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Mathematics LibreTexts

5.2: Fraction Concept

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    13997
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    Fraction Notation and Percent

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    Fraction Concept

    A fraction is a portion of a whole. Half a pie, two-thirds of a cake, a quarter of an hour.
    A whole can be any quantity or object. A whole pencil, a whole car, a whole inch.
    We shall use a basic rectangle (like a sheet of paper \(8 \times 12\) inches) as a whole.

    (0,8)(0,-5.9) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (12,-5)(0,1)5 (12,-7)(0,-1)5 (11,-6)12 in (4,-13)(1,0)4 (4,-13)(-1,0)4 (3,-14)\(8\) in (4,-12)(0,2)6(0,1)1 (0.2,-11.8)(0.15,0)24(0,1)11.6 (6,-6)\(\displaystyle \fbox{\)\(}\) (5.5,-5.7)(-1,0)1.2

    (0,1)(-15,-10.9) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-13)(1,0)4 (4,-13)(-1,0)4 (4,-14)\(8\) in (0,-4)(2,0)4(1,0)1 (0,-8)(2,0)4(1,0)1 (0.3,-11.8)(0.15,0)50(0,1)3.6 (3.2,-5.6)\(\displaystyle \fbox{\)\(}\) (3.5,-6.6)(0,-1)1.2 (15.5,-3)\(\displaystyle \frac{1}{2}\) means 1 part out of 2 (15.5,-8)\(\displaystyle \frac{1}{3}\) means 1 part out of 3 (10,-1.5)Num. (13,-1.4)(4,-1)2 (10,-4)Den. (13,-3.9)(4,1)1.5 (10,-6.5)Num. (13,-6.4)(4,-1)2 (10,-9)Den. (13,-8.9)(4,1)1.5

    (0,4)(0,1) (0,-8)

    A proper fraction is a fraction whose numerator is less than its denominator.

    (0,0)(1,0)10 (0,0)(0,1)6 (10,6)(-1,0)10 (10,6)(0,-1)6 (1,5)(2,0)5 (1,3)(2,0)5 (1,1)(2,0)5 (11,5)\(\displaystyle \frac{5}{15}=\frac{1}{3}\) (12,4)(-1,-1)2 (25,-6.9)

    A fraction whose numerator is equal to or more than its denominator is an improper fraction.

    (0,1)(-19,2.7) (0,0)(1,0)10 (0,0)(0,1)6 (10,6)(-1,0)10 (10,6)(0,-1)6 (1,5)(2,0)5 (1,3)(2,0)5 (1,1)(2,0)5 (-1.5,-2)\(\displaystyle \frac{20}{15}=\frac{4}{3}=1\frac{1}{3}\) (-1,0)(1,3)1 (6.8,-1.7)(4,1)6.5

    (0,0)(-31,0.0) (0,0)(1,0)10 (0,0)(0,1)6 (10,6)(-1,0)10 (10,6)(0,-1)6 (1,5)(2,0)5 (1,3)(2,0)5 (1,1)(2,0)5

    An improper fraction can be turned into a mixed number.

    Example \(1\)

    \(\displaystyle \frac{35}{6}=5+\frac{5}{6}=5\frac{5}{6}\) (simply divide 35 by 6.)
    There is an implied addition (plus) sign between the whole number and the proper fraction.
    Example \(2\)

    A mixed number can be turned into an improper fraction, like
    \(\displaystyle 7\frac{2}{5}=\frac{7}{1}+\frac{2}{5}=\frac{7\cdot 5}{5}+\frac{2}{5}=\frac{35}{5}+\frac{2}{5}=\frac{37}{5}\)
    Shortcut: \(\displaystyle 7\frac{2}{5}=\frac{7\cdot 5+2}{5}\)

    Equivalent Fractions

    (0,6)(1,-6) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-12)(0,2)6(0,1)1 (0.2,-11.8)(0.15,0)24(0,1)5.6 (4,-6)(2,0)2(1,0)1 (0.2,-5.8)(0.15,0)24(0,1)5.6 (6,-3)\(\displaystyle \frac{2}{4}\)

    (0,3)(-10,-16.5) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-4)(2,0)2(1,0)1 (4,-8)(2,0)2(1,0)1 (0.2,-11.8)(0.15,0)24(0,1)3.6 (0.2,-7.8)(0.15,0)24(0,1)3.6 (0.2,-3.8)(0.15,0)24(0,1)3.6 (4,-12)(0,2)6(0,1)1 (6,-2)\(\displaystyle \frac{3}{6}\)

    (0,3)(-22,-27.0) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-12)(0,2)6(0,1)1 (0.2,-2.8)(0.15,0)24(0,1)2.6 (0.2,-5.8)(0.15,0)24(0,1)2.6 (0.2,-8.8)(0.15,0)24(0,1)2.6 (0.2,-11.8)(0.15,0)24(0,1)2.6 (6,-2)\(\displaystyle \frac{4}{8}\) (4,-3)(2,0)2(1,0)1 (4,-6)(2,0)2(1,0)1 (4,-9)(2,0)2(1,0)1

    (0,3)(-33.5,-37.5) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-12)(0,2)6(0,1)1 (0.2,-2.2)(0.15,0)24(0,1)2.0 (0.2,-4.48)(0.15,0)24(0,1)2.0 (0.2,-6.9)(0.15,0)24(0,1)2.0 (0.2,-9.4)(0.15,0)24(0,1)2.0 (0.2,-11.8)(0.15,0)24(0,1)2.0 (6,-2.1)\(\displaystyle \frac{5}{10}\) (4,-2.4)(2,0)2(1,0)1 (4,-4.99)(2,0)2(1,0)1 (4,-7.26)(2,0)2(1,0)1 (4,-9.6)(2,0)2(1,0)1

    Since \(1\) is the identity for multiplication, any number (including a fraction) times 1 is that number.

    Example \(3\)

    \(\displaystyle \frac{1}{2}=\frac{1}{2}\cdot 1=\frac{1}{2}\cdot \frac{2}{2}=\frac{2}{4}\hspace{1.0in}\displaystyle \frac{1}{2}=\frac{1}{2}\cdot 1=\frac{1}{2}\cdot \frac{3}{3}=\frac{3}{6}\)
    \(\displaystyle \frac{1}{2}=\frac{1}{2}\cdot 1=\frac{1}{2}\cdot \frac{4}{4}=\frac{4}{8}\hspace{1.0in}\displaystyle \frac{1}{2}=\frac{1}{2}\cdot 1=\frac{1}{2}\cdot \frac{99}{99}=\frac{495}{990}\)
    The examples above illustrate that \[the numerator and the denominator of any fraction can be multiplied by the same number to get equivalent fractions.\] The key word here is multiplied.
    The process works in reverse. When we can factor out a \(1\), we obtain a reduced equivalent fraction. This leads to a simplified fraction.
    Example \(4\)

    \(\displaystyle\frac{2}{4}=\frac{1}{2}\cdot \frac{2}{2}=\frac{1}{2}\cdot 1= \frac{1}{2}\)
    \(\displaystyle \frac{3}{6}=\frac{1}{2}\cdot \frac{3}{3}=\frac{1}{2}\cdot 1=\frac{1}{2}\)
    \(\displaystyle \frac{4}{8}=\frac{1}{2}\cdot \frac{4}{4}=\frac{1}{2}\cdot 1=\frac{1}{2}\)
    \(\displaystyle \frac{495}{990}=\frac{1}{2}\cdot \frac{99}{99}=\frac{1}{2}\cdot 1=\frac{1}{2}\)
    The examples above illustrate that

    \[\fbox{\parbox[t]{3.5in}{the numerator and and the denominator of any fraction can be divided by the same number}}\]

    to get simplified equivalent fractions. This means we factor out a 1. The key word here is factored. Factoring means that the numerator and denominator must be factorable (transformed int a product).

    (0,5)(0,-2) (0,0)\(1=\displaystyle \frac{3+6}{9}\not=\frac{1+6}{3}=\frac{7}{3}=2\frac{1}{2}\) (2.4,1.4)(2,-1)1 (3.6,-0.2)(2,-1)1 (2.1,1.5)1 (3.1,-1.5)3 (1.0,2.6)(4,-1)15 (1.0,-1.3)(4,1)15 (8,-3.5)Be very careful with the word “cancel".

    (0,4)(0,1) (0,0)\(1=\displaystyle \frac{3+6}{9}=\frac{3(1+2)}{(3)(3)}=\frac{1+2}{3}\cdot \frac{3}{3}=\frac{1+2}{3}\cdot 1=\frac{3}{3}\cdot 1=1\)

    \[\fbox{\parbox[t]{3.2in}{If possible, always reduce fractions given as answers.}}\]

    Divisibility Tests

    \(\)
    The divisibility tests come in handy.

    A whole number is divisible by 2 if its last (ones place) digit is \(0, 2, 4, 6\), or \(8\). Such a number is an even number.

