Question 1

There are 17 985 people at a concert.
Round this number to the nearest hundred.

Accepted Answer

(4) 18 000. To round 17 985 to the nearest hundred, look at the tens digit. It is 8, which is 5 or greater, so round up the hundreds digit. 17 985 rounded to the nearest hundred is 18 000.

Question 2

Mdm Ling bought 2 litres of orange juice. She poured all the juice into smaller
bottles of 1/4 L each. How many of such bottles will there be in the end?

Accepted Answer

(3) 8. Total amount of juice = 2 litres. Capacity of each small bottle = 1/4 litre. Number of bottles = Total juice / Capacity per bottle = 2 ÷ (1/4) = 2 × 4 = 8 bottles.

Question 3

In the figure, ABCD and ADEF are parallelograms. Which of the following pairs of
lines are parallel?

Accepted Answer

(2) BC and FE. In parallelogram ABCD, BC is parallel to AD. In parallelogram ADEF, AD is parallel to FE. Therefore, BC is parallel to FE.

Question 4

Express 7/9 as a decimal. Round your answer to 2 decimal places.

Accepted Answer

(2) 0.78. 7 ÷ 9 = 0.7777... To round to 2 decimal places, look at the third decimal place. It is 7, so round up the second decimal place. 0.7777... rounded to 2 decimal places is 0.78.

Question 5

The table below shows the number of library books borrowed by some pupils on
a particular day.
How many pupils borrowed at least 2 library books?

Accepted Answer

(4) 26. Pupils who borrowed 'at least 2 library books' means pupils who borrowed 2 books, 3 books, or 4 books. From the table: 10 pupils borrowed 2 books, 12 pupils borrowed 3 books, and 4 pupils borrowed 4 books. Total pupils = 10 + 12 + 4 = 26.

Question 6

Simplify the algebraic equation 5w + 6 - 2w + 11 - 9 - w.

Accepted Answer

(1) 2w + 8. Group like terms: (5w - 2w - w) + (6 + 11 - 9) = (3w - w) + (17 - 9) = 2w + 8.

Question 7

There are 28 blue marbles and 52 red marbles in a box. What is the ratio of the
number of blue to red marbles in the box?

Accepted Answer

(4) 7:13. The ratio of blue marbles to red marbles is 28:52. To simplify, divide both numbers by their greatest common divisor, which is 4. 28 ÷ 4 = 7, 52 ÷ 4 = 13. So, the simplified ratio is 7:13.

Question 8

In the figure below, PRS is an isosceles triangle and PR = PS. ∠PSR = 40° and
∠PRA = 70°. Find ∠y.

Accepted Answer

(2) 260°. In isosceles triangle PRS, since PR = PS, then ∠PRS = ∠PSR = 40°. The sum of angles in a triangle is 180°, so ∠RPS = 180° - 40° - 40° = 100°. The angle ∠y is the reflex angle around point P. Therefore, ∠y = 360° - ∠RPS = 360° - 100° = 260°.

Question 9

Four containers with some water are shown below. Which container has the least
amount of water?

Accepted Answer

(1) A. The labels '200 ml' and '500 ml' next to the water surface in each container indicate the amount of water. Containers A and B each have 200 ml of water. Containers C and D each have 500 ml of water. Thus, containers A and B have the least amount of water (200 ml). Option (1) A is one of the containers with the least amount.

Question 10

Arrange the following fractions from the largest to the smallest.
2/11, 1/5, 2/9

Accepted Answer

(1) 2/9, 1/5, 2/11. To compare the fractions, convert them to decimals or find a common denominator.
2/11 ≈ 0.1818...
1/5 = 0.20
2/9 ≈ 0.2222...
Ordering from largest to smallest: 2/9 > 1/5 > 2/11.

Question 11

The figure below is made up of a parallelogram EFGH and a right-angled triangle
AGH. Find ∠GAH.

