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algebra question

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bmwhype2
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algebra question [#permalink] New post   02 Mar 2010, 09:37
$10 for 4 cans, $6 shipping. Each additional set of 4 cans adds $1 to shipping.

how do i set this up algebraically?



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Re: algebra question [#permalink] New post   02 Mar 2010, 09:43
bmwhype2 wrote:
$10 for 4 cans, $6 shipping. Each additional set of 4 cans adds $1 to shipping.

how do i set this up algebraically?


Let n represent number of sets of 4 cans each.

10n + 6 + (n-1)*1 should be the total cost, from my understanding of the statement.
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Re: algebra question [#permalink] New post   02 Mar 2010, 23:57
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Another equation -

Let n be the number of set of 4 cans.
6 + 10 + (10 +1)(n-1)
= 16 + 11(n-1)
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Re: algebra question [#permalink] New post   05 Mar 2010, 13:21
tarun wrote:
Another equation -

Let n be the number of set of 4 cans.
6 + 10 + (10 +1)(n-1)
= 16 + 11(n-1)



i set up the equation like this
let n=4

total = 10n + 6 + (n-1)
total = 11n + 5
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Re: algebra question [#permalink] New post   11 Mar 2010, 02:24
BarneyStinson wrote:
bmwhype2 wrote:
$10 for 4 cans, $6 shipping. Each additional set of 4 cans adds $1 to shipping.

how do i set this up algebraically?


Let n represent number of sets of 4 cans each.

10n + 6 + (n-1)*1 should be the total cost, from my understanding of the statement.


Why it is assumed that cans are going to be in the multiple of 4?
In case n is not divisible by 4, this equation wont hold good.
Should not it be like following,
price of 1 can = 10/4 = 2.5
If n is the no of cans, so total price is
2.5*n+6+⌊n/4⌋ - 1

where ⌊x⌋ means the floor of x, i.e. the largest integer less than or equal to x.



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Re: algebra question [#permalink]
11 Mar 2010, 02:24
Moderator:
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

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