Question

if the radius of cylinder is decreased by 50 percent and the height of cylinder is increased by 50 percent then find the percentage decreased in volume?

Answer 1

If the radius of cylinder is decreased by 50 percent and the height of cylinder is increased by 50 percent then find the percentage decreased in volume?

Here is your answer!

Let radius and height be r and h respectively.

Volume of a cylinder = pi*r²*h

According to question,

New radius = r - 50r/100

r'= r/2,

New height = h + 50h/100

h'= 3h/2,

New Volume = pi*r'²*h'

= pi*(r/2)² * 3h/2

= 3pi*r²*h/8

Decreased % = (Original V - new V) /original * 100%

= (V - 3V/8)/V * 100%

= 5V/8V * 100%

= 500/8 %

= 62.5%.

Answer 2

$\huge \bf{ \red{answer}}$

$\bf{62.5\%}$
$\bf{step \: by \: step \: explanation}$

$\sf{Given}$

The radius of a cylinder is decreased by 50 percent and the height increased by 50 percent. 

so the

volume before reducing and increasing was 

$\pi \: r^{2} h$

decrease in radius

=> r-50/100 r

=> r-1/2r

=> r-r/2

=> 2r-r/2

=> r/2

Now

increase in height

=> h+50/100h

=>h+1/2h

=> h+h/2

=> 2h+h/2

=> 3h/2

Now

new volume=

$\bf{\pi \: ( \frac{r}{2} )^{2} \frac{3h}{2} }$
$\pi \: \frac{ {r}^{2} }{4} \frac{3h}{2}$

$\pi \:\frac{3r^{2}h}{8}$
$\pi \:\frac{3}{8}{r}^{2}$ 

volume percent= orginal volume - new volume/original volume*100

=> 
$\frac{\pi \: r ^{2} h - \frac{3}{8} \pi \: r ^{2}h }{\pi \: r ^{2}h }\times 100$

=>
$\frac{ \frac{8\pi \: r ^{2} h - 3\pi \: r ^{2}h }{8} }{\pi \: r ^{2} h}\times{100}$

$\frac{ \frac{5\pi {r}^{2} h}{8} }{\pi {r}^{2} h}\times{100}$

$\bf{ \frac{5}{8} \times{100}}$

$\bf{ = > \: 62.5\%}$ 

$\huge{ \mathfrak{ \pink{ \: thanks}}}$