Water from 3 taps was used to fill an empty water tank. Tap X, Tap Y, Tap Z could fill the tank completely in 3 hours, 4 hours, 5 hours respectively.
(a) What fraction of the tank was filled with water after half an hour when all taps were turned on at the same time? Give your answer in the simplest form.
Now, only Tap X and Tap Y were both turned on to fill another empty water tank. Immediately after 20 minutes, Tap Z was also turned on. Both Tap X and Tap Y were turned off at the same time after turning on for an hour and only Tap Z was left to fill the tank completely.
(b) How much longer would Tap Z take to fill the tank completely with water after both Tap X and Tap Y were turned off? Express your answer in terms of hours in its simplest form.
(a) 47/120 (b) 1 & 5/12 h
X + Y + Z : 1 h > 1/3 + 1/4 + 1/5 = 47/60
1/2 h > 1/2 x 4/60 = 47/120
a) It was 47/120.
60  20 = 40 min
= 40/60 h
= 2/3 h
Z : 1 h > 1/5
2/3 h > 2/3 x 1/5 = 2/15
X + Y : 1 h > 1/3 + 1/4 = 7/12
2/15 + 7/12 = 43/60
1  43/60 = 17/60
60/60 > 5 h
17/60 > 17/60 x 5 = 17/12 h
= 1 5/12 h
b) It was 1 5/12 h.
Thursday, September 30, 2010
Wednesday, September 29, 2010
Ratio P6
Peter and Paul had a total of 63 marbles. During a game, each of them won 9 marbles and Peter then had 25% more marbles than Paul.
a) How many more marbles did Peter have than Paul after the game?
b) What was the ratio of the number of marbles Paul had to the number of marbles Peter had before the game?
Pe [][][][][] ] 63 + 9 + 9
Pa [][][][] ] = 81
9 u > 81
1 u > 81 / 9 = 9
a) Peter had 9 more marbles than Paul after the game. (before also  Constant Difference)
5 u > 5 x 9 = 45
45  9 = 36
4 u > 4 x 9 = 36
36  9 = 27
27 : 36 = 3 : 4 Be careful, the ratio required is Paul to Peter
b) It was 3 : 4.
a) How many more marbles did Peter have than Paul after the game?
b) What was the ratio of the number of marbles Paul had to the number of marbles Peter had before the game?
Pe [][][][][] ] 63 + 9 + 9
Pa [][][][] ] = 81
9 u > 81
1 u > 81 / 9 = 9
a) Peter had 9 more marbles than Paul after the game. (before also  Constant Difference)
5 u > 5 x 9 = 45
45  9 = 36
4 u > 4 x 9 = 36
36  9 = 27
27 : 36 = 3 : 4 Be careful, the ratio required is Paul to Peter
b) It was 3 : 4.
Whole Number P6
Mrs Jones has some almond and butter cookies.If she packs an almond cookie with a butter cookie,there will be 60 butter cookies left.If she packs every 2 almond cookies with 3 butter cookies,there will be 50 almond cookies left.How many cookies does she have in all?
A [ ]
B [ ][60]
< 1u >
A [ ][ ][50]
B [ ][ ][ ]
A [ ][ ][50]
B [ ][ ][50][60]
< 1 u >
1 u > 50 + 60 = 110
5 u > 5 x 110 = 550
550 + 50 = 600 cookies
She has 600 cookies.
A [ ]
B [ ][60]
< 1u >
A [ ][ ][50]
B [ ][ ][ ]
A [ ][ ][50]
B [ ][ ][50][60]
< 1 u >
1 u > 50 + 60 = 110
5 u > 5 x 110 = 550
550 + 50 = 600 cookies
She has 600 cookies.
Monday, September 27, 2010
Speed P6
Cynthia walked to the library from school and at the same time, Mrs Lee walked to school from the library. If they walked towards each other, they would meet 4 min later. If they walked in the same direction, it would take Cynthia 20 min to catch up with Mrs Lee whose walking speed was 40 m/min. How far was the library from the school? Assume that Mrs Lee's walking speed was constant at all times.
40 x 4 = 160 m
X 160
C[ ][ ]L
X + 160
C[ ][ ]
L[ ]
20 min > X + 160 Difference in distance
4 min > X  160 Difference in distance
20 min > 5 X  5 x 160 = 5 X  800
5 X  800 > X + 160
4 X > 800 + 160 = 960
X . 960/4 = 240 m
240 + 160 = 400 m
It was 400 m.
40 x 4 = 160 m
X 160
C[ ][ ]L
X + 160
C[ ][ ]
L[ ]
20 min > X + 160 Difference in distance
4 min > X  160 Difference in distance
20 min > 5 X  5 x 160 = 5 X  800
5 X  800 > X + 160
4 X > 800 + 160 = 960
X . 960/4 = 240 m
240 + 160 = 400 m
It was 400 m.
Speed P6
Last Monday, Danny walked from home to market. For the first 5 minutes, he was walking at an average speed of 35m/min. When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes. Find the distance between market and home.
5 min 4 min late
> 35 m/min X

H M
> 45 m/min 2 min early
X/35  X/45 > 4 +2
2/315 X > 6
X > 315/2 x 6 = 945 m
35 x 5 = 175 m
945 + 175 = 1120 m
= 1.12 km
It was 1.12 km.
5 min 4 min late
> 35 m/min X

H M
> 45 m/min 2 min early
X/35  X/45 > 4 +2
2/315 X > 6
X > 315/2 x 6 = 945 m
35 x 5 = 175 m
945 + 175 = 1120 m
= 1.12 km
It was 1.12 km.
Speed P6
Jacki’s mum leaves home at 12.35 pm every day and arrives 25 minutes later at his school to pick him up. She always cycles at the same speed along the same route. One day Jacki was dismissed early at 12.30 pm, so he started walking home at 6 km/h, hoping to meet her along the way. He met her at X and got onto her bicycle. They reached home 15 minutes earlier than before. How far did Jacki walked before his mum picked him up?
15 / 2 = 7. 5 min
25 : 7.5 = 10 : 3
12.35 am 7.5 min
C [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] 25 min
[ ][ ][ ]
7.5 min
6 km/h
< J
12.30 am
10 u > 25 min
7 u > 7/10 x 25 = 17.5 min
17.5 + 5 = 22.5 min
6 x 22.5/60 = 2.25 km
It was 2.25 km.
15 / 2 = 7. 5 min
25 : 7.5 = 10 : 3
12.35 am 7.5 min
C [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] 25 min
[ ][ ][ ]
7.5 min
6 km/h
< J
12.30 am
10 u > 25 min
7 u > 7/10 x 25 = 17.5 min
17.5 + 5 = 22.5 min
6 x 22.5/60 = 2.25 km
It was 2.25 km.
Speed P6
Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15 km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles reach the finishing line?
T [ ][15]
C [ ]
9.35 am time?
 