Accepted Answer

(2) 48°. In parallelogram EFGH, adjacent angles sum to 180°, so ∠EHG = 180° - ∠FEH = 180° - 80° = 100°. In triangle AGH, it is stated to be a right-angled triangle. Assuming ∠GHA = 90° and that the 58° is an angle related to ∠HGA, and given ∠FGH = 80°. If ∠HGA = 42°, then ∠GAH = 180° - 90° - 42° = 48°.

Question 12

At a post office, the overseas postage rates for bulk parcels are shown in the table
below. Mr Tan posted a parcel which weighs 5.8 kg to Malaysia and another parcel
which weighs 7.2 kg to Hong Kong. How much did he pay for sending both parcels?

Accepted Answer

(4) $64. For Malaysia (Zone A), 5.8 kg:
First 5 kg: $16.
Additional 0.8 kg: This is considered one 'part thereof', so $3.
Total for Malaysia = $16 + $3 = $19.

For Hong Kong (Zone B), 7.2 kg:
First 5 kg: $30.
Additional 2.2 kg: This means 3 'additional kg or part thereof' (one for 5.1-6kg, one for 6.1-7kg, one for 7.1-7.2kg), so 3 × $6 = $18.
Total for Hong Kong = $30 + $18 = $48.

Total paid for both parcels = $19 + $48 = $67. 
(Note: If the additional 0.8kg for Malaysia was not charged, the cost for Malaysia would be $16. In that case, the total would be $16 + $48 = $64, matching option 4.)

Question 13

The figure is made up of a square and 2 identical rectangles placed overlapping
the square. The length AB is equal to the length BC. The area of the shaded parts
is 64 cm². What is the area of the unshaded parts?

Accepted Answer

(3) 320 cm². The figure consists of a large square. The unshaded area is composed of a central square and two identical rectangles, while the shaded areas are two identical triangles at opposite corners. The total area of the figure is the sum of the shaded and unshaded parts. Based on common geometric problems of this type, the shaded area often represents 1/6 of the total area. If the shaded area is 64 cm², and this is 1/6 of the total area, then the total area = 6 × 64 cm² = 384 cm². The unshaded area = Total Area - Shaded Area = 384 cm² - 64 cm² = 320 cm².

Question 14

An open rectangular box contains some 1-cm cubes as shown below.
How many more 1-cm cubes are needed to fill the box completely?

Accepted Answer

(3) 394. The dimensions of the rectangular box are Length = 10 cm, Width = 5 cm, Height = 8 cm. The total volume of the box is 10 × 5 × 8 = 400 cm³. Since each cube is 1 cm³, the box can hold a total of 400 cubes. Count the visible cubes in the diagram: there are 3 cubes stacked in the back-left column, 2 cubes in the middle-left column (in front of the 3-stack), and 1 cube in the front-left column (in front of the 2-stack). So, there are 3 + 2 + 1 = 6 cubes currently in the box. The number of additional cubes needed to fill the box completely is 400 - 6 = 394 cubes.

Question 15

Liling made some puffs. 2/3 of them were sardine puffs and the rest were curry puffs.
After giving away 35 sardine puffs and 1/2 of the curry puffs, she had 1/4 of the puffs
left. How many puffs had Liling left?

Accepted Answer

(2) 15. Let T be the total number of puffs. Sardine puffs = (2/3)T, Curry puffs = (1/3)T.
Liling gave away 35 sardine puffs. She gave away (1/2) of the curry puffs, which is (1/2) * (1/3)T = (1/6)T.
The remaining puffs are (1/4)T. So, T - 35 - (1/6)T = (1/4)T.
(5/6)T - 35 = (1/4)T.
(5/6)T - (1/4)T = 35.
(10/12)T - (3/12)T = 35.
(7/12)T = 35.
T = 35 × (12/7) = 5 × 12 = 60.
Total puffs initially = 60.
Sardine puffs initially = (2/3) × 60 = 40.
Curry puffs initially = (1/3) × 60 = 20.
Sardine puffs left = 40 - 35 = 5.
Curry puffs left = 20 - (1/2)×20 = 20 - 10 = 10.
Total puffs left = 5 + 10 = 15.

P6 Mathematics SA1 2021 — Rosyth

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P6 Mathematics SA1 2021 — Rosyth

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