Timothy wants to buy a skateboard.It costs $17.25.Every week he saves $1.20 more than the week before.After saving for a number of weeks,he bought the skateboard. a)What is the least number of weeks he needs to save enough to buy the skateboard? b)How much of his savings did he have left after buying the skateboard?

Wk 1 --> $1.20

Wk 2 --> 1.20 + 1.20 x 2 --> $3.60

Wk 3 --> $3.60 + 1.20 x 3 = $7.20

Wk 4 --> $7.20 + 1.20 x 4 --> $12.00

Wk 5 --> 12.00 + 1.20 x 5 = $18.00 > $17.25

a) The least number of weeks he needs to save enough to buy the skateboard is 5.

18.00 - 17.25 = $0.75

b) He has $0.75 left.

## Monday, June 28, 2010

## Friday, June 25, 2010

### Annual Mathlympics 2009

4 men and 4 women are to be seated at a round table for dinner. If the men and women take alternate seats, how many arrangements are possible?

4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 576 ways

4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 576 ways

**The arrangement for men is 4 x 3 x 2 x 1 and the arrangement for women is also 4 x 3 x 2 x 1.**### Annual Mathlympics 2009

Janice was given a set of 5 different coloured beads to make a bracelet.If she has to use all of the 5 beads, how many different ways can Janice make the bracelet?

5 x 4 x 3 x 2 x 1 / 2 = 60 ways

5 x 4 x 3 x 2 x 1 / 2 = 60 ways

**There are 5 ways to pick the first bead, 4 ways to pick the second bead, 3 ways to pick the third bead, 2 ways to pick the fourth bead and 1 way to pick the last bead. As the bracelet can be turn over, the arrangement ABCDE is the same as EDCBA. So the number of ways is halved.**## Tuesday, June 22, 2010

### SAPMOPS 2006

Abel, Bob, Conan, Dave and Elijah divided a certain number of marbles amongst themselves in the following way: Abel took 1 marble and 20% of the remaining marbles, then Bob took 1 marble and 20% of the remaining marbles. Conan, Dave and Elijah did the same. At least how many marbles were there in tbe beginning?

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R)))))

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R))))

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + R))

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + R)

[1][][][][][] --> 1 + R

As R is a multiple of 5, try till you get the first expression as a whole number. Solve using Microsoft Office - Excel, the first whole number is 3121.

There were 3121 marbles in the beginning.

Alternative solution as suggested by a User.

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

Let x be the number of marbles in the beginning.

[ x - 1 ][1]

[ 4/5 x - 4/5 ][ A ]

[ 4/5 x - 4/5 - 1 ][1]

The remaining marbles after A has taken his share.

[ (4/5)^2 x - (4/5)^2 - 4/5 ][ B ]

[ (4/5)^2 x - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A and B have taken their shares.

[ (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5) ][ C ]

[ (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A, B and C have taken their shares.

[ (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) ][ D ]

[ (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A, B, C and D have taken their shares.

[(4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) ][ E ]

The remaining marbles, r --> (4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5)

(4/5) + (4/5)^2 + (4/5)^3 + (4/5)4 + (4/5)^5 = [(4/5)^6 - 1]/[4/5 - 1] - 1

= - 5(4/5)^6 + 5 - 1

= - 4 (5/4) (4/5)^6 + 4

= - 4 (4/5)^5 + 4

r --> (4/5)^5 x + 4 (4/5)^5 - 4 = (x + 4) (4/5)^5 - 4

For r to be a whole number, (x + 4) must be equal or greater than 5^5

For the smallest r, (x + 4) = 5 x 5 x 5 x 5 x 5

x = 3125 - 4

= 3121 marbles.

There were 3121 marbles in the beginning.

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R)))))

**units not drawn to scale**[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R))))

**units not drawn to scale**[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + R))

**units not drawn to scale**[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + R)

[1][][][][][] --> 1 + R

As R is a multiple of 5, try till you get the first expression as a whole number. Solve using Microsoft Office - Excel, the first whole number is 3121.

There were 3121 marbles in the beginning.

Alternative solution as suggested by a User.

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

Let x be the number of marbles in the beginning.

[ x - 1 ][1]

[ 4/5 x - 4/5 ][ A ]

[ 4/5 x - 4/5 - 1 ][1]

The remaining marbles after A has taken his share.

