Lesson Upper level word problems to solve using a single linear equation

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<H2>Upper level word problems to solve using a single linear equation</H2>

<H3>Problem 1</H3>Three shops sold the same number of lanterns during the lantern festival. 
Shop &nbsp;A &nbsp;sold &nbsp;1/2 &nbsp;or its lanterns, &nbsp;Shop &nbsp;B &nbsp;sold &nbsp;2/3 &nbsp;of its lanterns and 
Shop &nbsp;C &nbsp;sold &nbsp;3/4 &nbsp;of its lanterns. &nbsp;If the three shops had a total of &nbsp;330 &nbsp;lanterns left, 
how many lanterns did each shop sell?

<B>Solution</B>

<pre>
Let x be the same number of lanterns sold in each shop.

Shop 1 sold 1/2 of its lanterns; this 1/2 is x; 
       hence, the original amount of lanterns in shop 1 was 2x.
       2x-x = x is the amount of lanterns remained in shop 1.

Shop 2 sold 2/3 of its lanterns; this 2/3 is x; 
       hence, the original amount of lanterns in shop 2 was {{{(3/2)x}}}.
       {{{(3/2)x-x}}} = {{{(1/2)x}}} is the amount of lanterns remained in shop 2.

Shop 3 sold 3/4 of its lanterns; this 3/4 is x; 
       hence, the original amount of lanterns in shop 3 was {{{(4/3)x}}}.
       {{{(4/3)x-x}}} = {{{(1/3)x}}} is the amount of lanterns remained in shop 3.

The three shops had a total of 330 lanterns left. It gives us this equation

x + {{{(1/2)x}}} + {{{(1/3)x}}} = 330.

Simplify and find x

x + {{{x/2}}} + {{{x/3}}} = 330

{{{(6x)/6}}} + {{{(3x)/6}}} + {{{(2x)/6}}} = 330

{{{(11x)/6}}} = 330

x = {{{330*(6/11)}}} = 30*6 = 180.

Each shop sold 180 lanterns.    <U>ANSWER</U>
</pre>

<H3>Problem 2</H3>Jamie spent &nbsp;3/4 &nbsp;of her money on &nbsp;3 &nbsp;books of the same price. 
Grace spent &nbsp;5/9 &nbsp;of her money on &nbsp;2 &nbsp;diaries. 
They were each left with an equal amount of money. 
One book costs &nbsp;$3.30 &nbsp;more than one diary. 
How much money did they have altogether?

<B>Solution</B>

<pre>
Let x be the price of one diary, in dollars.

Then the price of one book is (x+3.30) dollars, according to the problem.

Jamie spent 3/4 of her money and bought 3 books.

Hence, one book costs 1/4 of Jamie's money.

But 1/4 of Jamie's money is exactly the amount which left in Jamie's possession..

So, we conclude that Jamie left (x+3.30) dollars.

+-----------------------------------------------------------------------------+
    |   OK, it is fine.  So, (x+3.30) dollars is the left side of the equation,   |
    |             which we are going to built to solve the problem.               |
    +-----------------------------------------------------------------------------+

Now let's calculate on the Grace side.

Grace spent 2x dollars, which is  5/9 of her starting money.  Hence, Grace's starting money was

{{{(9/5)*(2x)}}} = {{{(18/5)x}}}.

From it, we should take  4/9  of  {{{(18/5)x}}}  to calculate the part of Grace's money which left.  So,

Grace's money left =  {{{(4/9)*(18/5)x}}} = {{{((4*2)/5)x}}} = {{{(8/5)x}}}

+----------------------------------------------------+
    |   Thus, Grace's money left is  (8/5)*x  dollars.   |
    |   It is the right side of our equation for         |
    |                money left.                         |
    +----------------------------------------------------+

And now the equation for  <U>money left</U>  is

x + 3.30 = {{{(8/5)x}}}.    ( ! ! ! - here three exclamation signs express my delight)

To solve, multiply by 5 both sides of this equation

5x + 5*3.30 = 8x

5*3.30 = 8x - 5x

5*3.30 = 3x

x = {{{(5*3.30)/3}}} = 5*1.10 = 5.50.

So, one diary costs 5.50 dollars and one book costs $3.30 more,  or  5.50+3.30 = 8.80 dollars.

Now we are in the finish line.

Jamie spent 3*8.80 = 26.40 dollars.  It is 3/4 of her starting money.

Hence, Jamie's starting money was  {{{(4/3)*26.40}}} = 35.20 dollars.

Grace spent 2*5.50 = 11 dollars.  It is 5/9 of her starting money.

Hence, Grace's starting money was  {{{(9/5)*11}}} = 19.80 dollars.

Thus the starting money of two girls, altogether, was  35.20 + 19.80 = 55 dollars.

<U>ANSWER</U>.  Initially, the girls had  55 dollars, altogether.
</pre>

This problem is nice.  &nbsp;It teaches to think and forces to think.

My other additional lessons on solving single linear equations and word problems in one unknown in this site (section 2) are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/One-more-lesson-on-solving-problems-by-the-backward-method.lesson>One more lesson on solving problems by the backward method</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/Choose-an-unknown-variable-in-a-rational-way-for-your-problem.lesson>Choose an unknown variable in a rational way for your problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/Solving-problems-in-three-unknowns-using-only-one-equation.lesson>Entertainment problems on finding three unknowns using only one equation</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/OVERVIEW-of-my-additional-lessons-on-solving-single-linear-equations-and-word-problems-in-one-unknown.lesson>OVERVIEW of my additional lessons on solving single linear equations and word problems in one unknown</A>