NTSE 2015

Geometry Level 1

Two solid right circular cones have the same height. The radii of their bases are $a$ and $b$ . They are melted and recast into a cylinder of same height. The radius of the base of the cylinder is __ ?

This is a problem from the NTSE paper held on 8 November 2015.

\dfrac{a+b}3

\dfrac{a+b}{\sqrt 3}

\dfrac{\sqrt{a+b}}3

\sqrt{\dfrac{a^2+b^2}3}

1 solution

Sravanth C.
Nov 9, 2015

Let the height of the cones be $h$ . Therefore the volume of cylinder with base radius $a$ $=\dfrac 13\pi a^2h$ .

And that of the cylinder with base radius $b$ $=\dfrac 13\pi b^2h$ .

Let the base radius of the new cylinder be $x$ . Therefore it's volume is $\pi x^2h$ .

Now we can equate the volumes as follows: $\pi x^2h=\dfrac 13\pi a^2h+\dfrac 13\pi b^2h\\\pi x^2h=\dfrac 13\pi h(a^2+b^2)\\x^2=\dfrac 13(a^2+b^2)\\x=\sqrt{\dfrac{a^2+b^2}3}$

Moderator note:

Simple standard approach.

Hi shravanth...did u gave ntse this year..which state? How much are you expecting in it??

Mohit Gupta - 5 years, 7 months ago

Yeah I had given NTSE this year, but I did not do too well, I think I'll get 90 for sure, so I'm expecting around 100(SAT + MAT).

Sravanth C. - 5 years, 7 months ago

Which state and what is exp. cut off??