During a trip of 200 kilometres, Stella covers the first 80 kilometres

Question

During a trip of 200 kilometres, Stella covers the first 80 kilometres at an average speed of x kilometres per hour and the remaining distance at an average speed of (x + 15) kilometres per hour. If her average speed for the entire trip is 250/3 kilometres per hour, what is the value of x?

A. 223/3
B. 75
C. 455/6
D. 80
E. A unique value cannot be determined

A. 223/3
B. 75
C. 455/6
D. 80
E. A unique value cannot be determined

effatara · Answer

Let the total time for the entire trip be T hrs.

Average speed=200/T=250/3--->T=12/5 hrs

(Time for 1st 80 km) + (Time for remaining 120 km) = 12/5 
80/x + 120/(x+15) = 12/5
3x^2 - 205x - 1500 = 0
3x^2 - 225x +20x - 1500 = 0
3x(x - 75) + 25(x - 75) = 0
(x - 75)(3x + 25) = 0
x=75 (since value of 'x' can't be negative)

Ans: B

Subhrajyoti · Answer

A.s = Td/TT, now if we use the same analogy for the question, the end eq looks like this : 250/3= 80/80/x+ 120/120/x, when do the equation i am not getting the correct, can somebody suggest what am i doing wrong here

sumi747 · Answer

Subhrajyoti  The formula is total distance by total time
So it would look like: 200/{(80/x)+(120/x+15)}

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Subhrajyoti · Answer

Thanks, but can you advise why is the above equation incorrect i.e {250/3= 80/80/x+ 120/120/x}

IP1992 · Answer

Speed=DISATANCE/TIME

SO, 
250/3=200/((80/x)+(120/x+15))

On solving X=75

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ScottTargetTestPrep · Answer

Solution:
 
The time for the first 80 km of the trip is 80/x, and the time for the remaining 120 km of the trip is 120/(x + 15). We are given that the average speed of the entire trip is 250/3 kilometers per hour, Thus, the total time spent making the trip is 200/(250/3) = 200 x 3/250 = 4 x 3/5 = 12/5 hours. We can create the equation for the time of the trip as: 
80/x + 120/(x + 15) = 12/5

Multiplying the equation by 5x(x + 15), we have:

400(x + 15) + 600x = 12x(x + 15)

400x + 6000 + 600x = 12x^2 + 180x

12x^2 - 820x - 6000 = 0

3x^2 - 205x - 1500 = 0

(3x + 20)(x - 75) = 0

x = -20/3 or x = 75

Since x can’t be negative, x = 75.

Answer: B

Fdambro294 · Answer

Could we expect such a calculation-heavy type question on the GMAT?

IanStewart

Any insight would be greatly appreciated. Thank you.

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IanStewart · Answer

You could see a question in this general format on the test, but never with these numbers. GMAT questions are carefully designed, and the numbers carefully chosen, both so you get tractable algebra if you solve in a standard way (you would absolutely never need to factor a quadratic like 3x^2 - 205x - 1500 on the test), and so you can discover convenient shortcuts if you have a good conceptual understanding. So in a question like this, where it's clear the person will spend slightly more time driving at x+15 mph, you can be certain (because average speed is a weighted average, weighted by time) that 250/3 is slightly greater than the midpoint of x and x+15. But even trying to see what answers that eliminates takes far more time than it would on the GMAT because the answer choices are absurd here, and that still leaves you with A and B to consider. I'd be tempted at that point just to guess '75' is right (and confirm it, which isn't too bad using weighted average principles rather than algebra), because it's hard to imagine why anyone would design a question like this if the answer turned out to be 223/3, but since it's not a real GMAT question, it's hard to even know if a principle like that applies.

Kinshook · Answer

Given: During a trip of 200 kilometres, Stella covers the first 80 kilometres at an average speed of x kilometres per hour and the remaining distance at an average speed of (x + 15) kilometres per hour.

Asked:If her average speed for the entire trip is 250/3 kilometres per hour, what is the value of x?

Total time spent = 80/x + 120/(x+15)

Average speed = 250/3
Distance = 200 km
Time = 200/(250/3) = 2.4 hrs

80/x + 120/(x+15) = 2.4

IMO B

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During a trip of 200 kilometres, Stella covers the first 80 kilometres

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During a trip of 200 kilometres, Stella covers the first 80 kilometres

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