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i) What must be added to \[3\dfrac{3}{4}\] to get \[4\dfrac{2}{8}\]?
ii) What must be subtracted from \[4\dfrac{1}{4}\] to get \[2\dfrac{3}{{12}}\]?

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Last updated date: 21st Jul 2024
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Answer
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Hint: Here we have to find the sum and difference between the given data. Here the data is in the form of a fraction. Since the given data is in fraction the value of the denominator is different so we find the LCM for both the denominators and then we add the numbers. Hence, we obtain the required solution for the given question.

Complete step by step solution:
I.We apply the arithmetic operations on the fractions. Here in this question, we add and subtract the two fractions. The LCM is the least common multiple. The LCM will be common for both the numbers.
Now we will consider the first question
What must be added to \[3\dfrac{3}{4}\] to get \[4\dfrac{2}{8}\]
Let we consider the unknown number as x
The question is written in the form of algebraic equation as
\[ \Rightarrow 3\dfrac{3}{4} + x = 4\dfrac{2}{8}\]
Convert the mixed fractions into improper fractions we have
\[ \Rightarrow \dfrac{{15}}{4} + x = \dfrac{{34}}{8}\]
Take \[\dfrac{{15}}{4}\] to RHS we have
\[ \Rightarrow x = \dfrac{{34}}{8} - \dfrac{{15}}{4}\]
The LCM for the numbers 8 and 4 is 8.
So we have
\[ \Rightarrow x = \dfrac{{\dfrac{{34}}{8} \times 8 - \dfrac{{15}}{4} \times 8}}{8}\]
On simplifying we have
\[ \Rightarrow x = \dfrac{{34 - 30}}{8}\]
\[ \Rightarrow x = \dfrac{4}{8}\]
On dividing the both numerator and the denominator by 4 we have
\[ \Rightarrow x = \dfrac{1}{2}\]
So, the correct answer is “\[ x = \dfrac{1}{2}\]”.


II.Now we will consider the second question
What must be subtracted from \[4\dfrac{1}{4}\] to get \[2\dfrac{3}{{12}}\]
Let we consider the unknown number as x
The question is written in the form of algebraic equation as
\[ \Rightarrow 4\dfrac{1}{4} - x = 2\dfrac{3}{{12}}\]
Convert the mixed fractions into improper fractions we have
\[ \Rightarrow \dfrac{{17}}{4} - x = \dfrac{{27}}{{12}}\]
Take \[\dfrac{{17}}{4}\] to RHS we have
\[ \Rightarrow - x = \dfrac{{27}}{{12}} - \dfrac{{17}}{4}\]
The LCM for the numbers 12 and 4 is 12.
So we have
\[ \Rightarrow - x = \dfrac{{\dfrac{{27}}{{12}} \times 12 - \dfrac{{17}}{4} \times 12}}{{12}}\]
On simplifying we have
\[ \Rightarrow - x = \dfrac{{27 - 51}}{{12}}\]
\[ \Rightarrow - x = \dfrac{{ - 24}}{{12}}\]
On dividing the both numerator and the denominator by 12 we have
\[ \Rightarrow - x = - 2\]
On cancelling the minus we have
\[ \Rightarrow x = 2\]
So, the correct answer is “ x = 2”.

Note: While adding the two fractions we need to check the values of the denominator, if both denominators are having the same value then we can add the numerators. Suppose if the fractions have different denominators, we have to take LCM for the denominators and we simplify for further.