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Speed Dist DS

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Speed Dist DS

by Skywalker » Tue Aug 24, 2010 9:36 pm
3) X , Y and Z shared a 1500 mile drive. Which of the 3 drove the greatest distance on the trip ?

i) X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.
ii) Z drove for 9 hours at an avg rate of 50 miles per hour.

Note: How can we quickly calculate the details in option i) above, is what I am looking for. The OA is C I blv while I opted for E.
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by anticipation » Wed Aug 25, 2010 3:44 am
The correct answer indeed is E. Please check the source of OA.

1) We know that T and T+1 is time taken by Y and X respectively
And S and S-5 is the speed of Y and X

Insufficient
2) Z covered 450

450< 1500/3
So Insufficient

Combining both
(T+T+1)*(S+S-5)=1050

We cant solve this. So E.

If someone knows how to solve it combining both, they dont need an MBA, they deserve https://en.wikipedia.org/wiki/Fields_Medal
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by Gurpinder » Wed Aug 25, 2010 8:01 am
Hmm...... maybe I am wrong but C kind of makes sense....

Which of the 3 drove the greatest distance on the trip ?

i) X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

So this one is insufficient.

ii) Z drove for 9 hours at an avg rate of 50 miles per hour.

So Z drove for 450mph. We have 1050 mph to go. So CLEARLY Z did not drive the most. if we divide 1050/2 = 525. So even if x and Y drove the same, they still covered more distance than Z. So Z is out of the question.

But this one is also insufficient to say who drove the most.


Together
Z is out of the question. So we are left with X and Y

Notice that the question says X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

X drove an hour more than Y but at an AVERAGE speed of 5mph less than Y.

So logically, Y covered more distance than X because Y's AVG speed was higher.

To do this one algebraically, lets plug in values for y and x:

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

y1= 20mph (avg rate)
y2 = 5 hours

20x5=100 miles

x1 = 15mph (avg rate 5 less than y)
x2 = 6

15x6=90 miles

so y drove more than x. therefore, y drove the most

i hope this makes sense.......:D
Last edited by Gurpinder on Wed Aug 25, 2010 10:18 am, edited 1 time in total.
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by tomada » Wed Aug 25, 2010 10:09 am
Gurpinder,

Underneath the section titled Together, your third statement is incorrect. In the second sentence, you correctly reiterated that X drove 1 hour longer than Y, but the third statement reverses that, saying that X drove 1 hour less than Y.

This will most certainly impact the answer...

Gurpinder wrote:Hmm...... maybe I am wrong but C kind of makes sense....

Which of the 3 drove the greatest distance on the trip ?

i) X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

So this one is insufficient.

ii) Z drove for 9 hours at an avg rate of 50 miles per hour.

So Z drove for 450mph. We have 1050 mph to go. So CLEARLY Z did not drive the most. if we divide 1050/2 = 525. So even if x and Y drove the same, they still covered more distance than Z. So Z is out of the question.

But this one is also insufficient to say who drove the most.


Together
Z is out of the question. So we are left with X and Y

Notice that the question says X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

X drove an hour less than Y but at an AVERAGE speed of 5mph less than Y.

So logically, Y covered more distance than X because Y's AVG speed was higher.

To do this one algebraically, lets plug in values for y and x:

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

y1= 20mph (avg rate)
y2 = 5 hours

20x5=100 miles

x1 = 15mph (avg rate 5 less than y)
x2 = 6

15x6=90 miles

so y drove more than x. therefore, y drove the most

i hope this makes sense.......:D
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by Gurpinder » Wed Aug 25, 2010 10:18 am
tomada wrote:Gurpinder,

Underneath the section titled Together, your third statement is incorrect. In the second sentence, you correctly reiterated that X drove 1 hour longer than Y, but the third statement reverses that, saying that X drove 1 hour less than Y.

This will most certainly impact the answer...

Gurpinder wrote:Hmm...... maybe I am wrong but C kind of makes sense....

Which of the 3 drove the greatest distance on the trip ?

i) X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

So this one is insufficient.

ii) Z drove for 9 hours at an avg rate of 50 miles per hour.

So Z drove for 450mph. We have 1050 mph to go. So CLEARLY Z did not drive the most. if we divide 1050/2 = 525. So even if x and Y drove the same, they still covered more distance than Z. So Z is out of the question.

But this one is also insufficient to say who drove the most.


Together
Z is out of the question. So we are left with X and Y

Notice that the question says X drove 1 hour longer than Y but at an avg rate of 5 miles per hour less than Y.

X drove an hour less than Y but at an AVERAGE speed of 5mph less than Y.

So logically, Y covered more distance than X because Y's AVG speed was higher.

To do this one algebraically, lets plug in values for y and x:

distance for x: (y1-5)(y2+1) distance for y: (x1+5)(x2-1) ----> (var1 = rate, var2 = time)

y1= 20mph (avg rate)
y2 = 5 hours

20x5=100 miles

x1 = 15mph (avg rate 5 less than y)
x2 = 6

15x6=90 miles

so y drove more than x. therefore, y drove the most

i hope this makes sense.......:D
ahh sorry I got confused in all that jargon.

thanks
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- Alfred A. Montapert, Philosopher.
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