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Question

Question: How do you simplify the fraction \(\dfrac{{85}}{{110}}?\)...

How do you simplify the fraction 85110?\dfrac{{85}}{{110}}?

Explanation

Solution

A fraction is in its simplest form when the numerator and denominator do not have a common factor from which both numerator and denominator can be divisible.
The common factor is always a whole number, because having a common factor in decimal or fraction form will complicate the number more rather than simplifying it.

Complete step by step solution:
To simplify fractions we have to find Highest Common Factor (HCF) of
the numerator and denominator in order to divide them with that factor (HCF).

So let us find the highest common factor a between the numerator (85)(85) and the denominator (110)(110)
85=5×17 110=2×5×11  85 = 5 \times 17 \\\ 110 = 2 \times 5 \times 11 \\\
Therefore highest common factor of 8585 and 110110 comes out to be 55

Now dividing both the numerator and the denominator by 55,

85÷5110÷5=1722  \dfrac{{85 \div 5}}{{110 \div 5}} = \dfrac{{17}}{{22}} \\\ \\\

1722\therefore \dfrac{{17}}{{22}} is the simplest form of the fraction 85110\dfrac{{85}}{{110}}

Note: If numerator and denominator of a fraction are different prime numbers then it is already in its simplest form, because prime numbers have no common factors except one.

One is the common factor to all the numbers. Alternative method: You can convert a fraction into its simplest form by one more method that is by continue division of the numerator and the denominator with prime numbers (2,  3,  5,  7,  13.....)(2,\;3,\;5,\;7,\;13.....) until they does not have any common factor left except 11 Let us understand this by above example 85110\dfrac{{85}}{{110}}

Dividing 8585 and 110110 with prime numbers (2,  3,  5,  7,  13.....)(2,\;3,\;5,\;7,\;13.....) until they does not have any
common factor left.
85÷5110÷5=1722\dfrac{{85 \div 5}}{{110 \div 5}} = \dfrac{{17}}{{22}}

Now they do not have any common factor left except one.So the answer is 1722\dfrac{{17}}{{22}} This is a lengthy process, better go with the HCF process.