    \(\displaystyle \frac{1234}{2}=617\)
    A whole number is divisible by 3 if the sum of its digits is divisible by \(3\).

    \(1,249\) is not divisible by 3 because \(1+2+4+9=16\) which canot be divided by \(3\).

    \(\displaystyle \frac{12147}{3}=4,049\)

    because \(1+2+1+4+7=15\) which can be divided by 3.
    A whole number is divisible by 4 if the number represented by the last two digits is divisible by \(4\).

    \(\displaystyle \frac{1236}{4}=309\) because \(36\) in \(12\fbox{\)36\(}\) is divisible by \(4\).
    A whole number is divisible by 5 if its last (ones place) digit is \(0\) or \(5\).

    \(\displaystyle \frac{1235}{5}=247\)
    A whole number is divisible by 6 if it is divisible by 2 AND 3 simultaneously. ecause \(1,230\) is even and \(1+2+3+0=6\).
    For divisibility by \(7\), just perform a long division and check if there is a remainder. You can find methods on the internet, but the procedures take as long or longer than long division. Make sure you remember your multiplication table.
    A whole number is divisible by 8 if the number represented by the the last three digits is divisible by \(8\).

    \(\displaystyle \frac{1232}{8}=154\) because \(232\) in \(1\fbox{\)232\(}\) is divisible by \(8\).
    A whole number is divisible by 9 if the sum of its digits is divisible by \(9\).

    \(1,269\) is divisible by \(9\) because \(1+2+6+9=18\) which can be divided by \(9\).

    \(\displaystyle \frac{1269}{9}=141\)
    A whole number is divisible by 10 if its last (ones place) digit is \(0\).

    \(\displaystyle \frac{1,230}{10}=123\)
    A whole number is divisible by 11 if the difference of the sums of its even-positioned and odd-positioned digits is a multiple of \(11\) or \(0\).

    \(\displaystyle \frac{27,269}{11}=2,479\)

    (0,2)(0,0) (0,0)\(2\ 7,\ 2\ 6\ 9\\) (0.3,2)(1,0)6 (0.3,2)(0,-1)1.1 (3.0,2)(0,-1)1.1 (5.2,2)(0,-1)1.1 (1.2,-1.8)(1,0)5 (1.2,-1.8)(0,1)1.1 (3.2,-1.8)(0,1)1.1 (0.3,2)(1,0)6 (0.3,2)(0,-1)1.1 (7.3,2)\(2+2+9=13\) (8.9,-2)\(7+6=13\) (16.9,0)\(13-13=0\) (13.9,-1.2)(4,1)3.0 (14.7,1.8)(4,-1)2.4

    Example \(5\)

    Given \(\displaystyle \frac{3}{7}\) find the equivalent fraction whose denominator is \(21x^2\).

    Solution:

    \(21x^2=7\dot (3 x^2)\). Then \(\displaystyle \frac{3}{7}=\frac{3}{7}\cdot \frac{3x^2}{3x^2}=\frac{9x^2}{21x^2}\)
    Example \(6\)

    Reduce \(\left(\displaystyle \frac{2}{35}\right)\left(\displaystyle \frac{18}{11}\right)\left(\displaystyle \frac{12}{49}\right)\left(\displaystyle \frac{455}{432}\right)\)

    Solution:

    \(\begin{array}{cl lll} &\left(\displaystyle \frac{2}{35}\right)\left(\displaystyle \frac{18}{11}\right)\left(\displaystyle \frac{12}{49}\right)\left(\displaystyle \frac{455}{432}\right)\\[15pt] =&\left(\displaystyle \frac{2}{5\cdot 7}\right)\left(\displaystyle \frac{18}{11}\right)\left(\displaystyle \frac{12}{49}\right)\left(\displaystyle \frac{5\cdot 91}{432}\right)& \hbox{Factor\)\(}\\[15pt] =&\left(\displaystyle \frac{2}{7}\right)\left(\displaystyle \frac{18}{11}\right)\left(\displaystyle \frac{4\cdot 3}{49}\right)\left(\displaystyle \frac{91}{4\cdot 108}\right)& \hbox{Factor\)\(}\\[15pt] =&\left(\displaystyle \frac{2}{7}\right)\left(\displaystyle \frac{9\cdot 2}{11}\right)\left(\displaystyle \frac{3}{49}\right)\left(\displaystyle \frac{91}{9\cdot 12}\right)& \hbox{Factor\)\(}\\[15pt] =&\left(\displaystyle \frac{2}{7\cdot 1}\right)\left(\displaystyle \frac{2}{11}\right)\left(\displaystyle \frac{3}{49}\right)\left(\displaystyle \frac{7\cdot 13}{12}\right)& \hbox{Factor\)\(}\\[15pt] =&\left(\displaystyle \frac{2}{1}\right)\left(\displaystyle \frac{2}{11}\right)\left(\displaystyle \frac{3}{49}\right)\left(\displaystyle \frac{13}{2\cdot 2\cdot 3}\right)& \hbox{Factor\)\(}\\[15pt] =&\left(\displaystyle \frac{1}{1}\right)\left(\displaystyle \frac{1}{11}\right)\left(\displaystyle \frac{1}{49}\right)\left(\displaystyle \frac{13}{1\cdot 1\cdot 1}\right)\\[15pt] =&\displaystyle \frac{13}{11\cdot 49}=\displaystyle \frac{13}{539} \end{array}\)
    An enigma\(\)
    Each member of a class got a piece of cardboard \(8\times 12\)inches. The object was to cut the cardboard into two identical rectangular parts (halves).

    Some student ended up with a half measuring \(6 \times 8\) inches. Another student obtained a piece \(4 \times 12\).

    How can the halves be represented by the same fraction \(\displaystyle \frac{1}{2}\) and look so different?

    (0,15)(0,10) (0,0)(1,0)8 (0,0)(0,1)12 (8,12)(-1,0)8 (8,12)(0,-1)12 (5,6)\(12\) in (3,1)\(8\) in

    (0,0)(-12,7.5) (0,0)(1,0)8 (0,0)(0,1)12 (8,12)(-1,0)8 (8,12)(-1,0)8 (8,12)(0,-1)12 (4,12)(0,-1)12 (1.3,6)\(12\) in (1,0.5)\(4\) in

    (0,0)(-24,5) (0,0)(1,0)8 (0,0)(0,1)12 (0,6)(1,0)8 (8,12)(-1,0)8 (8,12)(0,-1)12 (5.5,3)\(6\) in (3,0.5)\(8\) in

    \(\)

    Addition of Fractions

    \(\)

    (0,0)(0,-1) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (10,-5)(0,1)5 (10,-7)(0,-1)5 (9,-6)12 in (4,-13)(1,0)4 (4,-13)(-1,0)4 (3,-14)\(8\) in (4,-12)(0,2)6(0,1)1 (0.2,-3.8)(0.15,0)24(0,1)3.6 (0.2,-7.7)(0.15,0)24(0,1)3.6 (0.2,-11.8)(0.15,0)24(0,1)3.8 (0,-4)(2,0)4(1,0)1 (0,-7.9)(2,0)4(1,0)1 (4.5,-17)\(\displaystyle \frac{1}{2}=\frac{3}{6}\)

    (0,0)(-13,-3.9) (0,0)(1,0)8 (0,0)(0,-1)12 (8,-12)(-1,0)8 (8,-12)(0,1)12 (4,-13)(1,0)4 (4,-13)(-1,0)4 (4,-14)\(8\) in (0,-4)(2,0)4(1,0)1 (0,-8)(2,0)4(1,0)1 (0.3,-11.8)(0.15,0)24(0,1)3.6 (4.3,-11.8)(0.15,0)24(0,1)3.6 (3.2,-17)\(\displaystyle \frac{1}{3}=\frac{2}{6}\) (4,-12)(0,2)6(0,1)1 (10,-5)(0,1)5.0 (10,-7)(0,-1)5.0 (9,-6)12 in

    (0,0)(-25.4,-4.0) (0,0)(1,0)4 (0,0)(0,-1)12 (4,-12)(-1,0)4 (4,-12)(0,1)12 (1,-14)\(4\) in (1.5,-3)

    A P P L E

    (0,0)(-31,-6.9) (0,-8)(1,0)8 (0,-8)(0,-1)4 (8,-12)(-1,0)8 (8,-12)(0,1)4 (4,-13)(1,0)4 (4,-13)(-1,0)4 (4,-14)\(8\) in (1.1,-10)

    ORANGE

    (9,-9)(0,1)1.0 (9,-11)(0,-1)1.0 (8.5,-10.5)4 in

    (0,0)(-32,-17.0) (0,-8)(1,0)4 (0,-8)(0,-1)4 (4,-12)(-1,0)4 (4,-12)(0,1)4 (2,-13)(1,0)2 (2,-13)(-1,0)2 (2,-14)\(4\) in (0.0,-10)

    BANANA

    (5,-9)(0,1)1.0 (5,-11)(0,-1)1.0 (4.7,-10.5)4 in

    We need to add \(\displaystyle \frac{1}{2}\) to \(\displaystyle \frac{1}{3}\).

