69 would be my answer.

I have a different answer. Pretty sure mine is right because the third time was the charmâ€¦

38

Thatâ€™s my answer too

And mine.

I get 38 too.

I got 67

Iâ€™m with CarolynA on 67

67 is a prime numberâ€¦ how could the answer possibly be 67?

I got 67 as wellâ€¦

Explain?

15 + 15 + 15 = 45

4 + 4 + 15 = 23

4 + 3 + 3 = 10

3 + 4 + (4 x 15) = 67

What did you get?

I think 67 too.

The symbols are not the same in all the equations.

If the 3 oâ€™clock clock is 3, then the 2 oâ€™clock clock shown in the last equation is probably 2 (but only if we assume it is a linear clock, so you would have to discount String theory, M theory, and Itzhak Barsâ€™ theory of time having two separate dimensions).

The bananas are also not the same. If the bunch of 4 bananas is 4, then the bunch of 3 bananas is 3.

And hexagon / pentagon / rectangle shape is also not the same in each equation.

This shape is certainly 15:

But assuming this shape is 11 simply by counting sides is not something that can be proven with given information.

Just because a hexagon / pentagon / rectangle is 15 (6 sided figure, 5 sided figure, and 4 sided figure), does not guarantee a hexagon / pentagon is 11 (6 sided figure, 5 sided figure). The shapes could be exponential representations for all we know.

There are 15 points in the first shape. There are 11 points in the final one. I think thatâ€™s just as sound as counting the bananas.

38 is simply one possible solution, and easily accepted because it is the easiest to compute.

Itâ€™s the answer that makes the most sense, unless you subscribe to the all clocks are worth 3, any bunch of bananas are worth 4, and all polygons are worth 15 theory. Or string theory, whatever that meansâ€¦

A hexagon/pentagon/rectangle is 15, sure. But that doesnâ€™t mean a hexagon/pentagon is 11.

You can cut an 8 in half to make it look like a 3, but that doesnâ€™t mean 1/2 of 8 is 3.

Well you can cut the bananas in half too. It is the only reasonable assumption. I applaud your creativity though.