T [ ][ ][ ][15][15][15]
C [ ][ ][ ]

time? = time?
15 x 3 = 45 km
45 / 90 = 1/2 h
= 1/2 x 60 min
= 30 min
25 min 5 min

9.35 am 10 am 10.05 am
It was 10.05 a.m.
T [ ][15]
C [ ]
9.35 am time?
 
T [ ][ ][ ][15][15][15]
C [ ][ ][ ]

time? = time?
15 x 3 = 45 km
45 / 90 = 1/2 h
= 1/2 x 60 min
= 30 min
25 min 5 min

9.35 am 10 am 10.05 am
It was 10.05 a.m.
Saturday, September 25, 2010
Percentage P6 2010 SA2 Nan Hua P2 Q15
Anna, Belinda and Clare bought a vase and shared the cost equally among themselves. Clare did not bring her money, so Anna and Belinda paid for the vase first. Anna paid 0.25 more than what Belinda had paid. The following day, Clare paid her share to both Anna and Belinda.
a) Clare paid Belinda $6.55. How much did Clare pay Anna?
b) Clare's brother bought a similar vase at the same shop during a sale and was given a discount of 20% of the price that Clare and her friends paid for. How much did Clare's brother pay for the Vase?
A [][][][][] 0.25 = 1/4
B [][][][] Highlighted  Clare's share
1 u > $6.55
2 u > 2 x 6.55 = $13.10
a) It was $13.10.
100% : 9 u > 9 x 6.55 = $58.95
80% > 90/100 x 58.95 = $47.16
b) It was $47.16.
a) Clare paid Belinda $6.55. How much did Clare pay Anna?
b) Clare's brother bought a similar vase at the same shop during a sale and was given a discount of 20% of the price that Clare and her friends paid for. How much did Clare's brother pay for the Vase?
A [][][][][] 0.25 = 1/4
B [][][][] Highlighted  Clare's share
1 u > $6.55
2 u > 2 x 6.55 = $13.10
a) It was $13.10.
100% : 9 u > 9 x 6.55 = $58.95
80% > 90/100 x 58.95 = $47.16
b) It was $47.16.
Labels:
M P6 NH,
M Percentage P6,
M Percentage P6 NH
Friday, September 24, 2010
Challenging P6
Tom bought 34 pens and books altogether. Each pen costs $0.40 and each book costs $2.20. The difference between the cost of the pens and the cost of the books is $9.80. How many pens and how many books did Tom buy?
34 x 2.20 = $74.80 34 books
74.80  9.80 = $65 Difference too much, need to be reduced by this amount.
Note: Actually which item cost more was not given.
See alternative method below.
2.20 + 0.40 = $2.60 Exchanging 1 book for 1 pen, difference reduces by $2.60
65 / 2.60 = 25 pens Need to exchange 25 books for 25 pens
34  25 = 9 books
Tom bought 25 pens and 9 books.
Alternative method

34 x 0.40 = $13.60
13.60  9.80 = $3.80 Assumption wrong!
2.20 + 0.40 = $2.60
3.80 / 2.60 = ?????????? Not a whole number.
13.60 + 9.80 = $23.40 Books should cost more than pens
2.20 + 0.40 = $2.60
23.40 / 2.60 = 9 books
34  9 = 25 pens
He bought 25 pens and 9 books.
34 x 2.20 = $74.80 34 books
74.80  9.80 = $65 Difference too much, need to be reduced by this amount.
Note: Actually which item cost more was not given.
See alternative method below.
2.20 + 0.40 = $2.60 Exchanging 1 book for 1 pen, difference reduces by $2.60
65 / 2.60 = 25 pens Need to exchange 25 books for 25 pens
34  25 = 9 books
Tom bought 25 pens and 9 books.
Alternative method

34 x 0.40 = $13.60
13.60  9.80 = $3.80 Assumption wrong!
2.20 + 0.40 = $2.60
3.80 / 2.60 = ?????????? Not a whole number.
13.60 + 9.80 = $23.40 Books should cost more than pens
2.20 + 0.40 = $2.60
23.40 / 2.60 = 9 books
34  9 = 25 pens
He bought 25 pens and 9 books.
Challenging P6
On a farm, there are 28 chickens and cows. The cows have 40 more legs than the chickens. How many cows and chickens are there?"
28 x 4 = 112 legs 28 cows
112  40 = 72 legs Difference too much (so needs to be reduced)
4 + 2 = 6 legs Exchanging 1 cow for 1 chicken (difference in legs increases by 6)
72 / 6 = 12 chickens 12 cows need to be exchanged for 12 chickens
28  12 = 16 cows
There are 16 cows and 12 chickens.
Alterrnative method

40 / 4 = 10 cows Remove the additonal legs from cows, so now the remaining cows and chickens have the same number of legs
28  10 = 18 Remaining cows and chickens
1 + 2 = 3 1 set [1 cow (4 legs) > 2 chickens (4 legs)]
18 / 3 = 6 sets
6 x 2 = 12 chickens
28  12 = 16 cows
There are 16 cows and 12 chickens.
Alternative method

4  2 = 2 legs 1 set [Difference between 1 cow and 1 chicken]
40 / 2 = 20 sets
20 + 20  28 = 12 Too many animals, need to reduce the number by 12.
2 + 1 = 3 1 group [To ensure that the difference in number of legs is maintained, whenever 1 cow reduced, 2 chickens would have to be reduced]
12 / 3 = 4 groups To be reduced
20  4 = 16 cows
20  4 x 2 = 12 chickens
There are 16 cows and 12 chickens.
28 x 4 = 112 legs 28 cows
112  40 = 72 legs Difference too much (so needs to be reduced)
4 + 2 = 6 legs Exchanging 1 cow for 1 chicken (difference in legs increases by 6)
72 / 6 = 12 chickens 12 cows need to be exchanged for 12 chickens
28  12 = 16 cows
There are 16 cows and 12 chickens.
Alterrnative method

40 / 4 = 10 cows Remove the additonal legs from cows, so now the remaining cows and chickens have the same number of legs
28  10 = 18 Remaining cows and chickens
1 + 2 = 3 1 set [1 cow (4 legs) > 2 chickens (4 legs)]
18 / 3 = 6 sets
6 x 2 = 12 chickens
28  12 = 16 cows
There are 16 cows and 12 chickens.
Alternative method

