# 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
7 years, 3 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


- 7 years, 3 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 $3 \times 5 \times 4 = 60$.

- 7 years, 3 months ago

My solution, word-for-word

- 7 years, 3 months ago

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

- 7 years, 3 months ago

(a)...60

- 7 years, 3 months ago

Good!

- 7 years, 3 months ago

The answer is definitely A) 60

- 7 years, 3 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


- 7 years, 3 months ago

(c)...70 looks correct here

- 7 years, 3 months ago

Yeah, I also got that.

- 7 years, 3 months ago

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

- 7 years, 3 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 $(1/3) *\sqrt{68}$

- 7 years, 3 months ago

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

- 7 years, 3 months ago

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

- 7 years, 3 months ago

Thanks!

- 7 years, 3 months ago

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

- 7 years, 3 months ago

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

- 7 years, 3 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 $3\sqrt{5} + 5$

- 7 years, 3 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=$\sqrt{5}$

- 7 years, 3 months ago

- 7 years, 3 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


- 7 years, 3 months ago

(d)..6

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

- 7 years, 3 months ago

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

- 7 years, 3 months ago

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

- 7 years, 3 months ago

i have edited it now..

- 7 years, 3 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 $155, 245, 335, 425, 515, 605$.

- 7 years, 3 months ago

It is no details. It is just an example!

- 7 years, 3 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


- 7 years, 3 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 $\frac{5}{108}$

- 7 years, 3 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

- 7 years, 3 months ago

Yeah, that's correct!

- 7 years, 3 months ago