NSM Seminar

DrYeap Ban Har
Consultant, New Syllabus Mathematics

Junyuan Secondary School, Singapore

1st Block 4 x 2 x 1
2nd Block 4 x 2 x 3
3rd Block
Victory Christian International School, Manila

10
76
10
3𝑥+10+𝑥+10=76

x
x
x
10
76
x
10
3𝑥+10+𝑥+10=76

4𝑥+20=76

x
x
x
10
76 – 20
x
10
3𝑥+10+𝑥+10=76

4𝑥+20=76

4𝑥=56

𝑥= 564=14

King Solomon Academy, London

Method 1
21 + 23 + 25 = 69
19 + 21 + 23 = 63
17 + 19 + 21 = 57
Method 2
Note that the mean of the three numbers is the middle number
57 : 3 = 19
The other numbers are 17 and 21.
Method 3
Make a list 1, 3, 5, 7, 9, …, 15, 17, 19, 21, 23, 25, 27, …
x
𝑥+𝑥+2+𝑥+4=57

Method 3
x
2
3𝑥+6=57

𝑥=513=17

3𝑥=57 −6=51

x
2
2
The numbers are 17, 19and 21.

𝑊h𝑒𝑛 𝑡=0, 20𝑡−5𝑡2=0

𝑡(20−5𝑡)=0

𝑡2−4𝑡+3=0

𝑊h𝑒𝑛 20𝑡−5𝑡2=0

𝑊h𝑒𝑛 20𝑡−5𝑡2=15

𝑡=0 𝑜𝑟𝑡=4

(𝑡−3)(𝑡−1)=0

𝑡=3 𝑜𝑟 𝑡=1

𝑊h𝑎𝑡 𝑖𝑠 𝑡h𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 20𝑡−5𝑡2?

189𝑥

189𝑥+6

189𝑥−189𝑥+6=960

21𝑥−21𝑥+6=160

21𝑥+6−21𝑥𝑥(𝑥+6)=160

126𝑥(𝑥+6)=160

7560=𝑥(𝑥+6)

𝑥2+6𝑥−7560=0

𝑥2+6𝑥−7560=0

𝑥−84𝑥+90=0

𝑥=84

x cannot be a negative value because it is an average speed.
87=189𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛=18987×60 𝑚𝑖𝑛

𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛=130min (𝑡𝑜 𝑡h𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑚𝑖𝑛𝑢𝑡𝑒)

Lesson 41
10.10.11

Lesson 41
10.10.11
a
b
To prove that angle at centre is equal to 2a + 2b
a

Northbrooks Secondary School, Singapore

Lesson 41
10.10.11
a
b
180o - 2a
2a
a

Lesson 41
10.10.11
a
b
2a
2b
b
a

Lesson 42
10.10.11
a
b
In the case of this type of figures, where is the angle at centre?
b
a
2b
2a

http://www.banhar.blogspot.com	180
http://banhar.blogspot.com	142
http://translate.googleusercontent.com	1
http://207.46.192.232	1
http://banhar.blogspot.jp	1

NSM Seminar

by Jimmy Keng on Oct 10, 2011

Accessibility

Categories

Tags

Upload Details

Usage Rights

5 Embeds 325

Statistics

NSM Seminar — Presentation Transcript