Saturday, October 22, 2011
Wednesday, October 12, 2011
Synthesis English P6
Try as I might, I was not able to write an answer that felt right.
Saturday, October 1, 2011
Friday, September 30, 2011
Decimal P6 2011 PSLE P2 Q
A fruit stall was selling pear at 70 cents each and apples at 40 cents each. Sally bought some pears and Tom bought some apples. Sally spent $1.10 more than Tom but had 7 less fruits than him.
a) How many pears did Sally buy?
b) How much did Tom spend?
0.40 x 7 = $2.80
1.10 + 2.80 = $3.90
0.70 - 0.40 = $0.30
3.90 ÷ 0.30 = 13 pears
a) It was 13 pears.
13 + 7 = 20
20 x 0.40 = $8
b) It was $8.
---------------------------------------
Alternative solution
---------------------------
0.70 x 7 = $4.90
4.90 + 1.10 = $6
0.70 - 0.40 = 0.30
6 ÷ 0.30 = 20
20 - 7 = 13 pears
a) It was 13 pears.
20 x 0.40 = $8
b) It was $8.
a) How many pears did Sally buy?
b) How much did Tom spend?
0.40 x 7 = $2.80
1.10 + 2.80 = $3.90
0.70 - 0.40 = $0.30
3.90 ÷ 0.30 = 13 pears
a) It was 13 pears.
13 + 7 = 20
20 x 0.40 = $8
b) It was $8.
---------------------------------------
Alternative solution
---------------------------
0.70 x 7 = $4.90
4.90 + 1.10 = $6
0.70 - 0.40 = 0.30
6 ÷ 0.30 = 20
20 - 7 = 13 pears
a) It was 13 pears.
20 x 0.40 = $8
b) It was $8.
Labels:
M Decimal P6,
M Decimal PSLE 2011,
M PSLE 2011
PSLE 2011 Incomplete questions and answers
Paper 1
-----------
Q13) Question on 2 Circles
Ans: 66 cm
Ans: Column A (Ans: 1)?? Remainder 3 --> Column C (Ans: 3)
Q) Eight right-angled triangles and square?
Ans: 17 cm
The dotted line area is 289?
Square root 289 = 17 cm
The dotted line area is 289?
Square root 289 = 17 cm
Q) Parallelogram and rhombus
Ans: 42 degrees
Q) How many pupils bought 3 books?
Ans: 22 pupils
28) Question on rice
Ans: 2.4 kg
Paper 2
-----------
Q6) 36/120 x 100%
Ans: 30%?? 20%??
Q13)
Ans: 66
Q) Tiles?
Ans: a) 5/9; b) 180m2
Whole Number P6 2011 PSLE P1 Q17
A school was holding an event and tickets were sold to the teachers and pupils. Each teacher had to pay $20 each while each pupil had to pay $8 each. 1/4 of the tickets were bought by the teachers and the rest bought by the pupils. The total cost of tickets bought by the pupils was $416 more than the total cost of the tickets bought by the teachers. What was the total cost of all the tickets sold?
1 x 20 = 20
3 x 8 = 24
24 – 20 = 4 u --> 416
44
44 u --> ----- x 416 = $4 576
4
It was $4 576.
1 x 20 = 20
3 x 8 = 24
24 – 20 = 4 u --> 416
44 u --> ----- x 416 = $4 576
4
It was $4 576.
Fraction P6 2011 PSLE P2 Q18
Faizal had a sum of money. He spent 1/4 of his money on a wallet and 2/5 of the remainder on a book. After that, his parents gave him $120. Finally, the ratio of the amount of money he had in the end to the amount of money he had at first was 5 : 4. What was the sum of money Faizal had at first?
1 3
1 - --- = ----
4 4
2 3
1 - --- = ----
5 5
3 3 9
--- x ---- = ----
5 4 20
9
---- + $120 --> 5 u
20
1 --> 4 u
9 9
---- --> ---- x 4 u = 1.8 u
20 20
5 u - 1.8 u --> $120
3.2 u --> $120
4
4 u --> ----- x 120 = $150
3.2
It was $150.
