Question

Gavin had 20 minutes to do a three-problem quiz. He spent 9 3/4 minutes on Problem 1 and 3 4/5 minutes on Problem 2. How much time did he have left for Problem 3? Write the answer in minutes and seconds.

PS: Solve the problem using the RDW strategy. Show all of your work.

Answer 1

Add the time for the first two problems:

9 3/4 + 3 4/5

To add fractions rewrite them so they have a common denominator:

3/4 = 15/20

4/5 = 16/20

Now you have:

9 15/20 + 3 16/20 =  12 31/20 = 13 11/20 minutes.

Now subtract that from the time allowed:

20 - 13 11/20 = 6 9/20 minutes.

Answer 2

Gavin has $6\dfrac{9}{20}$ minutes to solve the 3rd problem.

Given to us

• Total time with Gavin = 20 minutes,

• Time spent on problem 1 = $9\dfrac{3}{4}$ minutes,

• Time spent on problem 2 = $3\dfrac{4}{5}$ minutes,

Time spent on problem 1

Time spent on problem 1

$9\dfrac{3}{4} = \dfrac{(4\times 9)+3}{4}=\dfrac{39}{4}$  minutes

Time spent on problem 2

Time spent on problem 2

$3\dfrac{4}{5} =\dfrac{(3\times 5)+4}{5}=\dfrac{19}{5}$ minutes

Time left for problem 3

Time left for problem 3

    = Total time - time spent on problem 1 -time spent on problem 1

    $= 20 -\dfrac{39}{4} -\dfrac{19}{5}\\\\= \dfrac{20}{1} -\dfrac{39}{4} -\dfrac{19}{5}$

    taking LCM of denominators,

    $= \dfrac{20\times 20}{1\times 20} -\dfrac{39\times 5}{4\times 5} -\dfrac{19\times 4}{5\times 4}$

    $= \dfrac{400}{20} -\dfrac{195}{20} -\dfrac{76}{20}$

    $= \dfrac{400-195-76}{20}\\\\=\dfrac{129}{20}\\\\=6\dfrac{9}{20}\ \rm{minutes$

Hence, Gavin has $6\dfrac{9}{20}$ minutes to solve the 3rd problem.

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