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Mathematics 7 Online
OpenStudy (anonymous):

A group of n people decide to buy a $36,000 minibus. Each person will pay an equal share of the cost. If three additional people join the group, the cost per person will decrease by $1000. Find n.

OpenStudy (anonymous):

if we divide the cost of $36,000 amongst \(n\) people, each person must pay \(36000/n\) -- yes?

OpenStudy (anonymous):

now consider that with the addition of 3 more people, we now divide amongst \(n+3\) people i.e. \(36000/(n+3)\). the problem tells us that this new cost, \(36000/(n+3)\), is precisely $1000 less than \(36000/n\) (this should be intuitive, since with more people paying each individual only needs to contribute a smaller share)

OpenStudy (anonymous):

we write:$$\frac{36000}{n+3}=\frac{36000}n-1000$$can you solve for \(n\)?

OpenStudy (anonymous):

So, if you solve n+3, it would be -3?

OpenStudy (anonymous):

hmm?

OpenStudy (anonymous):

hint: multiply both sides by \(n\) and \(n+3\) to get rid of the denominators

OpenStudy (anonymous):

Cost per person with n people = 36000/n Cost per person with n + 3 people = 36000/(n+3) and this is 1000 less than (36000/n). 36000/(n+3) = (36000/n) - 1000 You can multiply both sides by n(n+3). On the left the n+3 cancels out and on the right the n cancels out. 36000n = 36000(n+3) - 1000n(n+3)

OpenStudy (anonymous):

get it?

OpenStudy (anonymous):

So, you divided by 1000?

OpenStudy (anonymous):

1 sec

OpenStudy (anonymous):

yes you divide by 1000

OpenStudy (anonymous):

It would be 36n+108+26n+n^2+3n?

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

combine like terms and solve using the quadratic formula

OpenStudy (anonymous):

n^2+3n-107+0

OpenStudy (anonymous):

yes!!

OpenStudy (anonymous):

(n+12) (n-9)=0 <---I think I am getting this wrong?

OpenStudy (anonymous):

no thts right :DD

OpenStudy (anonymous):

Finally! (/-.-)/ Well, I wish I can give you both best response, but I cannot. I became you guys fan. ^_^ And Thank you so much.

OpenStudy (anonymous):

haha no prob

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