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Mathematics LibreTexts

12.2: Basics

  • Page ID
    14040
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    Basics

    A coin (or a stamp) has a value. A dime is worth 10 cents.

    The total value of a number of coins (or stamps) is the product of the number and the value.

    Examples

    Some of the problems here are simple. The solution can be worked out fast by quick reasoning. The benefit of these problems is not to find the solution by reasoning, but by learning algebraic steps applicable to more challenging situations. Solving application problems is a skill. Cultivate it.

    My approach is not to read a problem first until I understand it. Many of my colleagues will disagree with me. I propose creating a preamble in which you write a symbol or set of symbols for each element in the problem. (Refer to exercise 10.2.9).

    Look at the end of the problem where you usually find what to solve for. Let \(x\) be the number we are looking for. Write this as the first step in the preamble (list of symbols in non-trivial exercises whose solution we seek).

    Next start reading the problem from the beginning and develop the preamble step by step. Write the symbol(s) for each step on a new line.

    Finish the preamble by converting all problem steps into symbols. Now look at your preamble and gain an overview of the your problem.

    It should be easier to obtain an equation using preamble symbols. Then solve the equation by the method introduced this far for linear equations. (A linear equation has a variable to the first degree (exponent).

    Example \(\PageIndex{1}\)

    A coin bank contains nickels, dimes and quarters.

    • There are \(8\) more dimes than quarters.
    • The number of nickels is \(8\) less than twice the number of dimes.
    • The total value of the coins is \(\$5.70\).

    Find the number of each coin.

    Solution:

    Preamble:

    Let \(x\) be the number of quarters (for example \(x=20\) quarters).
    There are \(8\) more dimes than quarters: \(x+8\) \(20+8=28\) dimes).
    Nickels: \(8\) less than \(2\times\) dimes: \(2(x+8)-8\) \(2(20+8)-8\)
    \(=2x+16-8\)
    \(=2x+8\)

    \(\begin{array}{|l|c|c |c|c|r |r|r|r|}\hline \hbox{Coin}&\hbox{Number of}&\hbox{Value of }&\hbox{Total value}\\ \hbox{}&\hbox{coins in cents}&\hbox{one coin}&\hbox{of each coins}\\ \hline \hbox{Nickels}&2x+8&5&5(2x+8)\\ \hline \hbox{Dimes}&x+8&10&10(x+8)\\ \hline \hbox{Quarters}&x&25&25x\\ \hline \end{array}\)

    Equation:

    \(\begin{array}{rcl lll} \hbox{Total value of coins}&=&\hbox{Total value of coins}\\[4pt] %\$5.70&=&\hbox{$=570$\hbox{ cents}}\\[4pt] 5(2x+8)+10(x+8)+25x&=&570\ \ \hbox{cents}\\[4pt] 10x+40+10x+80+25x&=&570\\[4pt] 45x+40+80&=&570\\[4pt] 45x+120&=&570\\[4pt] 45x+120-120&=&570-120\\[4pt] 45x&=&450\\[4pt] \displaystyle \frac{45x}{45}&=&\displaystyle \frac{450}{45}\\[10pt] x&=&10 \end{array}\)

    • Number of quarters: \(x=10\)
    • Number of dimes: \(x+8=10+8=18\)
    • Number of nickels: \(2x+8=2(10)+8=28\)

    Note: If \(x\) had been chosen to be the number of dimes or nickels, the development would have been more difficult.

    Example \(\PageIndex{2}\)

    A wallet contains \(42\) one-, five- and ten-dollar bills.
    The value of all the bills is \(\$135\).
    There are \(5\) times as many one-dollar bill as ten-dollar bills.
    Find the number of bills of each denomination.

    Solution:

    Preamble:

    Let \(x\) be the number of \(\$10\) bills (for example \(x=20\)).
    Then the number of \(\$1\) bills is \(5x\) \(5(20)\).
    [5pt] We need to become creative (more variables, more equations later),
    A wallet contains \(42\) bills (including \(???\) five-dollar bills):
    \(\begin{array}{rcl lll} x+5x+???&=&42\\[5pt] 6x+???&=&42\\[5pt] 6x-6x+???&=&42-6x\\[5pt] ???&=&42-6x \end{array}\)

    The number of \(\$5\) is \(???=42-6x\)

    \(\begin{array}{|r|c|c |c|c|r |r|r|r|}\hline \hbox{Bill}&\hbox{Number of}&\hbox{Value of }&\hbox{Total value}\\ \hbox{}&\hbox{bills}&\hbox{each bill}&\hbox{of the coins}\\ \hline \hbox{\)$10\(}&x&10&10x\\ \hline \hbox{\)$5\(}&42-6x&5&5(42-6x)\\ \hline \hbox{\)$1\(}&5x&1&5x\\ \hline \end{array}\)

    Equation:

    \(\begin{array}{rcl lll} \hbox{The total value of the coins}&=&\hbox{\)$135\(\hbox{ dollars}}\\[5pt] 10x+5(42-6x)+5x&=&135\\[5pt] 10x+210-30x+5x&=&135\\[5pt] 210-15x&=&135\\[5pt] 210-210-15x&=&135-210\\[5pt] -15x&=&-75\\[5pt] \displaystyle \frac{-15x}{-15}&=&\displaystyle \frac{-75}{-15}\\[10pt] x&=&5 \end{array}\)

    The number of ten-dollar bills is \(5\).
    The number of five-dollar bills is \(42-6x=42-6(5)=42-30=12\).
    The number of one-dollar bills is \(5(5)=25\).

