
How do you find the product of $\dfrac{1}{3} \times \dfrac{1}{4}$?
Answer
557.4k+ views
Hint: In this particular sum the student should combine both the fractions into a single fraction and then perform mathematical operations. After combining the fractions the student should now cancel out the common terms or multiples before finding the product of the fraction. After there is nothing left to cancel out the student should then multiply the remaining numbers to obtain the product.
Complete step by step solution:
In this sum the first step is to combine the fractions into a single fraction:
$\dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{{1 \times 1}}{{3 \times 4}}...........(1)$
From the above equation, we can see that there is nothing common between the numerator and the denominator, so we cannot cancel out anything. Thus the final step would be to multiply the given numbers.
$\dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{{12}}$
Thus the final answer for the given fraction is $\dfrac{1}{{12}}$.
Note: Since in this sum there was a multiplication sign the student could directly combine the fraction. But instead of there was a division sign the student would have to first invert the fraction and then combine the fraction. For example, if there was a sum $\dfrac{3}{{25}} \div \dfrac{9}{5}$, then in this sum the student will have to first invert the fraction $\dfrac{9}{5}$ to $\dfrac{5}{9}$ and then combine both the fractions and perform the rest of the operations.
Complete step by step solution:
In this sum the first step is to combine the fractions into a single fraction:
$\dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{{1 \times 1}}{{3 \times 4}}...........(1)$
From the above equation, we can see that there is nothing common between the numerator and the denominator, so we cannot cancel out anything. Thus the final step would be to multiply the given numbers.
$\dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{{12}}$
Thus the final answer for the given fraction is $\dfrac{1}{{12}}$.
Note: Since in this sum there was a multiplication sign the student could directly combine the fraction. But instead of there was a division sign the student would have to first invert the fraction and then combine the fraction. For example, if there was a sum $\dfrac{3}{{25}} \div \dfrac{9}{5}$, then in this sum the student will have to first invert the fraction $\dfrac{9}{5}$ to $\dfrac{5}{9}$ and then combine both the fractions and perform the rest of the operations.
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