[ (4/5)^2 x - (4/5)^2 - 4/5 ][ B ]

[ (4/5)^2 x - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A and B have taken their shares.

[ (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5) ][ C ]

[ (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A, B and C have taken their shares.

[ (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) ][ D ]

[ (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) - 1 ][1]

The remaining marbles after A, B, C and D have taken their shares.

[(4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) ][ E ]

The remaining marbles, r --> (4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5)

**The formula for goemetric series (which is taught at Sec 3) is as follows:****1 + z + z^2 + z^3 + . . . + z^n = [z^n+1 - 1]/[z - 1]**(4/5) + (4/5)^2 + (4/5)^3 + (4/5)4 + (4/5)^5 = [(4/5)^6 - 1]/[4/5 - 1] - 1

= - 5(4/5)^6 + 5 - 1

= - 4 (5/4) (4/5)^6 + 4

= - 4 (4/5)^5 + 4

r --> (4/5)^5 x + 4 (4/5)^5 - 4 = (x + 4) (4/5)^5 - 4

For r to be a whole number, (x + 4) must be equal or greater than 5^5

For the smallest r, (x + 4) = 5 x 5 x 5 x 5 x 5

x = 3125 - 4

= 3121 marbles.

There were 3121 marbles in the beginning.

**It is obvious that this cannot be P5/6 Stuff.**### Speed P6

Cars are travelling to and fro from Greens Town to Sins City. The cars leave the Towns at regular interval and a car will meet another car coming in the opposite direction at every 6 min. Ronald starts cycling from Greens Town to Sins City while Annie starts cycling in the opposite direction. Both leave the Towns at the same time. Ronald and Annie will meet a car coming in the opposite direction every 7 and 8 min respectively. They met each other after cycling for 56 min. What is the time taken by a car to travel from Greens Town to Sins City?

CG [ ][ ][ ][ ][ ][ ]

[ ][ ][ ][ ][ ][ ] CS

R [ 7 R ]

[ ][ ][ ][ ][ ][ ][ ] CS

CG [ ][ ][ ][ ][ ][ ][ ][ ]

[ 8 A ] A

7 R --> 5 C

If R cycled 56 min to travel 7 R to meet CS which travelled 7 C, to travel just 5 C. CS would take 5/7 x 56 = 40 min.

8 A --> 4 C

If A cycled 56 min to travel 8 A to meet CG which travelled 8 C. to travel just 4 C, CG would take 4/8 x 56 = 28 min.

40 + 28 = 68 min

Time taken by a car to travel from Green Town to Sins City was 68 minutes.

CG [ ][ ][ ][ ][ ][ ]

**Units show the respective**[ ][ ][ ][ ][ ][ ] CS

**distance travelled in the****time given in the question**R [ 7 R ]

[ ][ ][ ][ ][ ][ ][ ] CS

CG [ ][ ][ ][ ][ ][ ][ ][ ]

[ 8 A ] A

7 R --> 5 C

If R cycled 56 min to travel 7 R to meet CS which travelled 7 C, to travel just 5 C. CS would take 5/7 x 56 = 40 min.

8 A --> 4 C

If A cycled 56 min to travel 8 A to meet CG which travelled 8 C. to travel just 4 C, CG would take 4/8 x 56 = 28 min.

40 + 28 = 68 min

Time taken by a car to travel from Green Town to Sins City was 68 minutes.

## Thursday, June 17, 2010

### Speed P6

Toy Train A and Toy Train B are 15 m apart from each other. When the toy trains moved towards each other, they would meet in 5 min. However, if the toy trains moved in the same direction, toy Train A would overtake Toy Train B in 5/12 h. Find the speed of Toy Train A and Toy Train B.

5/12 h = 25 min

5 min

<><----- 25 min--->

B [ ][ ][ ][ ][ ][ ]

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

<15 m >

5 min

<------- 25 min ---------->

2 [ ] + [] --> 15 m

5 [ ] --> 3 [ ] + 4 []

2 [ ] --> 4 []

5 [] --> 15

[] --> 15 ÷ 5 = 3 m

[ ] --> 2 x 3 = 6 m

6 ÷ 5 = 1.2 m/min

9 ÷ 5 = 1.8 m/min

The speed of Toy Train A and Toy Train B are 1.8 m/min and 1.2 m/min respectively.

