Question

evaluate:(17/5+21/10) divided by (23/5-33/10)​

Answer 1

Answer:

13/55

Step-by-step explanation:

$( \frac{23}{5} - \frac{33}{10} ) \div ( \frac{17}{5} + \frac{21}{10} ) \\ = ( \frac{46 - 33}{10} ) \div ( \frac{34 + 21}{10} ) \\ = \frac{13}{55}$

Answer 2

★ How to do :-

Here, we are given with two set of fractions in which we are asked to divide them. So, first we should solve the two brackets which are given to us separately. The other concept used here is the LCM method. It's used to convert the unlike fractions to like fractions. We take the LCM of all denominators and then make them to the same as the LCM value. This is done because we cannot add the fractions having different denominators directly. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \bigg( \dfrac{23}{5} - \dfrac{33}{10} \bigg) \div \bigg( \dfrac{17}{5} + \dfrac{21}{10} \bigg)}$

First, let's solve the first bracket.

${\tt \leadsto \dfrac{23}{5} - \dfrac{33}{10}}$

LCM of 5 and 10 is 10.

${\tt \leadsto \dfrac{23 \times 2}{5 \times 2} - \dfrac{33}{10}}$

Multiply the numerator and denominator of first fraction.

${\tt \leadsto \dfrac{46}{10} - \dfrac{33}{10}}$

Subtract the obtained numerators.

${\tt \leadsto \dfrac{46 - 33}{10} = \dfrac{13}{10}}$

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Now, let's solve the second bracket.

${\tt \leadsto \dfrac{17}{5} + \dfrac{21}{10}}$

LCM of 5 and 10 is 10.

${\tt \leadsto \dfrac{17 \times 2}{5 \times 2} + \dfrac{21}{10}}$

Multiply the numerator and denominator of first fraction.

${\tt \leadsto \dfrac{34}{10} + \dfrac{21}{10}}$

Subtract the obtained numerators.

${\tt \leadsto \dfrac{34 + 21}{10} = \dfrac{55}{10}}$

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Now, let's divide both the obtained answers.

${\tt \leadsto \dfrac{13}{10} \div \dfrac{55}{10}}$

Take the reciprocal of second fraction and multiply both the fractions.

${\tt \leadsto \dfrac{13}{10} \times \dfrac{10}{55}}$

Now, multiply the numerator with numerator and the denominator with denominator.

${\tt \leadsto \dfrac{13 \times 10}{10 \times 55} = \dfrac{130}{550}}$

Write the obtained fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{130}{550} = \pink{\underline{\boxed{\tt \dfrac{13}{55}}}}}$

$\Huge\therefore$ The final answer is ${\tt \dfrac{13}{55}}$