Question

Abel was wrapping a gift for an upcoming party. He began with 20 1/2 feet of paper. He used 5 5/12 feet on his first gift and 3 1/3 feet on his second gift. If Abel bought 2 additional rolls of paper that each contained 10 1/2 feet, how many papers does he now have left for wraping.

Answer 1

Hey there!

$20 \dfrac{1}{2} - 5 \dfrac{5}{12} - 3 \dfrac{1}{3} + 2 \times 10 \dfrac{1}{2} \\ = \dfrac{41}{2} - \dfrac{65}{12} - \dfrac{10}{3} + 21 \\ = \dfrac{246}{12} - \dfrac{65}{12} - \dfrac{40}{12} + 21 \\ = \dfrac{141}{12} + 21 \\ = 32 \dfrac{3}{4}$

Answer : 32 3/4 feet or in decimal form, 32.75 feet.

Hope this helps. - M

Answer 2

$\text {At first = } 20\dfrac{1}{2} \text{ feet}$

$\text {Used } 5\dfrac{5}{12} \text{ on the first gift : }$
$20\dfrac{1}{2} - 5\dfrac{5}{12}$
$= 20\dfrac{6}{12} - 5\dfrac{5}{12}$
$= 15\dfrac{1}{12} \text{ feet}$

$\text {Used } 3\dfrac{1}{3} \text{ on the second gift : }$
$15\dfrac{1}{12} - 3\dfrac{1}{3}$
$= 14\dfrac{13}{12} - 3\dfrac{4}{12}$
$= 11\dfrac{9}{12}$
$= 11\dfrac{3}{4}\text{ feet}$

$\text {Bought additional 2 rolls of } 10\dfrac{1}{2} \text{ feet each : }$
$11\dfrac{3}{4} + 10\dfrac{1}{2} + 10\dfrac{1}{2}$
$= 11\dfrac{3}{4} + 10\dfrac{2}{4} + 10\dfrac{2}{4}$
$= 31\dfrac{7}{4}$
$= 32\dfrac{3}{4} \text{ feet}$

$\boxed{ \boxed{ \text{Answer: }32\dfrac{3}{4}\text{ feet}}}$