Question

$(2\frac{1}{3}+5\frac{1}{3})+3\frac{1}{3}$ is same as

• A. $2\frac{13}{3}+1$
• B. $\frac{13}{3}+1$
• C. $2(\frac{13}{3}+1)$
• D. $\frac{13}{3}+2$

Answer 1

The correct option is C

$2(\frac{13}{3}+1)$

$(2\frac{1}{3}+5\frac{1}{3})+3\frac{1}{3}$

= $(2\frac{1}{3}+3\frac{1}{3})+5\frac{1}{3}$ [$\because x+(y+z)=(x+y)+z$, for all numbers $x$, $y$ and $z$]

= $(\frac{7}{3}+\frac{10}{3})+5\frac{1}{3}$

= $(\frac{7}{3}+(\frac{7}{3}+1\left)\right)+5\frac{1}{3}$

We know that if we take any natural number and add the next natural number, or if we double the first number and add 1, we get the same number as the result.

So,

$(\frac{7}{3}+(\frac{7}{3}+1\left)\right)+5\frac{1}{3}$

= $(2\frac{7}{3}+1)+5\frac{1}{3}$

= $(\frac{13}{3}+1)+\frac{16}{3}$

= $(\frac{13}{3}+\frac{16}{3})+1$

= $(\frac{13}{3}+(\frac{13}{3}+1\left)\right)+1$

= $(2\frac{13}{3}+1)+1$

= $2\frac{13}{3}+2$

= $2(\frac{13}{3}+1)$