5/12 h = 25 min

5 min

<><----- 25 min--->

B [ ][ ][ ][ ][ ][ ]

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

<15 m >

5 min

**[ ][]**<------- 25 min ---------->

2 [ ] + [] --> 15 m

5 [ ] --> 3 [ ] + 4 []

2 [ ] --> 4 []

5 [] --> 15

[] --> 15 ÷ 5 = 3 m

[ ] --> 2 x 3 = 6 m

6 ÷ 5 = 1.2 m/min

9 ÷ 5 = 1.8 m/min

The speed of Toy Train A and Toy Train B are 1.8 m/min and 1.2 m/min respectively.

## Wednesday, June 16, 2010

### Challenging P6

Find the sum of the digits of the first 100 odd numbers.

First 100 odd numbers 1, 3, 5, ....., 199

Digits of first 100 numbers

1 , 3 , 5 , 7 , 9 --> 25

11 , 13 , 15 , 17 , 19 --> 25 + 1 x 5

21 , 23 , 25 , 27 , 29 --> 25 + 2 x 5

...

91 , 93 , 95 , 97 , 99 --> 25 + 9 x 5

101 , 103 , 105, 107 , 109 --> 25 + 1 x 5

111 , 113 , 115 , 117 , 119 --> 25 + 1 x 5 + 1 x 5

121 , 123 , 125 , 127 , 129 --> 25 + 1 x 5 + 2 x 5

....

191 , 193 , 195 , 197 , 199 --> 25 + 1 x 5 + 9 x 5

20 x 25 = 500

1 x 5 + 2 x 5 + .... + 9 x 5 = 5 x (1 + 2 + 9)

= 5 x (1+9)/2 x 9

= 225

225 x 2 = 450

10 x 1 x 5 = 50

500 + 450 + 50 = 1000

The sum of digits of the first 100 odd numbers is 1000.

First 100 odd numbers 1, 3, 5, ....., 199

Digits of first 100 numbers

1 , 3 , 5 , 7 , 9 --> 25

11 , 13 , 15 , 17 , 19 --> 25 + 1 x 5

21 , 23 , 25 , 27 , 29 --> 25 + 2 x 5

...

91 , 93 , 95 , 97 , 99 --> 25 + 9 x 5

101 , 103 , 105, 107 , 109 --> 25 + 1 x 5

111 , 113 , 115 , 117 , 119 --> 25 + 1 x 5 + 1 x 5

121 , 123 , 125 , 127 , 129 --> 25 + 1 x 5 + 2 x 5

....

191 , 193 , 195 , 197 , 199 --> 25 + 1 x 5 + 9 x 5

20 x 25 = 500

1 x 5 + 2 x 5 + .... + 9 x 5 = 5 x (1 + 2 + 9)

= 5 x (1+9)/2 x 9

= 225

225 x 2 = 450

10 x 1 x 5 = 50

500 + 450 + 50 = 1000

The sum of digits of the first 100 odd numbers is 1000.

## Sunday, June 13, 2010

### Whole Number P6

The value of $2, $5 and $10 that was saved by Rebecca was in the ratio of 8 : 10 : 3 respectively. After taking out 3/4 of the $2 notes to pay for the gift, and 18 of the $5 notes for the watch, she has $390 left. Find the value of the $10 notes.

Value

$2 $5 $10

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

8 : 10 : 3

8 ÷ 2 = 4

10 ÷ 5 = 2

3 ÷ 10 = 3/10

Number

$2 $5 $10

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

4 : 2 : 3/10

40u : 20u : 3u

-30u -18

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

10u 20u-18 3u

Value

$2 $5 $10

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

20u 100u - $90 30u

150u --> 390 + 90 = 480

1u --> 480 ÷ 150 = $3.20

30u --> 30 x 3.20 = $96

The value of the $10 notes was $96.

Value

$2 $5 $10

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

8 : 10 : 3

8 ÷ 2 = 4

10 ÷ 5 = 2

3 ÷ 10 = 3/10

Number

$2 $5 $10

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

4 : 2 : 3/10

40u : 20u : 3u

**Add u to differentiate it from 18 $5 taken out.**-30u -18

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

10u 20u-18 3u

Value

$2 $5 $10

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

20u 100u - $90 30u

150u --> 390 + 90 = 480

1u --> 480 ÷ 150 = $3.20

30u --> 30 x 3.20 = $96

The value of the $10 notes was $96.

