A piece of pasture grows at a constant rate every day. 200 sheep will eat up the grass in 100 days. 150 sheep will eat up the grass in 150 days. How many days does it take for 100 sheep to eat up the grass?

Number Number Pasture

of sheep of days

--------------------------------------------------------------------

200 100 S + 100 u

200 1 1/100 S + 1 u

150 150 S + 150 u

150 1 1/150 S+ 1 u

50 1 (1/100 - 1/150) S = 1/300 S

50 100 1/3 S

150 100 S

50 100 100 u

1/3 S --> 100 u

S --> 300 u

100 x 300 u + x u

100 1 1/150 S = 1/150 x 300 u

= 2 u

100 x 2 x u

2 x u --> 300 u + x u

x u --> 300 u

x --> 300 days

100 sheep will take 300 days to eat the grass.

## Friday, July 30, 2010

### Ratio P6 2009 SA2 P2 Nanyang Q12

Given that p : (q + r) = 1 : 3, r : (q + s) = 1 : 2 and p : s = 2 : 5.

(a) Find p : q : r : s.

(b) Find (r + q) : (p + s).

x 3

p : s = 2 : 5 = 6 : 15

x 2 x 3

p : (q + r) = 1 : 3 = 2 : 6 = 6 : 18

x 3

s + (q + r) = 5 + 6 = 11 ............... 33

x 11

r : (q + s) = 1 : 2 = 11 : 22

s = 15, (q + s) = 22, so q = 22 - 15 = 7

a) p : q : r : s = 6 : 7 : 11 : 15

b) (r + q) : (p + s) = (11 + 7) : (6 + 15)

= 18 : 21

= 6 : 7

(a) Find p : q : r : s.

(b) Find (r + q) : (p + s).

x 3

p : s = 2 : 5 = 6 : 15

x 2 x 3

p : (q + r) = 1 : 3 = 2 : 6 = 6 : 18

x 3

s + (q + r) = 5 + 6 = 11 ............... 33

x 11

r : (q + s) = 1 : 2 = 11 : 22

s = 15, (q + s) = 22, so q = 22 - 15 = 7

a) p : q : r : s = 6 : 7 : 11 : 15

b) (r + q) : (p + s) = (11 + 7) : (6 + 15)

= 18 : 21

= 6 : 7

## Thursday, July 29, 2010

### Speed P6

Mary and Ken travelled from Town A to Town B. Mary left Town A at 0915 and took 4 h to reach Town B. Ken left Town A at 1000 and took 2 h 45 min to reach Town B. At what time did Ken pass Mary?

9.15 am.

M [ DM ][ ] D

4 DM/D 4 h

10 am

K [ DM ][ ] D

2 3/4 DM/D 2 3/4 h

9 1/4 + 4 DM/D --> 10 + 2 3/4 DM/D

(4 - 2 3/4) DM/D --> 10 - 9 1/4 = 3/4

DM/D --> 3/4 x 4/5 = 3/5

3/5 x 4 = 2 2/5 h

= 2 h 24 min

2 h 24 min

|------------------|-------------|

9.15 am 11.15 am 11.39 am

or

3/5 x 2 3/4 = 1 13/20 h

= 1 h 39 min

1 h 39 min

|-------------------|-------------|

10 am 11 am 11.39 am

Ken passed Mary at 11.39 a.m.

9.15 am.

M [ DM ][ ] D

4 DM/D 4 h

10 am

K [ DM ][ ] D

2 3/4 DM/D 2 3/4 h

9 1/4 + 4 DM/D --> 10 + 2 3/4 DM/D

(4 - 2 3/4) DM/D --> 10 - 9 1/4 = 3/4

DM/D --> 3/4 x 4/5 = 3/5

3/5 x 4 = 2 2/5 h

= 2 h 24 min

2 h 24 min

|------------------|-------------|

9.15 am 11.15 am 11.39 am

or

3/5 x 2 3/4 = 1 13/20 h

= 1 h 39 min

1 h 39 min

|-------------------|-------------|

10 am 11 am 11.39 am

Ken passed Mary at 11.39 a.m.

### Speed P6

Alice is going up an escalator. When she takes 2 steps per second, she reaches the top in 24 steps. When she takes 1 step per second, she reaches the top in 16 steps. How long will she take to reach the top if she just stands on the escalator?