4  2 = 2 legs 1 set [Difference between 1 cow and 1 chicken]
40 / 2 = 20 sets
20 + 20  28 = 12 Too many animals, need to reduce the number by 12.
2 + 1 = 3 1 group [To ensure that the difference in number of legs is maintained, whenever 1 cow reduced, 2 chickens would have to be reduced]
12 / 3 = 4 groups To be reduced
20  4 = 16 cows
20  4 x 2 = 12 chickens
There are 16 cows and 12 chickens.
Thursday, September 23, 2010
Decimal P4
Jacob saved 60 cents a day. On the ninth day after he had started saving, Jason started saving $1 each day. How many days did Jason take to save the same amount of money as Jacob?
$0.60 x 8 = $4.80 Amount Jacob saved before Jason started saving
$1.00  $0.60 = $0.40 Difference in savings per day
4.80 ÷ 0.40 = 12 days Number of days for Jason to catch up
Jason took 12 days to save the same amount of money as Jacob.
$0.60 x 8 = $4.80 Amount Jacob saved before Jason started saving
$1.00  $0.60 = $0.40 Difference in savings per day
4.80 ÷ 0.40 = 12 days Number of days for Jason to catch up
Jason took 12 days to save the same amount of money as Jacob.
Wednesday, September 22, 2010
Whole Number P6
James had some money in his wallet with values between $60 to $150. If he buys the maximum number of water bottles he can afford at $7 each, he will have $1 left. If he buys the maximum number of towels at $9 each, he would have $3 left. What is the actual amount James had?
7 x (n + 1) + 1 > 9 x n + 3
2 n > 7 + 1  3 = 5 X
7 x (n + 2) + 1 > 9 x n + 3
2 n > 14 + 1  3 = 12
n > 12/2 = 6
9 x 6 + 3 = 57 Answer not within the range
9 x 7 = 63 9 units of 7 > 7 parts of 9
63 + 57 = $120 So need to add multiple of 63 (7 x 9 or 9 x 7)
It was $120.
Alternative method

[][][][][][][][ 7+1 ] [7][7][7][7][7][7][7][7][7]
[][][][][][][][][][3] [ 9 ][ 9 ][9 ][9 ][ 9 ][9 ][9 ]
2 u > 7 + 1  3 = 5 X
[][][][][][][][7 x 2 + 1]
[][][][][][][][][][ 3 ]
2 u > 7 x 2 + 1  3 = 12
1 u > 12/2 = 6
6 x 9 + 3 = 57 Not within range
7 x 9 = 63
63 + 57 = $120
It was $120.
7 x (n + 1) + 1 > 9 x n + 3
2 n > 7 + 1  3 = 5 X
7 x (n + 2) + 1 > 9 x n + 3
2 n > 14 + 1  3 = 12
n > 12/2 = 6
9 x 6 + 3 = 57 Answer not within the range
9 x 7 = 63 9 units of 7 > 7 parts of 9
63 + 57 = $120 So need to add multiple of 63 (7 x 9 or 9 x 7)
It was $120.
Alternative method

[][][][][][][][ 7+1 ] [7][7][7][7][7][7][7][7][7]
[][][][][][][][][][3] [ 9 ][ 9 ][9 ][9 ][ 9 ][9 ][9 ]
2 u > 7 + 1  3 = 5 X
[][][][][][][][7 x 2 + 1]
[][][][][][][][][][ 3 ]
2 u > 7 x 2 + 1  3 = 12
1 u > 12/2 = 6
6 x 9 + 3 = 57 Not within range
7 x 9 = 63
63 + 57 = $120
It was $120.
NMOS 2010 Prelim round Q15
Town A and Town B are 12 km apart. At 10 a.m., Tom and Harry travelled towards each other from Town A and Town B respectively. Tom walked at a speed of 20 m/min for 2 minutes after every 10 minutes of running at 165 m/min. Harry took a 4 minutes rest after each 15 minutes of running at 165 m/min. What time did they meet each other? [2 marks]
Tom
Distance (m)
1650 40 1650 40 1650 40 330
?
0 10 12 22 24 34 36 38
Time (min)
Harry
165 x 15 = 2475 m
Distance (m)
2475 0 2475 0
?
0 15 19 34 38
Time (min)
1650 x 3 + 40 x 3 + 330 = 5400 m
2475 x 2 = 4950 m
5400 + 4950 = 10350 m
12000  10350 = 1650 m
1650 /(165+165) = 5 min
38 + 5 = 43 min
43 min

10 am 10.43 am
It was 10.43 a.m.
Tom
Distance (m)
1650 40 1650 40 1650 40 330
?
0 10 12 22 24 34 36 38
Time (min)
Harry
165 x 15 = 2475 m
Distance (m)
2475 0 2475 0
?
0 15 19 34 38
Time (min)
1650 x 3 + 40 x 3 + 330 = 5400 m
2475 x 2 = 4950 m
5400 + 4950 = 10350 m
12000  10350 = 1650 m
1650 /(165+165) = 5 min
38 + 5 = 43 min
43 min

10 am 10.43 am
It was 10.43 a.m.
Angles P5 2009 SA2 ACS P2 Q11
180  60 = 120
120 / 2 = 60 A = B (folded)
180  90  60 = 30 C (Rightangled triangle)
180  30  30  20  20 = 80 degree D = C ; E = 20 (folded)
It is 80 degree.
Tuesday, September 21, 2010
Fraction P6 20010 SA2 ACS P2 Q13
Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal number of beads in each small box and equal number of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box? (Ans: 9)
15 x 5 = 75
5 x 14 u > 3 x (15 u + 75)
70 u > 45 u + 225
25 u > 225
1 u > 225/25 = 9 beads
There were 9 beads.
15 x 5 = 75
5 x 14 u > 3 x (15 u + 75)
70 u > 45 u + 225
25 u > 225
1 u > 225/25 = 9 beads
There were 9 beads.
Labels:
M Fraction P6,
M Fraction P6 ACS,
M P6 ACS
Volume P6 2010 SA2 ACS P2 Q16
A tank measuring 80 cm by 60 cm by 55 cm was 1/5 filled with water. A tap was turned on to fill it up with water at a rate of 9 litres per min. Every 30 seconds after the tap was turned on, an iron ball with a volume 500 cm3 was dropped into the tank. How many iron balls would there be in the tank when the water level reached the brim? (Ans 42 iron balls)
4/5 x 80 x 60 x 55 = 211200 cm3
60 s > 9000 cm3
30 s > 30/60 x 9000 = 4500 cm3
1 set > 4500 + 500 = 5000 cm3
211200 / 5000 = 42 sets (max) > 42 iron balls
There were 42 iron balls.
4/5 x 80 x 60 x 55 = 211200 cm3
60 s > 9000 cm3
30 s > 30/60 x 9000 = 4500 cm3
1 set > 4500 + 500 = 5000 cm3
211200 / 5000 = 42 sets (max) > 42 iron balls
There were 42 iron balls.
Percentage P6 2010 SA2 Catholic High P2 Q??
Daniel had 40% more stickers than Brandon. Daniel and Brandon each gave 20% of their stickers to Calvin. As a result, Calvin's stickers increased by 80%. If Daniel has 20 more stickers than Calvin in the end, how many stickers did Brandon have at first?
B > 100%
D > 140%
20/100 x (100 + 140)% = 48%
80% > 48%
100% > 100/80 x 48% = 60%
DN > 80/100 x 140% = 112%
CN > 60% + 48% = 108%
(112  108)% = 4%
4% > 20 stickers
100% > 100/4 x 20 = 500 stickers
a) He had 500 stickers at first.
B > 100%
D > 140%
20/100 x (100 + 140)% = 48%
80% > 48%
100% > 100/80 x 48% = 60%
DN > 80/100 x 140% = 112%
CN > 60% + 48% = 108%
(112  108)% = 4%
4% > 20 stickers
100% > 100/4 x 20 = 500 stickers
a) He had 500 stickers at first.
Labels:
M P6 CH,
M Percentage P6,
M Percentage P6 CH
Percentage P6 2010 SA2 Catholic High P2 Q??
John had $180 more than Raymond at first. Both bought a different printer with some of their money. In the end, Raymond was left with $20 more than John. If John's printer cost 50% more than Raymond's printer, how much does John's printer cost?
J [][ ][ ][ ]
R [][$20]<$180>[ ][ ]
1 u > 20 + 180 = 200
3 u > 3 x 200 = $600
It costs $600.
J [][ ][ ][ ]
R [][$20]<$180>[ ][ ]
1 u > 20 + 180 = 200
3 u > 3 x 200 = $600
It costs $600.
Labels:
M P6 CH,
M Percentage P6,
M Percentage P6 CH
Ratio P6 2010 SA2 Catholic High P2 Q??
At a conference made up of speakers and participants, there were 20% more men than women. The ratio of male speakers to female speakers was 8 : 5. There was an equal number of male and female participants.
a) Find the ratio of male speakers to male participants at the conference.
b) Halfway, 40 male participants left the conference and another 60 female participants joined the conference. In the end, there were 3/4 as many male participants as female participants remaining behind. How many speakers were there at the conference ?
M : F = 120 : 100
= 6 : 5 Multiply by 3 to get difference of 3
= 18 : 15 (equal to difference in speaker)
MS : FS = 8 : 5 Difference of 3 (8  5)
MP : FP = 1 : 1 Multiply by 10 so that totals tally
= 10 : 10
MS : MP = 8 : 10
= 4 : 5
a) It was 4 : 5.
MP FP

