# One secondary school question!

• Jul 4th 2006, 10:25 AM
laser2302
One secondary school question!
Hi guys, can anyone please tell me to solve this problem

Mr Sim pays $337.50 for a number of files. He can buy 10 more files with the same amount of money if each file costs 20 cents less. How much does he pay for a file and how many files does he buy? • Jul 4th 2006, 10:53 AM Quick Quote: Originally Posted by laser2302 Hi guys, can anyone please tell me to solve this problem Mr Sim pays$337.50 for a number of files. He can buy 10 more files with the same amount of money if each file costs 20 cents less. How much does he pay for a file and how many files does he buy?

Its me again,

We know the number of files, $\displaystyle n$, times the price of each file, $\displaystyle x$ is equal to $337.50 We also know that the number of files + 10,$\displaystyle n+10$, times the price of the files minus 20 cents,$\displaystyle x-0.20$, equals$337.50

so we solve the first equation for $\displaystyle n$
$\displaystyle nx=337.5$
$\displaystyle n=\frac{337.5}{x}$

and now we solve the second equation for $\displaystyle n$

$\displaystyle (n+10)(x-0.20)=337.5$

$\displaystyle (x-0.20)n+10x-2=337.5$

$\displaystyle (x-0.20)n=337.5+2-10x$

$\displaystyle n=\frac{339.5-10x}{x-0.20}$ so then we put the equations together...

$\displaystyle \frac{337.5}{x}=\frac{339.5-10x}{x-0.20}$ and solve for $\displaystyle x$
• Jul 4th 2006, 03:20 PM
Quick
Continuing
$\displaystyle \frac{337.5}{x}=\frac{339.5-10x}{x-0.20}$ turn them into recipricals...

$\displaystyle \frac{x}{337.5}=\frac{x-0.20}{339.5-10x}$ multiply both sides by $\displaystyle 339.5-10x$

$\displaystyle \frac{-10x^2+339.5x}{337.5}=x-0.20$subtract $\displaystyle x$ from both sides

$\displaystyle \frac{-10x^2+339.5x}{337.5}-x=-0.20$ multiply both sides by -1

$\displaystyle \frac{10x^2-339.5x}{337.5}+x=0.20$ multiply by 337.5

$\displaystyle 10x^2-339.5x+337.5x=67.5$ add the "x"s

$\displaystyle 10x^2+(-339.5+337.5)x=67.5$ add...

$\displaystyle 10x^2+-2x=67.5$ subtract 67.5 from both sides...

$\displaystyle 10x^2+-2x-67.5=0$ now we are left with a quadratic equation that equals zero, so to solve for $\displaystyle x$ we can use the quadratic formula...

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ substitute...

$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(10)(-67.5)}}{2(10)}$ simplify...

$\displaystyle x=\frac{2\pm\sqrt{4-(-2700)}}{20}$ simplify...

$\displaystyle x=\frac{2\pm\sqrt{2704}}{20}$ find the square root...

$\displaystyle x=\frac{2\pm52}{20}$ use only the principle square root...

$\displaystyle x=\frac{2+52}{20}$ add...

$\displaystyle x=\frac{54}{20}$ divide...

$\displaystyle x=2.7$ and voila! the price of each file is $2.70 now to find$\displaystyle n$we merely substitute 2.7 for$\displaystyle x\displaystyle n=\frac{337.5}{x}$substitute...$\displaystyle n=\frac{337.5}{2.7}$divide...$\displaystyle n=125$and that is your answer! (did I do more work than necessary?) • Jul 4th 2006, 09:54 PM Soroban Hello, laser2302! Quote: Mr Sim pays$337.50 for a number of files.
He can buy 10 more files with the same amount of money if each file costs 20 cents less.
How much does he pay for a file and how many files does he buy?

Change the money to cents to eliminate the decimals.

Let $\displaystyle N$ = number of files he bought.
Let $\displaystyle p$ = price per file.

Mr. Sim bought $\displaystyle N$ files at $\displaystyle p$ cents each . . . and paid 33750¢
. . $\displaystyle Np\:=\:33750\qiad\Rightarrow\quad p = \frac{33750}{N}$ [1]

For the same amount, he could buy $\displaystyle N+10$ files at $\displaystyle p - 20$ cents each.
. . $\displaystyle (N+10)(p - 20)\:=\:33750$ [2]

Substitute [1] into [2]: .$\displaystyle (N + 10)\left(\frac{33750}{N} - 20\right) \;= \;33750$

We have: .$\displaystyle 33750 - 20N + \frac{337500}{N} - 200\;=\;33750$

Multiply by $\displaystyle N:\;\;-20N^2 + 337500 - 200N\;=\;0$

Divide by $\displaystyle -20:\;\;N^2 + 10N - 16875\;=\;0$

This quadratic factors: .$\displaystyle (N - 125)(N + 135) \;= \;0$

. . and has the positive root: .$\displaystyle \boxed{N = 125}$

Substitute into [1]: .$\displaystyle p = \frac{33750}{125}\quad\Rightarrow\quad\boxed{p = 270}$

Therefore, he bought $\displaystyle 125$ files at $\displaystyle \$2.70$each. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Check He could buy$\displaystyle 135$files at$\displaystyle \$2.50$ each
. . It would cost: .$\displaystyle 135 \times \$2.50 \,=\,\$337.50$ . . . check!