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VIC - Best Way To Solve?

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by chipbmk » Tue Nov 10, 2009 2:21 pm
The given explanation is to use variables, is this the best way? Or is it quicker to plug in answers? Please let me know your thoughts.

Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

a) (x + y) / t
b) 2(x + t) / xy
c) 2xyt / (x + y)
d) 2(x + y + t) / xy
e) x(y + t) + y(x + t)

OA:C
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by truplayer256 » Tue Nov 10, 2009 8:08 pm
It'd be best to plug in numbers for this problem but in case you're wondering how to solve this problem algebraically:

We're told that Bob bikes to school at a steady rate of x miles per hour. On a particular day, his bike has a flat tire exactly half way to school. Let's assume that the distance from Bob's home to the school is d. On the day that his bike got a flat tire, Bob had already rode has bike for d/2x hours.
When he decided to walk to school in order to cover the other half distance, it took him d/2y hours. The problem tells us that it took Bob t hours to get school after he left his house. So:

d/2x+d/2y=t

solve for d:

t*4xy=d(2x+2y)

2xyt/x+y=d
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by heshamelaziry » Tue Nov 10, 2009 8:35 pm
truplayer256 wrote:It'd be best to plug in numbers for this problem but in case you're wondering how to solve this problem algebraically:

We're told that Bob bikes to school at a steady rate of x miles per hour. On a particular day, his bike has a flat tire exactly half way to school. Let's assume that the distance from Bob's home to the school is d. On the day that his bike got a flat tire, Bob had already rode has bike for d/2x hours.
When he decided to walk to school in order to cover the other half distance, it took him d/2y hours. The problem tells us that it took Bob t hours to get school after he left his house. So:

d/2x+d/2y=t

solve for d:

t*4xy=d(2x+2y)

2xyt/x+y=d

Why 2x and 2y ? i understand your algebraic solution but only if the rate to halfway is x "as told" and the rate for the second half is y "as told" ?
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by xcusemeplz2009 » Tue Nov 10, 2009 11:50 pm
heshamelaziry wrote:
truplayer256 wrote:It'd be best to plug in numbers for this problem but in case you're wondering how to solve this problem algebraically:

We're told that Bob bikes to school at a steady rate of x miles per hour. On a particular day, his bike has a flat tire exactly half way to school. Let's assume that the distance from Bob's home to the school is d. On the day that his bike got a flat tire, Bob had already rode has bike for d/2x hours.
When he decided to walk to school in order to cover the other half distance, it took him d/2y hours. The problem tells us that it took Bob t hours to get school after he left his house. So:

d/2x+d/2y=t

solve for d:

t*4xy=d(2x+2y)

2xyt/x+y=d

Why 2x and 2y ? i understand your algebraic solution but only if the rate to halfway is x "as told" and the rate for the second half is y "as told" ?
its halfway so distance is d/2 for which speed is x so time taken is d/2 divided x hence time for halfway with speed x is d/2x...
It does not matter how many times you get knocked down , but how many times you get up
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by heshamelaziry » Tue Nov 10, 2009 11:59 pm
Thanks. I understand yuor explanation. but why it is wrong to assign d to each half, since both halves are equal ? I bet you have solved problems in which a person goes to town a in 20 miles/hour and goes back home in 40 miles/hour and to calculate the total average speed, we say 2d/(d/20 + d/40) gives average speed.
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by xcusemeplz2009 » Wed Nov 11, 2009 12:06 am
heshamelaziry wrote:Thanks. I understand yuor explanation. but why it is wrong to assign d to each half, since both halves are equal ? I bet you have solved problems in which a person goes to town a in 20 miles/hour and goes back home in 40 miles/hour and to calculate the total average speed, we say 2d/(d/20 + d/40) gives average speed.
yes u are correct
even if each half is d
then t1=d/x
t2=d/y
avg speed= total distance/total time
ie 2d/(d/x+d/y)
which is same as 2xy/(x+y)

hence total dis is speed * time
which is 2xyt/(x+y

irresepective of d or d/2 ans is same. the only thing is changing is the total dist for d its 2d and for d/2 its d
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by palvarez » Wed Nov 11, 2009 2:36 pm
In this problem, i can easily pull this formula off of my mind. Avg speed = harmonic mean of x and y = 2xy(x+y)

distance = speed*time = 2xyt(x+y).

If I don't know this, i will just compute the distance with 'plugin numbers', and try to match up with the answer. That way, you would be SANE under time pressure.
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by chipbmk » Sun Nov 15, 2009 7:46 pm
I know this may seem like a really dumb question.

But can someone please show me how you would choose numbers and plug in for this question. Should you choose numbers for all the variables except for Distance, since that is what the question is asking us to solve for?

Thanks!
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by Testluv » Sun Nov 15, 2009 8:59 pm
chipbmk wrote:The given explanation is to use variables, is this the best way? Or is it quicker to plug in answers? Please let me know your thoughts.

Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

a) (x + y) / t
b) 2(x + t) / xy
c) 2xyt / (x + y)
d) 2(x + y + t) / xy
e) x(y + t) + y(x + t)

OA:C
You are most likely familiar with the average speed formula:

Average speed = total distance/total time

However, a very common occurence on the GMAT are round trip problems where an object goes...well for a round trip: from point A to B and back to A again. The significance of a round trip is that the distance travelled for each leg of the journey is equal. Here, the bike broke down halfway to school. So the distance travelled on bike is equal to the distance travelled on foot. Accordingly, we can use the special round trip formula:

average speed =(2 *s1*s2)/(s1+s2)

in which s1 is the speed for one half the journey (ie, point A to point B) and s2 the speed of the return trip (point B back to point A). In other words, the average speed is twice the product of the speeds divided by the sum of the speeds (the formula is derivative of the familiar average speed formula.)

Here, the two speeds are x and y.

So the average speed for the entire journey is:

Average speed for trip = 2xy/x+y

Because the problem asks for the distance, we will have to plug this speed expression into the distance = speed*time formula:

Distance = (t) * 2xy/(x+y)

or:

2xyt/(x+y)

Choose C

Note: Although I am calling it the "round trip" formula, it is applicable whenver you have an object whose journey is divdied into two legs of equal distance.
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