A Problem from the KANGAROO Contest // Un problème du concours KANGOUROU

 KANGAROO is a widely known mathematics competition. Here we discuss a Percentage Problem, taken from Book, which you will soon be able to find scanned here.

                          <<At "Frank Einstein" College, the number of students decreased by

                                 10% in one year, but the percentage of girls increased from 50%

                                  to 55%. The number of girls...

        A. grew up with 0,5%     B. grew up with 1%     C.  remained the same

        D. decreased by  1%     E.  decreased by 0,5%  >>

(page 11, Problem 22)

ANSWER CiP  :  D

A complete answer:

 the number of girls decreased by 1% and the number of boys decreased by 19%

                            Solution CiP 

                       Let's denote by  $b,\;f$  respectively the number of boys and girls at the beginning of the year and by $B,\;F$ their number at the end. From the text of the problem it follows that we have the equations:

$B+F=(1-10\text {%}) \cdot (b+f) \tag{1}$

$f=50\text{%} \cdot (b+f) \tag{2}$

$F=55\text{%} \cdot (B+F) \tag{3}$

From (2) we obtain  $f=b$  (which even grandma would have deduced). So from (1) we can write

$B+F=\frac{90}{100} \cdot 2f \tag{4}$

which substituted into (3) gives us

$F=\frac{55}{100} \cdot \frac{90}{100}\cdot 2f=\frac{99}{100} \cdot f=(1-1\text{%}) \cdot f$

So the number of girls decreased by 1%, hence the answer D.

     Further, from  $B+F\overset{(1)}{=}\frac{90}{100} \cdot 2f\;$, replacing  $f$  with  $b$  and  $F$  with  $\frac{99}{100} \cdot b$  we obtain  $B+\frac{99}{100}\cdot b=\frac{90}{100}\cdot b \Rightarrow $

$\Rightarrow B=\left (\frac{180}{100}-\frac{99}{100} \right )=\frac{81}{100}\cdot b=(1-19\text{%})\cdot b$

which completes the answer.

$\blacksquare$

         We are trying to formulate a more general problem :

    The total number of students decreased by  $t \text{%}$  and

 the percentage of girls increased from  $g\text{%}$  to  $h\text{%}$.      

                                      Determine the percentages by which the number of

                                   girls and boys changed.

  

ANSWER CiP (with the previous notations)

$F=\frac{(100-t)\cdot h}{g\cdot 100} \cdot f \tag{F}$

$B=\frac{(100-t)\cdot (100-h)}{(100-g) \cdot 100} \cdot b \tag {B}$

 

                         Solution CiP

$\blacksquare \;\blacksquare$

remark

(to be continue)