--------------------------------------------------------------------------
Alternative solution
------------------------
\<---------------- 4 p --------------->|
[ | | | | ] [ | | | | ] [ | | | |] [ | | | | ]
[ | | ][ | | ] [ | | ][ | | ][ | | ][ $120 ]
|<----------- 5 p ----------->|
5 p --> 9 u + 120
4 p --> 20 u
5
5 p --> ---- x 20 u = 25 u
4
25 u --> 9 u + 120
16 u --> 120
20
20 u --> ---- x 120 = $150
16
It was $150.
1 3
1 - --- = ----
4 4
2 3
1 - --- = ----
5 5
3 3 9
--- x ---- = ----
5 4 20
9
---- + $120 --> 5 u
20
1 --> 4 u
9 9
---- --> ---- x 4 u = 1.8 u
20 20
5 u - 1.8 u --> $120
3.2 u --> $120
4
4 u --> ----- x 120 = $150
3.2
It was $150.
--------------------------------------------------------------------------
Alternative solution
------------------------
\<---------------- 4 p --------------->|
[ | | | | ] [ | | | | ] [ | | | |] [ | | | | ]
[ | | ][ | | ] [ | | ][ | | ][ | | ][ $120 ]
|<----------- 5 p ----------->|
5 p --> 9 u + 120
4 p --> 20 u
5
5 p --> ---- x 20 u = 25 u
4
25 u --> 9 u + 120
16 u --> 120
20
20 u --> ---- x 120 = $150
16
It was $150.
Labels:
M Fraction P6,
M Fraction P6 PSLE 2011,
M PSLE 2011
Whole Number P6 2011 PSLE
The chairs in a hall were arranged in rows of 9 chairs each. For a concert to be held in the hall, Alex bought another 6 chairs and rearranged the chairs into rows of 7 chairs each. After the rearrangment, he found that he had 12 additional rows. How many chairs were there in the hall for the concert?
12 x 7 = 84
84 - 6 = 78
9 - 7 = 2
78 ÷ 2 = 39 rows originally
39 x 9 = 351
351 + 6 = 357 chairs
It was 357 chairs.
-----------------------------------------
Alternative solution
----------------------------
9 x 1 u --> 7 x 1 p - 6
{1 u -- number of rows at the start; 1 p -- number of rows in the end}
9 u --> 7 p - 6
1 u --> 1 p - 12
7 u --> 7 p - 7 x 12 = 7 p - 84
2 u --> 84 - 6 = 78
1 u --> 78 ÷ 2 = 39
39 x 9 = 351
351 + 6 = 357
It was 357 chairs.
12 x 7 = 84
84 - 6 = 78
9 - 7 = 2
78 ÷ 2 = 39 rows originally
39 x 9 = 351
351 + 6 = 357 chairs
It was 357 chairs.
-----------------------------------------
Alternative solution
----------------------------
9 x 1 u --> 7 x 1 p - 6
{1 u -- number of rows at the start; 1 p -- number of rows in the end}
9 u --> 7 p - 6
1 u --> 1 p - 12
7 u --> 7 p - 7 x 12 = 7 p - 84
2 u --> 84 - 6 = 78
1 u --> 78 ÷ 2 = 39
39 x 9 = 351
351 + 6 = 357
It was 357 chairs.
Fraction 2011 PSLE
Mr Lee has 185 more chicken pies than tuna pies. After he sold 3/5 of the chicken pies and half of the tuna pies there were 146 pies left. How many pies were sold?
5 C - 2 T --> 185
2 C + 1 T --> 146
4 C + 2 T --> 2 x 146 = 292
9 C --> 292 + 185 = 477
1 C --> 477 ÷ 9 = 53
1 T --> 146 - 2 x 53 = 40
3 C --> 3 x 53 = 159
159 + 40 = 199
It was 199 pies.
5 C - 2 T --> 185
2 C + 1 T --> 146
4 C + 2 T --> 2 x 146 = 292
9 C --> 292 + 185 = 477
1 C --> 477 ÷ 9 = 53
1 T --> 146 - 2 x 53 = 40
3 C --> 3 x 53 = 159
159 + 40 = 199
It was 199 pies.
