Form Four Science of a school has n students. A farewell dinner is planned at the end of the year at a total cost of RM252. It is found that 6 of the students are unable to attend the dinner and because of this, each of those attending dinner will have to pay RM1 extra. a) Show that n^2-6n-1512=0
@ganeshie8
can u clarify the meaning of RM? i assumed that meant room, but a room isn't a cost.
RM is just a currency unit like $ @britanyashton
There are \(n\) students and the total cost is \(252\), so each student has to contribute \(\dfrac{252}{n}\).
but 6 students are are unable to attend and that makes remaining \(n-6\) students to pay RM1 more. So each student now contributes \(\dfrac{252}{n}+1\)
\(n-6\) students contribting \(\dfrac{252}{n}+1\) each for a total of \(RM 252\) : \[(n-6)\left(\frac{252}{n}+1\right) = 252\] simplify ^
\[252+n-1512-6=252\]
@ganeshie8
\[(n-6)\left(\frac{252}{n}+1\right) = 252\] \[(n-6)\left(252+n\right) = 252n\] right ?
expand on left side and simplify
\[252n+n^2-1512-6n=252n\] \[n^2-6n-1512=0\]
thnx @ganeshie8
looks good!
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