A father shed a box of chocolates. He gave Simon, $\frac{1}{3}$ Rachel one quarter, Kerry one fifth and Darcy one sixth. he gave Judy six. How many chocolates were in the box?

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Q. A father shed a box of chocolates. He gave Simon,

\frac{1}{3}

Rachel one quarter, Kerry one fifth and Darcy one sixth. he gave Judy six. How many chocolates were in the box?

Define Total Chocolates as $x$ : Let's denote the total number of chocolates in the box as $x$ . According to the problem, the father gave away the following fractions of $x$ to his children: Simon got $\frac{1}{3}$ of $x$ , Rachel got $\frac{1}{4}$ of $x$ , Kerry got $\frac{1}{5}$ of $x$ , and Darcy got $\frac{1}{6}$ of $x$ . Judy got $x$ $1$ chocolates. We can write an equation that represents this situation: $\newline$ $x$ $2$
Convert Fractions to Common Denominator: To solve the equation, we need to find a common denominator for the fractions. The least common multiple of $3, 4, 5,$ and $6$ is $60$ . We will convert each fraction to have a denominator of $60$ : $(\frac{1}{3} \times x) \times (\frac{20}{20}) + (\frac{1}{4} \times x) \times (\frac{15}{15}) + (\frac{1}{5} \times x) \times (\frac{12}{12}) + (\frac{1}{6} \times x) \times (\frac{10}{10}) + 6 = x$
Combine and Add Fractions: Now, multiply each fraction by the equivalent of $1$ to get the common denominator: $\newline$ $(\frac{20}{60}) \times x + (\frac{15}{60}) \times x + (\frac{12}{60}) \times x + (\frac{10}{60}) \times x + 6 = x$
Isolate x: Combine the fractions: $(\frac{20}{60} + \frac{15}{60} + \frac{12}{60} + \frac{10}{60}) \cdot x + 6 = x$
Factor Out x: Add the fractions: $\newline$ egin{equation} $\newline$ \left(\frac{ $57$ }{ $60$ }\right) * x + $6$ = x $\newline$ \end{equation}
Simplify Expression: To isolate $x$ , we need to move the fraction to the other side of the equation: $\newline$ $x - \left(\frac{57}{60}\right) \cdot x = 6$
Multiply by $20$ : Factor out $x$ on the left side of the equation: $\newline$ $x \times (1 - \frac{57}{60}) = 6$
Calculate $x$ : Simplify the expression inside the parentheses: $\newline$ $x \times (\frac{60}{60} - \frac{57}{60}) = 6$ $\newline$ $x \times (\frac{3}{60}) = 6$
Calculate $x$ : Simplify the expression inside the parentheses: $\newline$ $x \times (\frac{60}{60} - \frac{57}{60}) = 6$ $\newline$ $x \times (\frac{3}{60}) = 6$ Simplify the fraction $(\frac{3}{60})$ to $(\frac{1}{20}):$ $\newline$ $x \times (\frac{1}{20}) = 6$
Calculate $x$ : Simplify the expression inside the parentheses: $\newline$ $x \times (\frac{60}{60} - \frac{57}{60}) = 6$ $\newline$ $x \times (\frac{3}{60}) = 6$ Simplify the fraction $(\frac{3}{60})$ to $(\frac{1}{20})$ : $\newline$ $x \times (\frac{1}{20}) = 6$ Multiply both sides of the equation by $20$ to solve for $x$ : $\newline$ $x = 6 \times 20$
Calculate $x$ : Simplify the expression inside the parentheses: $\newline$ $x \times (\frac{60}{60} - \frac{57}{60}) = 6$ $\newline$ $x \times (\frac{3}{60}) = 6$ Simplify the fraction $(\frac{3}{60})$ to $(\frac{1}{20}):$ $\newline$ $x \times (\frac{1}{20}) = 6$ Multiply both sides of the equation by $20$ to solve for $x$ : $\newline$ $x = 6 \times 20$ Calculate the value of $x$ : $\newline$ $x \times (\frac{60}{60} - \frac{57}{60}) = 6$ $0$