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Question: Two-thirds of a number is five-sixths. How do you find the number?...

Two-thirds of a number is five-sixths. How do you find the number?

Explanation

Solution

First consider the unknown number to be some variable and then express the given algebraic phrase in algebraic expression. After writing the phrase in the equation check all the constants should be at right side and variables at left side of the equation if not then send them with help of algebraic operations. And finally divide both sides with the coefficient of the variable to get the required number.

Complete step by step solution:
In order to find the number in the phrase “Two-thirds of a number is five-sixths”, we will first consider that number to be a variable.
Let us consider the required number be xx
Two-thirds can also be written as two parts of a whole divided in three parts and expressed algebraically as 23\dfrac{2}{3}
Similarly, five-sixth can also be written as five parts of a whole divided in six parts and expressed algebraically as 56\dfrac{5}{6}
Now we can express the given phrase “Two-thirds of a number is five-sixths” in algebraic expression as follows
23×x=56\Rightarrow \dfrac{2}{3} \times x = \dfrac{5}{6}
Dividing both sides with coefficient of xx, we will get
23÷23×x=56÷23 x=56÷23  \Rightarrow \dfrac{2}{3} \div \dfrac{2}{3} \times x = \dfrac{5}{6} \div \dfrac{2}{3} \\\ \Rightarrow x = \dfrac{5}{6} \div \dfrac{2}{3} \\\
It can also be written as
x=56×32 x=54  \Rightarrow x = \dfrac{5}{6} \times \dfrac{3}{2} \\\ \Rightarrow x = \dfrac{5}{4} \\\
Therefore the required number is equals to 54\dfrac{5}{4}

Note: Division sign between the two factors is converted into multiplication sign along with the multiplicative inverse of the second term, because as we know that division and multiplication are the inverse operation for each other. Also when solving this type of word problems where an unknown number is present, then your first step should be to consider that number to be a variable and the second step is to express the problem algebraically.