The smallest fraction which should be subtracted from the sum of $1 \frac{3}{4}, 2 \frac{1}{2}, 5 \frac{7}{12}, 3 \frac{1}{3}$ and $2 \frac{1}{4}$ to make the result a whole number is
(a)$\frac{5}{12}$
(b)$\frac{7}{12}$
(c)$\frac{1}{2}$
(d)7
Answer
Answer (as printed): A
Explanation
Sum of given fractions $=\frac{7}{4}+\frac{5}{2}+\frac{67}{12}+\frac{10}{3}+\frac{9}{4}$ $$=\left(\frac{21+30+67+40+27}{12}\right)=\frac{185}{12} .$$ The whole number just less than $\frac{185}{12}$ is 15. $$\text { Let } \frac{185}{12}-x=15 \text {. Then, } x=\left(\frac{185}{12}-15\right)=\frac{5}{12} \text {. }$$
Explanation as extracted from the printed page; notation may be imperfect.