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A father shed a box of chocolates. He gave Simon, 13\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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Full solution

Q. A father shed a box of chocolates. He gave Simon, 13\frac{1}{3} Rachel one quarter, Kerry one fifth and Darcy one sixth. he gave Judy six. How many chocolates were in the box?
  1. Define Total Chocolates as xx: Let's denote the total number of chocolates in the box as xx. According to the problem, the father gave away the following fractions of xx to his children: Simon got 13\frac{1}{3} of xx, Rachel got 14\frac{1}{4} of xx, Kerry got 15\frac{1}{5} of xx, and Darcy got 16\frac{1}{6} of xx. Judy got xx11 chocolates. We can write an equation that represents this situation:\newlinexx22
  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 3,4,5,3, 4, 5, and 66 is 6060. We will convert each fraction to have a denominator of 6060: (13×x)×(2020)+(14×x)×(1515)+(15×x)×(1212)+(16×x)×(1010)+6=x(\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
  3. Combine and Add Fractions: Now, multiply each fraction by the equivalent of 11 to get the common denominator:\newline(2060)×x+(1560)×x+(1260)×x+(1060)×x+6=x(\frac{20}{60}) \times x + (\frac{15}{60}) \times x + (\frac{12}{60}) \times x + (\frac{10}{60}) \times x + 6 = x
  4. Isolate x: Combine the fractions: (2060+1560+1260+1060)⋅x+6=x(\frac{20}{60} + \frac{15}{60} + \frac{12}{60} + \frac{10}{60}) \cdot x + 6 = x
  5. Factor Out x: Add the fractions: \newlineegin{equation}\newline\left(\frac{5757}{6060}\right) * x + 66 = x\newline\end{equation}
  6. Simplify Expression: To isolate xx, we need to move the fraction to the other side of the equation:\newlinex−(5760)⋅x=6x - \left(\frac{57}{60}\right) \cdot x = 6
  7. Multiply by 2020: Factor out xx on the left side of the equation:\newlinex×(1−5760)=6x \times (1 - \frac{57}{60}) = 6
  8. Calculate xx: Simplify the expression inside the parentheses:\newlinex×(6060−5760)=6x \times (\frac{60}{60} - \frac{57}{60}) = 6\newlinex×(360)=6x \times (\frac{3}{60}) = 6
  9. Calculate xx: Simplify the expression inside the parentheses:\newlinex×(6060−5760)=6x \times (\frac{60}{60} - \frac{57}{60}) = 6\newlinex×(360)=6x \times (\frac{3}{60}) = 6 Simplify the fraction (360)(\frac{3}{60}) to (120):(\frac{1}{20}):\newlinex×(120)=6x \times (\frac{1}{20}) = 6
  10. Calculate xx: Simplify the expression inside the parentheses:\newlinex×(6060−5760)=6x \times (\frac{60}{60} - \frac{57}{60}) = 6\newlinex×(360)=6x \times (\frac{3}{60}) = 6 Simplify the fraction (360)(\frac{3}{60}) to (120)(\frac{1}{20}):\newlinex×(120)=6x \times (\frac{1}{20}) = 6 Multiply both sides of the equation by 2020 to solve for xx:\newlinex=6×20x = 6 \times 20
  11. Calculate xx: Simplify the expression inside the parentheses:\newlinex×(6060−5760)=6x \times (\frac{60}{60} - \frac{57}{60}) = 6\newlinex×(360)=6x \times (\frac{3}{60}) = 6 Simplify the fraction (360)(\frac{3}{60}) to (120):(\frac{1}{20}):\newlinex×(120)=6x \times (\frac{1}{20}) = 6 Multiply both sides of the equation by 2020 to solve for xx:\newlinex=6×20x = 6 \times 20 Calculate the value of xx:\newlinex×(6060−5760)=6x \times (\frac{60}{60} - \frac{57}{60}) = 600

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