
What is $1\dfrac{1}{2}$ cup minus $1\dfrac{2}{3}$ cup?
Answer
462.9k+ views
Hint: We need to convert the given mixed fractions in the question into improper fractions. We then need to subtract these two terms. We then find the LCM for the two terms and simplify the subtraction of these two terms.
Complete step by step answer:
In order to solve this question, let us convert these given mixed fractions into improper fractions. Given the first term as $1\dfrac{1}{2}$ cups. We convert this quantity of cups to an improper fraction. The general way of doing this is by multiplying the denominator with the whole number and adding with the numerator. This can be shown as,
$\Rightarrow a\dfrac{b}{c}=\dfrac{c\times a+b}{c}$
This can be done as follows for the first term as,
$\Rightarrow 1\dfrac{1}{2}=\dfrac{2\times 1+1}{2}$
Simplifying the numerator by multiplying first and adding the terms,
$\Rightarrow \dfrac{3}{2}$
Therefore, the term $1\dfrac{1}{2}$ cups convert to $\dfrac{3}{2}$ cups. Next, we convert the second term to an improper fraction using the same method.
$\Rightarrow 1\dfrac{2}{3}=\dfrac{3\times 1+2}{3}$
Simplifying the numerator similarly,
$\Rightarrow \dfrac{5}{3}$
Therefore, the term $1\dfrac{2}{3}$ cups convert to $\dfrac{5}{3}$ cups. Subtracting the two improper fractions,
$\Rightarrow \dfrac{3}{2}-\dfrac{5}{3}$
Taking the LCM of the denominator, which is nothing but the LCM of 2 and 3 which is 6.
$\Rightarrow LCM\left( 2,3 \right)=6$
Multiplying both the numerator and denominator of the first term by 3 and the numerator and denominator of the second term by 2.
$\Rightarrow \dfrac{3\times 3}{2\times 3}-\dfrac{5\times 2}{3\times 2}$
Simplifying the multiplication of all the terms,
$\Rightarrow \dfrac{9}{6}-\dfrac{10}{6}$
Since the denominators are the same, we can now subtract the numerators,
$\Rightarrow \dfrac{9-10}{6}=-\dfrac{1}{6}$
Hence, $1\dfrac{1}{2}$ cup minus $1\dfrac{2}{3}$ cup is $-\dfrac{1}{6}$ cups.
Note: It is important to know the conversion of fractions from mixed to improper and vice-versa. We even need to know how to calculate LCM for the denominators. We need to simplify the terms in the numerator or denominator to its simplest form in case it is not in the final answer in its simplest form.
Complete step by step answer:
In order to solve this question, let us convert these given mixed fractions into improper fractions. Given the first term as $1\dfrac{1}{2}$ cups. We convert this quantity of cups to an improper fraction. The general way of doing this is by multiplying the denominator with the whole number and adding with the numerator. This can be shown as,
$\Rightarrow a\dfrac{b}{c}=\dfrac{c\times a+b}{c}$
This can be done as follows for the first term as,
$\Rightarrow 1\dfrac{1}{2}=\dfrac{2\times 1+1}{2}$
Simplifying the numerator by multiplying first and adding the terms,
$\Rightarrow \dfrac{3}{2}$
Therefore, the term $1\dfrac{1}{2}$ cups convert to $\dfrac{3}{2}$ cups. Next, we convert the second term to an improper fraction using the same method.
$\Rightarrow 1\dfrac{2}{3}=\dfrac{3\times 1+2}{3}$
Simplifying the numerator similarly,
$\Rightarrow \dfrac{5}{3}$
Therefore, the term $1\dfrac{2}{3}$ cups convert to $\dfrac{5}{3}$ cups. Subtracting the two improper fractions,
$\Rightarrow \dfrac{3}{2}-\dfrac{5}{3}$
Taking the LCM of the denominator, which is nothing but the LCM of 2 and 3 which is 6.
$\Rightarrow LCM\left( 2,3 \right)=6$
Multiplying both the numerator and denominator of the first term by 3 and the numerator and denominator of the second term by 2.
$\Rightarrow \dfrac{3\times 3}{2\times 3}-\dfrac{5\times 2}{3\times 2}$
Simplifying the multiplication of all the terms,
$\Rightarrow \dfrac{9}{6}-\dfrac{10}{6}$
Since the denominators are the same, we can now subtract the numerators,
$\Rightarrow \dfrac{9-10}{6}=-\dfrac{1}{6}$
Hence, $1\dfrac{1}{2}$ cup minus $1\dfrac{2}{3}$ cup is $-\dfrac{1}{6}$ cups.
Note: It is important to know the conversion of fractions from mixed to improper and vice-versa. We even need to know how to calculate LCM for the denominators. We need to simplify the terms in the numerator or denominator to its simplest form in case it is not in the final answer in its simplest form.
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