    \(\displaystyle \frac{1}{2}\) of the whole (labeled APPLE) is
    a rectangle with dimensions \(12\) by \(4\) inches.

    \(\displaystyle \frac{1}{3}\) of the whole (labeled ORANGE) is
    a rectangle with dimensions \(8\) by \(4\) inches.

    We cannot add apples to bananas (unless you make a cocktail salad).

    We can turn the APPLE into \(3\) BANANAs (dimensions \(4\) by \(4\) inches).

    We can turn the ORANGE into \(2\) BANANAs.

    Adding an APPLE to an ORANGES is equivalent to adding \(3\) BANANAs to \(2\) BANANAs.

    Mathematically, we obtain a common denominator and then add the numerators.

    \(\begin{array}{rcl lll lll} \displaystyle \frac{1}{2}&=&\displaystyle \frac{1}{2}\cdot \frac{3}{3}&=&\displaystyle \frac{3}{6}\\[15pt] \displaystyle \frac{1}{3}&=&\displaystyle \frac{1}{3}\cdot \frac{2}{2}&=&\displaystyle \frac{2}{6}\\[10pt]\) \(\\[-9pt]\hline\) \(\\[-9pt] & & &=&\displaystyle \frac{3+2}{6}=\frac{5}{6}\\ \end{array}\)

    A geometrical approach follows.

    If our reasoning is correct, then sum of the areas of the APPLE and the ORANGE should equal the area of \(5\) BANANAs.

    \(4\cdot 12 + 8\cdot 4 = 48+32=80\) in\(^2\) (APPLE+ORANGE)
    \(5\cdot (4\cdot 4)= 5\cdot 16 = 80\) in\(^2\) (\(5\) BANANAs)
    The original whole has \(12 \times 8 = 96\) in\(^2\).
    \(\displaystyle \frac{5}{6}\cdot 96\!=\!\frac{5}{6}\cdot \frac{96}{1}\!=\!\frac{5\cdot 12\cdot 8}{6}\!=\!\frac{5\cdot 2 \cdot 8}{1}\!=\!(5\cdot 2)(8)\!=\!10\cdot 8\!=\!80\)
    in\(^2\).

    Least Common Multiple (Denominator)

    Let’s add \(\displaystyle \frac{2}{15}+\frac{3}{10}\).

    Before discussing the LCD, the procedure in the previous section leads to

    \(\begin{array}{rcl lll lll} \displaystyle \frac{2}{15}&\!\!=\!\!&\displaystyle \frac{2}{15}\cdot \frac{10}{10}&\!\!=\!\!&\displaystyle \frac{20}{150}\\[15pt] \displaystyle \frac{3}{10}&\!\!=\!\!&\displaystyle \frac{3}{10}\cdot \frac{15}{15}&\!\!=\!\!&\displaystyle \frac{45}{150}\\[10pt]\) \(\\[-9pt]\hline\) \(\\[-4pt] &&&=&\displaystyle \frac{20+45}{150}\!\!\!=\!\frac{65}{150}\!\!=\!\!\frac{5\cdot 13}{3\cdot 5\cdot 10}\!\!=\!\!\frac{13}{30}&\!\!\hbox{(factor\)=1)\(}\\ \end{array}\)
    The denominator \(150\) in this problem is larger than needs to be. The largest denominator we need is the Least Common Multiple of the denominators (called Least Common Denominator — LCD) is obtained after writing the prime factorization of each denominator. The divisibility tests are handy in this procedure.

    (0,3)(-1,-2) (-0.5,0)\(\ \ 15\) (-0.3,-3.5)\(3\ \ \ 5\) (8.8,0)\(\ 10\) (8.8,-3.5)\(2\ \ \ 5\) (1,-0.5)(-1,-2)1 (1,-0.5)(1,-2)1 (10,-0.5)(-1,-2)1 (10,-0.5)(1,-2)1

    (0,3)(1,-2) (1,0)\(\begin{array}{ll|l| l|l|l lll lll} 15&=&&3&5\\[5pt] 10&=&2&&5\\[5pt] \hline \hbox{LCD}&=&2&3&5&=30\\ \end{array}\) (20,2)Same factors in the same column. (20,0.5)Different factors not in same column. (20,-1.0)No exponents. You need not remember (20,-2.5)to use smallest or largest exponent.

    \(\begin{array}{rcl lll lll} \displaystyle \frac{2}{15}&=&\displaystyle \frac{2}{15}\cdot \frac{2}{2}&=&\displaystyle \frac{4}{30}\ \hspace{0.5in} \hbox{\)2\(is missing row 1 above.}\\[15pt] \displaystyle \frac{3}{10}&=&\displaystyle \frac{3}{10}\cdot \frac{3}{3}&=&\displaystyle \frac{9}{30}\ \hspace{0.5in} \hbox{\)3\(is missing in row\)2\(above.}\\[10pt]\) \(\\[-9pt]\cline{1-5}\) \(\\[-9pt] & &&=&\displaystyle \frac{4+9}{30}=\frac{13}{30}\ \hbox{ Fraction can't be reduced further.}\\ \end{array}\)
    Example \(7\)

    The numbers are large, but so is the reward.

    Find the LCD of the denominators \(264\) and \(756\).

    (0,5)(-1,-4) (-0.1,0)\(\ 264\) (-0.6,-4.0)4 66 (18.8,0)\(\ 756\) (18.0,-4.0)4 189 (1.1,-0.5)(-1,-2)1 (2.5,-0.5)(1,-2)1 (20,-0.5)(-1,-2)1 (21.3,-0.5)(1,-2)1 (3.9,-4.2)(-1,-2)1 (3.9,-4.2)(1,-2)1 (-0.4,-4.5)(-1,-2)1 (-0.3,-4.5)(1,-2)1 (-2.3,-8.1)2 2 (2.5,-8.0)6 11 (2.8,-8.5)(-1,-2)1 (2.8,-8.5)(1,-2)1 (0.9,-12.0)2 3 (18.0,-4.5)(-1,-2)1 (18,-4.5)(1,-2)1 (16.2,-7.7)2 2 (21.7,-7.7)9 21 (23.0,-4.5)(-1,-2)1 (23.5,-4.5)(1,-2)1 (21.5,-7.9)(-1,-2)1 (21.7,-7.9)(1,-2)1 (25.6,-7.9)(-1,-2)1 (25.6,-7.9)(1,-2)1 (19.7,-11.4)3 3 (24.1,-11.4)7 3

    (0,3)(-5,-2) (0,0)\(\begin{array}{ll|l| l|l|l| l|l|l| l|ll } 264 &=&2&2&2&3& & & &11&\\[5pt] 756 &=&2&2& &3&3&3&7& &\\[5pt] \cline{1-10} \hbox{LCD}&=&2&2&2&3&3&3&7&11&\\[5pt] \end{array}\)

    The LCCD is \(756(2)(11)=16,632\)

    Just as a quick check, \(\displaystyle \frac{16,632}{756}=22\) and \(\displaystyle \frac{16,632}{264}=63\)

    Were you taught to find the LCM by using sets of multiples?

    Find the LCM of \(2\) and \(3\):

    Multiples of \(2\): \(\{2, 4, 6, 8, 10, 12, \cdots\}\)

    Multiples of \(3\): \(\{3, 6, 9, 12, \cdots\}\)

    \(6, 12, 18, \cdots\) are multiples of \(2\) and \(3\). \(6\) is the smallest (Least) multiple common to both \(2\) and \(3\).

    Do you see the difficulty in using this procedure to find the LCM of \(264\) and \(756\)?

    Multiples of \(264\): \(\{264,528,792, \cdots\}\)

    Multiples of \(756\): \(\{756,1512,2268, \cdots\}\)

    Addition of Fractions

    Example \(8\)

    Add \(\displaystyle \frac{7}{264}+\frac{5}{756}\)

    Solution:

    A common denominator is needed when adding fractions. We found that the LCD of \(264\) and \(756\) is \(16,632\).

    Note from the LCD array above that \(264\) needs to be multiplied by \((3)(3)(7)=(9)(7)=63\) to get \(16,632\).

    Similarly \(756\) needs to be multiplied by \((2)(11)=22\) to get \(16,632\).

    \(\begin{array}{c rcl lll} &\displaystyle \frac{7}{264}&=&\displaystyle \frac{7}{264}\cdot \frac{63}{63}&=&\displaystyle \frac{441}{16,632}&\hbox{Quick check: }=0.0265152\\[15pt] +&\displaystyle \frac{5}{756}&=&\displaystyle \frac{5}{756}\cdot \frac{22}{22}&=&\displaystyle \frac{110}{16,632}&\hbox{Quick check: }=0.0066138\\[10pt]\) \(\\[-9pt] \hline\) \(\\[-4pt] &&=&\displaystyle \frac{441+110}{16,632}&=&\displaystyle \frac{551}{16,632}&\hbox{Quick check: }=0.03312890\\ \end{array}\)

    \(0.0265152+0.0066138=0.033129\) (Nice!)

