Lesson Trap problems

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<H2>Trap problems</H2>

<H3>Problem 1</H3>A bottle filled with varnish cost &nbsp;$1.10. &nbsp;If the varnish alone is worth &nbsp;$1.00 &nbsp;more 
than the bottle, &nbsp;find the value of the bottle.

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This problem and many other similar problems is/are well known &nbsp;" trap " &nbsp;problems.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Those, who do not know that it is a trap, &nbsp;usually fall in it, &nbsp;giving wrong answer &nbsp;$0.10 &nbsp;for the bottle and &nbsp;$1.00 &nbsp;for the varnish alone,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;especially those of them who are in hurry to answer without thinking.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This answer is, &nbsp;OBVIOUSLY, &nbsp;incorrect, &nbsp;because the difference of &nbsp;$1.00 &nbsp;and &nbsp;$0.10 &nbsp;is not &nbsp;$1.00.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The correct answer is &nbsp;$0.05 &nbsp;for the empty bottle and &nbsp;$1.05 &nbsp;for the varnish alone.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;One way to get the correct answer is to get it mentally, &nbsp;after thinking couple seconds.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Another way is to solve it formally, &nbsp;as follows.

<pre>
Let x  be the cost for varnish in the bottle (in dollars);

Then the cost of the bottle is (x-1) dollars.

Total cost equation

x + (x-1) = 1.10  dollars

x + x = 1.10 + 1

2x   =   2.10

x   =   2.10/2 = 1.05,

giving the above mentioned correct answer.
</pre>

<H3>Problem 2</H3>Consider a population that grows according to the recursive rule &nbsp;P(n) = P(n-1) + 35, 
with initial population &nbsp;P(0)=20. 
 
Find an explicit formula for the population. 
 
Use your explicit formula to find &nbsp;P(100).

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Be careful : &nbsp;&nbsp;this problem is a trap.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The trap is that this sequence is &nbsp;SPECIAL : &nbsp;&nbsp;it &nbsp;IS &nbsp;an arithmetic sequence, 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;but starts from the zeroth term &nbsp;20.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Therefore, &nbsp;the standard formula for the n-th term of an arithmetic progression must be modified.

<pre>
The standard formula {{{P[n]}}} = 20 + (n-1)*35  does  {{{highlight(highlight(NOT))}}}  work now - it should be modified.

+----------------------------------------------+
    |   The correct formula is P[n] = 20 + n*35.   |
    +----------------------------------------------+

Again, your correct explicit formula is  {{{P[n]}}} = 20 + n*35.

You can  {{{CHECK}}}  it:  {{{P[1]}}} = 20 + 1*35 = 20 + 35 = 55.

In particular,  {{{P[100]}}} = 20 + 100*35 = 3520.
</pre>

<H3>Problem 3</H3>What is the square root of &nbsp;{{{9y^6}}} ?

<B>Solution</B>

Very often, unexperienced students give the answer &nbsp;&nbsp;{{{3y^3}}}.

It is incorrect.

The correct answer is  {{{3*abs(y)^3}}}  with the use of the absolute value sign.

It is because value  ' y '  in the original expression can be negative.

Then the mentioned wrong answer returns the negative value of the square root, which contradicts 
to the common agreement about the square root values.

My formula works universally for positive and negative values of &nbsp;' y ' - - - &nbsp;it is why it is preferable
and why it is the uniquely correct form.

The problems of this kind are a standard TRAP to catch unprepared/undertrained novices.
</pre>

<H3>Problem 4</H3>Simplify this expression &nbsp;&nbsp;{{{sqrt(75*a^2)/sqrt(5)}}}.

<B>Solution</B>

<pre>
The right way to simplify is THIS

{{{sqrt(75*a^2)/sqrt(5)}}} = {{{sqrt(15a^2)}}} = {{{abs(a)*sqrt(15)}}} = {{{sqrt(15)*abs(a)}}}.    <U>ANSWER</U>

According to commonly accepted agreements about square root, 
the expression &nbsp;' {{{a^2}}} '  &nbsp;always goes from the square root as  &nbsp;|a|, &nbsp;the absolute value of  &nbsp;'a'.

In this form, &nbsp;this transformation works universally for positive and negative values of &nbsp;' a '.

