Question 1

In the number line below, what is the value of X?

Accepted Answer

(3) 1.1. The difference between 1.3 and 0.5 is 0.8. There are 8 equal intervals between 0.5 and 1.3, so each interval represents 0.8 / 8 = 0.1. Point X is 6 intervals to the right of 0.5. Therefore, X = 0.5 + (6 * 0.1) = 0.5 + 0.6 = 1.1.

Question 2

Find the value of 4/7 + 2/3

Accepted Answer

(4) 1 5/21. To add the fractions, find a common denominator, which is 21. 
4/7 = (4*3)/(7*3) = 12/21
2/3 = (2*7)/(3*7) = 14/21
12/21 + 14/21 = 26/21. 
Converting to a mixed number, 26/21 = 1 5/21.

Question 3

Joan had some rubber bands. 40% of her rubber bands were blue and the remaining rubber bands were red. She gave away 75% of her red rubber bands. What percentage of her rubber bands were given away?

Accepted Answer

(3) 45%. If 40% of rubber bands were blue, the remaining red rubber bands were 100% - 40% = 60%. Joan gave away 75% of her red rubber bands, so the percentage given away is 75% of 60%. 
(75/100) * 60% = (3/4) * 60% = 45%.

Question 4

The square grid below shows Triangle A. What type of triangle is Triangle A?

Accepted Answer

(3) Right-angled triangle. By visual inspection of the triangle on the square grid, one of the angles appears to form a 90-degree corner, making it a right-angled triangle. (Note: Precise coordinate geometry calculations suggest it's acute, but for primary school questions involving grids, visual estimation of perpendicularity is often intended for 'right-angled'.)

Question 5

In the figure below, PQRS is a parallelogram. Which one of the following is false?
(1) PS = QR
(2) ∠SPQ = ∠QRS
(3) ∠QPS + ∠PSR = 180°
(4) PQ // PS

Accepted Answer

(4) PQ // PS. Properties of a parallelogram include: opposite sides are equal (PS=QR), opposite angles are equal (∠SPQ=∠QRS), and consecutive angles are supplementary (∠QPS+∠PSR=180°). Also, opposite sides are parallel (PQ // SR and PS // QR). The statement PQ // PS implies adjacent sides are parallel, which is false for a parallelogram.

Question 6

What is the area of triangle ABC shown below?

Accepted Answer

(1) 6 cm². The figure shows a right-angled triangle ABC with the right angle at A. The base and height for calculating the area are the two sides forming the right angle: AC = 3 cm and AB = 4 cm. 
Area = (1/2) × base × height = (1/2) × 3 cm × 4 cm = (1/2) × 12 cm² = 6 cm².

Question 7

Find the area of a circle of radius 10 cm. Leave your answer in terms of π.

Accepted Answer

(4) 100 π cm². The formula for the area of a circle is A = πr², where r is the radius. Given radius r = 10 cm. 
Area = π × (10 cm)² = π × 100 cm² = 100π cm².

Question 8

Which of the following is likely to be the length of the whiteboard in the classroom?

Accepted Answer

(2) 2 m. A whiteboard in a classroom is typically several feet long. 2 cm is too short, 20 m is too long for a single classroom, and 2 km is an extremely large distance. 2 meters (approximately 6.5 feet) is a reasonable length for a classroom whiteboard.

Question 9

Which month had the most amount of rainfall?

Accepted Answer

(1) March. From the table, compare the 'Amount of rainfall (mm)' for each month:
March: 744 mm
April: 162 mm
May: 696 mm
June: 6 mm
The highest amount of rainfall is 744 mm, which occurred in March.

Question 10

The amount of rainfall in July was a 50% decrease from the amount in June. How much was the amount of rainfall in July?

Accepted Answer

(1) 3 mm. From the table, the amount of rainfall in June was 6 mm. A 50% decrease means the amount in July was 50% of June's rainfall. 
50% of 6 mm = (50/100) × 6 mm = 0.5 × 6 mm = 3 mm.

Question 11

Anna, Beth and Cate had some stamps. Anna had 3/5 of what Cate had and 3/7 of what Beth had. Find the ratio of the number of stamps Beth had to the number of stamps Cate had.

Accepted Answer

(4) 7:5. Let A, B, and C be the number of stamps for Anna, Beth, and Cate, respectively.
From 'Anna had 3/5 of what Cate had': A = (3/5)C => C = (5/3)A.
From 'Anna had 3/7 of what Beth had': A = (3/7)B => B = (7/3)A.
We need to find the ratio of Beth's stamps to Cate's stamps, B:C.
B:C = (7/3)A : (5/3)A
Multiply both sides by 3 to remove the denominator: 7A : 5A
Divide both sides by A: 7:5.

Question 12

The figure shown below is a semicircle of diameter 40 cm. What is the perimeter of the figure? (Take π = 3.14)

Accepted Answer

(2) 102.8 cm. The perimeter of a semicircle consists of its curved arc length and its straight diameter.
Diameter = 40 cm.
Circumference of a full circle = π × diameter = 3.14 × 40 cm = 125.6 cm.
Length of the curved arc of the semicircle = (1/2) × 125.6 cm = 62.8 cm.
Perimeter of the figure = curved arc length + diameter = 62.8 cm + 40 cm = 102.8 cm.

Question 13

Nabil had 3460 g of cookies to sell at a carnival. He packed them into as many packets of 100 g as possible and had some cookies left unpacked. How much more cookies would he need so that he could pack 1 more packet of cookies of exactly 100 g?

Accepted Answer

(3) 40 g. Divide the total cookies by the packet size: 3460 g ÷ 100 g/packet = 34 packets with a remainder of 60 g.
This means Nabil had 60 g of cookies left unpacked. To pack 1 more packet of 100 g, he needs an additional 100 g - 60 g = 40 g of cookies.

Question 14

Express 8p + 6 - p + 3p - 2 in the simplest form.

Accepted Answer

(4) 10p + 4. Combine the 'p' terms: 8p - p + 3p = (8 - 1 + 3)p = 10p.
Combine the constant terms: 6 - 2 = 4.
So, the expression simplifies to 10p + 4.

Question 15

At a community event, there were some children and 156 adults. 2/3 of the children and 1/6 of the adults received a gift. Sixty people received a gift. How many children were there at the event?

Accepted Answer

(2) 51. Number of adults = 156.
Number of adults who received a gift = (1/6) × 156 = 26.
Total people who received a gift = 60.
Number of children who received a gift = Total gifts - Adult gifts = 60 - 26 = 34.
We know that 2/3 of the children received a gift. Let C be the total number of children.
(2/3) × C = 34.
C = 34 × (3/2) = 17 × 3 = 51.
There were 51 children at the event.

P6 Mathematics SA1 2019 — Nanyang

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