# system of equations word problem

• May 10th 2008, 09:38 PM
cityismine
system of equations word problem
• May 10th 2008, 10:08 PM
TheEmptySet
Quote:

Originally Posted by cityismine

We have three unknowns so we will need three equations to solve this system.

let C be the number correct; I be the number incorrect; and S the score

Mr Day: $\displaystyle 5c+10=s$
Ms. Knight: $\displaystyle 6c-1.5I=s$
question on quiz: $\displaystyle c+I=20 \iff I=20-c$

setting the first to equal we get

$\displaystyle 5c+10=6c-1.5I$

Now we can sub in I to get

$\displaystyle 5c+10=6c-1.5(20-c) \iff 5c+10=6c-30+1.5c \iff 40=2.5c \iff 16=c$

We can now back sub in to find the other values.

$\displaystyle I=20-16=4$ and $\displaystyle s=5c+10=5(16)+10=90$

Yeah :)
• May 11th 2008, 12:22 AM
cityismine
Hey, I was trying to solve this using two equations, but I guess you need three. Thanks for explaining it so thoroughly.
• May 11th 2008, 02:07 PM
Soroban
Hello, cityismine!

Quote:

Mr. Day and Ms. Knight gave the same test consisting of 20 short-answer questions,
. . but marked their test papers using different grading systems.
Mr. Day gave 5 points for each correct answer, made no deduction for an incorrect
. . or omitted answer, and then added 10 points to the total.
Ms. Knight gave 6 points for each correct answer and subtracted 1.5 points for each
. . incorrect or omitted answer.
Two students in different classes answered the same number of questions correctly

The student in Mr. Day's class got $\displaystyle x$ answers right.
He got 5 points per correct answer plus 10 points.
His grade was: .$\displaystyle 5x + 10$

The student in Ms. Knight's class got $\displaystyle x$ right and $\displaystyle 20-x$ wrong.
She got 6 points per correct answer: .$\displaystyle 6x$
and lost 1.5 point for each wrong answer: .$\displaystyle 1.5(20-x)$
Her grade was: .$\displaystyle 6x - 1.5(20-x) \:=\:7.5x - 30$

The two grades were equal: .$\displaystyle 5x + 10 \:=\:7.5x - 30 \quad\Rightarrow\quad 2.5x \:=\:40 \quad\Rightarrow\quad x \:=\:16$

Their grade was: .$\displaystyle 5(16) + 10 \:=\:7.5(16) - 30 \:=\:\boxed{90}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Graph the lines: .$\displaystyle \begin{array}{ccc}y &=& 5x+10 \\ y&=&7.5x - 30\end{array}$

They intersect at: $\displaystyle (16,\,90)$