The given problem is a standard &nbsp;TRAP &nbsp;to catch &nbsp;{{{highlight(highlight(millions))}}} &nbsp;of those, &nbsp;who are unfamiliar with this trap for advance.

My post shows and teaches a standard way  to avoid this trap.
</pre>

<H3>Problem 5</H3>Simplify &nbsp;&nbsp;{{{sqrt(4x^3*9y^5)}}}.

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This problem is a TRAP, and many students Fall in it.

<pre>
To solve the problem correctly, we should reveal a context accurately.

The domain, i.e. the area where this expression is defined, is the set of points (x,y)
in the 1st quadrant or in the 3rd quadrant.  It means that the domain is the union of both quadrants.

In other words x >= 0, y >= 0 are allowed under the square root, or x <= 0, y <= 0 simultaneously.
But x and y of different signs are not allowed.

Second note is that when square root is considered as a function or as an expression,
by a commonly accepted agreement, it is considered as non-negative value or expression.

Again, it is the common agreement for square root, when it is considered 
    as an expression of variables, that its output values are non-negative.

Next, the first move is to write  {{{sqrt(4x^3*9y^5)}}} = {{{2*3*x*y^2*sqrt(xy)}}}.    (1)

But this expression is only half-right.  On the other half it is WRONG.

It is wrong in 3rd quadrant.  Numerical values under the square root are non-negative there. 
It is good, since we can extract the square root then.

But the values of the simplified expression (1) will be negative in 3rd quadrant, 
which is not consistent with the agreement mentioned above.

Therefore, we should write (1) in MODIFIED form

{{{sqrt(4x^3*9y^5)}}} = {{{2*3*abs(x)*y^2*sqrt(xy)}}} = {{{6*abs(x)*y^2*sqrt(xy)}}},   (2)

placing x under the absolute value sign.

In this form (2),  it is consistent with (1) in the first quadrant and provides non-negative output
in the third quadrant.

Therefore, the commonly accepted <U>ANSWER</U> is THIS:

{{{sqrt(4x^3*9y^5)}}}}  is  {{{6*abs(x)*y^2*sqrt(xy)}}}  in the 1st and 3rd quadrants.
</pre>

For your info: practically all without exceptions Fall in this trap,
if they are unfamiliar with this logic/reasoning for advance, until somebody explains a right way reasoning.

So, to get the right answer to this problem and to many other similar problems,
you need accurately determine the domain, accurately reveal the context and accurately 
adopt the output expression in order for to have positive values of the final square root expression over the entire domain.

It is the KEY to solving such problems.

The key point to understand is that when square root of a non-negative number is considered, 
it has two values - one positive and one negative, opposite to the positive value.

On contrary, when square root of a function or of an expression is considered,
the common agreement is that in this situation the output (the result of applying square root) must be non-negative.

It requires special cares in the solution.

I think that 99% (or 99.99%) of contemporary people/students/teachers don't know this wisdom.

<H3>Problem 6</H3>If the surface of a cube is &nbsp;6x^2 -36x +54, &nbsp;what is the expression for the volume of the cube?

<B>Solution</B>

<pre>
The surface area of the cube is given by the polynomial expression  6x^2 - 36x + 54.

It is the combined area of 6 identical faces of the cube.  
The area of each face separately is

(6x^2 - 36x + 54)/6 = x^2 - 12x + 9 = (x-3)^2.

Notice that the given polynomial is always non-negative, 
so its values make sense as the face area for all values of x =/= 3.

The length of the side of this cube is  {{{sqrt((x-3)^2)}}} = |x-3|.

Notice that I use the absolute value  |x-3|  here, 
as the length is always assumed to be non-negative.

The volume of the cube is then  {{{abs(x-3)^3}}}.

This formula works for all values  x =/= 3, including  x < 3, 
while the formulas without the absolute value sign do not work at x < 3
giving negative edge length and negative volume.

So, my solution is MORE ACCURATE than the @mananth' solution and/or than the @Theo' solution.
</pre>

Actually, this seemingly simple task has a &nbsp;{{{highlight(highlight(well))}}} &nbsp;{{{highlight(highlight(hidden))}}} &nbsp;murderous &nbsp;{{{highlight(highlight(trap))}}}.

My solution discloses the trap, teaches you on how to avoid this trap 
and also teaches you to be always aware and always accurate in Math.

