Tuesday, 4 June 2019

SMO 2019 (Junior Section) Answers

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

62 comments:

Anonymous said...

How would u rate this year’s smo
Harder or easier than 2018
Will 12 get me bronze or silver

Anonymous said...

About the same difficulty. 12 should be a bronze

This is a meme said...

Hey I got 3 will I get into round 2?

Unknown said...

Will I get into round 2 with 13

Anonymous said...

U r meme

Anonymous said...

What do u think is this yr cutoff point

ROUND 2 PLX said...

Im not sure i got 2.5 marks

Ravisankar said...

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

Anonymous said...

I got 19, will i get gold or silver?

Anonymous said...

hey what will 5 get me

Anonymous said...

Is 20 enough to get gold

I'm failing math said...

is 0/25 able to get silver?

Unknown said...

I believe that is good.

Unknown said...

Gold definitely.

Anonymous said...

Would 15 or 16 get me to round 2?

Anonymous said...

The cutoff point is 16 last year.

uh said...

would 5 get me an honourable mention

Anonymous said...

would 15 be enough for round 2?
so many carelesssssssssssssssss

Anonymous said...

hey i got 7 what will i get

Anonymous said...

Q12 should be 11
A is $8 with 4 toys
B is $3 with 3 toys

Anonymous said...

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

Anonymous said...

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

Anonymous said...

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

Anonymous said...

Would 11 get me a bronze?

Anonymous said...

A and B have the same amount of toys.

Anonymous said...

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

Anonymous said...

How did you get the answers?

Anonymous said...

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

Anonymous said...

Qn 10 answer shld be 15 right?
15*2+3*4+7*5=77
15+3+7=25

Anonymous said...

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

Anonymous said...

is 6 enough for honourable mention?

Anonymous said...

When will the result be posted?

Anonymous said...

Q10: can't have only 3 $4 gifts

Anonymous said...

isn't the answer to q6 540

Anonymous said...

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

Unknown said...

will 17/25 get me into r2

Anonymous said...

Can anyone help to solve question 20? Thanks

Anonymous said...

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

Anonymous said...


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

Anonymous said...

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

Unknown said...

Is the SMO 2019 result list out?

Anonymous said...

The smo results are out!

Anonymous said...

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

Anonymous said...

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

Anonymous said...

The cut off was 14.....

Anonymous said...

How do you do question 24?

Anonymous said...

The question is not the same its like 1 word different

Sean said...

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.

Ava said...

@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.

Sean said...

@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.

Sean said...

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).

Anonymous said...

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

Anonymous said...

are results out?

Sean said...

Results are out!!!!

Anonymous said...

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

Anonymous said...

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.

Anonymous said...

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

Anonymous said...

So what did u get?

Anonymous said...

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

Anonymous said...

i just passed bruh that sucks

Anonymous said...

That is shitty low quality of question and answer bruh
The thing don't even give steps to answer lol
Proof

Anonymous said...

May I know for the first round, how many marks will lead to bronze silver and gold