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.

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.

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.

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.