2 steps SE steps

per second per second

24 steps

|-------------------|------------------|

2 steps --> 1 s

24 steps --> 24/2 = 12 s

1 step SE steps

per second per second

16 steps

|-----------------------|--------------|

1 step --> 1 s

16 steps --> 16 s

24 + 12 SE --> 16 + 16 SE

4 SE --> 24 - 16 = 8

SE --> 8/4 = 2 steps per second

24 + 2 x 12 = 48 steps [or 16 + 2 x 16 = 48 steps]

48 / 2 = 24 s

She took 24 seconds.

2 steps SE steps

per second per second

24 steps

|-------------------|------------------|

2 steps --> 1 s

24 steps --> 24/2 = 12 s

1 step SE steps

per second per second

16 steps

|-----------------------|--------------|

1 step --> 1 s

16 steps --> 16 s

24 + 12 SE --> 16 + 16 SE

4 SE --> 24 - 16 = 8

SE --> 8/4 = 2 steps per second

24 + 2 x 12 = 48 steps [or 16 + 2 x 16 = 48 steps]

48 / 2 = 24 s

She took 24 seconds.

### Time P6

Yao has 2 clocks in his house; a cuckoo clock and a pendulam clock. Unfortunately, neither is in good working condition.The cuckoo clock gains 5 seconds for every minute it runs (which means it gets faster and faster) whilst the pendulam clock loses 5 seconds every minute it runs (which means it gets slower and slower).Yao resets both at 6 o'clock in the morning. When the time shown on the pendulam clock is 7 a.m., what time is shown on the cuckoo clock?

C: 1 min --> 1 s

1 h --> 60 x 5 = 300 s

= 5 min

P: 1 min --> - 1 s

60 min --> - 60 x 5 = 300 s

= 5 min

55 min --> 5 + 5 = 10 min

60 min --> 60/55 x 10

= 10 min 55 s (rounded off to nearest s)

10 min 55 s

|----------------------------------|

7.00 a.m. 7:10:55 a.m.

(Pendulum clock) (Cuckoo clock)

The time shown on the cuckoo clock is 7:10:55 a.m.

C: 1 min --> 1 s

1 h --> 60 x 5 = 300 s

= 5 min

**7.05 a.m.**P: 1 min --> - 1 s

60 min --> - 60 x 5 = 300 s

= 5 min

**6.55 a.m.**55 min --> 5 + 5 = 10 min

**difference**60 min --> 60/55 x 10

= 10 min 55 s (rounded off to nearest s)

10 min 55 s

|----------------------------------|

7.00 a.m. 7:10:55 a.m.

(Pendulum clock) (Cuckoo clock)

The time shown on the cuckoo clock is 7:10:55 a.m.

## Wednesday, July 21, 2010

### Percentage P5 2009 SA2 Catholic High

18. Mrs Lim bought some bowls and plates.If she buys another 5 bowls, there are 60% as many bowls as plates. If she buys another 9 plates, there will be thrice as many plates as bowls. How many bowls did she buy?

B [ ][5]

[ ][ ][ ]

P [ ][ ][ ][ ][ ]

B [ ]

[ ][ ][ ][ ][ ][ ][ ][ ][ ]

[ ][ ][ ]< 5 x 3 >

P [ ][ 9 ]

4 u --> 9 + 5 x 3 = 24

1 u --> 24 ÷ 4 = 6

3 x 6 - 5 = 13 bowls

She bought 13 bowls.

B P

+ 5

------------

60 : 100 = 3 : 5

------------

5 x (B + 5) --> 3 x P

5 B + 25 --> 3 P

B P

+ 9

----------

1 : 3

----------

3 B --> P + 9

3 P --> 3 x (3 B - 9) = 9 B - 27

9 B - 27 --> 5 B + 25

4 B --> 25 + 27 = 52

B --> 52 ÷ 4 = 13 bowls

She bought 13 bowls.

B [ ][5]

[ ][ ][ ]

P [ ][ ][ ][ ][ ]

B [ ]

[ ][ ][ ][ ][ ][ ][ ][ ][ ]

**Compare**[ ][ ][ ]< 5 x 3 >

P [ ][ 9 ]

**Compare**4 u --> 9 + 5 x 3 = 24

**[9 u - 5 x 3 --> 5 u + 9]**1 u --> 24 ÷ 4 = 6

3 x 6 - 5 = 13 bowls

She bought 13 bowls.