    Subtraction of Fractions

    Rewrite subtraction as addition of opposites.

    Multiplication of Fractions

    What is two-thirds of a quarter?

    (0,5)(-2,-3) (-4,-5.5)The original whole (-4,-6.8)the whole pie. (0,0) (0,0)(-1,0)2 (0,0)(1,0)2 (0,0)(0,1)2 (0,0)(0,-1)2 (7.5,0)(4,4)[tr] (7.5,0)(0,1)2 (7.5,0)(1,0)2 (4,-3)\(\displaystyle \frac{1}{4}\) (4,-2.5)(-1,1)2.7 (5,-2.5)(1,1)2.5 (11,-5.5)The new whole (11,-6.8)is a quarter pie. (13.5,0)(4,4)[tr] (13.5,0)(0,1)2 (13.5,0)(1,0)2 (17.5,0)(4,4)[tr] (17.5,0)(0,1)2 (17.5,0)(1,0)2 (17.5,0)(2,1)1.8 (17.5,0)(1,2)0.9 (16,-3)\(\displaystyle \frac{1}{3}\) of the new whole (18,-2.0)(1,2)1 (14.5,-4.5)(0,1)4 (30,0) (30,0)(-1,0)2 (30,0)(1,0)2 (30,0)(0,1)2 (30,0)(0,-1)2 (30,0)(-2,-1)1.7 (30,0)(-1,-2)0.9 (30,0)(2,1)1.7 (30,0)(1,2)0.9 (30,0)(2,-1)1.7 (30,0)(1,-2)0.9 (30,0)(-2,1)1.7 (30,0)(-1,2)0.9 (23,0)(1,0)2 (23,0)(2,1)1.7 (25,0)(-1,4)0.25 (19,5)\(\displaystyle \frac{1}{12}\) of the original whole (24,4.7)(-1,-1)4.2 (24,4.7)(0,-1)3.0 (24,4.7)(2,-1)7.0 (19.8,0.3)(1,0)3.2 (28.7,0.3)(-1,0)4.2 (18.1,-2.0)(3,1)5.2

    The set of pictures is intended to show that \(\displaystyle \frac{1}{3}\) of \(\displaystyle \frac{1}{4}\) is \(\displaystyle \frac{1}{12}\)
    that is
    \(\displaystyle \frac{1}{3}\cdot \displaystyle \frac{1}{4}=\displaystyle \frac{1}{12}\).

    The last statement suggests that we can multiply numerators and denominators of the product of fractions. Reduce fractions before multiplying first, if possible.
    \[\displaystyle \frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}\]

    Division of Fractions

    Divide \(\begin{array}{c} \displaystyle \frac{2}{3}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{5}{7} \end{array}\)

    Multiply the fraction by \(\begin{array}{c} \displaystyle \frac{7}{5}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{7}{5} \end{array}=\) 1

    Note that the denominator is \(1\) also now, which an be omitted.

    \(\begin{array}{c} \displaystyle \frac{2}{3}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{5}{7} \end{array}\) \(\cdot\) \(\begin{array}{c} \displaystyle \frac{7}{5}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{7}{5} \end{array}\) \(=\) \(\begin{array}{c} \displaystyle \frac{2}{3}\cdot \frac{7}{5}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{1}{1} \end{array}\) \(=\displaystyle \frac{2}{3}\cdot \frac{7}{5}\)

    This example illustrates that division by a fraction is equivalent to multiplication by the reciprocal of the original divisor (denominator).

    missing

    Mixed Numbers

    A mixed number is a “mixed up" number. It cannot decide whether it is an improper fraction or a whole number. It is a whole number added to a proper positive fraction, with the addition sign omitted.

    (0,3)(0,5) (0,0)(1,0)9 (0,0)(0,1)7 (9,7)(-1,0)9 (9,7)(0,-1)7 (1.5,1)(2,0)4 (1.5,3)(2,0)4 (1.5,5)(2,0)3 (7.5,5) (10,5)

    The numerator of a proper fraction, like \(\displaystyle \frac{11}{12}\), is less than its denominator.

    (0,3)(-0,8) (0,0)(1,0)9 (0,0)(0,1)7 (9,7)(-1,0)9 (9,7)(0,-1)7 (1.5,1)(2,0)4 (1.5,3)(2,0)4 (1.5,5)(2,0)3 (7.5,5)

    (0,0)(-10.6,4.9) (0,0)(1,0)9 (0,0)(0,1)7 (9,7)(-1,0)9 (9,7)(0,-1)7 (1.5,1)(2,0)4 (3.5,3)(2,0)3 (1.5,5)(2,0)4 (1.5,3) (7.5,3) (9.8,6)

    The numerator of an improper fraction, like
    \(\displaystyle 1\frac{5}{12}=\frac{17}{12}\), is equal to or more
    than its denominator.

    Here is how to convert an improper fraction to a mixed number. Divide the numerator of an improper fraction by its denominator. The quotient is the whole number. The remainder is the numerator of the proper fraction.

    \(\displaystyle \frac{41}{12}=3\frac{5}{12}\)

    Now convert \(\displaystyle 5\frac{4}{7}\) to an improper fraction.

    \(\begin{array}{rcl lll} \displaystyle 5\frac{4}{7} %&=&\displaystyle 5\frac{4}{7}\\[15pt] &=&\displaystyle \frac{5}{1}+\frac{4}{7}\\[15pt] &=&\displaystyle \frac{5\cdot 7}{7}+\frac{4}{7}\\[15pt] &=&\displaystyle \frac{35}{7}+\frac{4}{7}\\[15pt] &=&\displaystyle \frac{35+4}{7}\\[15pt] &=&\displaystyle \frac{39}{7}\\[10pt] \end{array}\)

    Shortcut: \(\displaystyle 5\frac{4}{7}=\frac{5\cdot 7+4}{7}=\frac{39}{7}\)

    Arithmetic of Mixed Numbers

    Addition of Mixed Numbers

    Example \(9\)

    Add \(\displaystyle 4\frac{3}{5}+2\frac{1}{3}\)

    Solution:

    \(\hspace{-10pt}\begin{array}{c ccc ccc ccl lll lll} &\displaystyle 4\frac{3}{5}&\!=\!&\displaystyle \frac{4(5)+3}{5}&\!=\!&\displaystyle \frac{23}{5}&\!=\!&\displaystyle \frac{23}{5}\cdot \frac{3}{3}&\!=\!&\displaystyle \frac{69}{15}\\[15pt] \!\!+\!\!&\displaystyle 2\frac{1}{3}&\!=\!&\displaystyle \frac{2(3)+1}{3}&\!=\!&\displaystyle \frac{7}{3}&\!=\!&\displaystyle \frac{7}{3}\cdot \frac{5}{5}&\!=\!&\displaystyle \frac{35}{15}\\[10pt] \cline{7-10} \\[5pt] &&&&&&&&\!=\!&\!\!\!\displaystyle \frac{\!69\!+\!35}{15}\!=\!\frac{104}{15}\!=\!6\frac{14}{15} \end{array}\)

    Example \(10\)

    Add \(\displaystyle 87,654,321\frac{4}{5}+12,345,678\frac{2}{3}\)

    Solution:

    \(\begin{array}{c lll lll lll lll lll} &\displaystyle \frac{4}{5}&=&\displaystyle \frac{4}{5}&=&\displaystyle \frac{4\cdot 3}{5\cdot 3}&=&\displaystyle \frac{12}{15}\\[15pt] +&\displaystyle \frac{2}{3}&=&\displaystyle \frac{2}{3}&=&\displaystyle \frac{2}{3}\cdot \frac{5}{5}&=&\displaystyle \frac{10}{15}\\[10pt] \cline{4-8} \\[5pt] &&&&&&=&\displaystyle \frac{27}{15}=1+\frac{12}{15}=1+\frac{4}{5} \end{array}\)

    The method used in example \(5\) can be used, but it leads to very large numbers. Instead we add the whole numbers and the proper fractions individually.\(\)
    The sum is \(87,654,321+12,345,678+1+\displaystyle \frac{4}{5}=100,000,000\frac{4}{5}\).

    Subtraction of Mixed Numbers

    Example \(11\)

    Subtract \(\displaystyle 59\frac{1}{6}-3\frac{4}{5}\)

    Solution:

    The method of example \(11\) is straightforward, but suppose the numbers are large. Then the following technique is handy.

