SMO 2019 Junior Rd 2 Namelist

SMO 2019 Junior Rd 2 Questions

SMO 2019 Junior Questions

SMO 2019 (Junior Section) Answers

1. E

2. E

3. E

4. C

5. D

6. 360

7. 120

8. 9

9. 99

10. 14

11. 8

12. 12

13. 48

14. 201

15. 50

16. 8

17. 1010

18. 86

19. 6

20. 13

21. 73

22. 8

23. 12

24. 105

25. 20

## 61 comments:

How would u rate this year’s smo

Harder or easier than 2018

Will 12 get me bronze or silver

About the same difficulty. 12 should be a bronze

Hey I got 3 will I get into round 2?

Will I get into round 2 with 13

U r meme

What do u think is this yr cutoff point

Im not sure i got 2.5 marks

Hey, would 14 get me a silver? Also, what do you think the cut off is going to be this year?

I got 19, will i get gold or silver?

hey what will 5 get me

Is 20 enough to get gold

is 0/25 able to get silver?

I believe that is good.

Gold definitely.

Would 15 or 16 get me to round 2?

The cutoff point is 16 last year.

would 5 get me an honourable mention

would 15 be enough for round 2?

so many carelesssssssssssssssss

hey i got 7 what will i get

Q12 should be 11

A is $8 with 4 toys

B is $3 with 3 toys

They said that there were same amount of toys in A and B

I honestly don't know what you will attain but I personally considered it to be slightly more perplexing. I established way too many careless mistakes I attained 8 correct answers, could have procured 21 correct answers if they weren't 1 digit off. : C

Is 8 enough to get bronze or go to second round?

Would 11 get me a bronze?

A and B have the same amount of toys.

When do the results come out? Would 18 be enough to get into the next round?

How did you get the answers?

I got 16 or 17. Can I get into the next round? Can I get a silver?

Qn 10 answer shld be 15 right?

15*2+3*4+7*5=77

15+3+7=25

More than 3 gifts for each type so you have to buy at least four $4 gifts.. I read the question wrongly too hehe

is 6 enough for honourable mention?

When will the result be posted?

Q10: can't have only 3 $4 gifts

isn't the answer to q6 540

q6 is 360 as the sum of interior angles in the pentagon is 540 degrees, which is 180*5 (5 triangles) minus the 10 unknown angles, which will result in 360 degrees

will 17/25 get me into r2

Can anyone help to solve question 20? Thanks

May I ask if I can get a Bronze with 9? Thank you! (I'm so desperate to get a Bronze)

Question 20

Y= (x^2 + 2^2)^1/2 + ((12-x)^2 + 3^2)^1/2

Minimum Y = ((2 + 3)^2 + (12-x+x)^2)^1/2

= (5^2 + 12^2)^1/2

= 13

Will 9 give me bronze or honourable mention? And shouldnt q2 be c, last year smo junior paper 2nd question is the same as this year second question? Thank you for answering

Is the SMO 2019 result list out?

The smo results are out!

What is the cut-off mark to the 2nd round in this year?

I got 15/16 but did not get to the 2nd round. What is the cut off point this year?

The cut off was 14.....

How do you do question 24?

The question is not the same its like 1 word different

I just took the SMO Round 2. It was very hard. Can someone help me crack question 2 and 4 on the Round 2 exam! Thanks in advance.

@Sean

For question 2, the answer is that he would not be able to separate it into 315 piles but my friends and i had different workings. my

working was that 315 = 256 + 32 + 16 + 8 + 2 + 1. But since any pile would not be able to begin with one as 1 is an odd number, he cannot divide it into 315 piles

What were your answers for Q1, 3, and 5.

I was not able to do Q4 either.

@Ava

My solution for Q1 involves coordinate geometry, so I probably got deducted some points. For 3, I just checked n>m and n<m and n=m using inequalities. got (5,3), (3,5) and (1,1) as my answers. For q5 I got n^2.

I got it! The following is the solutions for Q4:

(2)^1/2*a^3 + 3/(ab - b^2) is larger or equal to (2)^1/2*a^3 + 3/(a^2/4)=

(2)^1/2*a^3 + 12/a^2 = ((2)^1/2)/2*a^3 + ((2)^1/2)/2*a^3 + 4/a^2 + 4/a^2 + 4/a^2 is larger or equal to five times the fifth root of each term in the preceding sum, which is equal to 5(32)^1/5 = 10. Thus, (2)^1/2*a^3 + 3/(ab - b^2) is always larger or equal to 10. Equality occurs when all the terms are equal, which is the solution to ((2)^1/2)/2*a^3 = 4/a^2 which is a = root 2. Because b must be so that the original sum is as low as possible, b = a/2 = (root 2)/2 = 1/(root 2).

Q1: the centroid G of triangle ABC is also the orthocentre of triangle ACE. Thus, the line GE meets AC at 90 degrees, so you can deduce that GE // BC. Therefore, using similar triangles, AE : EB = AG : GD = 2 : 1

are results out?

Results are out!!!!

I cannot find the results here: http://sms.math.nus.edu.sg/Competitions/Results.aspx

Where is it?

My solution for Q2 at round 2:

To be divisible to unit (one marble) piles, the number of marble in a pile should be a power of 2, e.g. 2, 4, 8, 16, etc.

Since initially, all three piles have uneven number of marbles, in the first step we can only merge two of them. There three ways to do that, e.g. 81+115 and 119, 81 and 115+119, 81 + 119 and 115.

In each way, the two piles obtained after the first step have a common divisor differ from a power of 2 (2^n). For example, in the first way, we have 196 and 119, and common divisor is 7.

What ever we will do (merge or divide by 2), the number of marbles in piles will be a multiplication of this common divisor (7 in previous example), what is indivisible into units piles.

are round 1 results out(hm,bronze,silver,etc)

So what did u get?

omg is this even sec 1???so hard leh

i just passed bruh that

sucksThat is

shittylow quality of question and answer bruhThe thing don't even give steps to answer lol

Proof

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