Laurel, Yanny, Cookies, Bananas, and Clocks

Today (May 19, 2018) the universe was telling me to write this section.

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.

Cookies, Bananas, and Clocks

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

cookie banana clock math problem

As in the Laurel/Yanny scenario, I got multiple ‘answers’:

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: $\,25\,$

cookie banana clock math problem Treat this as:
a single cookie icon
a single banana icon
a single clock icon

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

Let:

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.

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

cookie banana clock math problem Treat this as:
a single cookie icon
two different banana icons
a single clock icon

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:

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:

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: $$t + b_1 + b_1\cdot c \ \ =\ \ 3 + 1 + 1\cdot 10 \ \ =\ \ 14$$

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

cookie banana clock math problem Treat this as:
two different cookie icons
two different banana icons
a single clock icon

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:

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:

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: $$t + b_1 + b_1\cdot c_7 \ \ =\ \ 3 + 1 + 1\cdot 7 \ \ =\ \ 11$$

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

cookie banana clock math problem Treat this as:
two different cookie icons
two different banana icons
two different clock icons

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:

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:

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: $$t_2 + b_1 + b_1\cdot c_7 \ =\ 2 + 1 + 1\cdot 7 \ =\ 10$$

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!

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