    \(\hspace{-0.5in}\begin{array}{c ccc ccc cll } &&&\displaystyle \frac{1}{6}&=&\displaystyle \frac{1\cdot 5}{6\cdot 5}&=&\displaystyle \frac{5}{30}&\hbox{Keep the\)-\(sign at every}\\[15pt] &&&\displaystyle \frac{-4}{5}&=&\displaystyle \frac{-4}{5}\cdot \frac{6}{6}&=&\displaystyle \frac{-24}{30}&\hbox{ step so that you do not add}\\[15pt] \cline{4-8}\\[-7pt] &&&&&&=&\displaystyle \frac{-19}{30}&\hbox{unintentionally instead of subtracting.} \end{array}\)
    The difference is \(59-3+\displaystyle \frac{-19}{30}=56+\frac{-19}{30}\). If the proper
    fraction were positive, we would be finished. But here we shall rewrite \(56\) as \(55+1\) and convert the 1 to an equivalent fraction with denominator 30. This is called borrowing.

    \(\begin{array}{lll lll lll lll} 59-3+\displaystyle \frac{-19}{30}&=&\displaystyle 56+\frac{-19}{30}\\[11pt] &=&\displaystyle 55+1+\frac{-19}{30}\\[11pt] &=&\displaystyle 55+\frac{30}{30}+\frac{-19}{30}\\[11pt] &=&\displaystyle 55+\frac{30-19}{30}\\[11pt] &=&\displaystyle 55+\frac{11}{30} \end{array}\)
    \(\displaystyle 59\frac{1}{6}-3\frac{4}{5}=55\frac{11}{30}\)

    Multiplication of Mixed Numbers

    Convert the mixed numbers to improper fractions, then multiply the fractions. Do not forget about reducing fractions. When finished with the product, convert that product to a mixed number since you started with a mixed number.

    Division of Mixed Numbers

    Convert the mixed numbers to improper fractions, then divide the fractions (multiply by the reciprocal of the divisor (denominator)). Do not forget about reducing fractions. When finished with the product (which was originally a quotient), convert that product to a mixed number.

    Complex Fractions

    A complex fraction is a fraction whose numerator and/or denominator is a fraction.

    Example \(12\)

    Simplify \(\ \ \begin{array}{c} \displaystyle \frac{1}{2}-\frac{3}{5}\\\) \(\\[-6pt] \hline\) \(\\[-6pt] \displaystyle \frac{1}{4}-\frac{9}{25} \end{array}\)

    Solution:

    One method is to combine the fractions in the numerator and/or denominator into single fractions. Then multiply the numerator by the reciprocal of the denominator.

    Another method is to multiply both numerator and denominator of all the fractions by the LCD of all the fractions.

    The LCD of \(2\), \(5\), \(4\), and \(25\) is \(100\).
    \(\ \ \begin{array}{c} \displaystyle \frac{1}{2}-\frac{3}{5}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{1}{4}\!-\!\frac{9}{25} \end{array}\) \(\cdot\) \(\displaystyle \frac{100}{100}\) \(=\) \(\ \ \begin{array}{c} \displaystyle \frac{1(100)}{2}\!-\!\frac{3(100)}{5}\\\) \(\\[-8pt] \hline\) \(\\[-8pt] \displaystyle \frac{1(100)}{4}\!-\!\frac{9(100)}{25} \end{array}\) \(=\) \(\displaystyle \frac{1(50)\!-\!3(20)}{1(25)-9(4)}\),
    \(\displaystyle =\frac{50-60}{25-36}=\frac{10}{11}\)
    Do not confuse complex fractions with complex numbers. You will learn about complex numbers toward the end of this course.

    Convert a Decimal with Repeating Digits to a Rational Number

    \(\)
    We illustrate the technique by an example.

    Example \(13\)

    Convert \(N=0.3575757\cdots 57\cdots=0.3\overline{57}\) to a rational number (a ratio of two integers).

    Solution:

    The procedure is to rewrite \(N\) such that the decimal point occurs before the first group of repeating decimals. Call this number \(N_2\).

    Then rewrite \(N\) as another number \(N_1\) such that the decimal point occurs immediately after the first group of repeating decimals.

    Subtract the \(N_1-N_2\). The decimal points line up and the difference \(N_1-N_2\) will contain infinitely many zeros after the decimal point. One dision (and possible reduction) will leave you with a rational number.

    missing

    Check:

    \(\displaystyle \frac{354}{990}=0.35757575757575757575757575757576\)

    The rightmost 6 is due to calculator rounding.

    Percent

    “of" frequently means multiplication in mathematics; “per" similarly means division.

    Cent is found in words like century, cents, percent. It identifies a group of \(100\).

    “Percent" \(\Rightarrow\) Per Cent refers to division by \(100\).

    Thus \(5\) percent means \(\displaystyle \frac{5}{100}\) written \(5\)%.

    Is percent useful?

    You invest \(\$655\) in a bank and earn \(\$21.29\) in simple interest at the end of the year.

    Your friend invests \(\$967\) on the same day in another bank and earns \(\$28.53\) at the end of the year.

    Is your friend getting a better deal because his return is higher than yours?

    It is impossible to decide without further computations. Suppose you and your friend invest the same amount, say \(\$100\) at the same time in these banks. Suppose you earn \(\$3.25\) while your friend earns \(\$2.95\). Your bank has the better yield. Your bank pays \(3.25\)% while your friend’s bank returns \(2.95\)%.

    Basing every comparison on \(100\) is quite useful, for the sake of consistency.
    In the next example we convert a number to a percent,

    Note that \(1=\displaystyle 100\cdot \frac{1}{100}=100\%\)

    Example \(14\)

    \(2.7=(2.7)(1)=(2.7)(100)\%=270\%\)

    In the following example we convert a percent to a decimal number,

    Example \(15\)

    \(2.7\%=\displaystyle (2.7)\frac{1}{100}=0.027\)
    The following formula is extremely useful in dealing with percent

    missing

    Example \(16\)

    What is \(20\%\) of \(80\)?

    Solution:

    \(20\%=0.2\)

    (0,11)(0,-10) (2,0)\(A=\ P_d\ \cdot \ B\) (0,-3.3)What is \(20\%\) of \(80\%\)? (3.5,-0.7)(0,-1)1.5 (7.6,-0.7)(0,-1)1.5 (3.5,-4.0)(1,-4)0.5 (8,-4.0)(1,-2)0.7 (1,-7)\(\begin{array}{rcl lll} A&=&\ (0.2) \cdot (80)=16 \end{array}\) (0,-10)\(16\) is \(20\%\) of \(80\).

    Example \(17\)

    \(80\) is \(20\%\) of what number?

    Solution:

    \(20\%=0.2\)

    (0,13)(0,-12) (1.5,0)\(A=\ P_d\ \cdot \ B\) (1,-3.5)\(80\) is \(20\%\) of what number? (3.2,-0.7)(0,-1)1.5 (7.1,-0.7)(0,-1)1.5 (3.5,-4.0)(1,-2)0.7 (6.2,-4.0)(3,-2)2.4 (0.5,-9.5)\(\begin{array}{rcl lll} 80&=&\ (0.2) \cdot B\\[10pt] \displaystyle \frac{80}{0.2}&=&\ \left(\displaystyle \frac{0.2}{0.2}\right) \cdot B\\[10pt] \displaystyle \frac{800}{2}&=&\ B\\[10pt] B&=&400 \end{array}\) (0,-15.5)\(80\) is \(20\%\) of \(400\).

    Example \(18\)

    \(20\) is what percent of \(80\)?

    Solution:

    (0,13)(0,-12) (1.5,0)\(A= P_d\ \cdot \ B\) (1,-4)\(20\) is \(P_d\) of \(80\)? (3,-0.7)(0,-1)2 (6.5,-0.7)(0,-1)2 (3.0,-4.1)(1,-2)0.9 (7,-4.4)(1,-2)0.7 (0.8,-11)\(\begin{array}{rcl lll} 20&=& P_d \cdot 80\\[10pt] \displaystyle \frac{20}{80}&=&\ \left(\displaystyle \frac{80}{80}\right) \cdot P_d\\[10pt] \displaystyle \frac{1}{4}&=&\ P_d\\[10pt] P_d&=&0.25=(0.25)(1)=(0.25)(100\%)=(0.25)(100)\%=25\% \end{array}\) (0,-17)\(20\) is \(25\%\) of \(80\).

    You can also use \(\displaystyle \frac{P}{100}=\frac{A}{B}\). The danger is that you interchange
    the numbers for \(A\) and \(B\). The danger of error is lessened if you remember that \(B\) is the number following the word “of".

    Exercise 5

    1. Find \(3\) fractions equivalent to \(\displaystyle \frac{8}{11}\) with numerators \(n\le 35\).

    2. Reduce \(\displaystyle \frac{630}{1,764}\)

    3. Find the units (ones) digit \(d\) such that \(49,17``d"\) is divisible by \(2\).

    4. Find the units (ones) digit \(d\) such that \(49,17``d"\) is divisible by \(3\).

    5. If possible, construct a number that is divisible by all of the following numbers: \(2, 3, 4, 5, 6, 7, 8, 9\), and \(10\).

