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Valentin [98]
3 years ago
9

A school fair ticket costs $8 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who

went to the fair was 30, and the total money collected was $100. Which of the following options represents the number of children and the number of adults who attended the fair that day, and the pair of equations that can be solved to find the numbers? (4 points)
Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
7 0
A+c=30
(8)a+(1)c=100

a=10 adults
c=20 children
Nadusha1986 [10]3 years ago
5 0
$88 adults and $12 children
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<h3>Answer:  5 cakes</h3>

================================================

Explanation:

Let's start off converting the mixed number 12 & 1/4 to an improper fraction.

a \frac{b}{c} = \frac{a*c+b}{c}\\\\12 \frac{1}{4} = \frac{12*4+1}{4}\\\\12 \frac{1}{4} = \frac{49}{4}\\\\

Do the same for the other mixed number 2 & 1/3.

a \frac{b}{c} = \frac{a*c+b}{c}\\\\2 \frac{1}{3} = \frac{2*3+1}{3}\\\\2 \frac{1}{3} = \frac{7}{3}\\\\

-----------------------

From here, we divide the two fractions. I converted them to improper fractions to make the division process easier.

\frac{49}{4} \div \frac{7}{3} = \frac{49}{4} \times \frac{3}{7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{49\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 7\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 3}{4}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{21}{4}\\\\

The last step is to convert that result to a mixed number.

\frac{21}{4} = \frac{4*5+1}{4}\\\\\frac{21}{4} = \frac{4*5}{4}+\frac{1}{4}\\\\\frac{21}{4} = 4+\frac{1}{4}\\\\\frac{21}{4} = 5 \frac{1}{4}\\\\

Note that 21/4 = 5.25 and 1/4 = 0.25 to help check the answer.

-----------------------

Therefore, she can make 5 cakes. The fractional portion 1/4 is ignored since we're only considering whole cakes rather than partial ones.

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