## Saturday, June 12, 2010

### Speed P6

Car A and Car B left Town Y at the same time, heading in the opposite direction. Car A headed for Town Z while Car B left for Town X. The speed of Car B was 20 km/h faster than Car A. After 1/2 h, Car A had completed 2/3 of its journey while Car B had completed 1/2 of its journey. The two cars were also 110 km apart.

a) Calculate the speed of Car A.

b) How far was Car B from Town X when Car A reached its destination ?

Car A Car B

Z Y X

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

<---- 110 km ------>

20 x 1/2 = 10 km [][]

4u --> 110 - 10 = 100

1 u --> 100 / 4 = 25

2 u --> 2 x 25 = 50 km

50/ (1/2) = 50 x 2

= 100 km/h

a) The speed of Car A was 100 km/h.

2/3 --> 1/2 h

3/3 --> 3/2 x 1/2 = 3/4 h

1/2 h --> 50 + 10 = 60 km

3/4 h --> 3/4 x 2 x 60 = 90 km

60 x 2 - 90 = 30 km

b) Car B was 30 km from Town X.

a) Calculate the speed of Car A.

b) How far was Car B from Town X when Car A reached its destination ?

Car A Car B

Z Y X

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

<---- 110 km ------>

20 x 1/2 = 10 km [][]

4u --> 110 - 10 = 100

1 u --> 100 / 4 = 25

2 u --> 2 x 25 = 50 km

50/ (1/2) = 50 x 2

= 100 km/h

**Speed of car**a) The speed of Car A was 100 km/h.

2/3 --> 1/2 h

3/3 --> 3/2 x 1/2 = 3/4 h

1/2 h --> 50 + 10 = 60 km

3/4 h --> 3/4 x 2 x 60 = 90 km

60 x 2 - 90 = 30 km

b) Car B was 30 km from Town X.

## Thursday, June 10, 2010

### Speed P6

A driver and a motorcyclist both started at the same time to travel to Town B, 540 km from Town A. The car was travelling at an average speed 50 km/h faster than the motorcycle for the first 1.5 h. After that, the motorcyclist doubled his speed to catch up with the car. However, he arrived at Town B slightly more than 6 minutes later than the car. If the motocyclist were to arrive at the same time as the driver, at what speed should he increase to instead?

SM1 SM2 SC TC TM2 Remarks

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

50 100 100 Not possible as SM2 = SC

60 120 110 4 10/11 3.75 3.75 + 1.5 - 4 10/11 = 0.34 h > 20 min

70 140 120 4.5 3 3/70 3 3/70 + 1.5 - 4.5 = 3/70 h > 6 min

70 x 1.5 = 105 km

540 - 105 = 435 km

540 / 120 = 4.5 h

4.5 - 1.5 = 3 h

435/3 = 145 km/h

He should increase the speed to 145 km/h.

SM1 SM2 SC TC TM2 Remarks

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

50 100 100 Not possible as SM2 = SC

60 120 110 4 10/11 3.75 3.75 + 1.5 - 4 10/11 = 0.34 h > 20 min

70 140 120 4.5 3 3/70 3 3/70 + 1.5 - 4.5 = 3/70 h > 6 min

70 x 1.5 = 105 km

540 - 105 = 435 km

540 / 120 = 4.5 h

4.5 - 1.5 = 3 h

435/3 = 145 km/h

He should increase the speed to 145 km/h.

### Challenging P6

6,15,27,42,.... --> 6, (6+9), (6+9+12), (6+9+12+15),.... What's the formula?

1 --> 6 = 3 + 3 x 1

= 3 x 1 + 3 x 1

2 --> 15 = 6 + 9

= (3 + 3 x1) + (3 + 3 x 2)

= 3 x 2 + 3 x 1 + 3 x 2

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

3 --> 27 = 15 + 12

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

= 3 x 3 + 3 x 1 + 3 x 2 + 3 x 3

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

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

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

= 3 x N x (3 + N)/2

Testing

-------

1 --> 3 x 1 x (3 + 1)/2 = 3 x 2

= 6 (correct)

2 --> 3 x 2 x (3 + 2)/2 = 15 (correct)

3 --> 3 x 3 x (3 + 3)/2 = 27 (correct)

4 --> 3 x 4 x (3 + 4)/2 = 42 (correct)

5 --> 3 x 5 x (3 + 5)/2 = 60

6 --> 3 x 6 x (3 + 6)/2 = 81

...

