Question

Suppose that the first team member in a 3-person relay race must run 2 1/4 laps, the second team member must run 1 1/2 laps, and the third team member must run 3 5/8 laps. How many laps in all must each team run?

Answer 1

You need to add up all 3 mixed numbers to find the total laps.

2 + 1/4 + 1 + 1/2 + 3 + 5/8 =

(2 + 1 + 3) + (1/4 + 1/2 + 5/8) =

6 + (2/8 + 4/8 + 5/8) =

6 + (11/8) =

6 + (1 + 3/8) =

7 3/8 laps

Answer 2

Answer: There are $7\dfrac{3}{8}$ laps in all.

Step-by-step explanation:

Since we have given that

Number of laps must run by first team = $2\dfrac{1}{4}=\dfrac{9}{4}$

Number of laps must run by second team = $1\dfrac{1}{2}=\dfrac{3}{2}$

Number of laps must run by third team = $3\dfrac{5}{8}=\dfrac{29}{8}$

So, total number of laps in all must run is given by

$\dfrac{9}{4}+\dfrac{3}{2}+\dfrac{29}{8}\\\\=\dfrac{18+12+29}{8}\\\\=\dfrac{59}{8}\\\\=7\dfrac{3}{8}$

Hence, there are $7\dfrac{3}{8}$ laps in all.