Question

There are two types of mixed nuts – Type A and Type B. \[\dfrac{1}{3}\] of Type A nuts are peanuts, and \[\dfrac{2}{5}\] of Type B nuts are peanuts. A mixture of \[2\dfrac{1}{2}\] kg of Type A nuts and \[\dfrac{1}{2}\] kg of Type B nuts are mixed together. What fraction of this mixture are peanuts?

Answer 1

Hint:
Here, we need to find the fraction of peanuts in the final mixture. We will use multiplication of fractions to find the fraction of peanuts in the two types of peanuts added in the mixture. Then, we will use addition of fractions to find the fraction of peanuts in the mixture.

Complete step by step solution:
We will use multiplication of fractions, and then addition of fractions to find the fraction of peanuts in the mixture.
First, we will convert the mixed fraction \[2\dfrac{1}{2}\] to an improper fraction.
Therefore, we get
\[\begin{array}{l} \Rightarrow 2\dfrac{1}{2} = \dfrac{{2 \times 2 + 1}}{2}\\ \Rightarrow 2\dfrac{1}{2} = \dfrac{5}{2}\end{array}\]
Now, it is given that \[\dfrac{1}{3}\] of the Type A nuts are peanuts.
This means that \[\dfrac{1}{3}\] of the \[\dfrac{5}{2}\] kg Type A nuts are peanuts.
We will multiply the fractions to get the fraction of peanuts in \[\dfrac{5}{2}\] kg of Type A nuts.
Thus, we get
Fraction of peanuts in \[\dfrac{5}{2}\] kg of Type A nuts \[ = \dfrac{1}{3} \times \dfrac{5}{2} = \dfrac{5}{6}\] kg
Therefore, the fraction of peanuts in \[\dfrac{5}{2}\] kg of Type A nuts is \[\dfrac{5}{6}\] kg.
Next, it is given that \[\dfrac{2}{5}\] of the Type B nuts are peanuts.
This means that \[\dfrac{2}{5}\] of the \[\dfrac{1}{2}\] kg Type B nuts are peanuts.
We will multiply the fractions to get the fraction of peanuts in \[\dfrac{1}{2}\] kg of Type B nuts.
Thus, we get
Fraction of peanuts in \[\dfrac{1}{2}\] kg of Type B nuts \[ = \dfrac{2}{5} \times \dfrac{1}{2} = \dfrac{2}{{10}} = \dfrac{1}{5}\] kg
Therefore, the fraction of peanuts in \[\dfrac{1}{2}\] kg of Type B nuts is \[\dfrac{1}{5}\] kg.
Finally, we will find the fraction of peanuts in the final mixture by adding the fraction of peanuts in \[\dfrac{5}{2}\] kg of Type A nuts, and the fraction of peanuts in \[\dfrac{1}{2}\] kg of Type B nuts is \[\dfrac{1}{5}\] kg.
Thus, we get
Fraction of peanuts in the mixture \[ = \left( {\dfrac{5}{6} + \dfrac{1}{5}} \right)\] kg
The L.C.M. of the denominators 6 and 5 is 30.
Rewriting the fractions with denominator 30, we get
Fraction of peanuts in the mixture \[ = \left( {\dfrac{{25}}{{30}} + \dfrac{6}{{30}}} \right)\] kg
Adding the fractions, we get
Fraction of peanuts in the mixture \[ = \dfrac{{25 + 6}}{{30}} = \dfrac{{31}}{{30}}\] kg
Converting this to improper fraction, we get
Fraction of peanuts in the mixture \[ = \dfrac{{30 \times 1 + 1}}{{30}} = 1\dfrac{1}{{30}}\] kg

Therefore, the fraction of peanuts in the mixture is \[\dfrac{{31}}{{30}}\] kg, or \[1\dfrac{1}{{30}}\] kg.

Note:
We added and multiplied fractions in the solution. A fraction is a number which represents a part of a group. It is written as \[\dfrac{a}{b}\], where \[a\] is called the numerator and \[b\] is called the denominator. The group is divided into \[b\] equal parts. The fraction \[\dfrac{a}{b}\] represents \[a\] parts out of \[b\] equal parts of the group. For example: 105 out of 200 children can be represented as the fraction \[\dfrac{{105}}{{200}}\].
We converted a mixed fraction to improper fraction. An improper fraction is a fraction whose numerator is larger than its denominator. A mixed fraction is a fraction in the form \[a\dfrac{b}{c}\]. Every mixed fraction \[a\dfrac{b}{c}\] can be converted to an improper fraction \[\dfrac{{c \times a + b}}{c}\].