**Alternative method**B P

+ 5

------------

60 : 100 = 3 : 5

------------

5 x (B + 5) --> 3 x P

5 B + 25 --> 3 P

B P

+ 9

----------

1 : 3

----------

3 B --> P + 9

3 P --> 3 x (3 B - 9) = 9 B - 27

9 B - 27 --> 5 B + 25

4 B --> 25 + 27 = 52

B --> 52 ÷ 4 = 13 bowls

She bought 13 bowls.

## Tuesday, July 20, 2010

### Challenging

There is a big fruit tree in a garden. It bears fruits at a constant rate everyday. Paul will pick the fruits and sell. Animals will pluck and eat the fruits. 25 men can pick all the fruits in 15 days and 66 animals can eat all the fruits in 10 days. If 1 man can pick twice the number of fruits as 1 animal, find the number of days it take for 10 men to pick and 10 animals to eat all the fruits together.

25M (15 days) [ ][ ][ ] [][][][][][][][][][][][][][][]

50A (15 days)

66A (10 days) [ ][ ][ ] [][][][][][][][][][]

50A (10 days) [ ][ ] [][][][][][][][][][]

16A (10 days) [ ]

32A (10 days) [ ][ ]

18A (10 days) [][][][][][][][][][] --> 10 u

16A (10 days) --> 16/18 x 10 u = 80/9 u

[ ][ ] --> 2 x 80/9 u = 160/9 u

[ ][ ][ ] --> 3 x 80/9 = 80/3 u

25M (10 days) --> 160/9 + 10 = 250/9 u

10M (10 day) --> 10/25 x 250/9 = 100/9 u

10A (10 days) --> 10/18 x 10 = 50/9 u

10M + 10A (10 days) --> 100/9 + 50/9 = 50/3 u

10M + 10A (x days) --> x/10 x 50/3 = 5/3 x

5/3 x --> 80/3 + x

2/3 x --> 80/3

x --> 3/2 x 80/3 = 40 days

It takes 40 days for 10 men to pick and 10 animals to eat all the fruits.

25M (15 days) [ ][ ][ ] [][][][][][][][][][][][][][][]

50A (15 days)

66A (10 days) [ ][ ][ ] [][][][][][][][][][]

**Number of fruits at the beginning**50A (10 days) [ ][ ] [][][][][][][][][][]

16A (10 days) [ ]

32A (10 days) [ ][ ]

18A (10 days) [][][][][][][][][][] --> 10 u

16A (10 days) --> 16/18 x 10 u = 80/9 u

[ ][ ] --> 2 x 80/9 u = 160/9 u

[ ][ ][ ] --> 3 x 80/9 = 80/3 u

25M (10 days) --> 160/9 + 10 = 250/9 u

10M (10 day) --> 10/25 x 250/9 = 100/9 u

10A (10 days) --> 10/18 x 10 = 50/9 u

**[which is half of 10M (10 days)]**10M + 10A (10 days) --> 100/9 + 50/9 = 50/3 u

10M + 10A (x days) --> x/10 x 50/3 = 5/3 x

5/3 x --> 80/3 + x

**[ ][ ][ ][ x u ]**2/3 x --> 80/3

x --> 3/2 x 80/3 = 40 days

It takes 40 days for 10 men to pick and 10 animals to eat all the fruits.

## Sunday, July 18, 2010

### Percentage P6

In a stadium, 20 % of the people are performers for the concert and the rest are spectators. 72 % of the spectators are males and there are 220 more male spectators than female spectators. How many male spectators must leave the stadium so that 37.5 % of the people in the stadium are male spectators?

P S

------------------------

20 : 80 = 1 : 4 =

M F

----------

72 :

72 - 28 = 44

44% --> 220

1% --> 220 ÷ 44 = 5

72% --> 72 x 5 = 360 male spectators

53% --> 53 x 5 = 265

100 - 37.5 = 62.5

62.5% --> 265

37.5% --> 37.5/62.5 x 265 = 159 male spectators

360 - 159 = 201 male spectators

201 male spectators must leave the stadium.

P S

------------------------

20 : 80 = 1 : 4 =

**: 100***25*M F

----------

72 :

*28*72 - 28 = 44

44% --> 220

1% --> 220 ÷ 44 = 5

72% --> 72 x 5 = 360 male spectators

**25 + 28 = 53**53% --> 53 x 5 = 265

**Before**100 - 37.5 = 62.5

62.5% --> 265

**After (number the same, percentage has gone up)**37.5% --> 37.5/62.5 x 265 = 159 male spectators

360 - 159 = 201 male spectators

201 male spectators must leave the stadium.

