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A train traveled the first d miles of its journey it an average speed

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A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 09 Sep 2015, 00:50
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A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 09 Sep 2015, 01:58
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Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

Kudos for a correct solution.


Solution:
Total distance = 3d.
Total Time =d(1/60 + 1/y + 1/160)
Total Average = Total distance / Total Time ==> 96 = 3d/d(1/60 + 1/y + 1/160) ==> (1/60 + 1/y + 1/160) = 1/32 ==> y = 120 miles per hour.

Option, D
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 09 Sep 2015, 04:07
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Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

Kudos for a correct solution.


Average speed=Total distance traveled /Total time taken
3d/d/60+d/y+d/160=96
Solving for d and y,
15y=11y+480
4y=480
y=120
Answer D
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 14 Sep 2015, 06:24
Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

Kudos for a correct solution.


KAPLAN OFFICIAL SOLUTION:

When faced with an average rate problem, remember that average speed over an entire journey is equal to total distance divided by total time, and not the average of the individual speeds. In this problem you should also notice that no times or distance are given; we only know rates. When this happens, your best strategy is to pick a number for the distance and calculate the time based on that number.

We should make sure that the distance we choose will evenly divide by both 60 and 160, in order to only deal with integers. Thus, we should make d = 480. For the first d miles we travel at 60 miles per hour. Distance divided by rate is time, therefore 480/60 = 8 hours. For the final d miles we travel at 160 miles per hour; 480/160 = 3 hours. For the middle d hours, we travel at y miles per hour, so we can express the time as 480/y.

Our total distance will be 3 x 480, which equals 1440. Therefore total distance divided by total time is 1440/(8 + 3 + 480/y), which we are told must equal 96. From here you are doing algebra, and this can be simplified in the following manner:

1440/(11 + 480/y) = 96

1440/(11y/y + 480/y) = 96

1440/((11y + 480)/y) = 96

Keep in mind that in order to divide fractions, you multiply the numerator by the reciprocal of the denominator.

1440/1 x y/(11y + 480) = 96

1440y/(11y + 480) = 96

1440y = 96(11y + 480)

15y = 11y + 480

4y = 480

y = 120

which is choice (D).

You can set this up entirely algebraically from the beginning as well, and if you are extremely skilled with algebra, and fast and accurate at setting it up, then that is a valid approach as well. But for most, on a complicated algebraic equation such as this, you’ll have less room for error and be able to complete the question more quickly if you choose a number for the distance and move confidently from there.
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A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 27 Apr 2017, 08:00
d=LCM(96,160,60)=8*5*4*3
96=8*4*3 then d/96=5
60=8*4*5 then d/60=8
160=8*4*5 then d/160=3

3d/(d/60 + d/y +d/160)=3d/(11+y/d)=96
3d/96= 11 +y/d
15=11+ y/d
y/d=4 then y=8*5*3 y=120
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 02 May 2017, 16:43
Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135


We are given that a train traveled the first d miles of its journey at an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. Since time = distance/rate:

The time of the first d miles = d/60

The time of the next d miles = d/y

The time of the final d miles = d/160

We are also given that the overall average rate was 96 mph. We can use the average rate formula to determine y.

average rate = (total distance)/(total time)

96 = (d + d + d)/(d/60 + d/y + d/160)

96 = 3d/(d/60 + d/y + d/160)

We can divide the numerator and denominator by d:

96 = 3/(1/60 + 1/y + 1/160)

Now we get the common denominator of 480y:

96 = 3/(8y/480y + 480/480y + 3y/480y)

96 = 3/[(11y + 480)/480y]

96 = 3(480y)/(11y + 480)

96(11y + 480) = 1440y

11y + 480 = 15y

480 = 4y

y = 120

Answer: D
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 25 Jan 2019, 03:30
Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

Kudos for a correct solution.


I solved this using the inline logic,

1/60 + 1/y + 1/160 = 1/96

If LHS has to be equal to RHS, you can take the LCM of LHS.

So i just did that, looked for a value which can satisfy the remaining 2 values 60 & 160, value can be 480

Which will correspond with 120.

Answer D
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 25 Jan 2019, 05:43
Bunuel wrote:
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 96 miles per hour, what is the value of y?

(A) 68
(B) 84
(C) 90
(D) 120
(E) 135

Kudos for a correct solution.



When travelling 3 equal distances,

Average Speed = 3abc/(ab + bc + ca)

96 = 3*60*y*160 / (60y + 160y + 60*160)

32 * (220y + 9600) = 60*y*160 = 15 * 4 * y * 32*5

220y + 9600 = 15*4*5y

y = 960/8 = 120

Answer (D)

This and other formulas are discussed here on our blog:
https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 26 Jan 2019, 00:26
There is a way of solving this without involving too much calculation.
Let us assume d=480 ( LCM of 60 and 160) for ease of calculation.
Let the time taken for the 2nd leg be 't' hrs. So total time for the entire (480 x 3) miles is (8+3+t)hrs. Average speed for the entire trip is 'Total Distance' divided by the 'Total Time Taken'. Thus Ave. Speed for the entire trip is (480 X 3)miles/(11+t)hrs which is equal to 96 mph. Therefore, (480 x 3)/(11+t)=96. Thus, 't'=4. Thus the time taken to cover the 2nd leg of 480 miles is 4 hrs so 'y'=120 mph.
ANS: D
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Re: A train traveled the first d miles of its journey it an average speed  [#permalink]

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New post 07 Feb 2019, 10:10
KanishkM wrote:

1/60 + 1/y + 1/160 = 1/96

If LHS has to be equal to RHS, you can take the LCM of LHS.

So i just did that, looked for a value which can satisfy the remaining 2 values 60 & 160, value can be 480

Which will correspond with 120.

Answer D


Can you please show all the passages for this immediate and useful technique? kudos guaranteed :thumbup:
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Re: A train traveled the first d miles of its journey it an average speed   [#permalink] 07 Feb 2019, 10:10
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