10 : 10
 40 + 60

3 : 4

4 x (10 u  40) > 3 x (10 u + 60)
40 u  160 > 30 u + 180
10 u > 180 + 160 = 340
1 u > 340/10 = 34
13 u > 13 x 34 = 442 speakers
a) There were 442 speakers.
a) Find the ratio of male speakers to male participants at the conference.
b) Halfway, 40 male participants left the conference and another 60 female participants joined the conference. In the end, there were 3/4 as many male participants as female participants remaining behind. How many speakers were there at the conference ?
M : F = 120 : 100
= 6 : 5 Multiply by 3 to get difference of 3
= 18 : 15 (equal to difference in speaker)
MS : FS = 8 : 5 Difference of 3 (8  5)
MP : FP = 1 : 1 Multiply by 10 so that totals tally
= 10 : 10
MS : MP = 8 : 10
= 4 : 5
a) It was 4 : 5.
MP FP

10 : 10
 40 + 60

3 : 4

4 x (10 u  40) > 3 x (10 u + 60)
40 u  160 > 30 u + 180
10 u > 180 + 160 = 340
1 u > 340/10 = 34
13 u > 13 x 34 = 442 speakers
a) There were 442 speakers.
Speed P6 2010 SA2 Catholic High P2 Q15
At 9am John was cycling from Point A to Point B. At the same time, Mary was cycling from point B to point A using the same route as John. Cycling at a speed of 4km/h faster than Mary, they would pass each other 300 meter away from the mid point.
a) What time did they pass each other?
b) If John took another 3 minutes to reach Point B, what time would Mary reach Point A?
J
> S + 4 km/h
9 am 300 m 300 m <3 min>

A X m X B
9 am
S km/h <
M
< 9 min >
4000 m > 60 min
600 m > 600/4000 x 60 = 9 min
9 min

9 am 9.09 am
a) It was 9.09 am.
3 X > X + 300 + 300 3 min versus 9 min
2 X > 600
X > 600/2 = 300 m
300 m > 9 min
1200 m > 1200/300 x 9 = 36 min
36 min

9 am 9.36 am
b) It was 9.36 am.
a) What time did they pass each other?
b) If John took another 3 minutes to reach Point B, what time would Mary reach Point A?
J
> S + 4 km/h
9 am 300 m 300 m <3 min>

A X m X B
9 am
S km/h <
M
< 9 min >
4000 m > 60 min
600 m > 600/4000 x 60 = 9 min
9 min

9 am 9.09 am
a) It was 9.09 am.
3 X > X + 300 + 300 3 min versus 9 min
2 X > 600
X > 600/2 = 300 m
300 m > 9 min
1200 m > 1200/300 x 9 = 36 min
36 min

9 am 9.36 am
b) It was 9.36 am.
Thursday, September 16, 2010
Whole Number P6
A total of 548 adults and teenagers collected newspapers for a recycling project. Each adult collected 12 kg of newspaper. The teenagers worked in groups of 4 and each group collected 9 kg of newspapers. At the end of the project, a total 5250 kg of newspaper were collected. How many teenagers took part in the recycling project?
4 adults > 12 x 4 = 48 kg
4 teenagers > 9 kg
548 teenagers > 548/4 x 9 = 1233 kg
5250  1233 = 4017 kg
48  9 = 39 kg
4017 / 39 = 103 groups of adults
103 x 4 = 412 adults
548  412 = 136 teenagers
136 teenagers took part.
Check

136/4 x 9 + 412 x 12 = 5250 kg (correct)
4 teenagers > 9 kg
4 adults > 4 x 12 = 48 kg
548 adults > 548 x 12 = 6576 kg
6576  5250 = 1326 kg
48  9 = 39 kg
1326 / 39 = 34 groups of teenagers
34 x 4 = 136 teenagers
136 teenagers took part.
4 adults > 12 x 4 = 48 kg
4 teenagers > 9 kg
548 teenagers > 548/4 x 9 = 1233 kg
5250  1233 = 4017 kg
48  9 = 39 kg
4017 / 39 = 103 groups of adults
103 x 4 = 412 adults
548  412 = 136 teenagers
136 teenagers took part.
Check