Labels:
M Fraction P6,
M Fraction P6 PSLE 2011,
M PSLE 2011
Speed P6 2011 PSLE
25 m --> 1 min
300
300 m --> ----- x 1 = 12 min
25
190 x 12 = 2280 m
2280 ÷ 400 = 5.7
≈ 5 complete rounds
It was 5 complete rounds.
Labels:
M PSLE 2011,
M Speed P6,
M Speed PSLE 2011
Average 2011 PSLE
The average of A, B and C is 5y. C has 2 y more than B and A has y more than B.
a) Find B. Give your answer in terms of y.
b) If y = 5, what is B?
A [][ 2 y ] )
B [] ) 5 y x 3 = 15 y
C [][ y ] )
3 u --> 15 y - 2 y - y = 12 y
1 u --> 12 y ÷ 3 = 4 y
a) It was 4 y.
4 x 5 = 20
b) It was 20.
a) Find B. Give your answer in terms of y.
b) If y = 5, what is B?
A [][ 2 y ] )
B [] ) 5 y x 3 = 15 y
C [][ y ] )
3 u --> 15 y - 2 y - y = 12 y
1 u --> 12 y ÷ 3 = 4 y
a) It was 4 y.
4 x 5 = 20
b) It was 20.
Labels:
M Average P6,
M Average PSLE 2011,
M PSLE 2011
Whole Number P6 2011 PSLE P1 Q15
Sally has a book which has 525 pages. From Monday to Thursday she reads 15 pages per day and Friday to Sunday she reads 30 pages per day. What is the least number of days she needs to finish reading the whole book?
Friday to Sunday à 30 x 3 = 90
Monday to Thursday à 15 x 4 = 60
1 set (7 days) --> 90 + 60 = 150
525 ÷ 150 = 3 sets remainder 75 pages
75 ÷ 30 = 2 r 15 days
{So can start on Thursday, Friday or Saturday}
3 x 7 = 21
21 + 3 = 24 days
It is 24 days.
Friday to Sunday à 30 x 3 = 90
Monday to Thursday à 15 x 4 = 60
1 set (7 days) --> 90 + 60 = 150
525 ÷ 150 = 3 sets remainder 75 pages
75 ÷ 30 = 2 r 15 days
{So can start on Thursday, Friday or Saturday}
3 x 7 = 21
21 + 3 = 24 days
It is 24 days.
Thursday, September 29, 2011
Wednesday, September 28, 2011
Challenging P6 2011 Nanyang SA2 P2 Q14
(a) 16, 17, 18, 19, 20, 21
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= (1 + 10)/2 x 10 = 55
{Aveage x Number of numbers}
{Average = Sum of first and last numbers divided by 2}
(b) It is 55
1 + 2 + 3 + ..... + 99 + 100
= (1 + 100)/2 x 100 = 5050
5050 - 99 = 4951
c) It is 4951.
Percentage P6 2011 Nanyang SA2 P2 Q17
17. Packet A, Packet B and Packet C each contained some salt. At first, there were 200 g more salt in Packet A than Packet B. Packet C had 3/4 of the amount of salt in Packet A. After 1/8 of the amount of salt in Packet A was transferred to Packet B, there was 82.15 g of salt in Packet B. How many percent less salt were there in Packet C than Packet A at first? [5 marks] {Should be be at the end instead of at first?}
Answer given: 25%
1 - 3/4 = 1/4
1/4 x 100% = 25%
It was 25%.
Answer given: 25%
1 - 3/4 = 1/4
1/4 x 100% = 25%
It was 25%.
Ratio P6 2011 Nanyang SA2 P2 Q18
18. In a cinema, the ratio of the number of girls to the number of boys was 3 : 2. The ratio of the number of women to the number of boys was 5 : 4. The ratio of the number of girls to the number of men was 2 : 5. During the movie, 6 women and 27 men left the cinema. The ratio of the number of women to the number of men then became 1 : 2.
(a) Express the number of women as a fraction of the number of men at first. Leave your answer in its simplest form.
(b) How many children were there in the cinema?
Answer given: (a) 1/3 and (b) 30
x 3
2 : 5 = 6 : 15
5 /15 = 1/3
a) It was 1/3.
