Laurel, Yanny, Cookies, Bananas, and Clocks
 

Today (May 19, 2018) the universe was telling me to write this section.
Hope it's fun for you!

Laurel versus Yanny

I zipped onto Facebook.
An audio clip — Do you hear ‘Laurel’ or ‘Yanny’? — had gone viral.

I listened. It was definitely Yanny. No hint of Laurel.

Ten minutes later, my husband came in. I wanted to know what he heard.
I located the same post on Facebook.

I listened again. It was definitely Laurel. No hint of Yanny.

How could I have heard Yanny before?
Same computer, same location, same Facebook post, within ten minutes.

Here's a tool so you can hear both Yanny and Laurel.

Cookies, Bananas, and Clocks

Then, I came across a viral picture-math-puzzle involving cookies, bananas, and clocks:


As in the Laurel/Yanny scenario, I got multiple ‘answers’:
First look: the question mark is 25.
Another look ... the question mark is 14.
Another look ... the question mark is 11.
Another look ... the question mark is 10.
Follow the links to see my thought processes for getting each ‘answer’.
You need to read them in order, since they build on each other.

A Moral

I was led to these thoughts:

People hear things differently.
People see things differently.
Even the same person.

Emphasize understanding over right/wrong.

Cookie/Banana/Clock: $\,25\,$

With my mathematical background, I first saw a system of equations
(three equations, three unknowns).

Let:
  • $\,c := \,$ cookie icon
  • $\,b := \,$ banana icon
  • $\,t := \,$ clock (timepiece) icon
The problem becomes the system: $$ \begin{gather} c + c + c = 30\cr b + b + c = 14\cr b + t + t = 8 \end{gather} $$ or, more simply, $$ \begin{gather} 3c = 30\cr 2b + c = 14\cr b + 2t = 8 \end{gather} $$ with solution $\,c = 10\,$, $\,b = 2\,$ and $\,t = 3\,$. Thus: $$t + b + b\cdot c \ \ =\ \ 3 + 2 + 2\cdot 10 \ \ =\ \ 25$$ Be sure to recall PEMDAS: multiplication gets done before addition.

treat this as:
one cookie icon
one banana icon
one clock icon

Cookie/Banana/Clock: $\,14\,$

New perspective:
The banana icon in the bottom row is different.
It's a single banana—the others were double bananas.

With the paradigm ‘each unique icon represents a unique (unknown) number’,
there is now insufficient information to solve the mathematics problem, as follows:

We have:

  • $\,c := \,$ cookie icon
  • $\,b_2 := \,$ double banana icon
  • $\,b_1 := \,$ single banana icon
  • $\,t := \,$ clock (timepiece) icon
The system is: $$ \begin{gather} 3c = 30\cr 2b_2 + c = 14\cr b_2 + 2t = 8 \end{gather} $$ with solution $\,c = 10\,$, $\,b_2 = 2\,$ and $\,t = 3\,$.
Since $\,b_1\,$ is unknown, we are unable to evaluate the expression $\,t + b_1 + b_1\cdot c\,$.

However, it is reasonable to formulate the problem slightly differently, as follows.
Agree to let the double banana icon represent twice the single banana icon, so we have:

  • $\,c := \,$ cookie icon
  • $\,b_1 := \,$ single banana icon
  • $\,t := \,$ clock (timepiece) icon
  • agree that the double banana icon is $\,2b_1\,$
Now, the system is: $$ \begin{gather} 3c = 30\cr 2b_1 + 2b_1 + c = 14\cr 2b_1 + 2t = 8 \end{gather} $$ with solution $\,c = 10\,$, $\,b_1 = 1\,$ and $\,t = 3\,$. Thus: $$t + b_1 + b_1\cdot c \ \ =\ \ 3 + 1 + 1\cdot 10 \ \ =\ \ 14$$


treat this as:
one cookie icon
two banana icons
one clock icon

Cookie/Banana/Clock: $\,11\,$

Additional new perspective:
The cookie icon in the bottom row is different.
It has seven chips—the others have ten chips.