    Example \(\PageIndex{3}\)

    An order of \(\$0.44\) stamps and \(\$1.29\) custom photo postcards cost \(\$14.87\). The order consists of \(28\) items.

    Find the number of custom photo postcards.

    Solution:

    Preamble:

    Let \(x\) be the number of stamps (for example \(x=20\)).

    Then the number postcards is \(28-x\) \(28-20\).

    \(\begin{array}{|l|c|c |c|c|r |r|r|r|}\hline \hbox{Stamps}&\hbox{Number of}&\hbox{Value of }&\hbox{Total value}\\ \hbox{or postcards}&\hbox{items}&\hbox{one item}&\hbox{of the items}\\[5pt] \hline \hbox{stamp}&x&0.44&0.44x\\ \hline \hbox{postcard}&28-x&1.29&1.29(28-x)\\ \hline \end{array}\)

    Equation:

    \(\begin{array}{rcl lll} \hbox{The whole order costs }&=&\hbox{\)$14.87\(}\\[5pt] 0.44x+1.29(28-x)&=&14.87\\[5pt] 0.44x+36.12-1.29x&=&14.87\\[5pt] -0.85x+36.12&=&14.87\\[5pt] -0.85x+36.12-36.12&=&14.87-36.12\\[5pt] -0.85x&=&-21.25\\[5pt] \displaystyle \frac{-0.85}{-0.85}x&=&\displaystyle \frac{-21.25}{-0.85}\\[10pt] x&=&25 \end{array}\)

    The order consists of \(25\) stamps and \(28-25=3\) postcards.

    Exercises 12

    1. A collection of stamps from a foreign country consists of five-franc stamps and two-franc stamps.
      The number of two-franc stamps is \(12\) less than twice the number of five-franc stamps.
      The total value of the stamps is \(\$381\). Find the number of stamps.

    2. A cash register contains five-franc, ten-franc and twenty-five-franc coins.
      The number of five-franc coins is twice the number of ten-franc coins.
      The number of twenty-five-franc coins is \(4\) more than the number of five-franc coins.
      The total value of the coins is \(940\) francs. How many coins of each denomination are there?

    1. A collection of stamps from a foreign country consists of five-franc stamps and two-franc stamps.
      The number of two-franc stamps is \(12\) less than twice five-franc stamps.
      The total value of the stamps is \(381\) francs. Find the number of stamps.

      Solution:

      Preamble:

      Let \(x\) be the number of five-franc stamps (for example \(x=20\)).
      Two-franc stamps: \(12\) less than \(2\times\) five-franc stamps: \(2x-12\) \(2(20)-12\)).

      \(\begin{array}{|l|c|c |c|c|r |r|r|r|}\hline \hbox{Stamp}&\hbox{Number of}&\hbox{Value of }&\hbox{Total value}\\ &\hbox{stamps}&\hbox{each stamp}&\hbox{of the stamps}\\ \hline \hbox{five-franc}&x&5&5x\\ \hline \hbox{two-franc}&2x-12&2&2(2x-12)\\ \hline \end{array}\)

      Equation:

      \(\begin{array}{rcl lll} \hbox{The total value of the stamps }&=&\hbox{\)381\(francs}\\[5pt] 5x+2(2x-12)&=&381\\[5pt] 5x+4x-24&=&381\\[5pt] 9x-24&=&381\\[5pt] 9x-24+24&=&381+24\\[5pt] 9x&=&405\\[9pt] \displaystyle \frac{9}{9}x&=&\displaystyle \frac{405}{9}\\[11pt] x&=&45 \end{array}\)

      The number of five-franc stamps is \(45\).

      The number of two-franc stamps is \(2(45)-12=90-12=78\).

    2. A cash register contains five-franc, ten-franc and twenty-five-franc coins.

      The number of five-franc coins is twice the number of ten-franc coins.

      The number of twenty-five-franc coins is \(4\) more than the number of five-franc coins.

      The total value of the coins is \(940\) francs. How many coins of each denomination are there?

      Solution:

      Preamble:

      Let \(x\) be the number of ten-franc coins (for example \(x=20\)).

      Number of five-franc coins: \(2\times\) number of ten-franc coins: \(2x\) \(2(20)\).

      twenty-five-franc coins: \(4\) more than five-franc coins: \(2x+4\) \(2(20)+4.\)

      \(\begin{array}{|l|c|c |c|c|r |r|r|r|}\hline \hbox{Coin}&\hbox{Number of}&\hbox{Value of }&\hbox{Total value}\\ \hbox{}&\hbox{coins}&\hbox{one coin}&\hbox{of the coins}\\ \hline \hbox{five-franc}&2x&5&5(2x)\\ \hline \hbox{ten-franc}&x&10&10x\\ \hline \hbox{twenty-five}&2x+4&25&25(2x+4)\\ \hline \end{array}\)

      Equation:

      \(\begin{array}{rcl lll} \hbox{The total value of the coins}&=&\hbox{\)$940\(\hbox{ francs}}\\[5pt] 5(2x)+10x+25(2x+4)&=&940\\[5pt] 10x+10x+50x+100&=&940\\[5pt] 70x+100&=&940\\[5pt] 70x+100-100&=&940-100\\[5pt] 70x&=&840\\[5pt] \displaystyle \frac{70x}{70}&=&\displaystyle \frac{840}{70}\\[10pt] x&=&12 \end{array}\)

      The number of five-franc coins is \(2(12)=24\).
      The number of ten-franc coins is \(12\).
      The number of twenty-five-franc coins is \(2(12)+4=28\).