Question 1

In 9 783 124, the digit 7 stands for 7 × _____.

Accepted Answer

D. 100 000. The digit 7 in the number 9 783 124 is in the hundred thousands place. Therefore, it stands for 7 × 100 000.

Question 2

Spencer spent 40% of his money on a wallet and had $240 left. How much did the wallet cost?

Accepted Answer

A. $160. If Spencer spent 40% of his money, he had 100% - 40% = 60% of his money left.
This 60% corresponds to $240.
60% = $240
1% = $240 / 60 = $4
The wallet cost 40% of his money.
40% = 40 × $4 = $160.

Question 3

Simplify 6 + 9r - 2 + 2r.

Accepted Answer

C. 11r + 4. To simplify the expression, combine like terms:
(9r + 2r) + (6 - 2) = 11r + 4.

Question 4

Express 1.4 hours in hours and minutes.

Accepted Answer

C. 1 h 24 min. 1.4 hours can be split into 1 hour and 0.4 hours.
To convert 0.4 hours to minutes, multiply by 60:
0.4 hours × 60 minutes/hour = 24 minutes.
So, 1.4 hours is 1 hour 24 minutes.

Question 5

What is the digit in the tenths place in the sum of 44.2 and 0.81?

Accepted Answer

B. 0. First, find the sum of 44.2 and 0.81:
44.20 + 0.81 = 45.01
The digit in the tenths place of 45.01 is 0.

Question 6

Ahmad, Brayden and Kelly shared a packet of candies in the ratio of 2:3:4. Kelly had 24 candies. How many candies were there in the packet?

Accepted Answer

B. 54. The ratio of candies for Ahmad : Brayden : Kelly is 2 : 3 : 4.
Kelly's share is 4 units, which corresponds to 24 candies.
4 units = 24 candies
1 unit = 24 / 4 = 6 candies
The total number of units in the packet is 2 + 3 + 4 = 9 units.
Total candies in the packet = 9 units × 6 candies/unit = 54 candies.

Question 7

How many members were there in 2016?

Accepted Answer

B. 750. Locate the year 2016 on the horizontal axis of the line graph. Follow the vertical line up to the data point for 2016. Read the corresponding value on the vertical axis (Number of members). The point for 2016 is halfway between 600 and 900, which is 750.

Question 8

Between which 2 years was there a 100% increase in membership?

Accepted Answer

A. 2014 and 2015. A 100% increase means the number of members doubled from the first year to the second year.
Let's check the options using the data from the graph:
- For 2014 and 2015: Members in 2014 = 300, Members in 2015 = 600. Increase = 600 - 300 = 300. Percentage increase = (300 / 300) × 100% = 100%. This is the correct option.

Question 9

Which figure below does not have any line of symmetry?

Accepted Answer

D. (4). A line of symmetry divides a figure into two identical halves that are mirror images of each other.
- Figure (1) has a diagonal line of symmetry.
- Figure (2) has a diagonal line of symmetry.
- Figure (3) has both horizontal and vertical lines of symmetry (and diagonal ones too).
- Figure (4) is an asymmetrical arrangement of shaded squares and does not have any line of symmetry.

Question 10

The solid shown is made up of identical unit cubes. Which of the following shows the front view the solid?

Accepted Answer

C. (3). The 'Front view' arrow indicates the perspective from which to observe the solid. When looking at the solid (which forms the letter 'H') from the front:
- The left vertical leg appears as a column of 3 shaded squares.
- The right vertical leg appears as a column of 3 shaded squares.
- The horizontal bar connecting the legs appears as a single shaded square in the bottom-center position.
- The gaps above the horizontal bar in the middle are unshaded.
This description matches option (3), which shows a 3x3 grid with the first and third columns fully shaded, and only the bottom square of the middle column shaded.

Question 11

Peter is 13 years old now. His father is 3 times as old as he. What is their total age in 5 years' time?

Accepted Answer

D. 62 years. Peter's current age = 13 years.
Father's current age = 3 × 13 = 39 years.
In 5 years' time:
Peter's age = 13 + 5 = 18 years.
Father's age = 39 + 5 = 44 years.
Their total age in 5 years' time = 18 + 44 = 62 years.

Question 12

Jeff and 3 of his classmates scored an average of 80 marks in their Chinese test. If Jeff had scored 79 marks, their average score would have been 83 marks. How many marks did Jeff score?

Accepted Answer

A. 67. 1. There are 4 students (Jeff + 3 classmates). Their current average score is 80 marks.
   Total score of the 4 students = 4 × 80 = 320 marks.
2. Let J be Jeff's actual score and S_C be the sum of the scores of the 3 classmates.
   So, J + S_C = 320.
3. The problem states a hypothetical scenario: "If Jeff had scored 79 marks, their average score would have been 83 marks."
   This means if Jeff's score was 79, and the classmates' total score (S_C) remained the same, the new average for the 4 students would be 83.
   So, (79 + S_C) / 4 = 83.
4. From this, 79 + S_C = 4 × 83 = 332.
5. Calculate S_C: S_C = 332 - 79 = 253 marks.
6. Now, substitute S_C back into the first equation to find Jeff's actual score (J):
   J + 253 = 320
   J = 320 - 253 = 67 marks.

Question 13

Mdm Kalsa sold 120 cupcakes in the morning and 40% of the remaining cupcakes in the afternoon. The number of cupcakes left in the end was 1/3 of what she had at first. How many cupcakes were sold in the afternoon?

Accepted Answer

B. 60. 1. Let T be the total number of cupcakes Mdm Kalsa had at first.
2. Cupcakes sold in the morning = 120.
3. Remaining cupcakes after morning sales = T - 120.
4. Cupcakes sold in the afternoon = 40% of the remaining = 0.4 × (T - 120).
5. Cupcakes left at the end = (T - 120) - 0.4 × (T - 120) = 0.6 × (T - 120).
6. The problem states that cupcakes left at the end = 1/3 of T.
7. So, we can set up the equation: 0.6 × (T - 120) = T/3.
8. Convert 0.6 to a fraction: (3/5) × (T - 120) = T/3.
9. Multiply both sides by the least common multiple of 5 and 3 (which is 15) to clear denominators:
   15 × (3/5) × (T - 120) = 15 × (T/3)
   9 × (T - 120) = 5T
10. Expand the equation: 9T - 1080 = 5T.
11. Solve for T: 4T = 1080 => T = 1080 / 4 = 270.
12. Total cupcakes at first = 270.
13. Cupcakes remaining after morning sales = T - 120 = 270 - 120 = 150.
14. Cupcakes sold in the afternoon = 40% of 150 = 0.4 × 150 = 60.

Question 14

Some students are folding paper cranes during a craft lesson. In 9 minutes, 5 students can fold 10 paper cranes. At this rate, how long does it take for 3 students to fold 54 paper cranes altogether?

Accepted Answer

D. 81 min. 1. First, find the rate of one student:
   5 students fold 10 cranes in 9 minutes.
   This means 1 student folds 10 cranes in 9 × 5 = 45 minutes.
2. Next, find the time it takes for one student to fold one crane:
   1 student folds 1 crane in 45 / 10 = 4.5 minutes.
3. Now, consider 3 students. They work together.
   To fold 54 cranes, if one student were to do it alone, it would take 54 × 4.5 minutes = 243 minutes.
   With 3 students working together, the time taken will be divided by 3:
   Time = 243 minutes / 3 = 81 minutes.

P6 Mathematics SA2 2019 — Ai Tong

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P6 Mathematics SA2 2019 — Ai Tong

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