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Math Help - One secondary school question!

  1. #1
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    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?
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  2. #2
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    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, 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
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  3. #3
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    Continuing

    \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-0.20subtract 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?)
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  4. #4
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    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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