Challenging P5 2009 SA1 P2 CHIJ Q16

16. Look at the pattern below.

a) At what position does the number 12 first appear? [3]

b) What is the sum of the first 32 numbers? [2]

1st group -- > 1 , 2 , 3

2nd group -- > 2 , 3 , 4

3rd group -- > 3 , 4 , 5

….

10th group -- > 10 , 11 , 12

10 x 3 = 30 -- > 30th position

a) The number 12 first appears at the 30th position.

1, 2, 3, … 10

(1 + 10) x 10 ÷ 2 = 55

2, 3, 4, … 11

(2 + 11) x 10 ÷ 2 = 65

3, 4, 5, .. 12

(3 + 12) x 10 ÷ 2 = 75

55 + 65 + 75 + 11 + 12 = 218

or

3, 4, 5, … 10

(3 + 10) x 8 ÷ 2 = 52

(1 + 2 + 2 + 3 x 52 + 11 + 11 + 12) + (11 + 12) = 218

b) The sum of the first 32 numbers is 218.

Fraction P5 2009 SA1 P2 ACS Q15

15. The volume of water in bottle P was 3/5 of that in bottle Q. After 60 ml of water was added to bottle P and 20 ml was poured away from bottle Q, the amount of water in bottle P was 3/4 that in bottle Q. What was the amount of water in each bottle at first?

60 ÷ 3 = 20

20 x 4 = 80

P  [                     ][                     ][                     ] [      60     ]

Q [                     ][                     ][                     ][                     ][                       ] 
                                                                                                 [       80        ][20]

1 u --> 80 + 20 = 100

or

P  [                   ][                   ][                   ][      60     ]

Q [                   ][                   ][                   ][                      ][                      ]

   [                   ][20][                   ][20][                   ][20][                   ][20][20]

1 u --> 5 x 20 = 100

P: 3 u --> 3 x 100 = 300 ml

Q: 5 u --> 5 x 100 = 500 ml

The amount of water in P was 300 ml and the amount of water in Q was 500 ml.

Alternative method

                 P                   Q

   x 4                                         x 3

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

    12         3        :          5          15

+ 240   + 60                - 20       - 60

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

    12         3        :          4          12

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

12 u + 240 -- > 15 u – 60

3 u -- > 240 + 60 = 300

1 u -- > 300 ÷ 3 = 100

P: 3 u --> 3 x 100 = 300 ml

Q: 5 u --> 5 x 100 = 500 ml

The amount of water in P was 300 ml and the amount of water in Q was 500 ml.

Fraction P5 A004

Leah had $22 less than Emma at first. Leah spent 1/6 of her money while Emma spent 1/4 of her money at fair. If Emma had spent $8 more than Leah, how much money did Leah spend?

Given 1/6 and 1/4, make them into like fractions (denominators the same).
1/6 x 2/2 = 2/12. 1/4 x 3/3 = 3/12.

22 ÷ 4 = $5.50

E [][][][][][][][][][][][][5.50][5.50][5.50][5.50]

L [][][][][][][][][][][][]<------- 22 ------------->

3 u + 5.50 -- > 2 u + 8

1 u -- > 8 – 5.50 = 2.50

2 u -- > 2.50 x 2 = $5.00

Leah spent $5.

Alternative method 1 - model drawing 2

E [][8][][8][][8][][8]
L [][][][][][]<-22-->

4 u + 8 x 4 -- > 6 u +22
2 u -- > 32 – 22 = 10
1 u -- > 10 ÷2 = 5

Leah spent $5.

Alternative method 2

     L                    E

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

   1 u              1 u + $22

- 1/6 u     - (1/4 u + $5.50)

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

   5/6 u       3/4 u - $16.50        It’s ok if you do not show these.

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

1/4 u + 5.50 – 1/6 u -- > 8

1/12 u -- > 8 – 5.50 = 2.50

1/6 u -- > 2 x 2.50 = $5.00

Leah spent $5.

Whole Number P6 2009 SA1 P2 Rulang Q8

8. Uma had 6 times as many white beads as red beads. For every 5 white beads she used to make a necklace, she used 3 red beads. When she used up all the red beads, she had 52 white beads left. How many red beads did she have? [3]

                 <--------- 52 --------->

W  [][][][][][][][][][][][][][][][][][]

R   [][][]

13 u -- > 52

1 u -- > 52 ÷ 13 = 4

3 u -- > 3 x 4 = 12 red beads

She had 12 red beads.