136/4 x 9 + 412 x 12 = 5250 kg (correct)
4 teenagers > 9 kg
4 adults > 4 x 12 = 48 kg
548 adults > 548 x 12 = 6576 kg
6576  5250 = 1326 kg
48  9 = 39 kg
1326 / 39 = 34 groups of teenagers
34 x 4 = 136 teenagers
136 teenagers took part.
Wednesday, September 15, 2010
Volume P6
A container filled with some water as shown in Figure A below has a rectangular base area of 10.4cm2. Figure B shows the same container being turned upside down.
(a) Find the capacity of the container in cm3, correct to 2 decimal places.
(b) If the water in the container is transferred to another cubeshaped container of capacity 90cm3, find the height of the water level in the cubeshaped container.
(a) Find the capacity of the container in cm3, correct to 2 decimal places.
(b) If the water in the container is transferred to another cubeshaped container of capacity 90cm3, find the height of the water level in the cubeshaped container.
Assume 4.3 cm. [When assume other number < 4.6 cm, different volume will be obtained.]
4.6 x 10.4 = 47.84
4.3 x 10.4 = 44.72
47.84  44,72 = 3.12
7.8 x 10.4 = 81.12
81.12 + 3.12 = 84.24 cm3
a) It was 84.24 cm3. < 128.96 cm3
90^(1/3) = 4.48 Not sure if this is covered in P6
47.84/90 x 4.48 = 2.38 cm
b) It was 2.38 cm.
Speed P6
At 0800, Noel left Town A for Town B while Jane left Town B for Town A. At 1600 their cars passed each other. 5 h later Jane reached Town A but Noel was 240 km away from Town B.
(a) How long after they started would the two drivers passed each other?
(b) Find the distance between the two towns.
8 h

0800 1600
a) It was 8 h.
N [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] units [8 u : 5 u]
J [ 240 km ][][][][][][] [][][][][][] [] parts [240 + 5 p : 8 p]
5 x (5 p + 240) > 8 x 8 p
64 p  25 p > 1200
39 p > 1200
13 p > 13/39 x 1200 = 400
400 + 240 = 640 km
b) It was 640 km.
(a) How long after they started would the two drivers passed each other?
(b) Find the distance between the two towns.
8 h

0800 1600
a) It was 8 h.
N [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] units [8 u : 5 u]
J [ 240 km ][][][][][][] [][][][][][] [] parts [240 + 5 p : 8 p]
5 x (5 p + 240) > 8 x 8 p
64 p  25 p > 1200
39 p > 1200
13 p > 13/39 x 1200 = 400
400 + 240 = 640 km
b) It was 640 km.
Speed P6
At 0954 Mary left Everton Park and cycled towards Serene Park while Joy left Serene Park and cycled towards Everton Park. At 1004, the two cyclists passed each other. 5 minutes later, Mary reached Serene Park but Joy was 780 m from Everton Park.
(a) At what speed were the two cyclists approaching each other?
(b) Find the distance (in km) between Everton Parkf and Serene Park.
TM1 : TM2 = 10 : 5 = 2 : 1 DM1 : DM2 = 20 : 10
JM1 : JM2 = 10 : 5 = 2 : 1 DJ1 : DJ2 = 2 : 1 = 10 : 5
(DM2 = DJ1)
DM1  DJ2 = 20  5 = 15 u > 780 m
M [][][][][][][][][][][][][][][][][][][][][][][][][][][][][][]
J [][][][][][][][][][][][][][][] [][][][][][][][][][][][][][][]
< 780 m >
15 u > 780
1 u > 780/15 = 52
30 u > 30 x 52 = 1560 m
= 1.56 km
1560/10 = 156 m/min
a) It was 156 m/ min. Approaching speed (Combined speed)
b) It was 1.56 km.
(a) At what speed were the two cyclists approaching each other?
(b) Find the distance (in km) between Everton Parkf and Serene Park.
TM1 : TM2 = 10 : 5 = 2 : 1 DM1 : DM2 = 20 : 10
JM1 : JM2 = 10 : 5 = 2 : 1 DJ1 : DJ2 = 2 : 1 = 10 : 5
(DM2 = DJ1)
DM1  DJ2 = 20  5 = 15 u > 780 m
M [][][][][][][][][][][][][][][][][][][][][][][][][][][][][][]
J [][][][][][][][][][][][][][][] [][][][][][][][][][][][][][][]
< 780 m >
15 u > 780
1 u > 780/15 = 52
30 u > 30 x 52 = 1560 m
= 1.56 km
1560/10 = 156 m/min
a) It was 156 m/ min. Approaching speed (Combined speed)
b) It was 1.56 km.
Saturday, September 11, 2010
Whole Number P6 2010 SA2 CHIJ P2 Q11
There were some magnets in a box for sale. Mrs Frank bought ½ of the magnets in the box and was given 5 magnets free of charge. Next, Mrs Samson bought ½ of the remaining magnets and was given 3 magnets free of charge.Later, Mrs Parker bought ½ of the remaining magnets and was given 2 magnets free of charge.If there were a dozen magnets left in the box, how many magnets were there at first? [4 marks][Answer given: 134]
12 + 2 = 14
14 x 2 = 28
28 + 3 = 31
31 x 2 = 62
62 + 5 = 67
67 x 2 = 134 magnets
12 + 2 = 14
14 x 2 = 28
28 + 3 = 31
31 x 2 = 62
62 + 5 = 67
67 x 2 = 134 magnets
Labels:
M Whole Number P6,
M Whole Number P6 CHIJ
Volume P6 2010 SA2 CHIJ P2 Q12
The base of an empty tank measures 64 cm by 25 cm. It was being filled with water from Tap A and Tap B at a rate of 4 litres per minute and 10 litres per minute respectively. Both taps were turned on at the same time and they took 8 minutes to fill up the tank completely. (a) What was the height of the tank? (b) How much longer would it take if Tap A alone is used to fill up the empty tank? [4 marks][Answer given: 70 cm, 20 min]
4 + 10 = 14 litres/min
14 x 8 = 112 litres
112 x 1000 / (64 x 25) = 70 cm
a) It was 70 cm.
112/4 = 28
28  8 = 20 min
b) It was 20 min longer.
4 + 10 = 14 litres/min
14 x 8 = 112 litres
112 x 1000 / (64 x 25) = 70 cm
a) It was 70 cm.
112/4 = 28
28  8 = 20 min
b) It was 20 min longer.
Circle P6 2010 SA2 CHIJ P2 Q13
1/4 x 2 x 3.14 x 4 = 6.28
1/2 x 2 x 3.14 x 3 = 9.42
14 + 6.28 + 9.42 = 29.7 cm
a) It was 29.7 cm.
1/4 x 3.14 x 4 x 4 = 12.56 A + B
1/2 x 3.14 x 3 x 3 = 14.13 C + B
14.13  12.56 = 1.57 cm2.
Labels:
M Circle P6,
M Circle P6 CHIJ,
M P6 CHIJ
Speed P6 2010 SA2 CHIJ P2 Q15
Kitty Town and Melody Town are 234 km apart. Danny left Kitty Town for Melody Town at 8.42 a.m. travelling at an average speed of 85 km/h. At the same time, Jasmine left Melody Town for Kitty Town. They met each other at 10.30 a.m. (a) What was Jasmine’s average speed when she met Danny? (b) If Jasmine were to increase her speed by 26 km/h before meeting Danny, how much less time would she take before meeting him? [4 marks][Answer given: 45 km/h, 18 min
1 h 18 min 30 min

