One secondary school question!

**laser2302** · July 4th 2006, 10:25 AM

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?

**Quick** · July 4th 2006, 10:53 AM

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

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

and now we solve the second equation for $n$

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

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

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

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

$\frac{337.5}{x}=\frac{339.5-10x}{x-0.20}$ and solve for $x$

**Quick** · July 4th 2006, 03:20 PM

$\frac{337.5}{x}=\frac{339.5-10x}{x-0.20}$ turn them into recipricals...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$x=2.7$ and voila! the price of each file is $2.70

now to find $n$ we merely substitute 2.7 for $x$

$n=\frac{337.5}{x}$ substitute...

$n=\frac{337.5}{2.7}$ divide...

$n=125$

and that is your answer!

(did I do more work than necessary?)

**Soroban** · July 4th 2006, 09:54 PM

Hello, laser2302!

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

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

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

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

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

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

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

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

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

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

Therefore, he bought $125$ files at $\$2.70$ each.

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

Check

He could buy $135$ files at $\$2.50$ each
. . It would cost: . $135 \times \$2.50 \,=\,\$337.50$ . . . check!

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