## Wednesday, July 14, 2010

### Challenging P6

Row 1 2

Row 2 4 6 8

Row 3 10 12 14 16 18

Row 4 20 22 24 26 28 30 32

a) What is the largest number on Row 10?

b) 288 is the largest number in a certain row. Which row is it?

Row 1 --> 1 = 1 x 2 - 1

Row 2 --> 3 = 2 x 2 - 1

Row 3 --> 5 = 3 x 2 - 1

Row 4 --> 7 = 4 x 2 - 1

Row N --> N x 2 - 1 = 2 N - 1

Row 1 --> 2 x 1

Row 2 --> .......... 2 x (1 + 3)

Row 3 --> .......... 2 x (1 + 3 + 5)

Row 4 --> .......... 2 x (1 ++ 3 + 5 + 7)

Row N --> .......... 2 x [1 + 3 + 5 + 7 + .... (2 N - 1)]

= 2 x {[1 + (2 N - 1)] x N/2}

= 2 N x N

Row 10 --> 2 x 10 x 10 = 200

a) The largest number in Row 10 is 200.

2 N x N --> 288

N x N--> 288/ 2 = 144

= 12 x 12 -- Row 12

b) 288 is the largest number in Row 12.

Row 2 4 6 8

Row 3 10 12 14 16 18

Row 4 20 22 24 26 28 30 32

a) What is the largest number on Row 10?

b) 288 is the largest number in a certain row. Which row is it?

Row 1 --> 1 = 1 x 2 - 1

Row 2 --> 3 = 2 x 2 - 1

Row 3 --> 5 = 3 x 2 - 1

Row 4 --> 7 = 4 x 2 - 1

Row N --> N x 2 - 1 = 2 N - 1

Row 1 --> 2 x 1

Row 2 --> .......... 2 x (1 + 3)

Row 3 --> .......... 2 x (1 + 3 + 5)

Row 4 --> .......... 2 x (1 ++ 3 + 5 + 7)

Row N --> .......... 2 x [1 + 3 + 5 + 7 + .... (2 N - 1)]

= 2 x {[1 + (2 N - 1)] x N/2}

= 2 N x N

Row 10 --> 2 x 10 x 10 = 200

a) The largest number in Row 10 is 200.

2 N x N --> 288

N x N--> 288/ 2 = 144

= 12 x 12 -- Row 12

b) 288 is the largest number in Row 12.

### Ratio P6

In the figure below, which is not drawn to scale, 1/3 of the circle is shaded, 1/5 of the triangle is shaded and 1/4 of the rectangle is shaded. The total area of the circle and triangle is 7/8 of the rectangle. Given that the total area of the whole figure is 260 cm2, find the area of the rectangle.

C + T : R = 7 : 8 = 14 : 16

CS : C = 1 : 3 = 3 : 9

TS : T = 1 : 5

RS : R = 1 : 4 = 4 : 16

C + T + R --> 14 + 16 - 3 - 1 = 26

26 u --> 260

1 u --> 260/26 = 10

16 u --> 16 x 10 = 160 cm2.

The area of the rectangle is 160 cm2.

Alternative method

C + T : R = 7 : 8

RS : R = 1 : 4 = 2 : 8

7 + 8 - 2 = 13

13 u --> 260

1 u --> 260/13 = 20

8 u --> 8 x 20 = 160 cm2.

The area of the rectanglee is 160 cm2.

C + T : R = 7 : 8 = 14 : 16

**Multiply by 2 to make (C + T) the same****7 x ? = 3 x ? + 5 x ? [7 x 2 = 3 x 3 + 5 x 1 Guess and Check]**CS : C = 1 : 3 = 3 : 9

**Multiply by 3 to make (C + T) the same**TS : T = 1 : 5

RS : R = 1 : 4 = 4 : 16

**Multiply by 4 to get R = 16**C + T + R --> 14 + 16 - 3 - 1 = 26

26 u --> 260

1 u --> 260/26 = 10

16 u --> 16 x 10 = 160 cm2.

The area of the rectangle is 160 cm2.

