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If master chief has three and a half cookies, and the arbiter has five and three quarters of a cookie, how many cookies do they

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Answer:

$9\dfrac{1}{4}$

$4\dfrac{5}{8}$

Step-by-step explanation:

The master chief has three and a half cookies i.e. $3\dfrac{1}{2} = \dfrac{7}{2}$ numbers of cookies.

Again the arbiter has five and three quarters of a cookie i.e. $5\dfrac{3}{4} = \dfrac{23}{4}$ number of cookies.

Therefore, in total there are $(\dfrac{7}{2} +\dfrac{23}{4} ) = \dfrac{37}{4} = 9\dfrac{1}{4}$ numbers of cookies.

If we divide the total number of cookies in to equal parts and distribute to then then each of them will get $\dfrac{37}{4 \times 2} = \dfrac{37}{8} = 4\dfrac{5}{8}$ numbers of cookies. (Answer)