Question

A person of weight $70kg$ wants to loose $7kg$ by going up and down $12m$ high stairs. Assume he burns twice as much fat while going up than going down. If $1kg$ of fat is burnt on expending $9000k$-cal. How many times must he go up and down to reduce his $7kg$ weight? (Take $g=10m{s}^{-2}$ )

• A. $18\times 10^3$ times
• B. $24\times 10^3$ times
• C. $30\times 10^3$ times
• D. $21\times 10^3$ times

Answer 1

Answer: (D)

Given, $m=70kg,g=10m{s}^{-2},h=12m$ In going up and down once.
Number of k- cals burnt $=\left(mgh+\frac{mgh}{2}\right)$
$=\frac{3}{2}mgh$
$=\frac{3}{2}\times \frac{70\times 10\times 12}{4.2\times 1000}$
$=3k-cal$
Total no. of $k$-cal to be burnt for loosing $7kg$ of weight
$=7\times 9000=63000k-cal$
$\therefore$ Number of times the person has to go up and down the stairs $=\frac{63000}{3}=21000$
$=21\times 10^3$ times