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# On his trip to Alaska, Joe traveled y percent of the total

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On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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31 Oct 2012, 11:46
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Difficulty:

35% (medium)

Question Stats:

75% (02:52) correct 25% (03:04) wrong based on 208 sessions

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On his trip to Alaska, Joe traveled y percent of the total distance at an average speed of 50 miles per hour and the rest of the distance at an average speed of 70 miles per hour. What was Joe’s average speed in miles per hour for the whole trip in terms of y?

(A) 70(50 – y)

(B) (y + 60)/4

(C) 10,000 – 199y

(D) 17,500/(y + 250)

(E) y/2 + 35

If the first part is at 50 mile/h and the second part is 70 miles/h the overall speed is the toral distance / total time.

Pick a number for distance we can use 350. so the time for first part is 7 and for the rest is 5.

in the end we will have 26. something, but testing answer choices nothing comes up (

where I'm wrong

PS: I could use a bit of logic and come to the answer easier, but .............

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Posts: 59561
Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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01 Nov 2012, 06:27
6
5
carcass wrote:
On his trip to Alaska, Joe traveled y percent of the total distance at an average speed of 50 miles per hour and the rest of the distance at an average speed of 70 miles per hour. What was Joe’s average speed in miles per hour for the whole trip in terms of y?

(A) 70(50 – y)

(B) (y + 60)/4

(C) 10,000 – 199y

(D) 17,500/(y + 250)

(E) y/2 + 35

If the first part is at 50 mile/h and the second part is 70 miles/h the overall speed is the toral distance / total time.

Pick a number for distance we can use 350. so the time for first part is 7 and for the rest is 5.

in the end we will have 26. something, but testing answer choices nothing comes up (

where I'm wrong

PS: I could use a bit of logic and come to the answer easier, but .............

Say y=100%, so say Joe traveled entire trip at an average speed of 50 miles per hour. So, in this case we'd have that his average speed is simply 50 miles per hour.

Now, plug y=100 into the answer choices and see which one yields 50. Only answer choice D fits.

Or: say y=0%. In this case we'd have that Joe's average speed is 70 miles per hour.

Now, plug y=0 into the answer choices and see which one yields 70. Again, only answer choice D fits.

Hope it helps.
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Joined: 27 Feb 2012
Posts: 111
Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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31 Oct 2012, 12:41
3
1
As you need a quick logical answer let me try
Lety distance be X
so Time 1 = XY/100*50
Time 2 = X(100-Y)/100*70

Avg Speed = Total Distance / Total Time
= X/ Time 1 + Time 2

Ans D
reason is simple...X cancels and Y reamins in denominator
only 1 option has the same
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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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16 Nov 2012, 03:22
1
rajathpanta wrote:
i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.

If you plug y=0 or y=100, then you won't need to plug anything for the distance. Check here: on-his-trip-to-alaska-joe-traveled-y-percent-of-the-total-141661.html#p1138511
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Joined: 27 Aug 2012
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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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06 Nov 2012, 09:01
BangOn wrote:
As you need a quick logical answer let me try
Lety distance be X
so Time 1 = XY/100*50
Time 2 = X(100-Y)/100*70

Avg Speed = Total Distance / Total Time
= X/ Time 1 + Time 2

Ans D
reason is simple...X cancels and Y reamins in denominator
only 1 option has the same

I deducted the above within 30 secs but took another 3 mins to solve the whole mess to get to the ans choice.
As you rightly said if we notice Y remains in deno for only one choice, life becomes easier.

I tend to go the conventional way every time not the GMAT way

Any other tricks like this which could help save TIME?
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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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06 Nov 2012, 09:10
Another way - less than 2 mins
Take Total distance = 100
1) y% of 100 at 50 mi/hr = distance is y miles -- So its y miles at 50 mi/hr
2) remaining (100-y) miles at 70 mi/hr

Avg spd = Total Dist/Total Time = 100/(y/50 + (100-y)/70) -> leads to D.(in case there are more choices with y in deno)
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Posts: 59561
Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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06 Nov 2012, 09:38
HiteshPunjabi wrote:
Another way - less than 2 mins
Take Total distance = 100
1) y% of 100 at 50 mi/hr = distance is y miles -- So its y miles at 50 mi/hr
2) remaining (100-y) miles at 70 mi/hr

Avg spd = Total Dist/Total Time = 100/(y/50 + (100-y)/70) -> leads to D.(in case there are more choices with y in deno)

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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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15 Nov 2012, 10:50
i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.
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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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17 Nov 2012, 16:59
rajathpanta wrote:
i just plugged in some random nos-

say Total distance=100
y=30%
Then ball parked to find that D is the ans.

No need to assume the value of y. Assume total distance = 100. Therefore total time taken = (y/50) + (100-y)/70 = (y + 250)/175

Average speed = 100/( (y + 250)/175) = 175000/ (y + 250)
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Re: On his trip to Alaska, Joe traveled y percent of the total  [#permalink]

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13 Apr 2019, 12:50
I recently encountered one such questions and couldn't come up with the logic to start the solution.
After solving a bunch of these, the quickest method (at least for me) is to calculate the individual times and then divide the total distance (Which I assume 100 in most cases) with the total time. A little bit of calculation but it works better than plugging in numbers for me.
Re: On his trip to Alaska, Joe traveled y percent of the total   [#permalink] 13 Apr 2019, 12:50
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