Alternative method

**[Given informations 1/3 and 1/5 are redundant]**C + T : R = 7 : 8

RS : R = 1 : 4 = 2 : 8

7 + 8 - 2 = 13

**Minus 2 to remove double-counting of shaded areas.**13 u --> 260

1 u --> 260/13 = 20

8 u --> 8 x 20 = 160 cm2.

The area of the rectanglee is 160 cm2.

## Sunday, July 11, 2010

### HCI's S'pore MOPS 2004 - First round Q19

Aden and John started jogging along a circular track. Aden started at Point X while John started at Point Y where the line XY formed the diameter of the circle. Aden and John jogged towards each other along the circular track from their respective starting point and first met at Point W which was 80 m from Point X. After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Z which was 60 m from Point Y. Find the distance of the circular track.

X

A1 [ 80 ]

[ ] J1

Y

A2 [ ]

[ 60 ]

[ ] Z

[ ] J2

DA1 --> 80 m

DA2 --> 2 x 80 m = 160 m

160 - 60 = 100 m

100 + 80 = 180 m

180 x 2 = 360 m

The distance of the circular track was 360 m.

X

A1 [ 80 ]

**Half a circle to meet the first time**[ ] J1

Y

A2 [ ]

[ 60 ]

[ ] Z

**A circle to meet the second time**[ ] J2

DA1 --> 80 m

DA2 --> 2 x 80 m = 160 m

160 - 60 = 100 m

**W to Y**100 + 80 = 180 m

**XY (1/2 circular track)**180 x 2 = 360 m

The distance of the circular track was 360 m.

## Saturday, July 10, 2010

### Volume P6 2009 SA2 MGS Q 14

14. Jeffrey wants to pack some identical cuboids into the container as shown below. What is the maximum number of cuboids Jeffrey can pack into the container?

(1 ) 77

(2) 70

(3) 63

(4) 60

14 ÷ 2 = 7

6 ÷ 2 = 3

11 ÷ 3 = 3 r 2

**More cuboids can be packed for the r 2.**7 x 3 x 3 = 63 cuboids

6 ÷ 3 = 2

1 x 7 x 2 = 14 cuboids

63 + 14 = 77 cuboids (1)

## Friday, July 9, 2010

### Percentage P6 2009 SA2 P2 Catholic High Q15

Tom had $144 more than Ali at first. After Tom spent 25% of his money and Ali spent 1/3 of his money, Tom had $122 more than Ali. How much did Tom have at first? [5]

25/100 x 144 = $36

144 - 36 = $108

T [][][][][][][][][][][][][108][36]

A [][][][][][][][][][][][]

TN [][][][][][][][][][108]

AN [][][][][][][][]<122>

1 u --> 122 - 108 = 14

12 u --> 12 x 14 = 168

168 + 144 = $312

Tom had $312 at first.

Alternative method

T [ ][ ][ ] [ ] [][22][][22][][22][][22]

A [][][]< 144 > [][][]<-------144------->

<122>22

TN [ ][ ][ ]

AN [][] <122>

1 u --> 144 - 4 x 22 = 56

4 u --> 4 x 56 = 224

224 + 4 x 22 = $312

Tom had $312 at first.

Alternative method (Algebra)

4 T - 3 A --> $144

3 T - 2 A --> $122

1 T - 1 A --> 144 - 122 = 22

2 T - 2 A --> 2 x 22 = 44

1 T --> 122 - 44 = 78

4 T --> 4 x 78 = $312

Tom had $312 at first.

Alternative method

25/100 x 144 = $36

144 - 36 = $108

T [][][][][][][][][][][][][108][36]

**25% = 1/4, Make 1/4 and 1/3 like fractions**A [][][][][][][][][][][][]

**1/4 = 3/12 and 1/3 = 4/12**TN [][][][][][][][][][108]

AN [][][][][][][][]<122>

1 u --> 122 - 108 = 14

12 u --> 12 x 14 = 168

168 + 144 = $312

Tom had $312 at first.

Alternative method

T [ ][ ][ ] [ ] [][22][][22][][22][][22]

A [][][]< 144 > [][][]<-------144------->

<122>22

TN [ ][ ][ ]

**Comparing highlight**AN [][] <122>

**[ ] --> [][22]**1 u --> 144 - 4 x 22 = 56

4 u --> 4 x 56 = 224

224 + 4 x 22 = $312

Tom had $312 at first.

