NMOS 2016 Special Round answers

NMOS 2016 Results

1. 15
2. 126
3. 10
4. 671
5. 470
6. 184
7. 144
8. 72
9. 370
10. 180
11. 17
12. 800
13. 5800
14. 689
15. 272
16. 96 (*amended)
17. 100
18. 64
19. 12
20. 57

The answers are provided by LL. If you have different answers and you are sure you are correct, please leave a message in the cbox.

2013 NMOS and 2013 SPSO Top 20 Winners

These Primary 5 students may be the future Math and Science olympians of Singapore. They have been shortlisted for a training programme conducted by NUS High School, in preparation for the International Mathematics & Science Olympiad (IMSO). The final squad, which consists of 6 official participants and two reserves from each training squad (NMOS and SPSO, a total 12 official participants), will be selected to represent Singapore in IMSO 2013. 4 students (with background colour, below) have made it into both lists, which means they have 2 chances to make it into the national team representing Singapore in IMSO 2013, which will be held from 25 - 29 November 2013, in the Philippines.

National Mathematical Olympiad of Singapore (NMOS) 2013 Top 20 (ranked by merit):

HO LI XIONG TIMOTHY, ANGLO-CHINESE SCHOOL (PRIMARY)
LU MING YUAN, RULANG PRIMARY
ETHAN GOH CHEE KIAT, TAO NAN SCHOOL
NG YANGYI ALOYSIUS, HENRY PARK PRIMARY SCHOOL
PAN YIFAN, NANYANG PRIMARY SCHOOL
CHEN XINYI, NAN HUA PRIMARY SCHOOL
ZHANG YU CHI, HENRY PARK PRIMARY SCHOOL
TAN KIN HERN, CHONGFU SCHOOL
TAN WEI HAN MAX, NANYANG PRIMARY SCHOOL
WILLIAM YAO ZHANG, XINGNAN PRIMARY SCHOOL
CHEN LEIYU, CATHOLIC HIGH SCHOOL
SHEN XIN YI, ROSYTH SCHOOL
TAN WE KEAN, CATHOLIC HIGH SCHOOL
HAN JIATAO JERRY, NANYANG PRIMARY SCHOOL
DARREN LIM TECH WOON, NAN HUA PRIMARY SCHOOL
JOVAN LIAU YI JUN, NAN HUA PRIMARY SCHOOL
JERALD SIAH CHI MING, NANYANG PRIMARY SCHOOL
LI HANTAO, NAN HUA PRIMARY SCHOOL
FONG JUN YI, NANYANG PRIMARY SCHOOL
NATHANIEL ANG, ANGLO-CHINESE SCHOOL (PRIMARY)
BOO TSE YANG LUCAS, NANYANG PRIMARY SCHOOL

Singapore Primary Science Olympiad (SPSO) Top 20 (ranked by alphabetical order of schools):

JAYABASKARAN JAYANTH, ANGLO-CHINESE SCHOOL (PRIMARY)
LEONG EU-SHAUN, ANGLO-CHINESE SCHOOL (PRIMARY)
SEAN LEONG KAR WENG, ANGLO-CHINESE SCHOOL (PRIMARY)
TAY RAY EE, CATHOLIC HIGH SCHOOL
WONG ZHIAN, JOHN, CATHOLIC HIGH SCHOOL
ZHENG HONGYI, CATHOLIC HIGH SCHOOL
LIN ZIWEI, DEWEY, HENRY PARK PRIMARY SCHOOL
CHEN XINYI, NAN HUA PRIMARY SCHOOL
VICTOR NG WEI YUAN, NAN HUA PRIMARY SCHOOL
WU JIAYANG, NAN HUA PRIMARY SCHOOL
YAM HONG MENG, NAN HUA PRIMARY SCHOOL
CHARLES WINDLE, NANYANG PRIMARY SCHOOL
HAN JIATAO JERRY, NANYANG PRIMARY SCHOOL
TAN WEI HAN MAX, NANYANG PRIMARY SCHOOL
CHIA JING KANG, ROSYTH SCHOOL
LAI XIONG XING DANIEL, ROSYTH SCHOOL
ETHAN GOH CHEE KIAT, TAO NAN SCHOOL
GAN CHIU HOW, TAO NAN SCHOOL
JIANG BOWEN, TAO NAN SCHOOL
KOH JIN HAO, TAO NAN SCHOOL

Update (IMSO 2013 Results):

NMOS 2011 Round 1 Answers

This is posted by Lim Li
(Note: I did not take part in this competition as I am P6!)

NMOS 2011 R1 Answers
1. 220
2. 8
3. 30
4. 1006
5. 8
6. 4
7. 46
8. 46
9. 1003
10. 67
11. 210
12. 490
13. 30
14. 52 (answer given by someone from the c-box)
15. 20
16. 84
17. 125
18. 5
19. 80
20. 60
21. 21
22. 233
23. 2008
24. 19
25. 7
26. 3
27. 50
28. 15
29. 81
30. 72

NMOS 2010 Special Round Qn 5

Question 5: A company could use any of the 3 builder (Anthony, Bob and Charles) to build a wall. Anthony would take A days to build the wall. Likewise, Bob and Charles would take B days and C days to build the wall respectively.
Anthony and Bob take 16.8 days to complete a wall.
Anthony and Charles take 21 days to complete a wall.
Bob and Charles take 28 days to complete a wall.

Solution:
Anthony builds 1 wall in A days. Hence, Anthony builds 1/A wall in 1 day.
Similarly, Bob builds 1/B wall in 1 day, and Charles builds 1/C wall in 1 day.
In 1 day, Anthony and Bob together will build (1/A + 1/B) wall. In 16.8 days, Anthony and Bob will build [16.8x(1/A + 1/B)] wall which is 1 wall since Anthony and Bob complete the wall in 16.8 days. Hence, we get the following equations:

16.8(1/A + 1/B) = 1
21(1/A + 1/C) = 1
28(1/B + 1/C) = 1
Implying..
1/A + 1/B = 1/16.8 -------- (1)
1/A + 1/C = 1/21 ---------- (2)
1/B + 1/C = 1/28 ---------- (3)

(1)+(2)-(3)
2(1/A) = 1/16.8 + 1/21 - 1/28 = 1/14
Therefore, 1/A = 1/28, hence, A = 28.
From (1), 1/B = 1/16.8 - 1/28 = 1/42. Hence, B = 42.
From (2), 1/C = 1/21 - 1/28 = 1/84. Hence, C = 84.

Answer: A+B+C=154

NMOS 2010 Special Round Answers

1. 20
2. 2
3. 15
4. 216
5. 154
6. 54
7. 25
8. 12
9. 38
10. 75
11. 542
12. 48
13. 108
14. 12
15. 65
16. 24
17. 40
18. 25
19. 4
20. 62

Note: many answers were contributed by Lim Min and Lim Jeck.
Lim Li - if you have different answers and you are sure that you are correct, please let me know.