Lesson Solving entertainment problems on shortage of money

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Solving entertainment problems on shortage of money


Problem 1

Andrew and  Michael each wants to buy a sundae.  But  Andrew is  2 d ollars short,  while  Michael is  1  dollar short.
Combined,  they have exactly enough money to buy one sundae.  How many dollars does one sundae cost?

Solution

Let x represents the cost of one sundae in dollars.  Andrew is $2 short, so he has (x-2) dollars.  
Likewise, Michael has  (x-1) dollars.  Since together they have exactly enough to buy one sundae, x dollars, we can write:


    (x-2) + (x-1) = x.


Solve it for x

    2x - 3 = x,

    2x - x = 3,

       x   = 3.


ANSWER.  Each sundae costs 3 dollars.

Problem 2

Lola,  Michael and Joaquin each want to buy a sundae.  But Lola is  4  dollars short,  Michael is  3  dollars short,
and Joaquin is  1  dollar short.  Combined,  they have exactly enough money to buy one sundae.  How many dollars does one sundae cost?

Solution

Let x represents the cost of one sundae in dollars.  Lola is $4 short, so she must have (x-4) dollars.  
Likewise, Michael has  (x-3) dollars and Joaquin has (x-1).  Since together they have exactly enough to buy one sundae, x dollars, we can write:


  (x-4) + (x-3) + (x-1) = x.


Solve it for x



    3x - 8 = x

    3x - x = 8

       2x  = 8

        x  = 8/2 = 4.


ANSWER.  Each sundae costs 4 dollars.

Problem 3

Mr. Lee wants to buy  5  shirts.  If he buys  3  shirts,  he will have  $6  left.  If he gets all of them,  he will be  $10  short.
How much money does  Mr. Lee have?

Solution

In accordance with the context of the problem, all the shirts have the same price.


Let x be the price for one shirt.


Then Mr. Lee has, from one part of the condition,  3x + 6 dollars;

                  from the other part of the condition, he has  5x - 10 dollars.


It gives you THIS EQUATION


    3x + 6 = 5x - 10.


Simplify and find x


    6 + 10 = 5x - 3x

    16     = 2x

     x     = 16/2 = 8 dollars.


One shirt costs 8 dollars.


Hence, Mr Lee has  3x + 6 = 3*8 + 6 = 30 dollars.    ANSWER

Problem 4

Bob  and  John want to buy a gift.  But  Bob is  $20  short,  while  John is  $34  short.
Combined,  they are  $4  short.  What is the cost of gift ?

Solution

Let x be the price of the gift, in dollars.


Since Bob is 20 dollars short, he has  x-20 dollars;  

similarly, John has x-34 dollars.


Altogether, they have  (x-20) + (x-34) = 2x-54.


It is exactly x-4 dollars (since altogether they are 4 dollars short)

    2x- 54 = x - 4.


From this equation,

    x = 54 - 4 = 50.



ANSWER.  The gift costs 50 dollars.

Problem 5

A group of office workers had some prize money to distribute among themselves.
When all but one took  $9  each,  the last person only received  $5.
When they all took  $8  each,  there was  $12  left over.  How much had they won?

Solution

Let n be the number of workers.


We can express the total money in two different ways.


One way is      the total money = 9*(n-1) + 5

Another way is  the total money = 8*n + 12.


Since the total money is the same,


    9*(n-1) + 5 = 8n + 12


Simplify and find n


    9n - 9 + 5 = 8n + 12

    9n - 4 = 8n + 12

    9n - 8n = 12 + 4

       n    =   16.


So, there are 16 workers, and the total prize money is  8n+12 = 8*16 + 12 = 140 dollars.    ANSWER


My other lessons on solving single linear equations and word problems in one unknown are
    - HOW TO solve a linear equation
    - Simple word problems to solve using a single linear equation
    - More complicated word problems to solve using a single linear equation
    - Typical word problems to solve using a single linear equation
    - Typical problems on buying and selling items
    - Typical investment problems
    - Advanced word problems to solve using a single linear equation
    - HOW TO algebraize and solve these problems using one equation in one unknown
    - Challenging word problems to solve using a single linear equation
    - Selected word problems to solve by reducing to single linear equation
    - Solving some business-related problems
    - HOW TO solve these simple word problems MENTALLY without using equations
    - Using time equation to solve some Travel and Distance problems
    - Using price equation to solve some business related problems
    - Solving problems by the backward method
    - Solving more complicated problems by the backward method
    - OVERVIEW of lessons on solving linear equations and word problems in one unknown
in this site.


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