Alternative method (Algebra)

4 T - 3 A --> $144

3 T - 2 A --> $122

1 T - 1 A --> 144 - 122 = 22

2 T - 2 A --> 2 x 22 = 44

1 T --> 122 - 44 = 78

4 T --> 4 x 78 = $312

Tom had $312 at first.

Alternative method

## Thursday, July 8, 2010

### Whole Number P6

There were some boys and girls in a room. If 24 boys leave the room, the total number of students who remained in the room will be 7 times the number of boys who remained in the room. If 12 girls leave the room, the total number of students who remained in the room will be 4 times the number of boys. How many students were there in the classroom?

[24][1 u][ 6 u ]

[ ][ ][ ][ ][12]

[24][1 u][24][1 u][24][1 u]

3 u --> 24 x 3 + 12 = 84

1 u --> 84 ÷ 3 = 28

7 u --> 7 x 28 = 196

196 + 24 = 220 students

There were 220 students in the class.

[24][1 u][ 6 u ]

[ ][ ][ ][ ][12]

[24][1 u][24][1 u][24][1 u]

**Redraw the girl units (using boy unit)**3 u --> 24 x 3 + 12 = 84

**Compare the girls (in red)**1 u --> 84 ÷ 3 = 28

7 u --> 7 x 28 = 196

196 + 24 = 220 students

There were 220 students in the class.

### Decimal P5

Each day, Mike saves $0.50 while Carol saves $0.30 more than her. Carol starts saving 6 days after Mike. She has $17.40 more than Mike now. How much does Mike save now?

0.50 x 6 = $3.00

3.00 + 17.40 = $20.40

20.40 ÷ 0.30 = 68 days

68 + 6 = 74 days

74 x 0.50 = $37

Mike saves $37 now

0.50 x 6 = $3.00

**Amount saved by Mike in first 6 days**3.00 + 17.40 = $20.40

**Additional amount saved by Carol**20.40 ÷ 0.30 = 68 days

**Number of days Carol took to save $20.40 more**68 + 6 = 74 days

**Total number of days Mike saved**74 x 0.50 = $37

Mike saves $37 now

### Speed P6

Najip, Kumar and Gurmit started jogging at the same time from the same starting-point round a circular track. Najip and Kumar jogged in a clockwise direction and Gurmit jogged in an anti-clockwise direction. Gurmit took 5 minutes to complete each round. Gurmit met Najip after every 3 minutes. Gurmit met Kumar after every 2 minutes. The jogging speed of each person remained the same throughout.

(a) What was the ratio of Gurmit's speed to Najip's speed to Kumar's speed?

(b) When Gurmit and Najip met again at the starting-point after 15 minutes, Kumar had already jogged 3.6 km. What is the circumference of the circular track?

G [ ][ ][ ][ ][ ] 5 min, 1 round

3 min

G [ ][ ][ ] SG : SN = DG : DN

[ ][ ] N = 3 : 2

= 6 : 4

2 min

G [ ][ ] SG : SK = DG : DK

[ ][ ][ ] K = 2 : 3

= 6 : 9

SG : SN : SK = DG : DN : DK

= 6 : 4 : 9

a) The ratio of Gurmit's speed to Najip's speed to Kumar's speed was 6 : 4: 9.

DG : DN : DK = 6 : 4: 9

= 3 : 2 : 4.5

When Gurmit and Najip met again at the starting-point, Gurmit would have run 3 rounds while Najip 2 rounds and Kumar 4.5 rounds.

4.5 rounds --> 3.6 km

1 round --> 1/4.5 x 3.6 = 0.8 km

b) The circumference of the circular track was 0.8 km.

(a) What was the ratio of Gurmit's speed to Najip's speed to Kumar's speed?

(b) When Gurmit and Najip met again at the starting-point after 15 minutes, Kumar had already jogged 3.6 km. What is the circumference of the circular track?

G [ ][ ][ ][ ][ ] 5 min, 1 round

3 min

**Constant Time**G [ ][ ][ ] SG : SN = DG : DN

[ ][ ] N = 3 : 2

= 6 : 4

2 min

**Constant Time**G [ ][ ] SG : SK = DG : DK

[ ][ ][ ] K = 2 : 3

= 6 : 9

SG : SN : SK = DG : DN : DK

**Constant Time**= 6 : 4 : 9

a) The ratio of Gurmit's speed to Najip's speed to Kumar's speed was 6 : 4: 9.