5 : 15 10 {Use to 5 : 15 instead of 1 : 3}
- 6 - 27 - 12
----------------------------
1 : 2 2
----------------------------
(a) Express the number of women as a fraction of the number of men at first. Leave your answer in its simplest form.
(b) How many children were there in the cinema?
Answer given: (a) 1/3 and (b) 30
x 2
3 : 2 = 6 : 45 : 4
x 3
2 : 5 = 6 : 15
5 /15 = 1/3
a) It was 1/3.
W M W x 2
----------------------------5 : 15 10 {Use to 5 : 15 instead of 1 : 3}
- 6 - 27 - 12
----------------------------
1 : 2 2
----------------------------
15 u – 27 Ã 10 u – 12
5 u à 27 – 12 = 15
1 u à 15/5 = 3
10 u à 10 x 3 = 30
b) It was 30 children.
Volume P6 Nanyang 2011 SA2 P2 Q15
15. A rectangular tank measuring 3 m by 3 m by 1.5 m was 1/3 filled with water. A tap was turned on to fill up with water at a rate of 18 l/min. Every 2 minutes after the tap was turned on, 6 l of water was poured from a pail into the tank. How long did it take for the rectangular tank to be completely filled with water? Leave your answer as a mixed number in its simplest form. [4 marks]
Answer given: 428 2/3 min
Answer given: 428 2/3 min
2/3 x 300 x 300 x150 =9 000 000 cm3 = 9 000 litres
18 x 2 6 36 6
|---------|-----|----------|------|-----
2 min 0 2 min 0
2 min 0 2 min 0
1 cycle (2 min) Ã 18 x 2 + 6 =
9 000 litres à 9000/42 x 2 ≈ 214 cycles
9 000 - 214 x 42 =
12/18 = 2/3 min
214 x 2 = 428 = min
428 + 2/3 = 428 2/3 min
It was 428 2/3 min.
Tuesday, September 27, 2011
Monday, September 26, 2011
Monday, September 19, 2011
Fraction P6 Nan Hua 2011 SA2 P2 Q14
A train has a capacity of 154 seats. Tickets for seats are sold at $8 and $12. There are 1/5 more $8-seats than $12-seats on the train. During a trip, the amount collected from the sales of $8 tickets was twice the amount collected from the $12 tickets. The total amount collected was $540. How many $8-seats were not taken during the trip?
Note: Those in red are not necessary for the actual workings.
$8 $12 Total
------------------------------------------------------
Number of seats 6 u : 5 u 154
Amount collect 2 p : 1 p $540
------------------------------------------------------
11 u --> 154
1 u --> 154 ÷ 11 = 14
6 u --> 6 x 14 = 84
3 p --> 540
1 p --> 540 ÷ 3 = 180
2 p --> 2 x 180 = 360
360 ÷ 8 = 45
84 - 45 = 39 $8 tickets
It was 39 $8 tickets not taken up.
Note: Those in red are not necessary for the actual workings.
$8 $12 Total
------------------------------------------------------
Number of seats 6 u : 5 u 154
Amount collect 2 p : 1 p $540
------------------------------------------------------
11 u --> 154
1 u --> 154 ÷ 11 = 14
6 u --> 6 x 14 = 84
3 p --> 540
1 p --> 540 ÷ 3 = 180
2 p --> 2 x 180 = 360
360 ÷ 8 = 45
84 - 45 = 39 $8 tickets
It was 39 $8 tickets not taken up.
Monday, August 29, 2011
Ratio P5 Question
Mr Tang has some 20-cent and 50-cent coins.
The ratio of the value of the 20-cent coins to the value of the 50-cent coins is 7 : 6.
Find the ratio of the number of 20-cent coins to the number of 50-cent coins.
[Ans: 35 : 12]
The ratio of the value of the 20-cent coins to the value of the 50-cent coins is 7 : 6.
Find the ratio of the number of 20-cent coins to the number of 50-cent coins.
[Ans: 35 : 12]
Decimal P5 Question
Box A contained 1.36 kg of sand, box B contained 0.6 kg of sand, box C contained 1.68 kg of sand and box D contained 1.54 kg of sand.After some sand was transferred from box C to box A and from box D to box B, boxes A and B had the same amount of sand and box C had 5 times as much sand as box D.