8.42 am 9.42 am 10 am 10.30 am
1 h 48 min = 1.8 h
234/ 1.8 = 234 x 5/9
= 130 km/h
130  85 = 45 km/h
a) It was 45 km/h.
45 + 26 = 71 km/h
71 + 85 = 156 km/h
234/156 = 1.5 h
1.8  1.5 = 0.3 h
= 0.3 x 60 min
= 18 min
b) It was 18 min.
1 h 18 min 30 min

8.42 am 9.42 am 10 am 10.30 am
1 h 48 min = 1.8 h
234/ 1.8 = 234 x 5/9
= 130 km/h
130  85 = 45 km/h
a) It was 45 km/h.
45 + 26 = 71 km/h
71 + 85 = 156 km/h
234/156 = 1.5 h
1.8  1.5 = 0.3 h
= 0.3 x 60 min
= 18 min
b) It was 18 min.
Percentage P6 2010 SA2 CHIJ P2 Q16
Miss Flora had tulips, roses and carnations in her shop. In July, she sold a total of 1350 tulips, roses and carnations, of which 30% were tulips. There were as many roses as tulips sold. In August, the sale of carnations increased by 45%. This made up 27% of the total sales of tulips, roses and carnations. What is the percentage increase in the total sales of tulips and roses from July to August? (Leave your answer correct to 2 decimal places.)[5 marks][Answer given: 161.36%]
TJ > 30/100 X 1350 = 405
RJ > 405
405 + 405 = 810
CJ > 1350  810 = 540
CA > 145/100 x 540 = 783
27% > 783
73% > 73/27 x 783 = 2117
2117  810 = 1307
1307/810 x 100% = 161.36%
It was 161.36% increase.
TJ > 30/100 X 1350 = 405
RJ > 405
405 + 405 = 810
CJ > 1350  810 = 540
CA > 145/100 x 540 = 783
27% > 783
73% > 73/27 x 783 = 2117
2117  810 = 1307
1307/810 x 100% = 161.36%
It was 161.36% increase.
Ratio P6 2010 SA2 CHIJ P2 Q17
There were a total of 480 pink and black pearls in a box. The number of pink pearls was 1½ times that of black pearls. Some silver pearls and some white pearls were put into the box. For every 6 pink pearls that were already in the box, 13 white ones were added. Then the final number of black peals and silver pearls was 25% of the final number of pink pearls and white pearls. What is the ratio of the number of silver pearls to the number of white ones? [5 marks][Answer given: 3: 52]
5 u > 480
1 u > 480/5 = 96
P : 3 u > 3 x 96 = 288
B > 480  288 = 192
W > 288/6 x 13 = 624
P + W > 288 + 624 = 912
B + S > 25/100 x 912 = 228
S > 228  192 = 36
S : W = 36 : 624
= 3 : 52
It was 3 : 52.
5 u > 480
1 u > 480/5 = 96
P : 3 u > 3 x 96 = 288
B > 480  288 = 192
W > 288/6 x 13 = 624
P + W > 288 + 624 = 912
B + S > 25/100 x 912 = 228
S > 228  192 = 36
S : W = 36 : 624
= 3 : 52
It was 3 : 52.
Whole Number P6 2010 CHIJ P2 Q18
Hiroshi collects gamecards. Every day, he gets 10 additional new gamecards. On every third day, he gives three cards to his friend, Miki, If Hiroshi starts with eight cards on the first day, on which day will he have exactly 180 cards?[5 marks][20th day]
Day 1 > 8
Day 2 > 8 + 10 = 18
Day 3 > 18 + 10 = 28
Day 4 > 28 + 10  3 = 35
35  8 = 27
180  8 = 172
172/27 = 6 r 10
6 x 3 = 18
1 + 18 + 1 = 20th day
It was the 20th day.
Day 1 > 8
Day 2 > 8 + 10 = 18
Day 3 > 18 + 10 = 28
Day 4 > 28 + 10  3 = 35
35  8 = 27
180  8 = 172
172/27 = 6 r 10
6 x 3 = 18
1 + 18 + 1 = 20th day
It was the 20th day.
Labels:
M Whole Number P6,
M Whole Number P6 CHIJ
Volume P6 2010 SA2 RGS P2 Q11
A tank with a square base of side 60 cm, 80 cm tall, was 1/3 filled with water. (a) Find the volume of water in the tank (b) At 1 p.m., a tap was turned on to drain water out a rate of 10 litres per minute. 6 minutes later, another tap was turned on to fill the tank at 12 litres per minute. Find the volume of water in the tank at 1.10 p.m. [4 marks][Answer given: 96 litres or 96000 cm3, 44 litres]
60 x 60 x 80 = 288000 cm3
= 288 l
1/3 x 288 = 96 l
a) It ws 96 litres.
6 x 10 = 60
96  60 = 36 l
12  10 = 2 litres/min
4 x 2 = 8 l
36 + 8 = 44 l
b) It was 44 litres.
60 x 60 x 80 = 288000 cm3
= 288 l
1/3 x 288 = 96 l
a) It ws 96 litres.
6 x 10 = 60
96  60 = 36 l
12  10 = 2 litres/min
4 x 2 = 8 l
36 + 8 = 44 l
b) It was 44 litres.
Ratio P6 2010 SA2 RGS P2 Q14
Janice had three boxes, A, B and C, containing a total of 1512 pearls. The number of pearls in Box A to the total number of pearls was 2 : 7. Janice sold 190 pearls from Box B and sold ¼ of the pearls in Box C. The number of pearls left in Box B to the number of pearls left in Box C was 2 : 1. How many pearls were there in Box C at first? [5 marks][Answer given: 356 pearls]
5/7 x 1512 = 1080
B [][][][][][] [ 190 ] ] 1080
C [][][] [] ]
10 u > 1080  190 = 890
4 u > 4/10 x 890 = 356 pearls
It was 356 pearls.
5/7 x 1512 = 1080
B [][][][][][] [ 190 ] ] 1080
C [][][] [] ]
10 u > 1080  190 = 890
4 u > 4/10 x 890 = 356 pearls
It was 356 pearls.
Ratio P6 2010 SA2 RGS P2 Q16
Irene had a total of 1686 red, blue and green balloons on sale. The ratio of the number of red balloons to the number of blue balloons was 2:3. After Irene sold ¾ of the blue balloons, ½ of the green balloons and none of the red balloons, she had 922 balloons left. How many blue balloons did Irene have at first?[4 marks][Answer given: 948]
R [][][][][][][][] ]
B [][][][][][][][][][][][] ] 1686
G [ ][ ] ]
20 u + 2 p > 1686
R [][][][][][][][] ]
B [][][] ] 922
G [ ] ]
11 u + 1 p > 922
22 u + 2 p > 2 x 922 = 1844
2 u > 1844  1686 = 158
12 u > 12/2 x 158 = 948 blue balloons
It was 948 blue balloons.
R [][][][][][][][] ]
B [][][][][][][][][][][][] ] 1686
G [ ][ ] ]
20 u + 2 p > 1686
R [][][][][][][][] ]
B [][][] ] 922
G [ ] ]
11 u + 1 p > 922
22 u + 2 p > 2 x 922 = 1844
2 u > 1844  1686 = 158
12 u > 12/2 x 158 = 948 blue balloons
It was 948 blue balloons.
Speed P6 2010 SA2 RGS P2 Q17
Q17: Jason left Town A at 1 p.m. travelling at 64 km/h towards Town B. Half an hour later, Ben left Town A, travelling at 80 km/h towards Town B. (a) How far ahead was Jason when Ben left Town A? (b) At what time would Ben be 8 km ahead of Jason? [4mark][Answer given: 32 km, 4.00 pm
64 x 1/2 = 32 km
a) It was 32 km.
8 + 32 = 40 km
80  64 = 16 km/h
16 km > 1 h
40 km > 40/16 x 1 = 2 1/2 h
1/2 h 1 1/2 h