DG : DN : DK = 6 : 4: 9

= 3 : 2 : 4.5

When Gurmit and Najip met again at the starting-point, Gurmit would have run 3 rounds while Najip 2 rounds and Kumar 4.5 rounds.

4.5 rounds --> 3.6 km

1 round --> 1/4.5 x 3.6 = 0.8 km

b) The circumference of the circular track was 0.8 km.

## Wednesday, July 7, 2010

### Speed P6

Alfred and Peter jogged to and fro repeatedly along a straight path in a park between two points A and B. Alfred jogged at a uniform speed of 40m/min and Peter at 60m/min. They started jogging from the opposite directions at the same time as shown below.

Alfred-----> <-------Peter

A--------------------------X-----------------Y----------------B

Distance between X and Y: 200m

They first met one another at point X. The second time they met was at point Y.

a) Find the distance between A and B.

b) If they started jogging at 0715, when did they meet again for the third time?

A1 [ ][ ][ ][ ] X

P1 [ ][ ][ ][ ][ ][ ]

DA : DP = 4 : 6

= 2 : 3

A2 [ ]

[ a ]

[ 10 u - a ] Y

P2 [ ]

6 u + a : 4 u + (10 u - a) = 2 : 3

3 x (6 u + a) = 2 x (14 u - a)

3 a + 2 a = 28 u - 18 u

5 a = 10 u

a = 2 u

4 u --> 200 m

10 u --> 10/4 x 200 = 500 m

a) The distance between A and B was 500 m.

500 x 5 = 2500 m

40 + 60 = 100 m/min

2500 ÷ 100 = 25 min

25 min

|--------------|

0715 0740

b) They met again for the third time at 0740.

Alfred-----> <-------Peter

A--------------------------X-----------------Y----------------B

Distance between X and Y: 200m

They first met one another at point X. The second time they met was at point Y.

a) Find the distance between A and B.

b) If they started jogging at 0715, when did they meet again for the third time?

A1 [ ][ ][ ][ ] X

P1 [ ][ ][ ][ ][ ][ ]

DA : DP = 4 : 6

**Constant Time**= 2 : 3

A2 [ ]

[ a ]

[ 10 u - a ] Y

P2 [ ]

6 u + a : 4 u + (10 u - a) = 2 : 3

**Constant Time**3 x (6 u + a) = 2 x (14 u - a)

3 a + 2 a = 28 u - 18 u

5 a = 10 u

a = 2 u

4 u --> 200 m

10 u --> 10/4 x 200 = 500 m

a) The distance between A and B was 500 m.

500 x 5 = 2500 m

**Total distance travelled to meet the third time**40 + 60 = 100 m/min

**Combined speed**2500 ÷ 100 = 25 min

25 min

|--------------|

0715 0740

b) They met again for the third time at 0740.

## Tuesday, July 6, 2010

### Speed P6

A car and a motorcycle set off from Green Town and Pink Town respectively at the same time. The car, travelling at a constant speed, took 10 hours to reach Pink Town. The motorcycle, travelling at a constant speed, took 6 hours to reach Green Town. The vehicles passed by each other when they were 218 3/4 km from Pink Town. Find

a) the distance between the two towns and

b) the speed of the car.

C: 10 h --> 1 distance

1 h --> 1/10 distance

M: 6 h --> 1 distance

1 h --> 1/6 distance

Total: 1 h --> 1/10 + 1/6 = 4/15 distance

4/15 distance --> 1 h

15/15 distance --> 15/4 x 1 = 3.75 h

3.75 h --> 218 3/4 km

6 h --> 6/3.75 x 218 3/4= 350 km

a) The distance between the two towns was 350 km.

350/10 = 35 km/h

b) The speed of the car was 35 km/h.

Alternative method

For constant distance, SC : SM = TM : TC

= 3 : 5

For constant time, DC : DM = SC : SM

5 u --> 218 3/4 km

8 u --> 8/5 x 218 3/4 = 350 km

a) The distance between the two towns was 350 km.

350/10 = 35 km/h

b) The speed of the car was 35 km/h

a) the distance between the two towns and

b) the speed of the car.

C: 10 h --> 1 distance

1 h --> 1/10 distance

M: 6 h --> 1 distance

1 h --> 1/6 distance

Total: 1 h --> 1/10 + 1/6 = 4/15 distance

4/15 distance --> 1 h

15/15 distance --> 15/4 x 1 = 3.75 h

3.75 h --> 218 3/4 km

6 h --> 6/3.75 x 218 3/4= 350 km

a) The distance between the two towns was 350 km.