10 --> 3 x 10 + (3 + 10)/2 = 195

1st Fig is 1 hexagon - made up of 6 sticks

(Hexagon resting on one edge)

2nd Fig is 3 hexagon - made up of 15 sticks

(1 hexagon resting on two hexagons)

3rd Fig is 6 hexagon - made up of 27 sticks

(The two hexagons resting on 3 hexagons)

Based on figures given,

1 --> 6

2 --> 6 x (1 + 2) - 3 x 1 = 18 - 5

= 5

3 --> 6 x (1 + 2 + 3) - 3 x (1 + 2) = 36 - 9

= 27

N --> 6 x (1 + 2 + ...+ N) - 3 x (1 + 2 + ...N -1)

= 6 x (1 + N)/2 x N - 3 x (1 + N - 1)/2 x (N -1)

= 3 x N x (N + 1) - 3 x N x (N - 1)/2

if factorised will get the same formula as above

Testing

1 --> 3 x 1 x 2 - 3 x 1 x 0/2 = 6 (correct)

2 --> 3 x 2 x 3 - 3 x 2 x 1/2 = 15 (correct)

3 --> 3 x 3 x 4 - 3 x 3 x 2/2 = 27 (correct)

10 --> 3 x 10 x 11 - 3 x 10 x 9/2 = 195 (same as above)

1 --> 6 = 3 + 3 x 1

= 3 x 1 + 3 x 1

2 --> 15 = 6 + 9

= (3 + 3 x1) + (3 + 3 x 2)

= 3 x 2 + 3 x 1 + 3 x 2

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

3 --> 27 = 15 + 12

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

= 3 x 3 + 3 x 1 + 3 x 2 + 3 x 3

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

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

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

**You may stop here as the next step**= 3 x N x (3 + N)/2

**involves factorisation (Sec 1 level)**Testing

-------

1 --> 3 x 1 x (3 + 1)/2 = 3 x 2

= 6 (correct)

2 --> 3 x 2 x (3 + 2)/2 = 15 (correct)

3 --> 3 x 3 x (3 + 3)/2 = 27 (correct)

4 --> 3 x 4 x (3 + 4)/2 = 42 (correct)

5 --> 3 x 5 x (3 + 5)/2 = 60

6 --> 3 x 6 x (3 + 6)/2 = 81

...

10 --> 3 x 10 + (3 + 10)/2 = 195

1st Fig is 1 hexagon - made up of 6 sticks

(Hexagon resting on one edge)

2nd Fig is 3 hexagon - made up of 15 sticks

(1 hexagon resting on two hexagons)

3rd Fig is 6 hexagon - made up of 27 sticks

(The two hexagons resting on 3 hexagons)

Based on figures given,

1 --> 6

2 --> 6 x (1 + 2) - 3 x 1 = 18 - 5

= 5

3 --> 6 x (1 + 2 + 3) - 3 x (1 + 2) = 36 - 9

= 27

N --> 6 x (1 + 2 + ...+ N) - 3 x (1 + 2 + ...N -1)

= 6 x (1 + N)/2 x N - 3 x (1 + N - 1)/2 x (N -1)

= 3 x N x (N + 1) - 3 x N x (N - 1)/2

if factorised will get the same formula as above

Testing

1 --> 3 x 1 x 2 - 3 x 1 x 0/2 = 6 (correct)

2 --> 3 x 2 x 3 - 3 x 2 x 1/2 = 15 (correct)

3 --> 3 x 3 x 4 - 3 x 3 x 2/2 = 27 (correct)

10 --> 3 x 10 x 11 - 3 x 10 x 9/2 = 195 (same as above)

## Wednesday, June 9, 2010

### Speed P6

The distance around the perimeter of a reservoir is 5.52 km. Rick and Reuben started walking from the same point but in the opposite direction round the reservoir. Rick's speed was 25 m/min and he rested for 5 minutes after every 35 minutes. Reuben's speed was 35 m/min and he rested for 15 minutes after every 45 minutes. How long did it take both of them to meet again?

25 x 35 = 875 m

25 x 25 = 625 m

35 x 45 = 1575 m

Ri [ 875 ] [ 875 ] [625]

!____!__!____ !__!___!

Time 35 5 35 5 25 (105 min)

Re [ 1575 ] [ 1575 ]

!______!_____!______!

