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

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Upper level word problems to solve using a single linear equation


Problem 1

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

Solution

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 %283%2F2%29x.
       %283%2F2%29x-x = %281%2F2%29x 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 %284%2F3%29x.
       %284%2F3%29x-x = %281%2F3%29x 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 + %281%2F2%29x + %281%2F3%29x = 330.


Simplify and find x

    x + x%2F2 + x%2F3 = 330

    %286x%29%2F6 + %283x%29%2F6 + %282x%29%2F6 = 330

    %2811x%29%2F6 = 330

    x = 330%2A%286%2F11%29 = 30*6 = 180.

 
Each shop sold 180 lanterns.    ANSWER

Problem 2

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

Solution

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  

    %289%2F5%29%2A%282x%29 = %2818%2F5%29x.


From it, we should take  4/9  of  %2818%2F5%29x  to calculate the part of Grace's money which left.  So,

    Grace's money left =  %284%2F9%29%2A%2818%2F5%29x = %28%284%2A2%29%2F5%29x = %288%2F5%29x


    +----------------------------------------------------+
    |   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  money left  is

    x + 3.30 = %288%2F5%29x.    ( ! ! ! - 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 = %285%2A3.30%29%2F3 = 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  %284%2F3%29%2A26.40 = 35.20 dollars.



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

Hence, Grace's starting money was  %289%2F5%29%2A11 = 19.80 dollars.



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


ANSWER.  Initially, the girls had  55 dollars, altogether.

This problem is nice.  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
    - One more lesson on solving problems by the backward method
    - Choose an unknown variable in a rational way for your problem
    - Entertainment problems on finding three unknowns using only one equation
    - OVERVIEW of my additional lessons on solving single linear equations and word problems in one unknown


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