(a) How much sand was transferred from box D to box B? [Ans: 1.315 kg]
(b) How much sand was transferred from box C to box A? {Ans: 0.555 kg]
(a) How much sand was transferred from box D to box B? [Ans: 1.315 kg]
(b) How much sand was transferred from box C to box A? {Ans: 0.555 kg]
Fraction P5 Question
William had $6 and Xena had $40 at first.William save 90 cents a day and Xena saved 40 cents a day.
(a) How long did it take William to save $18? [20 days]
(b) How long did it take for William's total money to be 1 1/2 times of Xena's total money? [Ans: 180 days]
(a) How long did it take William to save $18? [20 days]
(b) How long did it take for William's total money to be 1 1/2 times of Xena's total money? [Ans: 180 days]
Fraction P5 Question
Amy saved 5/8 of her salary every month and she spent the rest.
Ben saved 3/7 of his salary every month and he spent the rest which amounted to $600.
Amy took 1 1/6 years less than Ben to save $12600.
(a) What was Ben's monthly salary? [Ans: $1050]
(b) What was Amy's monthly salary? [Ans: $1440]
Ben saved 3/7 of his salary every month and he spent the rest which amounted to $600.
Amy took 1 1/6 years less than Ben to save $12600.
(a) What was Ben's monthly salary? [Ans: $1050]
(b) What was Amy's monthly salary? [Ans: $1440]
Monday, August 8, 2011
Tuesday, July 26, 2011
Ratio P5
The ratio of the amount of money Jenny had to the amount of money Siti had was 3:4. The next day, their parent gave each of them some money. Siti received twice as much as Jenny. The ratio then became 5:8. If both girls had a total of $70 at first, how much did Siti receive?
J [][][] [][]
S [][][][] [][][][]
7 u --> $70
4 u --> 4/7 x 70 = $40
Siti received $40.
J [][][] [][]
S [][][][] [][][][]
7 u --> $70
4 u --> 4/7 x 70 = $40
Siti received $40.
Tuesday, July 19, 2011
Monday, June 27, 2011
Sunday, June 26, 2011
Fraction P6 2010 SA1 Rosyth Q18
18. There were 156 more pupils in the playground than in the library. 1/5 of the pupils in the library were girls and 5/6 of the pupils in the playground were boys. If there were 17 more girls in the playground than in the library, find the total number of pupils in the two places. [5]
Monday, May 30, 2011
??
Ali has $300. He lost 1/3 of it and ends up with $200. To have the $300 again, he needs to win $100 which is ½ of the $200. 1/3 is known as the Losing Fraction (LF) and ½ is known as the Winning Fraction (WF). ½ times 1/3 = 1/6. 1/6 is the product of the LF and WF, known as the PLW. Find the LF and the WF if the PLW is 81/400.
81 = 81 x 1
= 27 x 3
= 9 x 9
(a and c)
400 = 400 x 1
= 200 x 2
= 100 x 4
= 80 x 5
= 50 x 8
= 40 x 10
= 25 x 16
(b and d)
LF = a/b
WF = c/d
a/b = c/d (1 - a/b)
a d = c (b - a)
Using the above factors, trial and error,
a = 9, b = 25, c = 9 and =16
LF = 9/25 and WF = 9/16
81 = 81 x 1
= 27 x 3
= 9 x 9
(a and c)
400 = 400 x 1
= 200 x 2
= 100 x 4
= 80 x 5
= 50 x 8
= 40 x 10
= 25 x 16
(b and d)
LF = a/b
WF = c/d
a/b = c/d (1 - a/b)
a d = c (b - a)
Using the above factors, trial and error,
a = 9, b = 25, c = 9 and =16
LF = 9/25 and WF = 9/16
Monday, May 23, 2011
Fraction P6 2010 SA1 Rosyth P2 Q18
There were 156 more pupils in the playground than in the library. 1/5 of the pupils in the library were girls and 5/6 of the pupils in the playground were boys. If there were 17 more girls in the playground than in the library, find the total number of pupils in the two places.
