Lesson Entertainment problems on fractions

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Entertainment problems on fractions


Problem 1

A container can hold either  60  small boxes or  48  big boxes.
If there are already  20  small and  24  big boxes in the container,
how many more big boxes can the container still hold.

Solution 

20%2F60 = 1%2F3.  Hence, 1/3 of the container volume is just occupied by the 20 small boxes.


24%2F48 = 1%2F2.  Hence, 1/2 of the container volume is just occupied by the 24 big boxes.


Taking the sum, we see that 1/3 + 1/2 = 5/6 of the container volume is just occupied
by the 20 small and 24 big boxes; 1/6 of the container volume still remains free.


So, %28%281%2F6%29%29%2F%28%281%2F48%29%29 = 48%2F6 = 8 big boxes can be added to occupy free space.   ANSWER


Problem 2

Eleven fidget spinners cost less than  $12.  Twelve fidget spinners cost more than  $13.  How many cents does one fidget spinner cost?

Solution 

    11x < 12  ---->   x < 12%2F11 = 1.090909090. . . 


    12x > 13  ---->   x > 12%2F13 = 1.083333333. . . 


Comparing these decimals, and taking into account that the number of cents MUST BE integer,

you CONCLUDE that one fidget costs  1.09  dollars.    ANSWER


No,  it is  NOT  FOR  NOTHING  people invented decimal fractions  (!)


Problem 3

Brian had a sum of money.  He spent an equal amount of money each day.
After  6  days he had  3/5  of his money left.  3  days later he had  360  dollars left.
How much did he have at first?
Solve the problem mentally,  in your head,  without using equation  (use logical reasoning,  only).

Solution 

From the condition, you know that Brian spent  1 - 3/5 = 2/5  of his money in 6 day.


Keeping the same rate, in the next 3 days he will spend half of 2/5, which is 1/5.


So,  in 6 + 3 = 9 days  Brian will spend  2/5 + 1/5 = 3/5  of his money.


Hence, after 9 days Brian will still have  1- 3/5 = 2/5  of his money left.


Thus 2/5 of his money is 360 dollars.


Hence,  1/5  of his money is half of  360  dollars,  i.e.  180 dollars.


It means that  Brian had initially  5 times as much as 180 dollars, i.e.  900  dollars.    ANSWER

Problem 4

Find the sum of the series   18/12 + 18/20 + 18/30 + 18/42 + ... + 18/1260.

Solution

                                Step by step 

(1)  18/12 + 18/20 + 18/30 + 18/42 + ... + 18/1260 = 18*(1/12 + 1/20 + 1/30 + 1/42 + . . . + 1/1260).



(2)  Every addend in the parentheses is 

     1%2F12 = 1%2F%283%2A4%29 = 1%2F3 - 1%2F4

     1%2F20 = 1%2F%284%2A5%29 = 1%2F4 - 1%2F5

     1%2F30 = 1%2F%285%2A6%29 = 1%2F5 - 1%2F6

     1%2F42 = 1%2F%286%2A7%29 = 1%2F6 - 1%2F7

          . . . . . . . . . . . . . . . . 

     1%2F1260 = 1%2F%2835%2A36%29 = 1%2F35 - 1%2F36



(3)  Add fractions in n.2, on the left sides and on the right sides.

     On the left side, you will get the sum, which is in parentheses in n.1.

     On the right side, all interior term will cancel each other, and only two extreme terms will survive


        1/12 + 1/20 + 1/30 + 1/42 + . . . + 1/1260 = 1%2F3 - 1%2F36 = 12%2F36+-+1%2F36 = 11%2F36.



(4)  THEREFORE

        18/12 + 18/20 + 18/30 + 18/42 + ... + 18/1260 = 18%2A%2811%2F36%29 = 11%2F2.


ANSWER.  The requested sum is equal to 11%2F2 = 51%2F2 = 5.5.    ANSWER

Problem 5

The sum of the reciprocals of two numbers is  7.
The larger reciprocal exceeds the smaller one by  7/3.  Find the numbers.

Solutions

            Since the entire problem is formulated as the relations between reciprocals,
            it is natural to solve it for reciprocal.   Therefore,  I introduce reciprocal and work with them. 

Let  x and y be THE RECIPROCALS.


Then we have these equations


    x + y = 7

    x - y = 7/3.


Add equations.  You will get  2x = 7+%2B+7%2F3 = 21%2F3%2B7%2F3 = 28%2F3.

                Hence,  x = 14%2F3.



Subtract equations.  You will get  2y = 7+-+7%2F3 = 21%2F3-7%2F3 = 14%2F3.

                Hence,  y = 7%2F3.


Thus the numbers are  1%2Fx = 3%2F14  and  1%2Fy = 3%2F7.    ANSWER

Solved.

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The lesson to learn from this solution is  HOW  TO  select your starting unknowns to make the solution  AS  SIMPLE  as possible.


My other lessons on fractions in this site are
    - Using fractions to solve word problems
    - Using fractions to solve Travel problems
    - Calculations with fractions
    - Advanced problems on ratios
    - OVERVIEW of my lessons on fractions


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