Problem Solving

A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

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rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

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circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>
60*4*pi = 8*2*pi*r
=> r = 15 inches.
=> diameter = 2*r = 30 inches.

Hence, D

anshumishra wrote:

rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

12
15
25
30
32

circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>
60*4*pi = 8*2*pi*r
=> r = 15 inches.

Hence, B

and it's D for diameter, i.e. R*2=15*2=30

Night reader wrote:

anshumishra wrote:

rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

12
15
25
30
32

circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>
60*4*pi = 8*2*pi*r
=> r = 15 inches.

Hence, B

and it's D for diameter, i.e. R*2=15*2=30

Thanks night reader ! Missed that. Let me edit the first post.

Night reader wrote:

anshumishra wrote:

rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

12
15
25
30
32

circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>
60*4*pi = 8*2*pi*r
=> r = 15 inches.

Hence, B

and it's D for diameter, i.e. R*2=15*2=30

I must admit ambiguity in this problem... the third wheel has been left out

anshumishra wrote:

rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

12
15
25
30
32

circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>

How do you say they cover equal distance , can you please explain further?

Deepthi Subbu wrote:

anshumishra wrote:

rtaha2412 wrote:A machine consisting of 3 connected wheels is mixing concrete. The engine is running at a speed of 60 rpm and the mixer is mixing at a rate of 8 rpm. If the radius of the smaller (drive) wheel is 2 inches, what is the diameter of the mixing wheel? The smaller wheel is driven directly by the engine.

12
15
25
30
32

circumference of smaller wheel = 2*pi*2 = 4*pi inches
circumference of mixer wheel = 2*pi*r

In one minute they must have covered equal distance =>

How do you say they cover equal distance , can you please explain further?

This is equivalent to the scenario where a vehicle(may be a cycle in circus) with unequal radius moves. You will see that the one with the smaller radius has to rotate faster to match with the larger one. Put a mark on the tyre with larger radius, when one rotation completes the cycle has moved to 2*pi*R distance (right ?). Same logic for the smaller one (as to cover up this distance smaller has to rotate x times so that 2*pi*r*x = 2*pi*R, as eventually the cycle is same and it covered a fixed distance.)