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# Give my Round 1 Problems a Try!

Hi Brilliant!

I wrote the first Round of the South African Mathematics Olympiad today. I wanted to share the problems with the rest of you. I am eager to compare my answers.

STAR SOLVER: Shikhar Jaiswal, Michael Tong

Note by Mark Mottian
3 years, 7 months ago

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A four-digit even number ABCD is formed by using four of the digits 1,2,3,4,5,6 and 7 without repetition. How many even numbers less than 2014 can be formed in this way?

OPTIONS
A    60
B    90
C    120
D    150
E    180


- 3 years, 7 months ago

Clearly, $$A < 3$$. Consider $$A = 2$$. But then $$B > 0$$, so it is greater than $$2014$$. This leaves our one case, $$A = 1$$. To make our number even, we choose $$2, 4,$$ or $$6$$ as the last digit. Otherwise, we can choose whatever digit we want. This gives $$3 \times 5 \times 4 = 60$$.

- 3 years, 7 months ago

WOW Michael! Thanks for such clear answers with explanations. I think you also deserve STAR solver.

- 3 years, 7 months ago

My solution, word-for-word

- 3 years, 7 months ago

The answer is definitely A) 60

- 3 years, 7 months ago

(a)...60

- 3 years, 7 months ago

Good!

- 3 years, 7 months ago

Rectangle ABCD has sides AB = 5 and CB = 3. Points E and F lie on diagonal AC such that AE = EF = FC. What is the area of triangle BEF?

OPTIONS


A 3/2

B 5/8

C 5/2

D $$(1/2) *\sqrt{34}$$

E $$(1/3) *\sqrt{68}$$

- 3 years, 7 months ago

Note that $$[BAE] = [BEF] = [BFC]$$, since they have equal bases and the same altitude. Thus, $$[BEF] = \frac{1}{6} [ABCD] = \frac{5}{2}$$

- 3 years, 7 months ago

C)5/2 because the there are 3 same triangles which means same area.

- 3 years, 7 months ago

(c)...5/2...cause the 3 triangles have equal areas(length of altitude and base are same)

- 3 years, 7 months ago

Correct! I've awarded you the STAR SOLVER (see note description).

- 3 years, 7 months ago

Thanks!

- 3 years, 7 months ago

Points P and Q are on line segment AB on the same side of the midpoint of AB. Point P divides AB in the ratio 2:3 and Q divides AB in the ratio 3:4. If PQ = 2, then the length of the segment AB is:

OPTIONS
A    12
B    28
C    70
D    75
E    105


- 3 years, 7 months ago

$$|\frac{4}{7} - \frac{3}{5}|x = 2 \rightarrow x = 70$$

- 3 years, 7 months ago

(c)...70 looks correct here

- 3 years, 7 months ago

Yeah, I also got that.

- 3 years, 7 months ago

Three standard dice are rolled and the numbers thrown are added. The probability of getting a sum of 15 is:

OPTIONS
A    15/36
B    5/6
C    15/216
D    5/108
E    15/18


- 3 years, 7 months ago

Note the bijection between rolls that sum to $$15$$ and rolls that sum to $$6$$.

Thus, we count the probability of getting a $$6$$: Since each die must be at least one, we can count instead the number of ways to put $$3$$ balls in $$3$$ urns, namely $${5 \choose 3} = 10$$.

The total number of outcomes (assuming distinct dice) is $$6^3 = 216$$.

The desired probability is $$\frac{5}{108}$$

- 3 years, 7 months ago

(d)...5/108 looks like the correct one..

Total possibilities=6^3=216

Favourable possibilities:-(3,6,6)-'3" from here

                                            (4,5,6)-'6' from here

(5,5,5)-'1' from here


no of favourable possibilities=10

Probability=10/216=5/108

- 3 years, 7 months ago

Yeah, that's correct!

- 3 years, 7 months ago

When the tens digit of a three digit number "abc" is deleted, a two-digit number "ac" is formed. How many numbers "abc" are there such that "abc" = 9 x "ac" + 4c. For example, 245 = 9 x 25 + 4 x 5

OPTIONS
A    3
B    4
C    5
D    6
E    7


- 3 years, 7 months ago

The details of the problem are useless. We just need the algebra. $$100a + 10b + c = 9(10a + c) + 4c$$

$$100a + 10b + c = 90a + 13c$$

$$10(a+b) = 12c$$

Thus $$c = 5$$. $$a+b = 6$$, since $$a \neq 0$$ we get $$6$$ solutions, namely $$155, 245, 335, 425, 515, 605$$.

- 3 years, 7 months ago

It is no details. It is just an example!

- 3 years, 7 months ago

(d)..6

The numbers are-155,515,245,425,335,605

- 3 years, 7 months ago

I got 6 (i.e. option D) as my answer. The numbers I found were: 155, 245, 335, 425, 515 and 605.

- 3 years, 7 months ago

i have edited it now..

- 3 years, 7 months ago

oh yes my mistake! sorry....i was a bit careless there

- 3 years, 7 months ago

A right angled triangle has an area of 5. The altitude perpendicular to the hypotenuse has a length of 2. The perimeter of the triangle is:

A $$5\sqrt{5} + 5$$

B CUBEROOT(3) + 3

C $$5\sqrt{3} + 3$$

D $$3\sqrt{5} +$$$$5\sqrt{3}$$

E $$3\sqrt{5} + 5$$

- 3 years, 7 months ago

(E)....Take out length of hypotenuse using the information it comes out to be 5 units

Let one side be $$a$$

the other side=$$\sqrt{25-a^2}$$

Given:

$$\frac {1}{2}a\sqrt{25-a^2}=5$$

On solving...$$a=2\sqrt{5}$$

while the other side=$$\sqrt{5}$$

- 3 years, 7 months ago

- 3 years, 7 months ago

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