Answer key for the maths brevet France 2016

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The answer key to the June 2016 math patent topic.

Exercise 1:

$\frac{27}{100}$
2. $\frac{27}{27+38}=\frac{27}{65}$
Plant A: $\frac{27}{500}\times ,100=5,4$

Plant B: $\frac{38}{500}\times ,100=7,6$

Only plant A has a satisfactory control?

Exercise 2:

Program A :
Program B:
Program A = Program B

$x=-\frac{8}{5}=-1,6$

Exercise 3:

Figure 1: ABC is a right triangle and $AB=\sqrt{12^2-6^2}=10,4\,cm$ .

Figure 2: ABC is a right triangle and

$sin53%C2%B0=\frac{AB}{36}$ so $AB=36\times ,sin53=28,8\,cm$

Figure 3:

$P=\pi\times ,D=154$

$D=\frac{154}{\pi}=49\,cm$

Exercise 4:

$54\times ,(1-\frac{30}{100})=54*0,7=37,8$ €
a. ${\color{DarkRed},=(B_1*30)/100}$

b. ${\color{DarkRed},=B_1-B_2}$

3. $x\times ,0,7=42$

$x=\frac{42}{0,7}=60$ €

Exercise 5:

$A_{APS}=\frac{30\times ,18}{2}=15*18=270\,m^2$

It will take 2 bags or €

Property: (AS) // (RC) because two lines perpendicular to a third are parallel to each other.

2. We know that $A\in,(RP);S\in(RC);(AS)//\,(RC%C2%B0)))$

according to the direct part of Thales’ theorem, we have :

$\frac{PA}{PR}=\frac{AS}{RC}$

$\frac{30}{40}=\frac{18}{RC}$

so $RC=\frac{18*40}{30}=24\,m$

$A_{skatepark}=A_{PRC}-A_{PAS}\\=\frac{40*24}{2}-270\\=480-270\\=210\,m^2$

Exercise 6:

Part 1:

Piece 1 = 8 cm and piece 2 = 12 cm.a square of side 8:4=2cm and an equilateral triangle of side 12 : 3 = 4 cm.
$2*2=4\,cm^2$
$AH=\sqrt{4^2-2^2}=3,5\,cm$ with AH the height of the equilateral triangle.

We obtain $\frac{4*3,5}{2}=7\,cm^2$ .

Part 2:

Let x : the length of the piece 1.

$Aire_carre=(\frac{x}{4})^2=\frac{x^2}{16}$

2. a. $\simeq,3\,cm$

b. $\simeq,8,6\,cm$

Exercise 7:

$V_{vase}=(9-0,2*2)^2*(21,7-1,7)=1479,2\,cm^3$

$V_{150billes}=150\times ,\frac{4}{3}\times ,\pi\times ,0,9^3\simeq,458\,cm^3$

$V_{vide}=1479,2-458=1021,2cm^3=1,021L$

Yes, he can add a liter of colored water without risking an overflow.

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