Time 45 15 45 (105 min)

875 + 875 +625 + 1575 + 1575 = 5525 m

5525 - 5520 = 5 m

25 + 35 = 60 m/min

5 ÷ 60 = 1/12 min

= 5 s

105 min - 5 s = 104 min 55 s

= 1 h 44 min 55 s

It took 1h 44 min 55 s for both of them to meet again.

25 x 35 = 875 m

25 x 25 = 625 m

35 x 45 = 1575 m

Ri [ 875 ] [ 875 ] [625]

!____!__!____ !__!___!

Time 35 5 35 5 25 (105 min)

Re [ 1575 ] [ 1575 ]

!______!_____!______!

Time 45 15 45 (105 min)

875 + 875 +625 + 1575 + 1575 = 5525 m

5525 - 5520 = 5 m

25 + 35 = 60 m/min

5 ÷ 60 = 1/12 min

= 5 s

105 min - 5 s = 104 min 55 s

= 1 h 44 min 55 s

It took 1h 44 min 55 s for both of them to meet again.

## Sunday, June 6, 2010

### Whole Number P5

Mdm Lek had 57 apples and oranges altogether.There were 3 fewer oranges than apples.She gave away half as many oranges as apples to her neighbours. She was left with twice as many oranges as apples.How many apples did she give away?

A [ ][3] )

O [ ] ) 57

2 u --> 57 - 3 = 54

1 u --> 54 ÷ 2 = 27

A --> 27 + 3 = 30

[][][3][][3]

A [][ ][ ] 30

O [][][ ]

3 u --> 30 - 3 - 3 = 24

1 u --> 24 ÷ 3 = 8

30 - 8 = 22 apples

She gave away 22 apples.

A [ ][3] )

O [ ] ) 57

2 u --> 57 - 3 = 54

1 u --> 54 ÷ 2 = 27

A --> 27 + 3 = 30

[][][3][][3]

A [][ ][ ] 30

**A - O --> [ ] - [] = [3],**O [][][ ]

**27****so redraw [ ] as [][3]**3 u --> 30 - 3 - 3 = 24

**1 u is different from the earlier 1 u**1 u --> 24 ÷ 3 = 8

**If you like, you may use 1 su**30 - 8 = 22 apples

She gave away 22 apples.

### Whole Number P6

Alex is as many months old as his grandpa is in years and about as many days old as his dad is in weeks. If the sum of their 3 ages is 140 years, how old is each?

A [][][][][][][][][][][][] months --> [] years

G [][][][][][][][][][][][] years

A [12 x 4 x 7] days

D [12 x 4 x 7] weeks --> [][][][][][][] years

20 u --> 140 years

1 u --> 140 ÷ 20 = 7 years

12 u --> 12 x 7 = 84 years

7 u --> 7 x 7 = 49 years

Alex, his dad and grandpa are 7 years, 49 years and 84 years old respectively.

A [][][][][][][][][][][][] months --> [] years

G [][][][][][][][][][][][] years

A [12 x 4 x 7] days

**x 4 to convert to weeks, x 7 to convert to days**D [12 x 4 x 7] weeks --> [][][][][][][] years

**÷ 4 to convert to months, ÷ 12 to convert to years**20 u --> 140 years

1 u --> 140 ÷ 20 = 7 years

12 u --> 12 x 7 = 84 years

7 u --> 7 x 7 = 49 years

Alex, his dad and grandpa are 7 years, 49 years and 84 years old respectively.

## Thursday, June 3, 2010

### Speed P6

A Van left Port Penang for Town Lalu at 9 a.m. A Car started its journey along the same route as the Van from starting point, two hours later. The car overtook the Van at 3 p.m. The speed of the Car was 30 km/h faster than the Van. The Van took 7 hours to reach Town Lalu. Find the distance between Port Penang and Town Lalu.

3 h 3 h

!----------!-----------!

9 am 12 noon 3 am

6 - 2 = 4 h

30 x 4 = 120 km

120 ÷ 2 = 60 km/h

60 x 7 = 420 km

The distance between Port Penang and Town Lalu was 420 km.

3 h 3 h

!----------!-----------!

9 am 12 noon 3 am

6 - 2 = 4 h

30 x 4 = 120 km

**A****dditional distance travelled by Car****to catch up with Van in 4 h****= Distance travelled by Van in the first 2 h**120 ÷ 2 = 60 km/h

**Speed of van**60 x 7 = 420 km

The distance between Port Penang and Town Lalu was 420 km.