    6. Reduce \(\left(\displaystyle \frac{14}{35}\right)\left(\displaystyle \frac{36}{77}\right)\left(\displaystyle \frac{49}{24}\right)\)

    7. What is the prime factorization of \(264\)?

    8. What is the prime factorization of \(495\)?

    9. Find the Least Common Multiple of \(264\) and \(495\).

    10. Add \(\displaystyle \frac{5}{264}+\frac{7}{495}\).

    11. Mother has \(\displaystyle \frac{5}{6}\) yards of cloth. She wants to make napkins
      which require \(9\) inches of cloth each. How many napkins can she make?
      There are \(12\) inches in a ft and \(3\) ft equal \(1\) yard.

    12. A warehouse contains a number of boxes all the same size.
      \(\displaystyle \frac{2}{21}\) of the remaining boxes is removed on Monday.
      \(\displaystyle \frac{1}{6}\) of the remaining boxes is shipped on Tuesday.
      \(\displaystyle \frac{1}{2}\) of the boxes still left are trucked out on Wednesday.
      All of the remaining \(190\) boxes are sold on Thursday.
      How many boxes were in the warehouse on Monday?
      Hint: Assume you know the number of boxes in the warehouse on Monday.
      Hint: Let that number be locked in the \(x\)-box.

    13. Subtract \(\displaystyle \frac{5}{42}-\frac{2}{1155}\)

    14. Reduce \(\left(\displaystyle \frac{12}{35}\right)\left(\displaystyle \frac{21}{4}\right)\left(\displaystyle \frac{10}{9}\right)\left(\displaystyle \frac{11}{26}\right)\)

    15. A troupe of boy scouts come to camp with \(45\displaystyle \frac{2}{3}\) ft of rope. Each scout is to be cut \(\displaystyle 1\frac{3}{4}\) ft of rope to practice knots. The
      amount of rope is just sufficient for all the scouts (a little piece of rope may be left over). How many scouts are in the troupe?

    16. Convert \(\displaystyle \frac{150}{37}\) to a mixed number.

    17. Convert \(7\displaystyle \frac{9}{28}\) to an improper fraction.

    18. Add \(25\displaystyle \frac{6}{7}+12\frac{11}{14}\).

    19. Subtract \(38\displaystyle \frac{9}{14}-12\frac{11}{14}\).

    20. Multiply \(\left(\displaystyle 3 \frac{6}{7}\right)\left(\displaystyle 2 \frac{1}{3}\right)\)

    21. Convert \(81\displaystyle \frac{2}{3}\%\) to an exact fraction, no decimals.

    22. Convert \(\displaystyle \frac{2}{3}\) to an exact percent. Give your answer as a mixed number.

    23. A retailer buys a dozen coats for \(\$250\) each. The mark up rate (taken as a percent of cost) is \(30\%\). Find the sale price of a coat (cost plus mark up).
      A year later the retailer has one coat left and wants to give his customers a discount of \(30\%\) of the sale price to sell the coat. The discount is computed using the sale price above. How much does the customer have to pay after the discount?
      Is the discounted price equal to the original cost of \(\$250\) since the mark up rate was \(30\%\) and the discount rate was also \(30\%\)?
      If not, what is the retailer’s profit or loss?

    24. What is \(5\%\) of \(25\)?

    25. \(25\) is \(5\%\) of what number?

    26. \(5\) is what percent of \(25\)?

    The solutions follow:

    Keep them covered up till you are able to solve each of the problems above, without help from any source.

    1. Solution:\(\)
      \(\displaystyle \frac{8}{11}=\frac{8}{11}\cdot \frac{2}{2}=\frac{16}{22}\)
      \(\displaystyle \frac{8}{11}=\frac{8}{11}\cdot \frac{3}{3}=\frac{24}{33}\)
      \(\displaystyle \frac{8}{11}=\frac{8}{11}\cdot \frac{4}{4}=\frac{32}{44}\)
    2. Solution:
      \(\begin{array}{rcl lll} \displaystyle \frac{630}{1,764}&=&\displaystyle \frac{9\cdot 70}{9\cdot 196}&\hbox{Factor\)=1\(}\\[15pt] &=&\displaystyle \frac{70}{ 196}\\[15pt] &=&\displaystyle \frac{2\cdot 35}{2\cdot 98}&\hbox{Factor\)=1\(}\\[15pt] &=&\displaystyle \frac{35}{98}&\\[15pt] &=&\displaystyle \frac{7\cdot 5}{7\cdot 14}&\hbox{Factor\)=1\(}\\[15pt] &=&\displaystyle \frac{5}{14} \end{array}\)
    3. Solution:

      \(d\) can be \(0\) so the number is \(49,170\).
      \(d\) can also be \(2\), so the number is \(49,172\).
      \(d\) can further be \(4\), so the number is \(49,174\).
      \(d\) can be \(6\) as well, so the number is \(49,176\).
      [5pt] Finally \(d\) can be \(8\), so the number is \(49,178\).

    4. Solution:

      The sum of the digits must be divisible by \(3\). The sum is \(4+9+1+7+d=21+d\).

      \(d=0\), \(21\) is divisible by \(3\). The number is \(49,170\).

      \(d=3\), sum\(=21+3=24\), div. by \(3\). Number: \(49,173\).

      \(d=6\), sum\(=21+6=27\), div. by \(3\). Number: \(49,176\).

      \(d=9\), sum\(=21+9=30\), div. by \(3\). Number: \(49,179\).

    5. Solution:

      If the number is divisible by \(8\), it will automatically be divisible by \(2\) and \(4\).

      If the number is divisible by \(9\), it will automatically be divisible by \(3\).

      Smallest number: \(N=5\cdot 7\cdot 8 \cdot 9=40\cdot 63=2,520\).

      \(N\) can be any multiple of \(2,520\), namely \(5040\), \(7560\), \(10080\), \(12600\), \(15120\), \(\cdots\)

    6. Reduce \(\left(\displaystyle \frac{14}{35}\right)\left(\displaystyle \frac{36}{77}\right)\left(\displaystyle \frac{49}{24}\right)\)
      Solution:
      \(\begin{array}{rcl lll} \left(\displaystyle \frac{14}{35}\right)\left(\displaystyle \frac{36}{77}\right)\left(\displaystyle \frac{49}{24}\right) &=&\left(\displaystyle \frac{2\cdot 7}{5\cdot 7}\right)\left(\displaystyle \frac{2\cdot 2\cdot 3\cdot 3}{7\cdot 11}\right)\left(\displaystyle \frac{7\cdot 7}{2\cdot 2\cdot 2 \cdot 3}\right)\\[20pt] &=&\displaystyle \frac{2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 7\cdot 7\cdot 7} {2\cdot 2\cdot 2\cdot 3 \cdot5\cdot 7\cdot 7\cdot 11 }\\[15pt] &=&\displaystyle \frac{\not2\not 2\not 2\not 3\cdot 3\not 7\not 7\cdot 7} {\not 2\not 2\not 2\not 3 \cdot5\not 7\not 7\cdot 11 }\\[20pt] &=&\displaystyle \frac{3\cdot 7} { 5\cdot 11 }=\displaystyle \frac{21} {55}\\[20pt] \end{array}\)

    7. What is the prime factorization of \(264\)?
      Solution:
      \(264=2(132)=2(2)(66)=2(2)(2)(33)=2(2)(2)(3)(11)=2^3(3)(11)\)

    8. What is the prime factorization of \(495\)?
      Solution:
      \(495=(3)(165)=(3)(3)(55)=(3)(3)(5)(11)\)

    9. Find the Least Common Multiple of \(264\) and \(495\).
      Solution:

      (0,5)(1,-2) (0,0)\(\begin{array}{ll|l| l|l|l| l|l|l| lll } 264 &=&2&2&2&3& & &11&\\[5pt] 495 &=& & & &3&3&5&11 &\\[5pt] \hline \hbox{LCD}&=&2&2&2&3&3&5&11&=264(3)(5)=3,960\\ \end{array}\)

    10. Add \(\displaystyle \frac{5}{264}+\frac{7}{495}\).
      Solution:
      We found that the LCD of \(264\) and \(495\) is \(3,960\).

      Note from the LCD array above that \(264\) needs to be multiplied by \((3)(5)=15\) to get \(3,960\).

      Similarly \(495\) needs to be multiplied by \((2)(2)(2)=8\) to get \(3,960\).

      \(\hspace{-0.5in}\begin{array}{c rcl lll} &\displaystyle \frac{5}{264}&=&\displaystyle \frac{5}{264}\frac{15}{15}&=&\displaystyle \frac{75}{3,960}&\hbox{Quick check: }=0.0189394\\[15pt] +&\displaystyle \frac{7}{495}&=&\displaystyle \frac{7}{495}\frac{8}{8}&=&\displaystyle \frac{56}{3,960}&\hbox{Quick check: }=0.0141414\\[5pt]\) \(\\[-10pt] \hline\) \(\\[-5pt] &&=&\displaystyle \frac{56+75}{3,960}&=&\displaystyle \frac{131}{3,960}&\hbox{Quick check: }=0.033080808\\ \end{array}\)
      \(0.0189394+0.0141414=0.033080808\) (Nice!)

