Question

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

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