﻿ Laurel, Yanny, Cookies, Bananas, and Clocks

# 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.

Then, I came across a viral picture-math-puzzle involving cookies, bananas, and clocks: 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: $\,25\,$ Treat this as:
a single banana icon
a single clock icon

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.

## Cookie/Banana/Clock: $\,14\,$ Treat this as:
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:

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

## Cookie/Banana/Clock: $\,11\,$ Treat this as:
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:

• $\,c_{10} := \,$ 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: $$t + b_1 + b_1\cdot c_7 \ \ =\ \ 3 + 1 + 1\cdot 7 \ \ =\ \ 11$$

## Cookie/Banana/Clock: $\,10\,$ Treat this as:
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:

• $\,c_{10} = 10\,$
• $\,c_{7} = 7\,$
• $\,b_1 := \,$ 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 has numerical value $\,2\,$ ($\,t_2 = 2\,$)
• agree that the clock icon that reads $\,3\,$ o'clock has numerical value $\,3\,$ ($\,t_3 = 3\,$)
• 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: $$t_2 + b_1 + b_1\cdot c_7 \ =\ 2 + 1 + 1\cdot 7 \ =\ 10$$

## 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:

• For cookies, the number of chips determines its numerical value: $\text{ten chips} = 10\,,$ $\text{seven chips} = 7\,$
• For bananas, the number of bananas determines its numerical value: $\text{one banana} = 1\,,$ $\text{two bananas} = 2\,$
• For clocks, the hour time determines its numerical value: $\,2 \text{ o'clock} = 2\,,$ $\,3 \text{ o'clock} = 3\,$

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! 