How do you solve \[\dfrac{1}{2}.\dfrac{x}{5}=\dfrac{3}{10}?\]
Answer
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Hint: We are given two fractions as \[\dfrac{1}{2}\] and \[\dfrac{x}{5}.\] When they are multiplied to the product, it is \[\dfrac{3}{10}.\] We are asked to find the value of x. To do so we will first learn how to find the product of two fractions and then we will see some examples of such fractions, multiplication and then use this knowledge we will follow to solve our problem. We can also go by hit and trial method.
Complete step-by-step solution:
We are given two fractions \[\dfrac{1}{2}\] and \[\dfrac{x}{5}.\] A fraction is a number of the form \[\dfrac{a}{b}\] where \[b\ne 0.\] To multiply \[\dfrac{1}{2}\] and \[\dfrac{x}{5}\] we will start by learning how to find the product of two fractions. Consider if we have two fractions say \[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] then when we multiply \[\dfrac{a}{b}\times \dfrac{c}{d}\] and then we must check if there are terms common to numerator and denominator that can be cancelled. If there are terms then we can cancel it. After cancelling, we will multiply numerator to numerator and denominator to the denominator. For example, we have \[\dfrac{2}{9}\] and \[\dfrac{6}{5}.\] On multiplying, we get, \[\dfrac{2}{9}\times \dfrac{6}{5}\] Here we can see 3 is common in numerator and denominator. So, we can cancel it. So, we get,
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{2\times 2\times 3}{3\times 3\times 5}\]
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{2\times 2}{3\times 5}\]
Now we will multiply numerator with numerator and denominator with denominator. So, we get,
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{4}{15}\]
Now in our problem we have \[\dfrac{1}{2}\] and \[\dfrac{x}{5}\] whose product is given as \[\dfrac{3}{10}.\] Since x is known, so we cannot find the common in the numerator and denominator. So, multiplying directly numerator to numerator and denominator to the denominator, we get,
\[\dfrac{1}{2}\times \dfrac{x}{10}=\dfrac{1\times x}{2\times 5}=\dfrac{x}{10}\]
So, we get the product as \[\dfrac{x}{10}\] but we are given that their product is \[\dfrac{3}{10}.\] Hence, \[\dfrac{x}{10}\] must be equal to \[\dfrac{3}{10}.\] So,
\[\dfrac{x}{10}=\dfrac{3}{10}\]
On cross multiplying, we get,
\[\Rightarrow x=\dfrac{3\times 10}{10}\]
On simplifying, we get,
\[\Rightarrow x=3\]
So, the value of x is 3.
Note: We can also go by hit and trial. As we learn how to multiply, we will use various values of x and check that it satisfies the equation or not. Say if let x = 1, we get,
\[\dfrac{1}{2}\times \dfrac{x}{5}=\dfrac{1}{2}\times \dfrac{1}{5}\]
On simplifying, we get,
\[\Rightarrow \dfrac{1}{10}\ne \dfrac{3}{10}\]
So, x = 1 is not our solution.
If we put x = 3, we get,
\[\dfrac{1}{2}\times \dfrac{x}{5}=\dfrac{1}{2}\times \dfrac{1}{5}\]
On simplifying, we get,
\[\Rightarrow \dfrac{3}{10}=\dfrac{3}{10}\]
So, x = 3 is the solution to our problem.
Complete step-by-step solution:
We are given two fractions \[\dfrac{1}{2}\] and \[\dfrac{x}{5}.\] A fraction is a number of the form \[\dfrac{a}{b}\] where \[b\ne 0.\] To multiply \[\dfrac{1}{2}\] and \[\dfrac{x}{5}\] we will start by learning how to find the product of two fractions. Consider if we have two fractions say \[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] then when we multiply \[\dfrac{a}{b}\times \dfrac{c}{d}\] and then we must check if there are terms common to numerator and denominator that can be cancelled. If there are terms then we can cancel it. After cancelling, we will multiply numerator to numerator and denominator to the denominator. For example, we have \[\dfrac{2}{9}\] and \[\dfrac{6}{5}.\] On multiplying, we get, \[\dfrac{2}{9}\times \dfrac{6}{5}\] Here we can see 3 is common in numerator and denominator. So, we can cancel it. So, we get,
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{2\times 2\times 3}{3\times 3\times 5}\]
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{2\times 2}{3\times 5}\]
Now we will multiply numerator with numerator and denominator with denominator. So, we get,
\[\Rightarrow \dfrac{2}{9}\times \dfrac{6}{5}=\dfrac{4}{15}\]
Now in our problem we have \[\dfrac{1}{2}\] and \[\dfrac{x}{5}\] whose product is given as \[\dfrac{3}{10}.\] Since x is known, so we cannot find the common in the numerator and denominator. So, multiplying directly numerator to numerator and denominator to the denominator, we get,
\[\dfrac{1}{2}\times \dfrac{x}{10}=\dfrac{1\times x}{2\times 5}=\dfrac{x}{10}\]
So, we get the product as \[\dfrac{x}{10}\] but we are given that their product is \[\dfrac{3}{10}.\] Hence, \[\dfrac{x}{10}\] must be equal to \[\dfrac{3}{10}.\] So,
\[\dfrac{x}{10}=\dfrac{3}{10}\]
On cross multiplying, we get,
\[\Rightarrow x=\dfrac{3\times 10}{10}\]
On simplifying, we get,
\[\Rightarrow x=3\]
So, the value of x is 3.
Note: We can also go by hit and trial. As we learn how to multiply, we will use various values of x and check that it satisfies the equation or not. Say if let x = 1, we get,
\[\dfrac{1}{2}\times \dfrac{x}{5}=\dfrac{1}{2}\times \dfrac{1}{5}\]
On simplifying, we get,
\[\Rightarrow \dfrac{1}{10}\ne \dfrac{3}{10}\]
So, x = 1 is not our solution.
If we put x = 3, we get,
\[\dfrac{1}{2}\times \dfrac{x}{5}=\dfrac{1}{2}\times \dfrac{1}{5}\]
On simplifying, we get,
\[\Rightarrow \dfrac{3}{10}=\dfrac{3}{10}\]
So, x = 3 is the solution to our problem.
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