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% >>
(1991: Grades 6-7 Problem #29, page 9
and Grade 8 Problem #22, page 11)
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
From the problem statement we write the equations :
$B+F=\left (1-\frac{t}{100} \right ) \cdot (b+f) \tag {i}$
$f=\frac{g}{100} \cdot (b+f) \tag{ii}$
$F=\frac{h}{100} \cdot (B+F) \tag{iii}$
Equations (ii) and (iii) lead to the relations :
$(100-g)\cdot f=g \cdot b \tag{b,f}$
$(100-h) \cdot F=h \cdot B \tag{B,F}$
Substituting $B\underset{(B,F)}{=}\frac{100-h}{h}\cdot F$ and $b\underset{(b,f)}{=}\frac{100-g}{g}\cdot f$ in equation (i) we have
$\frac{100-h}{h}\cdot F +F=\frac{100-t}{100}\cdot \left ( \frac{100-g}{g}\cdot f+f\right )\;\Leftrightarrow$
$\Leftrightarrow\;\frac{100}{h}\cdot F=\frac{100-t}{100}\cdot \frac{100}{g} \cdot f\;\Rightarrow\; F=\frac{(100-t)\cdot h}{g\cdot 100} \cdot f$ that is, equation (F).
Substituting $F\underset{(B,F)}{=}\frac{h}{100-h}\cdot B$ and $f\underset{(b,f)}{=}\frac{g}{100-g}\cdot b$ in equation (i) we have
$B+\frac{h}{100-h} \cdot B=\frac{100-t}{100}\cdot \left (b+\frac{g}{100-g}\cdot b\right )\;\Leftrightarrow$
$\Leftrightarrow\;\frac{100}{100-h}\cdot B=\frac{100-t}{100}\cdot \frac{100}{100-g} \cdot b\;\Rightarrow\; B=\frac{(100-t)\cdot (100-h)}{(100-g)\cdot 100} \cdot b$ that is, equation (B).
$\blacksquare \;\blacksquare$
In the initial problem : $t=10,\;g=50,\;h=55$ so
$F=\frac{90\cdot 55}{100\cdot 50}\cdot f=\frac{99}{100}\cdot f=(1-1\text{%})\cdot f$,
$B=\frac{90\cdot 45}{100\cdot 50}\cdot b=\frac{81}{100}\cdot b=(1-19\text{%})\cdot b.$
Other EXAMPLES :
- If the total number of students decreases by 12% and the number of girls increases from 44% to 50% then
$F=\frac{(100-12)\cdot 50}{44\cdot 100}\cdot f=\frac{88\cdot 50}{44\cdot 100}\cdot f=f$,
so the percentage of girls remains the same, and
$B=\frac{(100-12)\cdot (100-50)}{(100-44)\cdot 100}\cdot b=\frac{44}{56}\cdot b\approx 0,79\cdot b=(1-21\text{%})\cdot b$
so the percentage of boys decreases by about 21%.
- If the total number of students decrease by 8% and the number of girls increases from 46% to 54% then
$F=\frac{(100-8)\cdot 54}{46\cdot 100}\cdot f=\frac{92\cdot 54}{46\cdot 100}\cdot f=\frac{108}{100}\cdot f=(1+8\text{%})\cdot f$
so the percentage of girls increases by 8%. ATTENTION !! : it is a coincidence that 54%-46%=8%.
$B=\frac{(100-8)\cdot (100-54)}{(100-46)\cdot 100}\cdot b=\frac{92\cdot 46}{54 \cdot 100}\cdot b=\frac{529}{675}\cdot b\approx 0,78 \cdot b=(1-22\text{%})\cdot b$
so the percentage of boys decreases by about 22%.
- If the total number of students decrease by 48% (a bit of a bizarre situation) and the number of girls increases from 52% to 64% then
$F=\frac{(100-48)\cdot 64}{52\cdot 100}\cdot f=\frac{52\cdot 64}{52\cdot 100}\cdot f=\frac{64}{100}\cdot f=(1-36\text{%})\cdot f$
so the percentage of girls decreases by 36%;
$B=\frac{(100-48)\cdot (100-64)}{(100-52)\cdot 100}\cdot b=\frac{52\cdot 36}{48\cdot 100}\cdot b=\frac{39}{100}\cdot b=(1-61\text{%})\cdot b$
so the percentage of boys decreases by 61%.
Remarks CiP
- If the total number of students increases instead of decreasing then we can juggle with negative percentage values, but the formulas remain the same (as in the second example, the number of girls)
- We have seen in some examples that the answer obtained is only approximate. It would be interesting to study the relationships between the quantities $t,\; g,\;h\;$ for which the answer is expressed as an integer percentage, i.e.
$\frac{(100-t)\cdot h}{g}\;\;,\;\;\;\frac{(100-t)\cdot (100-h)}{100-g}$
are integers.
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