Ratio P6 2009 SA1 P2 RGS Q1

1. Mrs Tan used syrup and water in the ratio of 2 : 7 to make a drink. How many litres of drink did she make if she used 4 litres of syrup?

2 u -- > 4 l

9 u -- > 9/2 x 4 = 18 l

Algebra P6 2009 SA1 P2 RGS Q2

2. John is (m + 2) years old now. 2 years ago, his cousin was twice his age. How old is his cousin now?

John's age 2 years ago.
(m + 2) – 2 = m

His cousin's age 2 years ago.
2 x m = 2m

His cousin's age now.
2m + 2

Speed 2009 SA1 P2 RGS Q3

3. At 10 am, Andy drove from Town A to Town B at an average speed of 100 km/h. At the same time, Jeremy drove from Town B towards Town A at an average speed of 95 km/h. They met at 12.30 pm. Find the distance between Town A and Town B.

          2 h           1/2 h

   !--------------!-------------!

10 am        12 pm      12.30 pm 

100 + 95 = 195 km/h

195 x 5/2 = 487.5 km

Ratio P6 2009 SA1 P2 RGS Q4

4. The ratio of Ann’s amount of money to Jane’s amount of money is 4 : 9. How much more money does Jane have if Ann has $240?

4 u -- > $240

9 – 4 = 5

5 u -- > 5/4 x 240 = $300

Percentage P6 2009 SA1 P2 RGS Q5

5. 2/5 of 80 km is the same as ____ % of 320 km.

2/5 x 80 = 32

32/320 x 100% = 10%

Percentage P6 2009 SA1 P2 RGS Q6

6. Dan and Ella shared a sum of money. When Dan’s share increased from $500 to $605, the amount Ella received decreased by 15%. How much did Ella receive at first? [3]

             100%

[  500  ][         ]

[   605    ][      ]

                85%

15% -- > 605 – 500 = 105

100% -- > 100/15 X 105 = $700

Ella received $700 at first.

Average P6 2009 SA1 P2 RGS Q7

7. The average marks received by Jane, Alice, Susan and Mabel in a recent Science test was 76 marks. Jane and Alice got 8y marks each while Susan got half of Mabel’s marks. How many marks did Susan score for the test? Express your answer in terms of y. [3]

76 x 4 = 304

3 u -- > 304 – 8y - 8y = 304 – 16y

1 u -- > (304 – 16y)/3

Susan scored (304 – 16y)/3 marks.

Whole Number P6 2009 SA1 P2 RGS Q8

8. Jason had $78 and Ben had $25. After Jason and Ben spent $53 altogether on some games, Jason had 3 times as much money as Ben. How much did Ben spend on the games? [3]

J  [][][][  SJ  ]  78

B      [][ SB ]  25

           $53

78 + 25 = 103

4 u -- > 103 – 53 = 50

1 u -- > 50 ÷ 4 = 12.50

25 – 12.50 = $12.50

Ben spent $12.50 on the games.

Fraction P6 2009 SA1 P2 RGS Q9

 9. Tom’s savings was 2/3 of Jane’s savings. After Jane had spent $180, her savings became ¾ of Tom’s savings. How much did Tom save? [3]


Tom's savings was 2/3 of Jane's savings.
After Jane had spent $180, her savings became 3/4 of Toms' savings.
After - Tom's savings was 4/3 of Jane's savings.
Before - Tom's savings unchanged at 4 units. So chaange 2/3 to 4/6.

2/3 x 2/2 = 4/6

6 – 3 = 3

3 u -- > $180

4 u -- > 4/3 x 180 = $240

Tom saved $240.

Circle P6 2009 SA1 P2 RGS Q10

10. The diagram below shows a square, a semicircle and a right-angled isosceles triangle. Given that the radius of the circle is 4 cm, find the area of the shaded part. Take π as 3.14. [3]

4 x 4 = 16

1/4 x 3.14 x 4 x 4 = 12.56

16 – 12.56 = 3.44

4 x 2 = 8

1/2 x 8 x 8 = 32

32 + 3.44 = 35.44 cm²

The area of the shaded part is 35.44 cm².

