Question

A student was asked to solve the fraction \[\dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}}\] and his answer was \[\dfrac{1}{4}\] . By how much was his answer wrong?

Answer 1

Hint: In this question, first of all we find out the correct value of the fraction by using the BODMAS rule. Then the difference between the correct answer and the student's answer is our final answer.
BODMAS Rule: is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. In certain regions, PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction) is used, which is the synonym of BODMAS.
It explains the order of operations to be performed while solving an expression. According to BODMAS rule, if an expression contains brackets ((), {}, []) we have first to solve or simplify the bracket followed by ‘of’ (that means powers and roots, etc.), then division, multiplication, addition and subtraction from left to right. Solving the problem in the wrong order will result in a wrong answer.

Complete step-by-step answer:
It is given that the fraction is \[\dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}}\].
First, we will solve the given fraction i.e., \[\dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}}\]
First we are going to convert the mixed fractions into improper fractions
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{7}{3} + \left( {\dfrac{3}{2} \times \dfrac{5}{3}} \right)}}{{2 + \dfrac{5}{3}}}\]
Now we are going to cancel 3 on both numerator and denominator the terms inside the open bracket.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{7}{3} + \dfrac{5}{2}}}{{2 + \dfrac{5}{3}}}\]
Now we are going to add the terms in both numerator and denominator separately. For this we are going to take LCM.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{{7 \times 2 + 5 \times 3}}{{3 \times 2}}}}{{\dfrac{{2 \times 3 + 5}}{3}}}\]
Now we are going to use the BODMAS rule to solve the above numerator fraction. For this first we multiply then we add.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{{(7 \times 2) + (5 \times 3)}}{{3 \times 2}}}}{{\dfrac{{(2 \times 3) + 5}}{3}}}\]
Now we multiply the terms within the bracket.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{{14 + 15}}{6}}}{{\dfrac{{6 + 5}}{3}}}\]
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{\dfrac{{29}}{6}}}{{\dfrac{{11}}{3}}}\]
Now we have fractions on both numerator and denominator so we make it to a proper fraction.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{29}}{6} \times \dfrac{3}{{11}}\]
Now we have simplified the fraction by using the BODMAS rule. No we can solve the fraction.
Now we are going to cancel the common terms to get the final answer. We can cancel 6 by 3. Then we get 2.
\[ \Rightarrow \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{29}}{{2 \times 11}}\]
\[\therefore \dfrac{{\dfrac{7}{3} + \left( {1\dfrac{1}{2} \times \dfrac{5}{3}} \right)}}{{2 + 1\dfrac{2}{3}}} = \dfrac{{29}}{{22}}\]
But the student got his answer as \[\dfrac{1}{4}\].
Now we have to find by how much he was wrong by which is given by the difference of correct answer to wrong answer.
\[ \Rightarrow \dfrac{{29}}{{22}} - \dfrac{1}{4} = \dfrac{{29 \times 2 - 1 \times 11}}{{44}}\]
Now we apply the BODMAS rule to solve the above fraction.
\[ \Rightarrow \dfrac{{29}}{{22}} - \dfrac{1}{4}\]\[ = \dfrac{{58 - 11}}{{44}} = \dfrac{{47}}{{44}}\]

Therefore, the student was wrong by a fraction of \[\dfrac{{47}}{{44}}\].

Note: Mixed fraction is nothing but a combination of a whole number and a fraction.
For example, $3\dfrac{1}{2}$ is a mixed fraction where 3 is a whole number and $\dfrac{1}{2}$ is a fraction. So we can convert any mixed fraction into an improper fraction by multiplying the denominator part of fraction and the whole number then we add it to the numerator part of the fraction. Hence $3\dfrac{1}{2} = \dfrac{7}{2}$