Question

Solve the following :
$(3\frac{1}{5}+2\frac{1}{3})-4\frac{1}{5}$

• A. $4$
• B.
  $1\frac{1}{3}$
• C. $1\frac{1}{5}$
  
• D. $\frac{4}{3}$

Answer 1

Answer: (B), (D)

$\frac{4}{3}$

The given expression is $(3\frac{1}{5}+2\frac{1}{3})-4\frac{1}{5}$.
To solve this mathematical expression, we use the PMDAS rule.
These rules start with parentheses. Then we perform operations on multiplication or division from left to right. Finally, operations on addition or subtraction are performed from left to right.
First, perform operation on the parentheses and add $(3\frac{1}{5}+2\frac{1}{3})$
$3\frac{1}{5}$ and $2\frac{1}{3}$ are mixed numbers.
$3\frac{1}{5}=3+\frac{1}{5}=3\times \frac{5}{1}\times 5+\frac{1}{5}=\frac{15}{5}+\frac{1}{5}=\frac{16}{5}$

$2\frac{1}{3}=2+7\frac{1}{3}=2\times \frac{3}{1}\times 3+\frac{1}{3}=\frac{6}{3}+\frac{1}{3}=\frac{7}{3}$
Now, $\frac{16}{5}$ and $\frac{7}{3}$ are unlike fractions.
Now, find the equivalent fractions of $\frac{16}{5}$ and $\frac{7}{3}$ with the same denominator.
$\frac{16\times 3}{5\times 3}=\frac{48}{15}$

$\frac{7\times 5}{3\times 5}=\frac{35}{15}$
Now, add $\frac{48}{15}+\frac{35}{15}=\frac{48+35}{15}=\frac{83}{15}$
Now, subtract $4\frac{1}{5}$ from $\frac{83}{15}$
$4\frac{1}{5}=4+\frac{1}{5}=\frac{4\times 5}{1\times 5}+\frac{1}{5}=\frac{20}{5}+\frac{1}{5}=\frac{21}{5}$
Now, $\frac{83}{15}$ and $\frac{21}{5}$ are unlike fractions.
So, find the equivalent fractions of $\frac{21}{5}$ with the denominator 15.
$\frac{21\times 3}{5\times 3}=\frac{63}{15}$

Now, subtract $\frac{83}{15}-\frac{21}{15}=\frac{83-63}{15}=\frac{20}{15}=\frac{4}{3}=1\frac{1}{3}$
Hence, options B and C are correct.