Fraction P6 2009 SA1 P2 RGS Q11

11. Jane was thinking of a fraction. The sum of its numerator and denominator was 34. After I added 98 to its denominator, the fraction became 1/11. What was the fraction that Jane was thinking of? [3]

 

N []          )

D [    ]      ) 34

N []         

D [][][][][][][][][][][]

   [    ][      98        ]

11 u - 98

12 u -- > 34 + 98 = 132

1 u -- > 132 ÷ 12 = 11

11 u -- > 11 x 11 = 121

121 – 98 = 23

11/23

The fraction that Jane was thinking of is 11/23.

Alternative method

N + D -- > 34

N/(D + 98) -- > 1/11

11 N -- > D + 98

11 N – D -- > 98

12 N -- > 34 + 98 = 132

1 N -- > 132 ÷ 12 = 11

D -- > 34 – 11 = 23

N/D -- > 11/23

The fraction that Jane was thinking of is 11/23.

Whole Number P6 2009 SA1 P2 RGS Q12

12. Alex, Ben and David went for a run but none of them completed the run.
Ben ran 5 times as far as Alex before he stopped.
David stopped 1 km before the finishing line and he ran 3 km less than twice the distance Ben ran. The three of them ran 21 km, how long was the run? [4]

A  [ ]                                  )

B  [ ][ ][ ][ ][ ]                      )  21 km

D  [                    ]<  3  >     )

     [                    ]<1> Finishing line  

2B [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ]

16 u -- > 21 + 3 = 24

1 u -- > 24 ÷ 16 = 1.5

10 u -- > 10 x 1.5 = 15

15 – 3 + 1 = 13 km

The run was 13 km long.

Fraction P6 2009 SA1 P2 RGS Q13

13a. The figure below is made up of unit squares. What fraction of the figure is shaded? [1]

1/2 x 3 x 4 = 6

½ x 2 x 1 = 1

1 + 6 = 7

5 x 4 = 20

7/20

13b. The figure below is not drawn to scale. Given that the area of WXYZ is 40 cm², find the area of the shaded part. [3]

The two shaded triangles may be adjusted to form a big triangle (base 12 cm and height 18 cm) such that they do not overlap.

1/2  x 12 x 18 = 108

108 – 40 = 68 cm²

The area of the shaded part is 68 cm².

Area P6 2009 SA1 P2 RGS Q14

14. The figure below is made up of two identical right-angled triangles, a small square and a big square. The perimeter of the shaded region is 50 cm, and the total area of the two unshaded squares is 254.5 cm. Find the total area of the two shaded right-angled triangles. [5]

Firstly, we may adjust the figure and add a red rectangle to form another square as follows:

2 L + 2 B

50 – 3 – 3 = 44

L + B

44 ÷ 2 = 22

Area of biggest square (L + B) x (L + B)

22 x 22 = 484

484 – 254.5 = 229.5

229.5 ÷ 2 = 114.75 cm²

The total area of the two shaded right-angled triangles is 114.75 cm².

Circle P6 2009 SA1 P2 RGS Q15

15. The figure is made up of 4 equal quadrants and one semicircle. AB = CD = EF = 8 cm. The total area of the unshaded parts is 33.5 cm². Take π as 3.14.

a) Find the total perimeter of the shaded parts. [1]

b) Find the total area of the shaded parts. [4]

2 x 3.14 x 8 = 50.24

50.24 + 8 + 8 = 66.24 cm

a) The total perimeter of the shaded parts is 66.24 cm.

1/2 x 3.14 x 4 x 4 = 25.12

33.5 – 25.12 = 8.38

3.14 x 8 x 8 = 200.96

200.96 – 8.38 – 8.38 = 184.2 cm²

The total area of the shaded parts is 184.2 cm².

Fraction P6 2009 SA1 P2 RGS Q16

16. In a math competition, participants can obtain 4 possible awards: Gold, Silver, Bronze and Participation. 3/7 of the participants obtained Gold awards, 1/4 of them obtained Silver awards, and 1/6 of them obtained Bronze awards. Given that there were less than 100 participants taking the competition,
a) How many participants obtained the awards for Participation? [3]
b) How many more participants obtained Gold awards than Bronze awards? [2]

3/7 x 12/12 = 36/84

1/4  x 3/3 x 7/7 = 21/84

1/6 x 2/2 x 7/7 = 14/84

84 – 36 – 21 – 14 = 13

a) 13 participants obtained the awards for Participation.

36 – 14 = 22

b) 22 more participants obtained Gold awards than Bronze awards.