    11. There are \(12\) inches in a ft and 3 ft equal 1 yard.

      Solution:

      For the first napkin,
      \(\displaystyle \frac{\!9 \hbox{ inches}}{1} \cdot \frac{1 \hbox{ ft}} {\!12\!\hbox{ inches}} \!=\!\frac{3\cdot 3 \hbox{ ft}} {3\cdot 4} \!=\!\frac{3\hbox{ ft}}{4} \!=\!\frac{3\hbox{ ft}}{4} \cdot \frac{ \hbox{ yards}} {3\hbox{ ft}}\!=\!\frac{1\hbox{ yard}}{4}\)
      of cloth is removed, cut from the cloth. For the second napkin, \(9\) inches \(=\displaystyle \frac{1}{4}\) yard of cloth is removed, subtracted, cut from the cloth.
      This repeated subtraction suggests division.
      The number of napkins is
      \(\displaystyle \frac{5\hbox{ yds}}{6}\div \frac{1\hbox{ yrd}}{4\hbox{ napkins}}=\displaystyle \frac{5\hbox{ yds}}{6}\cdot \frac{4\hbox{ napkins}}{1\hbox{ yrd}}=\displaystyle \frac{5}{3}\cdot \frac{2\hbox{ napkins}}{1}=3\) napkins with \(\displaystyle \frac{1}{3}\) of a napkin left over.

      Instead of using fractions, we could have solved the problem using inches.
      One napkin requires \(9\) inches. The available cloth length is
      \(\displaystyle \frac{5\hbox{ yd}}{6}\cdot \frac{3\hbox{ ft}}{1\hbox{ yd}}\cdot \frac{12\hbox{ in}}{1\hbox{ ft}}=\frac{5}{6}\cdot \frac{3}{1}\cdot \frac{6\cdot 2\hbox{ in}}{1}=\frac{5}{1}\cdot \frac{3}{1}\cdot \frac{1\cdot 2\hbox{ in}}{1}=30\) inches.

      The number of napkins is \(\displaystyle \frac{30\hbox{ inches }}{1}\cdot \frac{1 \hbox{ napkin}}{9\hbox{ inches}}=\frac{30}{9}=3\) napkins with \(3\) inches of cloth left over.

    12. A warehouse contains a number of boxes all the same size.
      \(\displaystyle \frac{2}{21}\) of the remaining boxes is removed on Monday.
      \(\displaystyle \frac{1}{6}\) of the remaining boxes is shipped on Tuesday.
      \(\displaystyle \frac{1}{2}\) of the boxes still left are trucked out on Wednesday.
      All of the remaining \(190\) boxes are sold on Thursday.
      How many boxes were in the warehouse on Monday?
      Hint: Assume you know the number of boxes in the warehouse on Monday.
      Hint: Let that number be locked in the \(x\)-box.
      Solution:
      Cut the number in the \(x\)-box into \(21\) parts and subtract \(2\) of those parts. That leaves \(1-\displaystyle \frac{2}{21}=\frac{19}{21}\) of the original number of boxes for Tuesday.
      Cut the \(\displaystyle \frac{19}{21}\) parts from Monday into six more parts to get
      \(\displaystyle \frac{19}{21}\cdot \frac{1}{6}=\frac{19}{126}\) parts. \(\displaystyle \frac{1}{6}\) of these parts is shipped out on
      Tuesday. This leaves \(\displaystyle \frac{19}{21}-\frac{19}{126}=\frac{114}{126}-\frac{19}{126}=\frac{95}{126}\) of
      the original boxes on Tuesday.
      Half of \(\displaystyle \frac{95}{126}\cdot \frac{1}{2}=\displaystyle \frac{95}{252}\) the original boxes are sold on
      Wednesday. That leaves \(\displaystyle \frac{95}{252}\) of the original boxes for Thursday.
      This leaves \(\displaystyle \frac{95}{252}x=190\) of the original number of boxes.
      The number of boxes is
      \(x=\displaystyle \frac{252}{95}\cdot 190=\frac{(252)(95)}{95}=\frac{(252)(2)}{1}=504\).
      W0W! You are exceptionally talented if you got the answer on your own by reasoning, not by guessing and trying.
      As a preview to algebra, let’s rework this problem. Give the name \(x\) to the number locked up number in the \(x\)-box.

      \(\begin{array}{lll lll lll lll lll} x&=&\displaystyle \frac{21}{21}\ \ \hbox{Original number}\\[15pt] \displaystyle \frac{2}{21}x& &\ \ \hbox{Number of boxes sold on Monday}\\[13pt] x-\displaystyle \frac{2}{21}x&=&\displaystyle \frac{19}{21}x\ \ \hbox{Number of boxes left on Monday}\\[13pt] \displaystyle \frac{1}{6} \cdot \frac{19}{21}x&=&\displaystyle \frac{19}{126}x\ \ \hbox{Number of boxes sold on Tuesday}\\[13pt] \displaystyle \frac{19}{21}x-\frac{19}{126}x&=&\displaystyle \frac{114}{126}x-\frac{19}{126}x=\frac{95}{126}x\ \ \hbox{Boxes left on Tuesday}\\[13pt] \displaystyle \frac{1}{2}\cdot \frac{95}{126}x&=&\displaystyle \frac{95}{252}x\ \ \hbox{Number of boxes sold on Wednesday} \end{array}\)

      \(\begin{array}{lll lll lll lll lll} \displaystyle \frac{95}{126}x-\frac{95}{252}x&=&\displaystyle \frac{95}{252}x\ \ \hbox{Number of boxes left on Wednesday}\\[15pt] \end{array}\)
      All of these boxes was sold on Thursday.
      \(\displaystyle \frac{95}{252}x=190\) so
      \(x=\displaystyle \frac{252\cdot 190}{95}=\frac{252\cdot 2}{1}=504\)
      Originally there were \(504\) boxes in the warehouse.
      Let’s check this number:
      \(\displaystyle \frac{2\cdot 504}{21}=48\) boxes were shipped out on Monday leaving
      \(504-48=456\) boxes.
      \(\displaystyle \frac{1\cdot 456}{6}=76\) boxes were shipped out on Tuesday leaving
      \(456-76=380\) boxes.
      \(\displaystyle \frac{1\cdot 380}{2}=190\) boxes were shipped out on Wednesday
      leaving \(380-190=190\) boxes.
      Does this problem wet your appetite for algebra?

    13. Subtract \(\displaystyle \frac{5}{42}-\frac{2}{1155}\)
      Solution:
      A common denominator is needed when adding fractions. We found that the LCD of \(42\) and \(1155\):

      (0,3)(-1,-2) (0,0)\(\ \ 42\) (-0.3,-3.5)6 7 (0,-3.5)(-1,-2)1 (0,-3.5)(1,-2)1 (0,0)\(\ \ 42\) (-1.3,-6.5)2 3 (8.8,0)\(\ \ 1155\) (8.8,-3.5)5 231 (12,-3.5)(-1,-2)1 (12,-3.5)(1,-2)1 (10.3,-6.5)3 77 (1,-0.5)(-1,-2)1 (1,-0.5)(1,-2)1 (10,-0.5)(-1,-2)1 (10,-0.5)(1,-2)1 (11.7,-9.5)7 11 (13,-6.7)(-1,-2)1 (13,-6.7)(1,-2)1

      (0,3)(1,-2) (0,0)\(\begin{array}{ll|l| l|l|l lll lll} 42 &=&2&3& &7\\[6pt] 1155 &=& &3&5&7&11\\ [6pt]\hline \hbox{LCD}&=&2&3&5&7&11\!=\!2310\\[6pt] \end{array}\) (19,2)Like factors in the same column. (19,0.5)Only like factors in same column. (19.0,-1.0)Avoid exponents. Why remember (19.2,-2.5)use of smallest or largest exponent?

      \(\begin{array}{rcl lll lll} \!\!\displaystyle \frac{5}{42}&\!\!=\!\!&\displaystyle \frac{5}{42}\cdot \frac{55}{55}&\!\!=\!\!&\displaystyle \frac{275}{2310}&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\hbox{\)511\(is missing in row 1. (see LCD array).}\\[10pt] \!\!\displaystyle \frac{2}{1155}&\!\!=\!\!&\displaystyle \frac{2}{1155}\cdot \frac{2}{2}&=&\displaystyle \frac{4}{2310}&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\hbox{\)2\(is missing in row 2 (see LCD array).}\\[10pt]\) \(\\[-9pt]\cline{1-5}\) \(\\[-9pt] &&&=&\displaystyle \frac{275+4}{2310}=\frac{279}{2310}&\hbox{Fraction can't be reduced further.}\\ \end{array}\)
      Quick check (with calculator approximations):
      \(\displaystyle \frac{5}{42}=0.11904762\)
      \(\displaystyle \frac{2}{1155}=0.00173160\)
      \(\displaystyle \frac{279}{2310}=0.12077922\)
      \(0.11904762+0.00173160=0.12077922\) (Nice!)

