

Patrick
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by dddanny2006 » Fri Mar 14, 2014 1:17 pm
Well,Yes I do understand it.Infact,I did change my post,but I guess you were already in to the reply when I re-edited it.Thanks anyway.

A small situation here Patrick.

I buy 100 pens from a dealer for $20 a piece

I sell 80 of those pens for $30 a piece,since the other 20 dont get sold I return them back to the manufacturer for a 50% refund thats $10

Now if Im asked to find the profit based on the initial investment of 100pens @ $20 a piece,is it safe for me to do it this way

(2400-2000) *100 equalling to 20%
2000

Is this right?Im a little confused about the 20 that he returned.Do we need to subtract them from our initial investment and then calculate the profit based on the re-adjusted investment(12000-Refund)

Thanks







Patrick_GMATFix wrote:
dddanny2006 wrote:You have reversed my ratio.The ratio I considered was 2:3,but you say its 3:2.Why?Also,I find it hard to identify and fill weights in the formulas.Leg 1 takes (20/50)hrs.So this leg 1 takes the 2/5.Can you please explain.Thanks
Daniel,

It is true that the correct ratio is the reverse of your original ratio, but that's just a coincidence. I didn't actually reverse your ratio. My post explains how I came up with the right ratio.

As for your 2nd question, the weights should not be the actual amounts of time, but rather the percentage distribution of those amounts.

I think my posts answers your questions. Maybe you read through it too quickly?

Patrick
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Last edited by dddanny2006 on Fri Mar 14, 2014 1:21 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Fri Mar 14, 2014 1:19 pm
dddanny2006 wrote:The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph

[spoiler]OA : D[/spoiler]
Here's another option:

The total distance is 40 miles, and we want the average speed to be 60 miles per hour.
Average speed = (total distance)/(total time)
So, we get: 60 = (40 miles)/(total time)
Solve equation to get: total time = 2/3 hours
So, the TIME for the ENTIRE 40-mile trip needs to be 2/3 hours.

driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour.
How much time was spent on this FIRST PART of the trip?
time = distance/speed
So, time = 20/50 = 2/5 hours

The ENTIRE trip needs to be 2/3 hours, and the FIRST PART of the trip took 2/5 hours

2/3 hours - 2/5 hours = 10/15 hours - 6/15 hours
= 4/15 hours
So, the SECOND PART of the trip needs to take 4/15 hours


The SECOND PART of the trip is 20 miles, and the time is 4/15 hours
Speed = distance/time
So, speed = 20/(4/15)
= (20)(15/4)
= 75

Answer: D

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by dddanny2006 » Fri Mar 14, 2014 1:27 pm
Thank you Brent.What do you make of this?
I buy 100 pens from a dealer for $20 a piece

I sell 80 of those pens for $30 a piece,since the other 20 dont get sold I return them back to the manufacturer for a 50% refund thats $10

Now if Im asked to find the profit based on the initial investment of 100pens @ $20 a piece,is it safe for me to do it this way

(2400-2000) *100 equalling to 20%
2000

Is this right?Im a little confused about the 20 that he returned.Do we need to subtract them from our initial investment and then calculate the profit based on the re-adjusted investment(12000-Refund)
Brent@GMATPrepNow wrote:
dddanny2006 wrote:The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph

[spoiler]OA : D[/spoiler]
Here's another option:

The total distance is 40 miles, and we want the average speed to be 60 miles per hour.
Average speed = (total distance)/(total time)
So, we get: 60 = (40 miles)/(total time)
Solve equation to get: total time = 2/3 hours
So, the TIME for the ENTIRE 40-mile trip needs to be 2/3 hours.

driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour.
How much time was spent on this FIRST PART of the trip?
time = distance/speed
So, time = 20/50 = 2/5 hours

The ENTIRE trip needs to be 2/3 hours, and the FIRST PART of the trip took 2/5 hours

2/3 hours - 2/5 hours = 10/15 hours - 6/15 hours
= 4/15 hours
So, the SECOND PART of the trip needs to take 4/15 hours


The SECOND PART of the trip is 20 miles, and the time is 4/15 hours
Speed = distance/time
So, speed = 20/(4/15)
= (20)(15/4)
= 75

Answer: D

Cheers,
Brent
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by Brent@GMATPrepNow » Fri Mar 14, 2014 1:35 pm
dddanny2006 wrote:Thank you Brent.What do you make of this?
I buy 100 pens from a dealer for $20 a piece

I sell 80 of those pens for $30 a piece,since the other 20 dont get sold I return them back to the manufacturer for a 50% refund thats $10

Now if Im asked to find the profit based on the initial investment of 100pens @ $20 a piece,is it safe for me to do it this way

(2400-2000) *100 equalling to 20%
2000

Is this right?Im a little confused about the 20 that he returned.Do we need to subtract them from our initial investment and then calculate the profit based on the re-adjusted investment(12000-Refund)
I'd say that your initial EXPENSE was $2000 (100 pens a $20 each)

Sold 80 pens for $30 apiece. This is $2400
You also "sold" (returned) 20 pens for $10 apiece. This is $200
TOTAL REVENUE = $2400 + $200 = $2600

TOTAL PROFIT = REVENUE - EXPENSE
= $2600 - $2000
= [spoiler]$600[/spoiler]

Cheers,
Brent
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by Patrick_GMATFix » Fri Mar 14, 2014 1:37 pm
You are over-thinking it Daniel. Profit is always total revenues - total expenses.

Think in real world terms. If you buy a toy for $100, and the manufacturer gives you a 50% refund, what did you actually spend? Only $50!. The problem with your formula is that at the end of the day, the merchant did not actually spend $2000. $200 was refunded, so he/she only spent $1800.

Profit is $2400 - $1800
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by dddanny2006 » Fri Mar 14, 2014 1:51 pm
Thanks Brent.
Brent@GMATPrepNow wrote:
dddanny2006 wrote:Thank you Brent.What do you make of this?
I buy 100 pens from a dealer for $20 a piece

I sell 80 of those pens for $30 a piece,since the other 20 dont get sold I return them back to the manufacturer for a 50% refund thats $10

Now if Im asked to find the profit based on the initial investment of 100pens @ $20 a piece,is it safe for me to do it this way

(2400-2000) *100 equalling to 20%
2000

Is this right?Im a little confused about the 20 that he returned.Do we need to subtract them from our initial investment and then calculate the profit based on the re-adjusted investment(12000-Refund)
I'd say that your initial EXPENSE was $2000 (100 pens a $20 each)

Sold 80 pens for $30 apiece. This is $2400
You also "sold" (returned) 20 pens for $10 apiece. This is $200
TOTAL REVENUE = $2400 + $200 = $2600

TOTAL PROFIT = REVENUE - EXPENSE
= $2600 - $2000
= [spoiler]$600[/spoiler]

Cheers,
Brent
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by dddanny2006 » Fri Mar 14, 2014 1:52 pm
Thanks Patrick
Patrick_GMATFix wrote:You are over-thinking it Daniel. Profit is always total revenues - total expenses.

Think in real world terms. If you buy a toy for $100, and the manufacturer gives you a 50% refund, what did you actually spend? Only $50!. The problem with your formula is that at the end of the day, the merchant did not actually spend $2000. $200 was refunded, so he/she only spent $1800.

Profit is $2400 - $1800
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