Challenging P6 2009 SA1 P2 RGS Q17

17. The cuboid below is made of 2-cm cubes. All 6 faces of the cuboid are painted.

Find the ratio of the number of cubes with 3 painted faces to the number of cubes with 2 painted faces to the number of cubes without any painted faces in its simples form. [4]

Number of cubes with 3 painted faces

[4 at Top corners + 4 at Bottom corners]

4 x 2 = 8 cubes

Number of cubes with 2 painted faces

[4 vertical on each edge x 4 vertical edges]

[4 horizontal front x 4 (front top, back top, front bottom, back bottom)]

[3 horizontal side x 4 (right top, left top, right bottom, left bottom)]

4 x 4 = 16

3 x 4 = 12

16 + 12 + 16 = 44

Number of cubes without any painted faces

The cuboid is 6 x 5 x 6. Remove all cubes by the sides, the cuboid left is 4 x 3 x 4, which gives the number of cubes not painted.

4 x 3 x 4 = 48 cubes

8 : 44 : 48 = 2 : 11 : 12

The ratio is 2 : 11 : 12.

Percentage P6 2009 SA1 P2 RGS Q18

18. Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles. How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red? [5]

Firstly list out the relationships given.

25 : 75 = 1 : 3              Container A – Red : Blue

75 : 25 = 3 : 1              Container B – Red : Blue

The total number of red marbles and total number of blue marbles remain constant since they are moved from Container A to Container B. (Before = After)

A + 3 B --> 250 + 600 = 850
3 A + B --> 200 + 150 = 350

4 A + 4 B --> 850 + 350 = 1200
1 A + 1 B --> 1200 ÷ 4 = 300

2 B --> 850 - 300 = 550
B --> 550 ÷ 2 = 275
4 B --> 4 x 275 = 1100              There is no need to solve for A.

1100 – 600 – 150 = 350 marbles             
Always remember to write down the unit of measurement.

350 red and blue marbles must be moved from Container A to Container B.

Alternative method 1 – Model Drawing
Firstly list out the relationships given.

250 : 200 = 5 : 4           Container A – Red : Blue (Before)

                                    Not necessary for Container B

                                         

25 : 75 = 1 : 3              Container A – Red : Blue (After)

75 : 25 = 3 : 1              Container B – Red : Blue (After)

Next, move 50 blue marbles (and shaded them) to Container B to achieve the end ratio of 3 : 1. [600 : 150 + 50]

Then shade away 3 units of RA for every 1 unit of BA, until you cannot get any more 3 : 1. [**]

RA [    ][    ] [    ] [    ] [    ]   Cut all units into 2 small units

BA [    ][    ] [    ] [    ]             

After that cut the units into smaller units (in this case – 1 unit into 2 smaller units).

Then repeat ** above, till RA : RB = 1 : 3
[Do you realise that there is no need to draw model for Container B at all]

RA [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ]   250

BA [ ][ ][ ][ ][ ][ ][ ][ ]          200

8 u --> 200

1 u --> 200 ÷ 8 = 25

14 u --> 14 x 25 = 350 marbles

350 red and blue marbles must be moved from Container A to Container B.

Alternative method 2

Firstly list out the relationships given.

250 : 200 = 5 : 4           Container A – Red : Blue (Before)

                                    Not necessary for Container B

                                         

25 : 75 = 1 : 3              Container A – Red : Blue (After)

75 : 25 = 3 : 1              Container B – Red : Blue (After)

Next, move 50 blue marbles (and shaded them) to Container B to achieve the end ratio of 3 : 1.  [600 : 150 + 50]

200 – 50 = 150

Then for Container A, -3 for Red every -1 for Blue (- 3 : - 1) to achieve end results of 1 : 3 for Red : Blue.

As we are finding the number of marbles move, multiply column (Red) by 3 to get same end result of 3 as blue. Then solve.

             Red          Blue

     x 3  ----------------------

750        250          150

- 9         - 3      :     - 1

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

  3     :     1      :       3

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

750 – 9 u --> 150 – 1 u

8 u --> 750 – 150 = 600

1 u --> 600 ÷ 8 = 75

4 u --> 4 x 75 = 300

Remember to include the 50 blue marbles moved earlier (to make the ratio of the Red to Blue marbles 3 : 1 in Container B.

300 + 50 = 350 marbles

350 red and blue marbles must be moved from Container A to Container B.