    14. Reduce \(\left(\displaystyle \frac{12}{35}\right)\left(\displaystyle \frac{21}{4}\right)\left(\displaystyle \frac{10}{9}\right)\left(\displaystyle \frac{11}{26}\right)\)
      Solution:
      \(\begin{array}{rcl lll} \!\!\!\!\!\!\!\left(\displaystyle \frac{12}{35}\right)\left(\displaystyle \frac{21}{4}\right)\left(\displaystyle \frac{10}{9}\right)\left(\displaystyle \frac{11}{26}\right)\!\!\!\!&=&\left(\displaystyle \frac{2\cdot 2\cdot 3}{5\cdot 7}\right)\left(\displaystyle \frac{3\cdot 7}{2\cdot 2}\right)\left(\displaystyle \frac{2\cdot 5}{3\cdot 3}\right)\left(\displaystyle \frac{11}{2\cdot 13}\right)\\[15pt] &=&\displaystyle \frac{\not 2\cdot \not 2\cdot \not 3\cdot \not 3\cdot \not 7\cdot \not 2\cdot \not 5\cdot11} {\not 5\cdot \not 7\cdot \not 2\cdot \not 2\cdot \not 3\cdot \not 3\cdot \not 2\cdot 13}\\[15pt] &=&\displaystyle \frac{11}{13} \end{array}\)

    15. A troupe of boy scouts come to camp with \(45\displaystyle \frac{2}{3}\) ft of rope. Each scout is to be cut \(\displaystyle 1\frac{3}{4}\) ft of rope to practice knots. The amount of rope is just sufficient for all the scouts (a little piece of rope may be left over). How many scouts are in the troupe?
      Solution:
      Subtract \(\displaystyle 1\frac{3}{4}\) ft of rope from \(45\displaystyle \frac{2}{3}\), then cut and subtract another \(\displaystyle 1\frac{3}{4}\) ft. Repeated subtraction means division.

      The rope can be cut into
      \(\begin{array}{rcl lll} 45\displaystyle \frac{2\hbox{ ft}}{3}\div \displaystyle 1\frac{3}{4}\frac{\hbox{ft}}{\hbox{scout}}&=&\displaystyle \frac{(45\cdot 3+2)\hbox{ ft}}{3}\div \displaystyle \frac{(4\cdot 1+3)}{4}\frac{\hbox{ft}}{\hbox{scout}}\\[15pt] &=&\displaystyle \frac{137\hbox{ ft}}{3}\div \displaystyle \frac{7\hbox{ ft}}{4\hbox{ scouts}}\\[15pt] &=&\displaystyle \frac{137\hbox{ ft}}{3}\cdot \displaystyle \frac{4\hbox{ scouts}}{7\hbox{ ft}}\\[15pt] &=&\displaystyle \frac{137\cdot 4\hbox{ scouts}}{3\cdot 7}\\[15pt] &=&\displaystyle \frac{548}{21}\hbox{ scouts}\\[15pt] &=&26\displaystyle \frac{2}{21}\hbox{ scouts} \end{array}\)

      There are \(26\) scouts in the troupe.

    16. Convert \(\displaystyle \frac{150}{37}\) to a mixed number.
      Solution:

      (0,0)(0,0) (0,0)\(\begin{array}{rcl lll} \displaystyle \frac{150}{37}&=&4\displaystyle \frac{2}{37}\\ \end{array}\) (10,0)\(37\ \ \ \ 1\ 5\ 0\) (12,1)(1,0)4 (12,1)(0,-1)1.3 (14,1.5)\(4\) (11.3,-1.5)\(-\ 1\ 4\ 8\) (12.0,-2.0)(1,0)4 (15,-3.0)\(2\)

      \(\)

    17. Solution:

      \(7\displaystyle \frac{9}{28}=\frac{7\cdot 28+9}{28}=\frac{196+9}{28}=\frac{205}{28}\).

    18. Add \(25\displaystyle \frac{6}{7}+12\frac{11}{14}\).
      Solution:
      \(\begin{array}{lll lll lll lll} &25+\displaystyle \frac{6}{7}&\Rightarrow &\displaystyle \frac{6}{7}&=&\displaystyle \frac{6}{7}\cdot \frac{2}{2}&=&\displaystyle \frac{12}{14}\\[15pt] +&\displaystyle 12+\frac{11}{14}&\Rightarrow &\displaystyle \frac{11}{14}&=&\displaystyle \frac{11}{14}&=&\displaystyle \frac{11}{14}\\[-6pt] &&&&\\ \hline \ &&&&&&\) \(\\[-5pt] &37+\Rightarrow &&&&&=&\displaystyle \frac{23}{14}=1+\frac{9}{14} \end{array}\)
      Answer: \(37+1+\displaystyle \frac{9}{14}=38\displaystyle \frac{9}{14}\)

    19. Solution:
      \(\begin{array}{lll lll lll lll} &38\displaystyle \frac{9}{14}&=& 38&+&\displaystyle \frac{9}{14}&=& 38&+&\displaystyle \frac{9}{14}\\[15pt] -&\displaystyle 12\displaystyle \frac{11}{14}&=&-\left(\displaystyle 12\begin{array}{c}\) \(\\\) \(\end{array}\right.&+&\left.\displaystyle \frac{11}{14}\right)&=&-12&+&\displaystyle \frac{-11}{14}\\[15pt] &&&&&&\) \(\\[-15pt] \hline &&&&&&\) \(\\[-11pt] &&&&&&=&26&+&\displaystyle \frac{-2}{14} \end{array}\)
      \(\begin{array}{lll lll lll lll lll} &26&+&\displaystyle \frac{-2}{14}\\[11pt] =&25&+&1&+&\displaystyle \frac{-2}{14}\\[11pt] =&25&+&\displaystyle \frac{14}{14}&+&\displaystyle \frac{-2}{14}\\[11pt] =&25&+&\displaystyle \frac{12}{14}\\[11pt] =&25&+&\displaystyle \frac{2\cdot 6}{2\cdot 7}\\[11pt] =&25\displaystyle \frac{6}{7} \end{array}\)
    20. Multiply \(\left(\displaystyle 3 \frac{6}{7}\right)\left(\displaystyle 2 \frac{1}{3}\right)\)
      Solution:
      \(\left(\displaystyle 3 \frac{6}{7}\right)\left(\displaystyle 2 \frac{1}{3}\right)=\left(\displaystyle \frac{27}{7}\right)\left(\displaystyle \frac{7}{3}\right)=\displaystyle \frac{27\cdot 7}{3\cdot 7}=\frac{27}{3}=9\)

    21. Convert \(81\displaystyle \frac{2}{3}\%\) to an exact fraction, no decimals.
      Solution:
      \(81\displaystyle \frac{2}{3}\%=\frac{243+2}{3}\%=\frac{245}{3}\cdot \frac{1}{100}=\frac{5\cdot 49}{3}\cdot \frac{1}{5\cdot 20}\)
      \(\displaystyle =\frac{49}{3\cdot 20}=\frac{49}{60}\)

    22. Solution:

      \(\displaystyle \frac{2}{3}=\frac{2}{3}\cdot 1= \frac{2}{3}\cdot 100\%=\frac{200}{3}\%=66\frac{2}{3}\%\)

    23. A retailer buys a dozen coats for $\(250\) each. The mark up rate (taken as a percent of cost) is \(30\)%. Find the sale price of a coat (cost plus mark up).
      A year later the retailer has one coat left and wants to give his customers a discount of \(30\)% of the sale price to sell the coat. The discount is computed using the sale price above. How much does the customer have to pay after the discount?
      Is the discounted price equal to the original cost of $\(250\) since the mark up rate was \(30\)% and the discount rate was also \(30\)%?
      If not, what is the retailer’s profit or loss?
      Note: Mark up is taken using cost (\(\$250\)), discount is taken using sales price before applying the discount, also referred to as regular price.

      Solution:

      Mark up is \(30\)% of cost.

      Mark up \(=(0.3)(250)=\$75\)

      Sale price of a coat = cost plus mark up.

      Sale price \(=250 + 75 = \$325\).

      Discount is \(30\)% of sale (regular) price.

      Discount \(=(0.3)(325)=\$97.50\)

      The customer has to pay \(325-97.50=\$227.50\).

      The retailer loses \(250-227.50=\$22.50\) on this last coat.

    24. Solution:

      \(A=(0.05)(25)\)

      \(A=1.25\)

    25. Solution:

      \(25=(0.05)B\)

      \(B=\displaystyle \frac{25}{0.05}\)

      \(B=\displaystyle \frac{2500}{5}=500\)

    26. \(5\) is what percent of \(25\)?
      Solution:
      \(5=x(25)\)
      \(x=\displaystyle \frac{5}{25}=\frac{20}{100}=20\%\)