1 pm 1.30 pm 4 pm
b) It was 4 pm.
64 x 1/2 = 32 km
a) It was 32 km.
8 + 32 = 40 km
80  64 = 16 km/h
16 km > 1 h
40 km > 40/16 x 1 = 2 1/2 h
1/2 h 1 1/2 h

1 pm 1.30 pm 4 pm
b) It was 4 pm.
Percentage P6 2010 SA2 RGS P2 Q18
There are more pupils in School A than School B. 30% of the pupils in School A is 45 more than 40% of the pupils in School B, If 10% of the pupils in School A leaves to join School B, there will be 200 more pupils in School A than School B. (a) How many pupils are there in School B? (b) How many percent less pupils are there in School B than School A? Leave your answer as fraction in the simplest form.[5 marks][Answer given: 1200, 31 3/7%]
3 A > 4 B + 45
24 A > 32 B + 8 x 45 = 32 B + 360
9 A > 10 B + 1 A + 200
8 A > 10 B + 200
24 A > 30 B + 3 x 200 = 30 B + 600
32 B + 360 > 30 B + 600
2 B > 600  360 = 240
10 B > 5 x 240 = 1200 pupils
a) It was 1200 pupils.
3 A > 2 x 240 + 45 = 525
10 A > 10/3 x 525 = 1750
1750  1200 = 550
550/1750 x 100% = 31 3/7%
b) It 31 3/7% less.
3 A > 4 B + 45
24 A > 32 B + 8 x 45 = 32 B + 360
9 A > 10 B + 1 A + 200
8 A > 10 B + 200
24 A > 30 B + 3 x 200 = 30 B + 600
32 B + 360 > 30 B + 600
2 B > 600  360 = 240
10 B > 5 x 240 = 1200 pupils
a) It was 1200 pupils.
3 A > 2 x 240 + 45 = 525
10 A > 10/3 x 525 = 1750
1750  1200 = 550
550/1750 x 100% = 31 3/7%
b) It 31 3/7% less.
Decimal P6 2010 SA2 Maha Bodhi P2 Q6
Mrs Gopal and Mrs Ari went to the supermarket. Both women decided to spend all their money on either apples or pears and they would not buy the same fruit.
If Mrs Gopal bought only pears, she would have 9 more fruit than Mrs Ari.
If she bought only apples, she would have 41 more fruit than Mrs Ari.
How much more money has Mrs Gopal than Mrs Ari?
40 : 60 = 2 : 3 Number bought 3 : 2 (inverse)
Number
G [ ][ ]
A [ ][ ][ ]<9>
2 u > 3 p + 9
6 u > 9 p + 3 x 9 = 9 p + 27
Number
G [ ][ ][ ]
A [ ][ ]< 41 >
3 u > 2 p + 41
6 u > 4 u + 2 x 41 = 4 p + 82
9 p + 27 > 4 p + 82
5 p > 82  27 = 55
1 p > 55/5 = 11
3 p > 3 x 11 = 33
2 u > 33 + 9 = 42
42 x 0.60 = $25.20
33 x 0.40 = $13.20
25.2  13.20 = $12
It was $12
Alternatively

40 : 60 = 2 : 3 Number bought 3 : 2 (inverse)
Number
G [ ][ ]
A [ ][ ][ ]<9>
Number
G [ ][ ][ ]
A [ ][ ]< 41 >
5 u > 5 p + 9 + 41 = 5 p + 50
5 u  5 p > 50
2 u  2 p > 2/5 x 50 = 20 Compare only number of pears bought
20 x 0.60 = $12
It was $12.
Rates P6 2010 SA2 Maha Bodhi P2 Q11
When both Machines P and Q are used, a book can be printed in 5 hours. When printing a book, Machine P broke down. As a result, it took 7 hours to complete the printing of the book. If it would take Machine Q 8 hours 20 minutes to print the book alone, (a) how long was Machine P used before it broke down? (b) how long would it take to print the book using only Machine P? [4 marks][Answer given: 2 hours before, 12½ h]
Q : 8 1/3 h > 1 book
7 h > 7 x 3/25 = 21/25 book
5 h > 5 x 3/25 = 3/5 book
1  21/25 = 4/25 book
1  3/5 = 2/5 book
P : 2/5 book > 5 h
4/25 book > 4/25 x 5/2 x 5 = 2 h
1 book > 5/2 x 5 = 12 1/2 h
a) It was 2 h.
b) It was 12 1/2 h.
Q : 8 1/3 h > 1 book
7 h > 7 x 3/25 = 21/25 book
5 h > 5 x 3/25 = 3/5 book
1  21/25 = 4/25 book
1  3/5 = 2/5 book
P : 2/5 book > 5 h
4/25 book > 4/25 x 5/2 x 5 = 2 h
1 book > 5/2 x 5 = 12 1/2 h
a) It was 2 h.
b) It was 12 1/2 h.
Decimal P6 2010 SA2 Maha Bodhi P2 Q12
Talik was given some pocket money for recess. He realised that if he had spent $0.80 on each meal, he would have 8 meals fewer than if he had spent $0.60 per meal. Talik spent $0.60 on 8 meals and spent $0.80 on the remaining meals. How many meals did the money last him? [4 marks][Answer given: 26 meals]
8 x 0.60 = $4.80
$0.80  0.60 = $0.20
4.80/0.20 = 24 meals  $0.80
24 x 0.80 = $19.20
19.20  4.80 = $14.40
14.40/0.80 = 18 meals
18 + 8 = 26 meals
It was 26 meals.
8 x 0.60 = $4.80
$0.80  0.60 = $0.20
4.80/0.20 = 24 meals  $0.80
24 x 0.80 = $19.20
19.20  4.80 = $14.40
14.40/0.80 = 18 meals
18 + 8 = 26 meals
It was 26 meals.
Volume P6 2010 Maha Bodhi P2 Q13
In every minute, three times as much water flows out of Tap A as Tap B into two empty containers, X and Y.
After 20 minutes, Container X is 2/5 filled with water and the height of water level in Container X is the same as that of Container Y as shown in the diagram above.
a) What is the length of the base of Container X?
b) How many litres of water can Container X hold?
50 x 45 = 2250
2250/3 = 750
750/25 = 30 cm
a) It is 30 cm.
Information not complete to solve (b).
Speed P6 2010 SA2 Maha Bodhi Q15
10/SF + 10/SS = 13 1/3
= 40/3
1/SF + 1/SS = 1/10 x 40/3
= 4/3
24/SF + 6/SS = 14
18 SF = 14  6 x 4/3
SF = 1/18 x 18/3
= 3
30/3 = 10 s
It was 10 s.
Percentage P6 2010 SA2 Maha Bodhi P2 Q16
Tom and Jerry shared a box of marbles. If Tom was given 10 fewer marbles, Jerry would have five times as many marbles as Tom. If Tom was given 10 more marbles, he would have ½ as many marbles as Jerry. What percentage of the marbles did Tom receive? [5 marks][Answer given: 25%]
T J
 10 + 10