You may start with
a) There were 156 more pupils in the playground than in the library, or
b) there were 17 more girls in the playground than in the library
1/5 of the pupils in the library were girls --> 5 units in library with 1 unit girls
5/6 of the pupils in the playground were boys --> 6 UNITS in playground, with 1 UNIT girls
Note: unit and UNIT are not the same
a)
There were 156 more pupils in the playground than in the library
Draw 6 UNITS for playground + 156 [divided in 6 parts]
Draw 6 UNITS for library, but with 1 UNIT divided into 5 parts
P [ ][ ][ ][ ][ ][ ][][][][][][]
L [ ][ ][ ][ ][ ][||||]<---156--->
156 ÷ 6 = 26
Compare the girls in the playground and library
26 [ ][ ]
[ ][|
17
26 - 17 = 9
1 u --> 9 x 5 = 45 Compare the 6 UNITS of playground with 6 UNITS of libary
6 u --> 6 x 45 = 170
270 + 156 + 270 = 696 pupils
There were 696 pupils.
Alternative solution (b)
----------------------------
there were 17 more girls in the playground than in the library
[ ][17]
[ ]
[][17] [][17] [][17] [][17] [] [17] [][17]
[] [] [] [] [] <------------156------------>
6 u + 17 x 6 --> 5 u + 156
1 u --> 156 - 102 = 54
5 u --> 5 x 54 = 270
270 + 270 + 156 = 696 pupils
There were 696 pupils.
You may start with
a) There were 156 more pupils in the playground than in the library, or
b) there were 17 more girls in the playground than in the library
1/5 of the pupils in the library were girls --> 5 units in library with 1 unit girls
5/6 of the pupils in the playground were boys --> 6 UNITS in playground, with 1 UNIT girls
Note: unit and UNIT are not the same
a)
There were 156 more pupils in the playground than in the library
Draw 6 UNITS for playground + 156 [divided in 6 parts]
Draw 6 UNITS for library, but with 1 UNIT divided into 5 parts
P [ ][ ][ ][ ][ ][ ][][][][][][]
L [ ][ ][ ][ ][ ][||||]<---156--->
156 ÷ 6 = 26
Compare the girls in the playground and library
26 [ ][ ]
[ ][|
17
26 - 17 = 9
1 u --> 9 x 5 = 45 Compare the 6 UNITS of playground with 6 UNITS of libary
6 u --> 6 x 45 = 170
270 + 156 + 270 = 696 pupils
There were 696 pupils.
Alternative solution (b)
----------------------------
there were 17 more girls in the playground than in the library
[ ][17]
[ ]
[][17] [][17] [][17] [][17] [] [17] [][17]
[] [] [] [] [] <------------156------------>
6 u + 17 x 6 --> 5 u + 156
1 u --> 156 - 102 = 54
5 u --> 5 x 54 = 270
270 + 270 + 156 = 696 pupils
There were 696 pupils.
Monday, January 3, 2011
Graph PSLE 2010 P2 Q15
Chris filled a tank with water using two taps. He turned on Tap A first and after 4 minutes, he also turned on Tap B. Both taps were turned off at the same time when the tank was completely filled without overflowing.
The graph below shows the amount of water in the tank over 16 minutes.
10 - 4 = 6
30/6 = 5
5/4 = 1.25
5 - 1.25 = 3.75 litres
Alternative method
----------------------
10 - 4 = 6
6/4 x 5 = 7.5
35 - 5 - 7.5 = 22.5
22.5/6 = 3.75 litres
b) In one minutes, 3.75 litres of water flowed from Tap B.
The graph below shows the amount of water in the tank over 16 minutes.
(a) What fraction of the tank was filled 4 minutes after Tap A was turned on? Express your answer in the simplest form.
(b) In one minute, how many litres of water flowed from Tap B?
5/35 = 1/7
a) 1/7 of the tank was filled 4 minutes after Tap A was turned on.
35 - 5 = 30
10 - 4 = 6
30/6 = 5
5/4 = 1.25
5 - 1.25 = 3.75 litres
Alternative method
----------------------
10 - 4 = 6
6/4 x 5 = 7.5
35 - 5 - 7.5 = 22.5
22.5/6 = 3.75 litres
b) In one minutes, 3.75 litres of water flowed from Tap B.
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