Verified answer

[tex] \sf \: Let r=10 units \: and h=10 units \\ \\ [/tex]

V = π²h = h=π×100×10=1000πunit³

Now,

r =50% of r=5 units and

h =160 % of h=16 unit

V =πr²2

h =π(5) ² ×16=400πunits³

[tex] \therefore \: \%\: decrease= \: = \frac{1000 \pi \: - 400\pi}{1000 \pi } \times 100 \\ \\ \\ = 60\%[/tex]

[tex]{ \color{blue}{ \fbox{60\% }}}[/tex]

hope its help u

Answer:

The volume of cylinder will be decreased by 60 %.

Option b) 60 %

Step-by-step-explanation:

Let the radius of the cylinder be r units.

And the height of the cylinder be h units.

∴ Volume of cylinder = π r² h

Original volume = π r² h cu.units

Now,

The radius of cylinder is decreased by 50%.

New radius ( R ) = r - 50 % r

⇒ New radius ( R ) = r - ( 50 / 100 ) r

⇒ New radius ( R ) = r - ( r / 2 )

⇒ New radius ( R ) = ( 2r - r ) / 2

⇒ New radius ( R ) = r / 2 units

Now,

The height of cylinder is increased by 60%.

New height ( H ) = h + 60 % h

⇒ New height ( H ) = h + ( 60 / 100 ) h

⇒ New height ( H ) = h + ( 3h / 5 )

⇒ New height ( H ) = ( 5h + 3h ) / 5

⇒ New height ( H ) = 8h / 5 units

Now,

New volume = π R² H

⇒ New volume = π ( r / 2 )² * 8h / 5

⇒ New volume = π r² / 4 * 8h / 5

⇒ New volume = 2 π r² h / 5 cu.units

Now,

Volume Decrease % = ( Decrease in volume / Original volume ) * 100

⇒ Volume Decrease % = [ ( Original volume - New volume ) / Original volume ] * 100

⇒ Volume Decrease % = { [ π r² h - ( 2 π r² h / 5 ) ] / π r² h } * 100

⇒ Volume Decrease % = [ ( 5 π r² h - 2 π r² h ) / 5 ] / π r² h } * 100

⇒ Volume Decrease % = ( 3 π r² h / 5 π r² h ) * 100

⇒ Volume Decrease % = ( 3 / 5 ) * 100

⇒ Volume Decrease % = 3 * 20

⇒ Volume Decrease % = 60

∴ The volume of cylinder will be decreased by 60 %.