<H3>Problem 7</H3>What is the exact value of &nbsp;{{{arccos(cos(3pi/2))}}} ?

<B>Solution</B>

<pre>
This problem is a  {{{highlight(highlight(TRAP))}}},  provoking you to answer "{{{3pi/2}}}".

This answer is incorrect.

To answer correctly, you should remember that function arccos has values in interval  [{{{-pi/2}}},{{{pi/2}}}].

Therefore, you should transform  {{{3pi/2}}}  to the equivalent angle in interval  [{{{-pi/2}}},{{{pi/2}}}].

This equivalent angle is  {{{-pi/2}}},  so the correct answer to the problem's question is  {{{-pi/2}}}.
</pre>

<H3>Problem 8</H3>Determine the amplitude, &nbsp;period and phase shift of &nbsp;&nbsp;y = {{{-2*cos(pi*x -3)}}}.

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In the &nbsp;Internet, &nbsp;where are many similar problems.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;But this one is &nbsp;SPECIAL: &nbsp;it is a &nbsp;TRAP, &nbsp;and its solution in the post by @lwsshar3 is incorrect.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Below is my correct solution with complete explanations.

<pre>
Regarding the amplitude, the question is trivial: I hope, 9 of 10 tutors will answer "amplitude is 2 units".

With the period, the question is trivial, too: I hope that many tutors will give a correct answer "the period is 2".

But regarding the phase shift, 9 of 10 tutors and 99 of 100 students will answer incorrectly, 
saying that the phase shift is  {{{3/pi}}}.

To answer correctly, we should remember that the parent function to compare with is cos(x).

When we consider y = {{{-2*cos(pi*x-3)}}}, we should first write it equivalently with the positive coefficient 
before cosine

y = {{{2*cos(pi*x-3-pi)}}},

shifting right by half-period {{{pi}}}.  ONLY AFTER THAT we can rewrite the function, extracting the shift explicitly

y = {{{2*cos(pi*(x - (3+pi)/pi))}}} = {{{2*cos(pi*(x-(3/pi+1)))}}}.

Now it becomes obvious that the shift is  {{{3/pi+1)}}}  units right, comparing with the parent function cos(x).

<U>ANSWER</U>.  The shift is  {{{3/pi+1)}}}  units right.
</pre>

REMEMBER: &nbsp;&nbsp;The sign &nbsp;" - " &nbsp;before the trigonometric functions "sine" and "cosine" 
is not a harmless symbol that can be ignored in such analysis.

Actually, &nbsp;this sign means the same as the half-period shift of the argument.

Therefore, &nbsp;it must be taken into account, &nbsp;and it completely changes the game, &nbsp;turning everything upside down.

My other additional lessons on &nbsp;Miscellaneous word problems &nbsp;(section 3) &nbsp;in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/More-complicated-problems-on-finding-number-of-elements-in-finite-subsets.lesson>More complicated problems on finding number of elements in finite subsets</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-problems-by-the-Backward-method.lesson>Solving problems by the Backward method</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/A-solid-cube-is-placed-into-a-cylindrical-tank-displacing-water.lesson>A solid cube is placed in a cylindrical tank displacing water in the tank</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Minimax-problems-that-to-be-solved-MENTALLY-based-on-common-sense.lesson>Minimax linear problems to solve MENTALLY based on common sense</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-linear-optimization-problems-by-reduction-to-linear-function.lesson>Solving linear optimization problems without LP-method by reduction to linear function</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-one-special-linear-minimax-problem-in-100-D-space.lesson>Solving one special linear minimax problem in 100-D space by the Linear Programming method</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Miscellaneous-logical-problems.lesson>Miscellaneous logical problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Math-Olympiad-level-problem-on-divisibility-numbers.lesson>Math Olympiad level problem on divisibility numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Find-a-sequense-of-transformations-of-a-given-number-to-get-a-desired-number.lesson>Find a sequence of transformations of a given number to get a desired number</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Upper-class-entertainment-Math-problems-for-all-ages.lesson>Upper class entertainment Math problems for all ages</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/OVERVIEW-of-my-additional-lessons-on-Miscellaneous-word-problems-section-3.lesson>OVERVIEW of my additional lessons on Miscellaneous word problems, section 3</A>

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-I.

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-II.