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
5 years, 8 months ago

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1 vote

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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

Mark Mottian - 5 years, 8 months ago

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Clearly, A<3A < 3. Consider A=2A = 2. But then B>0B > 0, so it is greater than 20142014. This leaves our one case, A=1A = 1. To make our number even, we choose 2,4,2, 4, or 66 as the last digit. Otherwise, we can choose whatever digit we want. This gives 3×5×4=603 \times 5 \times 4 = 60.

Michael Tong - 5 years, 8 months ago

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My solution, word-for-word

Sharky Kesa - 5 years, 8 months ago

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

Mark Mottian - 5 years, 8 months ago

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(a)...60

Shikhar Jaiswal - 5 years, 8 months ago

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Good!

Mark Mottian - 5 years, 8 months ago

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The answer is definitely A) 60

Kahsay Merkeb - 5 years, 8 months ago

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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

Mark Mottian - 5 years, 8 months ago

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(c)...70 looks correct here

Shikhar Jaiswal - 5 years, 8 months ago

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Yeah, I also got that.

Mark Mottian - 5 years, 8 months ago

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4735x=2x=70|\frac{4}{7} - \frac{3}{5}|x = 2 \rightarrow x = 70

Michael Tong - 5 years, 8 months ago

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

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

Mark Mottian - 5 years, 8 months ago

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

Shikhar Jaiswal - 5 years, 8 months ago

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

Mark Mottian - 5 years, 8 months ago

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@Mark Mottian Thanks!

Shikhar Jaiswal - 5 years, 8 months ago

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

Kahsay Merkeb - 5 years, 8 months ago

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Note that [BAE]=[BEF]=[BFC][BAE] = [BEF] = [BFC], since they have equal bases and the same altitude. Thus, [BEF]=16[ABCD]=52[BEF] = \frac{1}{6} [ABCD] = \frac{5}{2}

Michael Tong - 5 years, 8 months ago

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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 55+55\sqrt{5} + 5

B CUBEROOT(3) + 3

C 53+35\sqrt{3} + 3

D 35+3\sqrt{5} + 535\sqrt{3}

E 35+53\sqrt{5} + 5

Mark Mottian - 5 years, 8 months ago

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

Let one side be aa

the other side=25a2\sqrt{25-a^2}

Given:

12a25a2=5\frac {1}{2}a\sqrt{25-a^2}=5

On solving...a=25a=2\sqrt{5}

while the other side=5\sqrt{5}

Shikhar Jaiswal - 5 years, 8 months ago

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I got the same answer!

Mark Mottian - 5 years, 8 months ago

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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

Mark Mottian - 5 years, 8 months ago

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(d)..6

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

Shikhar Jaiswal - 5 years, 8 months ago

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

Mark Mottian - 5 years, 8 months ago

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

Shikhar Jaiswal - 5 years, 8 months ago

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@Mark Mottian i have edited it now..

Shikhar Jaiswal - 5 years, 8 months ago

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The details of the problem are useless. We just need the algebra. 100a+10b+c=9(10a+c)+4c100a + 10b + c = 9(10a + c) + 4c

100a+10b+c=90a+13c100a + 10b + c = 90a + 13c

10(a+b)=12c10(a+b) = 12c

Thus c=5c = 5. a+b=6a+b = 6, since a0a \neq 0 we get 66 solutions, namely 155,245,335,425,515,605155, 245, 335, 425, 515, 605.

Michael Tong - 5 years, 8 months ago

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It is no details. It is just an example!

Sagnik Saha - 5 years, 8 months ago

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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

Mark Mottian - 5 years, 8 months ago

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Note the bijection between rolls that sum to 1515 and rolls that sum to 66.

Thus, we count the probability of getting a 66: Since each die must be at least one, we can count instead the number of ways to put 33 balls in 33 urns, namely (53)=10{5 \choose 3} = 10.

The total number of outcomes (assuming distinct dice) is 63=2166^3 = 216.

The desired probability is 5108\frac{5}{108}

Michael Tong - 5 years, 8 months ago

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(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

Shikhar Jaiswal - 5 years, 8 months ago

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Yeah, that's correct!

Mark Mottian - 5 years, 8 months ago

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