Answer key for the maths mock exam 2020 to download in PDF

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1 Exercise 1:
2 Exercise 2:
3 Exercise 3:
4 Exercise 4:
5 Exercise 5:
6 Exercise 6:
7 Exercise 7:

The answer key for the subject of brevet de maths 2020 .This correction will allow you to spot your mistakes if you have done the subject in its entirety.

Exercise 1:

Question 1: Answer B.

Question 2: Answer C.

Question 3: Answer B.

Question 4: Answer C.

Question 5: Answer A.

Exercise 2:

1. 2530 is not divisible by 19 so the chocolatier cannot make 19 packs.

2. 2622=1×2530 +92

2530= 27×92+ 46

92=2×46 + 0

The gcd is the last non-zero remainder in Euclid’s algorithm so gcd(2622,2530)=46.

The chocolatier will be able to make a maximum of 46 packages.

2622 : 46 = 57 and 2530 : 46 = 55

Each package will consist of 57 Easter eggs and 55 chocolate fish.

Exercise 3:

There is 75% of sunshine so in June, July and August there are 30+31+31=92 days,

It will be sunny $\frac{75\times ,92}{100}=69$ days.

La paillotte:

69×500+23×50-2500×3 =28 150 €.

The store:

69×350+23×300-92×60=25 530 €.

Conclusion: the benefit of the straw hut on the beach being the highest, Peio will choose the beach.

Exercise 4:

1. $V_{SABC}=\frac{B\times ,h}{3}=\frac{1}{3}\times ,\frac{7,5\times ,7,5}{2}\times ,15=140,625\,cm^3$

The volume of the pyramid is 141 .

a.The section SMN is a reduction of the triangle ABC so it is a right triangle isosceles in S’.

b. The reduction coefficient is $k=\frac{6}{15}$ .

$S'N=k\times ,AC=\frac{6}{15}\times ,7,5=3\,cm$ .

3. Let’s calculate the volume of the plug:

$V_{SS'MN}=,(\frac{6}{15},,)^3\times ,V_{SABC}=\frac{8}{125}\times ,140,625=9\,cm^3$

therefore
the maximum volume of perfume is :
141-9=132 .

Exercise 5:

1. The probability that the candidate will reach the treasure room is $\frac{1}{5}$ .

2.b. There are 6 possibilities out of 8 that he will win at least 200 e; the probability is therefore $\frac{6}{8}=\frac{3}{4}$ .

3. He has 3 possibilities out of 8 to win nothing so the probability is $\frac{3}{8}$ .

Exercise 6:

Here is the figure built with Geogebra software.

a) The triangle ABC is rectangular: TRUE.
b) The segment [BC] measures 10 cm: FALSE (use the Pythagorean theorem).
c) The angle $\widehat{AOC}$ measures 60°: TRUE.
d) The area of triangle ABC is $18\sqrt{3}$ cm² : TRUE.
e) The angle $\widehat{BOC}$ measures 31° : FALSE.

Exercise 7:

Let x : the length of the small side of the hexagon and y : the length of the large side of the hexagon.

The sum of the perimeters of the three small triangles is equal to the perimeter of the remaining gray hexagon.

We thus obtain the equation :

3x+3y=9x

3y=9x-3x

3y=6x

y=(6/3)x

y=2x

So the length of the long side of the hexagon is twice the length of the short side of the hexagon and therefore twice the length of the side of the small equilateral triangles.

The length of the large equilateral triangle is 6 cm,

so we have :

2x+y=6

2x+2x=6

4x=6

x=6/4=1.5 cm.

Conclusion: the side measure of the small triangles is 1.5 cm.

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