Question

Plz, help me with this question too!
If you multiply two positive mixed fractions, the result could be
A) greater than both fractions
B) less than both fractions
C) equal to one
D) less than one

Answer 1

(A) Greater than both fractions

We know about mixed fractions. Positive mixed fractions are always greater than 1 and are written in the form of $a\dfrac{b}{c}=a+\dfrac {b}{c}$ for some positive integers $a,\ b,\ c$ where $b\ \textless\ c.$ Or we can say that a mixed fraction is a combination of a whole number and a proper fraction.

E.g.: $2\dfrac {3}{5}=2+\dfrac{3}{5}$

We can convert a mixed fraction into improper fraction as follows:

$a\dfrac {b}{c}\ \iff\ \dfrac {ac+b}{c}$

E.g.: $2\dfrac {3}{5}=\dfrac {2\times 5+3}{5}=\dfrac {13}{5}$

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Suppose I have two mixed fractions,

$a\dfrac {b}{c}\ \&\ p\dfrac {q}{r}$

for some positive integers $a,\ b,\ c,\ p,\ q,\ r$ where $b\ \textless\ c\ \&\ q\ \textless\ r.$

We can multiply these two fractions in two ways.

1. Converting the mixed fractions into improper ones and then multiply directly.
2. Direct multiplication of the mixed fractions by using the formula,

$\large(a+b)(c+d)=ac+ad+bc+bd$

I prefer second method for getting the answer for the question simply. So,

$\left (a+\dfrac {b}{c}\right)\left (p+\dfrac {q}{r}\right)\\\\\\=ap+\dfrac {aq}{r}+\dfrac {bp}{c}+\dfrac {bq}{cr}$

Now we compare each of the two mixed fractions $a\dfrac {b}{c}\ \&\ p\dfrac {q}{r}$ with the product.

Consider, $a\dfrac {b}{c}=a+\dfrac {b}{c}$

Since $p>0,$ the value of the mixed fraction increases when it is multiplied by $p.$ Thus we can say that,

$a+\dfrac {b}{c}\ \textless\ \left (a+\dfrac {b}{c}\right) p\\\\\\a+\dfrac {b}{c}\ \textless\ ap+\dfrac {bp}{c}$

When some positive quantities are added to the RHS of this inequality, the value also increases, doesn't it?

$a+\dfrac {b}{c}\ \textless\ ap+\dfrac {aq}{r}+\dfrac {bp}{c}+\dfrac {bq}{cr}$

Thus the product is greater than $a\dfrac {b}{c}.$

Similarly, consider $p\dfrac {q}{r}=p+\dfrac {q}{r}.$

Then we have,

$p+\dfrac {q}{r}\ \textless\ \left (p+\dfrac {q}{r}\right)a\quad [\text {Since a>0 }]\\\\\\p+\dfrac {q}{r}\ \textless\ ap+\dfrac {aq}{r}\\\\\\p+\dfrac {q}{r}\ \textless\ ap+\dfrac {aq}{r}+\dfrac {bp}{c}+\dfrac {bq}{cr}$

Thus the product is also greater than $p\dfrac {q}{r}.$

Hence the correct answer is,

(A) Greater than both fractions

#answerwithquality

#BAL