Question

How many silver coins ,1.75cm in diameter and of thickness 2mm, must be meltted to form a cuboid of dimensions 5.5cm*10cm*3.5cm?

Answer 1

GIVEN:-

For coin -

d = 1.75cm

r = $\small\sf{\frac{1.75}{2}cm}$

h= 2mm = 0.2cm

⠀⠀

$\therefore$ No. of silver coins = $\large\sf{\frac{Volume\:of\:Cuboid}{Volume\:of\:Coin}}$

$\implies$$\large\sf{\frac{5.5 \times 10 \times 3.5}{ \frac{22}{7} \times \frac{1.75}{2} \times \frac{1.75}{2} \times 0.2}}$

$\implies$$\large\sf{\frac{5.5 \times 10 \times 3.5 \times 7 \times 2}{2.2 \times 1.75 \times 1.75}}$

$\implies$$\large\sf{\frac{5.5 \times 10 \times 350 \times 700 \times 2}{2.2 \times 175 \times 175}}$

$\implies$$\large\sf{400coins}$

Answer 2

For a coin : $\sf{r=\frac{175}{200}=\frac{7}{8}cm,h=2mm=\frac{2}{10}cm}$

Let n be the number of coins. Then,

n × Volume of one coin = Volume of a cuboid

= $\sf{n×π{r}^{2}h=l×b×h}$

= $\sf{n \times \frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times \frac{2}{10} = 5.5 \times 10 \times 3.5}$

Hence, $\sf{n = \frac{55 \times 10 \times 35 \times 7 \times 8 \times 8 \times 10}{22 \times 7 \times 7 \times 2 \times 10 \times 10} = 400}$