1 : 5

5 T  50 > J + 10
J > 5 T  60
T J
+ 10  10

1 : 2

2T + 20 > J  10
J > 2 T + 30
5 T  60 > 2 T + 30
3 T > 30 + 60 = 90
T > 90/3 = 30
J > 2 x 30 + 30 = 90
90 + 30 = 120
30/120 x 100% = 25%
T J
 10 + 10

1 : 5

5 T  50 > J + 10
J > 5 T  60
T J
+ 10  10

1 : 2

2T + 20 > J  10
J > 2 T + 30
5 T  60 > 2 T + 30
3 T > 30 + 60 = 90
T > 90/3 = 30
J > 2 x 30 + 30 = 90
90 + 30 = 120
30/120 x 100% = 25%
Percentage P6 2010 SA2 Maha Bodhi P2 Q17
A group of 328 P5 children were divided into two teams at the P5 camp this year. There were 8 more children in Team B than in Team A. There were also 50% more boys in Team B than in Team A. If 3/8 of the girls were in Team B, (a) what percentage of the pupils in Team B are girls? (b) what percentage of the children are boys? Leave your answers as mixed numbers in the simplest form. [5 marks][Answers given: 28 and 4/7%, 60 and 40/41%]
328 + 8 = 336
336/2 = 168  Team B
328  168 = 160  Team A
A B

B G B G
 
2 : 3
5 : 3

2 B + 5 G > 160
6 B + 15 G > 3 x 160 = 480
3 B + 3 G > 168
6 B + 6 G > 2 x 168 = 336
9 G > 480  336 = 144
3 G > 3/9 x 144 = 48
48/168 x 100% = 28 4/7%
a) It was 28 4/7%.
3 B > 168  48 = 120
5 B > 5/3 x 120 = 200
200/328 x 100% = 60 40/41%
b) It was 60 40/41%.
328 + 8 = 336
336/2 = 168  Team B
328  168 = 160  Team A
A B

B G B G
 
2 : 3
5 : 3

2 B + 5 G > 160
6 B + 15 G > 3 x 160 = 480
3 B + 3 G > 168
6 B + 6 G > 2 x 168 = 336
9 G > 480  336 = 144
3 G > 3/9 x 144 = 48
48/168 x 100% = 28 4/7%
a) It was 28 4/7%.
3 B > 168  48 = 120
5 B > 5/3 x 120 = 200
200/328 x 100% = 60 40/41%
b) It was 60 40/41%.
Speed P6 2010 SA2 Maha Modhi P2 Q18
18. Tom is travelling on a bus that is moving at a uniform speed of 30 km/h. If he alights at Bus Stop A, he will walk home by Route A which is 800 m away from his flat.
He can also alight 1 km down the road at Bus Stop B and take the 730 m Route B home.
If Tom walks at a uniform speed of 40 m/min, find:
a) the diffference in time when he alights at the two different bus stops.
He can also alight 1 km down the road at Bus Stop B and take the 730 m Route B home.
If Tom walks at a uniform speed of 40 m/min, find:
a) the diffference in time when he alights at the two different bus stops.
b) the earliest time he can reach home if he boards the bus from the interchange 20 km away at 13 05.
Bus Stop A, Route A
800/40 = 20 min
Bus Stop B, Route B
730/40 = 18.25 min
1/30 x 60 = 2 min
18.25 + 2 = 20.25 min
20.25  20 = 0.25 min
= 0.25 x 60 s
=1 5 s
a) It was 15 s.
20/30 x 60 = 40 min
40 + 20 = 60 min
= 1 h
1 h

1305 1405
b) It was 14 05.
Ratio P6 2010 SA2 Nanyang P1 Q21
The cost of 3 mangoes is the same as the cost of 5 grapefruits. The cost of 3 mangoes is also the same as the cost of 10 pears. Find the ratio of the cost of a mango to the cost of a grapefruit to the cost of a pear.
(Ans : 10 : 6 : 3)
3 M > 5 G
M/G > 5/3 = 10/6
3 M > 10 P
M/P > 10/3
M : G : P = 10 : 6 : 3
Alternative method

Assume 3 mangoes = 5 grapefruits = 10 pears = $1
1 M > 1/3
1 G > 1/5
1 P > 1/10
M : G : P = 1/3 : 1/5 : 1/10
= 10 : 6 : 3 Multiply by 30
(Ans : 10 : 6 : 3)
3 M > 5 G
M/G > 5/3 = 10/6
3 M > 10 P
M/P > 10/3
M : G : P = 10 : 6 : 3
Alternative method

Assume 3 mangoes = 5 grapefruits = 10 pears = $1
1 M > 1/3
1 G > 1/5
1 P > 1/10
M : G : P = 1/3 : 1/5 : 1/10
= 10 : 6 : 3 Multiply by 30
Area P6 2010 SA2 Nanyang P1 Q30
The figure below is formed by stacking 4 pieces of square paper one on top of another. The papers have different prints and sizes of sides 3cm, 4cm, 5cm and 6cm. The 6cm piece is placed at the bottom of the stack, followed by the 5cm piece, then the 4cm piece and the 3cm piece is placed right on top.
Find the sum of the area A and B. (ans : 8 cm2)
6  1 = 5
5  3 = 2
6  4 = 2
2 x 2 = 4
4 x 2 = 8 cm2
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