Adjusting notation to account for the new information, let:

  • $\,c_{10} := \,$ cookie icon with ten chips
  • $\,c_{7} := \,$ cookie icon with seven chips
  • $\,b_1 := \,$ single banana icon
  • $\,t := \,$ clock (timepiece) icon
  • agree that the double banana icon is $\,2b_1$
The system is: $$ \begin{gather} 3c_{10} = 30\cr 2b_1 + 2b_1 + c_{10} = 14\cr 2b_1 + 2t = 8 \end{gather} $$ with solution $\,c_{10} = 10\,$, $\,b_1 = 1\,$ and $\,t = 3\,$.
Since $\,c_7\,$ is unknown, we are unable to evaluate the expression $\,t + b_1 + b_1\cdot c_7\,$.

However, it is reasonable to formulate the problem slightly differently, as follows.
Agree to let the number of chips on a cookie dictate its numerical value.
Thus, we have:

  • $\,c_{10} = 10\,$
  • $\,c_{7} = 7\,$
  • $\,b_1 := \,$ single banana icon
  • $\,t := \,$ clock (timepiece) icon
  • agree that the double banana icon is $\,2b_1$
Now, we have a true statement, and a system of two equations and two unknowns: $$ \begin{gather} 10 + 10 + 10 = 30 \quad \text{(true statement)}\cr 2b_1 + 2b_1 + 10 = 14\cr 2b_1 + 2t = 8 \end{gather} $$ with solution $\,b_1 = 1\,$ and $\,t = 3\,$. Thus: $$t + b_1 + b_1\cdot c_7 \ \ =\ \ 3 + 1 + 1\cdot 7 \ \ =\ \ 11$$


treat this as:
two cookie icons
two banana icons
one clock icon

Cookie/Banana/Clock: $\,10\,$

Additional new perspective:
The clock in the bottom row is different.
It reads 2 o'clock—the others read 3 o'clock.

Building on prior scenarios, and adjusting notation to account for the new information, let:

  • $\,c_{10} = 10\,$
  • $\,c_{7} = 7\,$
  • $\,b_1 := \,$ single banana icon
  • $\,t_2 := \,$ clock (timepiece) icon that reads 2 o'clock
  • $\,t_3 := \,$ clock (timepiece) icon that reads 3 o'clock
  • agree that the double banana icon is $\,2b_1$
We now have $$ \begin{gather} 10 + 10 + 10 = 30\cr 2b_1 + 2b_1 + 10 = 14\cr 2b_1 + 2t_3 = 8 \end{gather} $$ with solution $\,b_1 = 1\,$ and $\,t_3 = 3\,$.
Since $\,t_2\,$ is unknown, we are unable to evaluate the expression $\,t_2 + b_1 + b_1\cdot c_7\,$.

However, it is reasonable to formulate the problem slightly differently, as follows.
Agree to let the (hour) time on a clock dictate its numerical value.
Thus, we have:

  • $\,c_{10} = 10\,$
  • $\,c_{7} = 7\,$
  • $\,b_1 := \,$ single banana icon
  • agree that the clock icon that reads 2 o'clock ($\,t_2\,$) has numerical value $\,2\,$
  • agree that the clock icon that reads 3 o'clock ($\,t_3\,$) has numerical value $\,3\,$
  • agree that the double banana icon is $\,2b_1$
Now, we have two equations in a single unknown: $$ \begin{gather} 10 + 10 + 10 = 30 \quad \text{(true statement)}\cr 2b_1 + 2b_1 + 10 = 14\cr 2b_1 + 2\cdot 3 = 8 \end{gather} $$ Fortunately, the second and third equations are equivalent, each yielding $\,b_1 = 1\,$.
The desired expression is then: $$t_2 + b_1 + b_1\cdot c_7 \ \ =\ \ 2 + 1 + 1\cdot 7 \ \ =\ \ 10$$


treat this as:
two cookie icons
two banana icons
two clock icons

In Conclusion

In hindsight, we might merely want to present the following ‘solution’:

Use the first three equations to guide us to the following agreements:

With these agreements, the first three equations are true.
Thus, it's reasonable to think that this is what the author of the problem wanted.
Then, the final expression becomes $$2 + 1 + 1\cdot 7$$ and you need only remember that multiplication is done before addition, yielding $$2 + 1 + (1\cdot 7)\ \ =\ \ 2 + 1 + 7\ \ =\ \ 10$$

When you've had sufficient fun with this digression,
move on to:
Taking PEMDAS Too Literally: Don't Make This Mistake!