$A$ person of mass $60 \, kg$ wants to lose $5 \, kg$ by going up and down a $10 \, m$ high stairs. Assume he burns twice as much fat while going up than coming down. If $1 \, kg$ of fat is burnt on expending $7000 \, kcal$,how many times must he go up and down to reduce his weight by $5 \, kg$?

(D) Mass of the person $m = 60 \, kg$. Height of the stairs $h = 10 \, m$. Energy required to burn $1 \, kg$ of fat $= 7000 \, kcal = 7 \times 10^6 \, cal$. Total energy required to burn $5 \, kg$ of fat $= 5 \times 7 \times 10^6 = 35 \times 10^6 \, cal$.
Work done in going up $= mgh = 60 \times 9.8 \times 10 = 5880 \, J$.
Work done in coming down is half of going up (since fat burnt is twice going up),so work done in coming down $= 2940 \, J$.
Total work done in one round trip $= 5880 + 2940 = 8820 \, J$.
Converting to calories $(1 \, cal = 4.2 \, J)$: Total energy per trip $= 8820 / 4.2 = 2100 \, cal$.
Number of trips $n = \frac{\text{Total energy required}}{\text{Energy per trip}} = \frac{35 \times 10^6}{2100} = 16666.67$.
Rounding to the nearest integer,the person must go up and down $16667$ times.