Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q10

In the figure shown below, SUVX is a square. STU is an equilateral triangle and TXW is a straight line.
a) Find the value of Angle STX.
b) Find the value of Angle WVX.

Solution

a)
Line TZ passes through V, while line TY is passes through the centre of Line SU.
Angle STX is 1/4 of Angle STU.
Angle STU is 60 degrees (Triangle STU is equilateral)

Angle STX --> (1/4) x 60 degrees = 15 degrees

Answer: 15 degrees

b)
Angle SXT = 15 degrees (Triangle STX is isosceles)
Angle TXV --> (90 - 15) degrees = 75 degrees
Angle WXV --> (180 - 75) degrees = 105 degrees

Angle WVX
--> (180 - 105) degrees divided by 2 = 37.5 degrees
(Triangle WVX is isosceles)

Answer: 37.5 degrees

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q9

Nigel had a total of 227 durians and pears. He sold half of the durians and bought another 40 pears. As a result, he had an equal number of durians and pears.
a) How many durians were there at first?
b) How many pears were there at first?

Solution

(a)

3 units --> 267
1 unit --> 267 divided by 3 = 89

(Durians) 2 units --> 2 x 89 = 178

Answer: 178 durians


(b)
(Pears) 1 unit - 40
--> 89 - 40 = 49

Answer: 49 pears

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q8

Wilson and Yi Lin had $71 altogether. Yi Lin and Patrick had $105 altogether. Wilsom hand 3/5 of the money that Patrick had. How much money did Yi Lin have?

Solution

Yi Lin + Patrick
? + 5 units --> $105

Yi Lin + Wilson
? + 3 units --> $71

5 units - 3 units --> $105 - $71
2 units --> $34
1 unit --> $34 divided by 2 = $17

Yi Lin + Wilson
? + 3 units --> $71
? + (3 x $17) --> $71
? + $51 --> $71
? --> $71 - $51 = $20

Answer: $20

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q6

(a) Arif is 2x years old. His father is 4 times as old as he. His mother is 3 years younger than his father. What is their total age in terms of x?

(b) If x is = 4, find their total age.

Solution

(a)
Arif --> 2x
Father --> 4 x 2x = 8x
Mother --> 8x - 3

Total --> 2x + 8x + 8x - 3 = 18x - 3

Answer: (18x - 3) years

(b)
18(4) - 3
= 72 - 3
= 69

Answer: 69 years

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q5

A container with Bottle A placed in it has a mass of 4.27 kg. An identical container with Bottle B placed in it has a mass of 6.58 kg. The mass of Bottle A is 1/3 that of Bottle B. What is the mass of the Bottle A? Give your answer correct to 2 decimal places.

Solution

Bottle A --> 1 unit
Bottle B --> 3 unit

Container + Bottle A (1 unit) --> 4.27 kg
Container + Bottle B (3 units) --> 6.58 kg

3 units - 1 unit --> 6.58 kg - 4.27 kg
2 units --> 2.31 kg
1 unit --> 2.31 kg divided by 2 units = 1.155 kg
1.155 kg --> 1.16 kg correct to 2 decimal places

Answer: 1.16 kg

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q4

Box A contains only one-dollar coins. Box B contains only fifty-cent coins and Box C contains only twenty-cent coins. Box A has 5 times as many coins as Box C. Box B contains 12 coins fewer than Box A. Box C contains half the number of coins in Box B. How much money is there in Box B?

Solution

2 units --> 5 units - 12
3 units --> 12
1 unit --> 12 divided by 3 = 4

(Box B)
5 units - 12
-->(5 x 4) - 12
= 20 - 12
= 8 (coins)

8 coins x $0.50 = $4

Answer: $4

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q3

The ratio of the number of Chloe's stickers to the number of Faith's stickers is 3:5. The ratio of the number of Faith's stickers to Melissa's stickers is 6:7. If Melissa has 204 stickers more than Chloe, how many stickers do they have altogether?

Solution

Melissa - Chloe --> 204

35 units - 18 units --> 204

17 units --> 204

1 unit --> 204 divided by 17 = 12

All stickers

Chloe + Faith + Melissa

18 + 30 + 35 = 83

83 units --> 83 x 12 = 996

Answer: 996 stickers

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q1

LMNO is a square. PQN and PLQ are isosceles triangles. Angle QNM is 23 degrees. Find Angle NPQ.

Solution

Angle PNO --> 23 deg (mirror image of Angle MNQ)

Angle OPN --> 90 deg - 23 deg = 67 deg

Angle LPQ --> 45 deg (Triangle LQP is isosceles and Angle LPQ is a rt angle)

Angle NPQ --> (180 - 45 - 67) deg = 68 deg

Answer: 68 degrees

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q18

Ai Tong School organized a 2-day camp. On the first day, the number of boys was 600 more than the girls. On the second day, the number of boys decreased by 10% but the number of girls increased by 10%. If there were 2540 children on the second day, how many children were there on the first day?

Solution

2nd Day

Boys --> 0.9 unit + 540

Girls --> 1.1 unit

Total --> 2540

0.9 unit + 540 + 1.1 unit --> 2540

2 units + 540 --> 2540

2 units --> 2540 - 540 = 2000

1 unit --> 2000 divided by 2 = 1000

1st Day

2 units + 600

(2 x 1000) + 600

= 2000 + 600

= 2600

Answer: 2600 children

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q17

75% of the children in the stadium were girls. After 52 girls and 4 boys left, the remaining children formed groups of 8. In each group, there were 3 boys. How many children were there in the stadium at first?

Solution

Note - "Before" for girls is 3 parts, while "Before" for boys is 1 part. This is to represent 75% girls and 25% boys, before 52 girls and 4 boys left.

From the table above

4 units --> 40

1 unit --> 40 divided by 4 = 10

Total boys (after)

3 units --> 3 x 10 = 30

Total girls (after)

5 units --> 5 x 10 = 50

Total children (after)

30 + 50 = 80

Total children at first

80 + 52 + 4 = 136

Answer: 136 children

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