350/10 = 35 km/h

b) The speed of the car was 35 km/h.

Alternative method

For constant distance, SC : SM = TM : TC

**(S = D/T)****(total distance)**= 6 : 10= 3 : 5

For constant time, DC : DM = SC : SM

**(D = S x T)****(when they met)**= 3 : 55 u --> 218 3/4 km

8 u --> 8/5 x 218 3/4 = 350 km

a) The distance between the two towns was 350 km.

350/10 = 35 km/h

b) The speed of the car was 35 km/h

### Challenging P6

6, 24, 60, 120, 210, ?

What follows after 210 ?

Factors of 6 --> 1, 2, 3, 6

1st number --> 6= 2 x 3

= (1 x 2) x (1 + 2)

Factors of 24 --> 1, 2, 3, 4, 6, 8, 12, 24

2nd number --> 24 = 4 x 6

= (2 x 2) x (1 + 2 + 3)

Factors of 60 --> 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

3rd number --> 60 = 6 x 10

= (3 x 2) x (1 + 2 + 3 + 4)

Factors of 120 --> 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

4th number --> 120 = 8 x 15

= (4 x 2) x (1 + 2 + 3 + 4 + 5)

Factors of 210 --> 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210

5th number --> 210 = 10 x 21

= (5 x 2) x (1 + 2 + 3 + 4 + 5 + 6)

6th number --> (6 x 2) x (1 + 2 + 3 + 4 + 5 + 6 + 7)

= 12 x (7 x 8)/2

= 336

The next number is 336.

What follows after 210 ?

Factors of 6 --> 1, 2, 3, 6

1st number --> 6= 2 x 3

= (1 x 2) x (1 + 2)

**= 2 x (2 x 3)/2 = 1 x 2 x 3**Factors of 24 --> 1, 2, 3, 4, 6, 8, 12, 24

2nd number --> 24 = 4 x 6

= (2 x 2) x (1 + 2 + 3)

**= 4 x (3 x 4)/2 = 2 x 3 x 4**Factors of 60 --> 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

3rd number --> 60 = 6 x 10

= (3 x 2) x (1 + 2 + 3 + 4)

**= 6 x (4 x 5)/2 = 3 x 4 x 5**Factors of 120 --> 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

4th number --> 120 = 8 x 15

= (4 x 2) x (1 + 2 + 3 + 4 + 5)

**= 8 x (5 x 6)/2 = 4 x 5 x 6**Factors of 210 --> 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210

5th number --> 210 = 10 x 21

= (5 x 2) x (1 + 2 + 3 + 4 + 5 + 6)

**= 10 x (6 x 7)/2 = 5 x 6 x 7**6th number --> (6 x 2) x (1 + 2 + 3 + 4 + 5 + 6 + 7)

= 12 x (7 x 8)/2

**= 6 x 7 x 8**= 336

The next number is 336.

## Friday, July 2, 2010

### Challenging P6

Candle A and candle B were placed on a table. Candle A was 1.5 cm longer than candle B. Candle A and candle B were lighted at 0630 h and 0800 h respectively. They burnt down to the same length at 1030 h. At 1200 h, candle B was burnt out while A only burnt out at 1230 h. Given that the rate of burning of each candle was constant throughout, find the original length of each candle.

[][][] [][][] [][][]

A [ ][ ][ ]

B [][][] [][][] [][]1.5cm

1 u --> 1.5 cm

A: 9 u --> 9 x 1.5 = 13.5 cm

B: 8 u --> 8 x 1.5 = 12 cm

The original lengths of candle A and candle B are 13.5 cm and 12 cm respectively.

**2 h 4 h****| -------| ------------- |****1230 1030 0630**[][][] [][][] [][][]

A [ ][ ][ ]

**2 h -- 1 Bu; 4 h -- 2 Bu**B [][][] [][][] [][]1.5cm

**1.5 h -- 3 u; 2.5 h -- 5 u****1200 1030 0800****[1 Bu = 3 u, same length]****|______|__________ |****1.5 h 2.5 h**1 u --> 1.5 cm

A: 9 u --> 9 x 1.5 = 13.5 cm

B: 8 u --> 8 x 1.5 = 12 cm

The original lengths of candle A and candle B are 13.5 cm and 12 cm respectively.

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