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How do you simplify $6.4 \div ( - 0.08)$?

seo-qna
Last updated date: 27th Jul 2024
Total views: 384k
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Answer
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Hint: Here we are given an expression in the division form and the numbers contain decimal points in it. First of all remove the decimal point placing appropriate number of zeros below it and then frame the expression in the simplified form and at last perform division or make it more simpler removing common factors from the numerator and the denominator.

Complete step-by-step solution:
Take the given expression: $6.4 \div ( - 0.08)$
It can be expressed as $6.4 \div ( - 0.08) = - \dfrac{{6.4}}{{0.08}}$
Remove decimal point in the above expression, place ten for the one digit after decimal point and hundred for two digits after decimal point.
$6.4 \div ( - 0.08) = - \dfrac{{\dfrac{{64}}{{10}}}}{{\dfrac{{008}}{{100}}}}$
Numerator’s denominator goes to the denominator and the denominator's denominator goes to the numerator.
$6.4 \div ( - 0.08) = - \dfrac{{64 \times 100}}{{10 \times 008}}$
Find the factors in the above expression.
$6.4 \div ( - 0.08) = - \dfrac{{8 \times 8 \times 10 \times 10}}{{10 \times 8}}$
Common factors from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
$6.4 \div ( - 0.08) = - 8 \times 10$
Simplify the above equation by finding the product on the right hand side of the equation. Also, the product of one positive term and the one negative term gives the resultant value in negative.
$6.4 \div ( - 0.08) = - 80$
This is the required solution.

Note: Be good in multiples and always remember that common factor from the numerator and the denominator cancel each other. Also be careful while removing the decimal point. Always count the number of digits after the decimal point and then put zeros under it or ten for one digit, hundred for two digits and so on. Always remember that the product of one positive term and the one negative term gives the resultant value in negative whereas the product of two negative terms gives us the positive value.