### Algebra P6

If p is directly proportional to the square root of q.The sum of p when q=49 and q=784 is 10.

p ^ 2 = c ^ 2 x q

p1 x p1 = c x c x 49

= c x c x 7 x 7

p1 = 7c

p2 x p2 = c x c x 784

= c x c x 28 x 28

p2 = 28c

p1 + p2 = 7c + 28c

= 35c

35c = 10

c = 10 ÷ 35 = 2/7

p ^ 2 = 2/7 x 2/7 x q

If q = 196, p x p = 2/7 x 2/7 x 196

= 4 x 4

p = 4

**This may not be a P6 Question as square root is not cover in Primary School.**p ^ 2 = c ^ 2 x q

p1 x p1 = c x c x 49

= c x c x 7 x 7

p1 = 7c

p2 x p2 = c x c x 784

= c x c x 28 x 28

p2 = 28c

p1 + p2 = 7c + 28c

= 35c

35c = 10

c = 10 ÷ 35 = 2/7

p ^ 2 = 2/7 x 2/7 x q

If q = 196, p x p = 2/7 x 2/7 x 196

= 4 x 4

p = 4

### Whole Number P6

Before the Great Singapore Sale, Loft Departmental Store and Gallery Departmental Store were selling a similar bag at the same price. During the sale, such a bag was sold at the 2 depatmental stores at the discounted price of $279 and $313 respectively. Mdm Lim bought 2 such bags from each store during the sale and saved $190. What was the discount given for one such bag by Loft Departmental Store during the sale?

L [279][][1u][279][][1u]

G [ 313 ][1u][ 313 ][1u]

313 - 279 = 34

4 u --> 190 - 34 - 34 = 122

1 u --> 122 ÷ 4 = 30.50

30.50 + 34 = $64.50 [][1 u]

The discount given for one such bag by Loft Departmental Store during the sale was $64.50.

L [279][][1u][279][][1u]

G [ 313 ][1u][ 313 ][1u]

313 - 279 = 34

**The difference in price for 1 bag. []**4 u --> 190 - 34 - 34 = 122

1 u --> 122 ÷ 4 = 30.50

30.50 + 34 = $64.50 [][1 u]

The discount given for one such bag by Loft Departmental Store during the sale was $64.50.

**There is no need to convert to percentage discount as Question asks specifically for Discount only.**## Tuesday, June 1, 2010

### Speed P6

A train took 12 hours to travel from Singapore to Ipoh. Its average speed for the first 1/3 of the journey was 1/2 the average speed for the remaining journey. Find the time taken for the first 1/3 of the journey.

Distance [ ][ ][ ]

Speed 1/2 S< S >

Time [ ][ ] [ ] [ ] 12 h

4 u --> 12 h

1 u --> 12 ÷ 4 = 3 h

2 u --> 2 x 3 = 6 h

The time taken for the first 1/3 of the journey was 6 h.

Alternative method

12 h

1/3 D 2/3 D

!-----!-----!-----!

!____!________!

1/2 S S

T1 = 1/3 D ÷ 1/2 S

= 2/3 DS

T2 = 2/3 D ÷ S

= 2/3 DS

T1 + T2 = 12 h

2/3 DS + 2/3 DS = 12 h

T1 = T2 = 12 ÷ 2 = 6 h

Time taken for the first 1/3 of the journey was 6 h.

Distance [ ][ ][ ]

Speed 1/2 S< S >

Time [ ][ ] [ ] [ ] 12 h

**[Each distance is 1 unit, so Time inversely proportional to Speed]**4 u --> 12 h

1 u --> 12 ÷ 4 = 3 h

2 u --> 2 x 3 = 6 h

The time taken for the first 1/3 of the journey was 6 h.

Alternative method

12 h

1/3 D 2/3 D

!-----!-----!-----!

!____!________!

1/2 S S

T1 = 1/3 D ÷ 1/2 S

= 2/3 DS

T2 = 2/3 D ÷ S

= 2/3 DS

T1 + T2 = 12 h

2/3 DS + 2/3 DS = 12 h

T1 = T2 = 12 ÷ 2 = 6 h

Time taken for the first 1/3 of the journey was 6 h.

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