Question

The sum of three fractions is $\dfrac{59}{24}$. If the greatest fraction is divided by the smallest fraction, the result is $\dfrac{7}{6}$, which is greater than the middle fraction by $\dfrac{1}{3}$ .Find the fractions. \[\]
A.$\dfrac{3}{7},\dfrac{4}{5},\dfrac{2}{3}$\[\]
B. $\dfrac{7}{8},\dfrac{5}{6},\dfrac{3}{4}$\[\]
C. $\dfrac{7}{9},\dfrac{2}{5},\dfrac{2}{3}$\[\]
D. $\dfrac{3}{7},\dfrac{4}{5},\dfrac{2}{3}$\[\]

Answer 1

Hint: We assume the three fractions in descending order as $x,y,z$. We use the data in the question and design the equations$x+y+z=\dfrac{59}{24},\dfrac{x}{z}=\dfrac{7}{6},\dfrac{7}{6}=y+\dfrac{1}{3}$. We express $x$ in terms of $z$ from the second equation and find $y$ from the third equation. We put $x,y$ in first equation to get $z$ and then $x,y$.\[\]

Complete step by step answer:
We are given the question that The sum of three fractions is $\dfrac{59}{24}$. Let us assume that the three fractions as $x,y,z$ in decreasing or descending order. So we have
\[x+y+z=\dfrac{59}{24}........\left( 1 \right)\]
We are further given the question that the greatest fraction is divided by the smallest fraction, the result is $\dfrac{7}{6}$. The greatest fraction according to our assumption is $x$ and the smallest is $z$. So we have,
\[\dfrac{x}{z}=\dfrac{7}{6}\]
We multiply both sides of the above equation by $z$to express $x$ in terms of $z$ as,
\[x=\dfrac{7}{6}z.....\left( 2 \right)\]
We are also given in the question the sum of the greatest and smallest fraction that is $\dfrac{7}{6}$is greater than the middle fraction by $\dfrac{1}{3}$. The middle fraction according to our assumption is $y$. So we have,
\[\dfrac{7}{6}=y+\dfrac{1}{3}\]
We subtract both sides by $\dfrac{1}{3}$ to have,
 \[\begin{align}
  & \dfrac{7}{6}-\dfrac{1}{3}=y \\
 & \Rightarrow y=\dfrac{7}{6}-\dfrac{1}{3} \\
 & \Rightarrow y=\dfrac{7-2\times 1}{6}=\dfrac{5}{6}.....\left( 3 \right) \\
\end{align}\]
We put the values of $x$ and $y$ obtained from equation (2) and (3) in equation (1) to have,
\[\begin{align}
  & x+y+z=\dfrac{59}{24} \\
 & \Rightarrow \dfrac{7z}{6}+\dfrac{5}{6}+z=\dfrac{59}{24} \\
\end{align}\]
Let us subtract $\dfrac{5}{6}$ from both sides of the equation and then take $z$ common in the left hand side to have,
\[\begin{align}
  & \Rightarrow \dfrac{7z}{6}+z=\dfrac{59}{24}-\dfrac{5}{6} \\
 & \Rightarrow z\left( \dfrac{7}{6}+1 \right)=\dfrac{59-5\times 4}{24}=\dfrac{39}{24} \\
 & \Rightarrow z\left( \dfrac{7+6}{6} \right)=\dfrac{39}{24} \\
 & \Rightarrow z\left( \dfrac{13}{6} \right)=\dfrac{39}{24} \\
\end{align}\]
We multiply both side of the equation to obtain the value of $z$ as
\[\Rightarrow z=\dfrac{39}{24}\times \dfrac{6}{13}=\dfrac{3}{4}\]
We put the value of $z$ in the equations (2) to get $x$ as
\[x=\dfrac{7z}{6}=\dfrac{7}{6}\times \dfrac{3}{4}=\dfrac{7}{8}\]
We obtain the values of $x,y,z$ are $\dfrac{7}{8},\dfrac{5}{6},\dfrac{3}{4}$ respectively.

So, the correct answer is “Option C”.

Note: We can verify the results by converting the fractions $\dfrac{7}{8},\dfrac{5}{6},\dfrac{3}{4}$ into equivalent fractions where we convert equate the denominators by finding the least common multiple (LCM) of denominators and then compare the numerators. We can alternatively solve by expressing $z$ in terms of $x$ and then solving for $x$.