Popular Maths Questions over the years!

Primary 3 Question 1

Mr Li and Mr Siew had $1592 altogether. When Mr Siew received $69 from Mr Li, both of them had the same amount of money. How much money did each man have at first?

$1592 ÷ 2 = $796

Each man had $796 in the end.

This was after Mr Li had given $69 to Mr Siew.

$796 + $69 = $865

**Mr Li** had **$865 **at first.

$796 - $69 = $727

**Mr Siew** had **$727 **at first.

Primary 3 Question 2

A salesman sold pens at $10 each for a company. Every time he collected $200 from the sale of the pens, the company paid him $40. In a certain week, he received $120 from the company. How many pens did he sell that week?

200 ÷ 10 = 20

He needed to sell 20 pens to collect $200 and earned $40 from the company.

(Note this step carefully.)

$120 ÷ $40 = 3 (or 40 + 40 + 40 = 120)

There are 3 groups of $40 in $120.

$40 (1 group) ---> 20 pens

$120 (3 groups) ---> 3 x 20 = 60 pens

He sold **60 pens **that week.

Primary 3 Question 3

Alan and Bala have 481 marbles altogether.Bala and Cindy have 625 marbles altogether. How many marbles does Bala have if the 3 children have 749 marbles altogether?

A + B ---> 481 ―—― (1)

B + C ---> 625 ―—― (2)

A + B + C ---> 749 ―—― (3)

Add (1) and (2):

A + B + B + C ---> 481 + 625 = 1106 ―—― (4)

Subtract (3) from (4):

B ---> 1106 – 749 = 357

Ans: **357 marbles**

Primary 4 Question 1

33 children were standing in a queue. The first child in the queue was a boy. There were 3 girls standing between 2 boys.

(a) How many boys were there in the queue?

(b) What fraction of the children in the queue were girls?

(a) Look for the pattern.

B G G G B G G G B G G G B …

33 – 1 = 32

32 ÷ 4 = 8

Number of boys = 8 + 1 = 9

Ans: **9 boys**

(b) Number of girls = 33 – 9 = 24

Ans:

Primary 4 Question 2

Sally had $2.70 more than Betty at first. When Sally received another $8.90, she had 5 times as much as Betty. What was the total amount the two girls had at first?

$2.70 + $8.90 = $11.60

4 units = $11.60

1 unit = $11.60 ÷ 4 = $2.90

6 units = $2.90 x 6 = $17.40

$17.40 - $8.90 = $8.50

Ans: **$8.50**

Primary 4 Question 3

Jasper and Joyce had a total of $177.50. After Joyce gave Jasper $8.60, Jasper had 4 times as much money as Joyce. How much money did Jasper have at first?

The total amount of money remained uncharged.

After

$177.50 ÷ 5 = $35.50 (Joyce)

4 x $35.50 = $142 (Jasper)

Before

142 - $8.60 = $133.40 (Jasper)

Ans: **$133.40**

Primary 4 Question 4

A bag of balloons was supposed to be shared equally among 38 children. 3 children did not want any balloons and their share was given equally to the remaining children. As a result, each of the remaining child received 6 more balloons. How many balloons were there in the bag?

38 – 3 = 35 children (3 children did not want any balloons)

35 x 6 = 210 balloons (each of the remaining child received 6 more balloons)

These three children originally owned 210 balloons.

210 ÷ 3 = 70 balloons (each child received 70 balloons)

70 x 38 = 2660 (total)

Ans:** 2660 balloons**

Assumption Method - Type 3

Mr. Tan prepared some sweets to distribute to his class. He had a total of 68 students. He gave each girl 3 sweets and each boy 5 sweets. If the total number of sweets that the boys received was 156 more than the total number of sweets the girls received, how many girls were there in the class?

**Ans**: There were **23** girls in the class.

Grouping Concepts 2 - Type 4

Damien had some $2 and $5 notes. He had 3 more $2 notes than $5 notes. The value of the $2 notes was $9 less than the value of $5 notes. How many $5 notes did Damien have?

**Ans**: Damien had **5** $5 notes.

Simultaneous Equation (Replacement) - Type 2

A cabinet can be packed from end to end with 20 big boxes or 30 small boxes. Jody had already packed the shelf with 2 big boxes and 10 small boxes. At most, how many more big boxes can Jody pack the cabinet with?

**Ans**: At most, Jody can pack the cabinet with **11** more big boxes.

(a) Fill in the table above. (1m)

(b) Find the total number of grey and white triangles for Figure 250. (1m)

(c) Find the percentage of grey triangles in Figure 250. (3m)

(a)

(b) The total number of triangles is just figure number -> n x n

Figure 250 -> 250 x 250 = 62 500

(c) Let’s extend the table in (a).

From (a), notice a pattern!

In order to find the number of grey triangles, we use this formula -> (N÷2) x (N+1)

Percentage of grey triangles in Figure 250: 31 375 ÷ 62 500 x 100% = 50.2%

4 similar golden rings overlap each other as shown below. Each ring has a diameter of 9 cm and a thickness of 1 cm.

(a) Find the length of the 4 overlapping rings.

(b) If the length of the first ring to the last ring is 198 cm, how many rings are there?

(a) 9 + (3 x 7) = 30 cm The length of 4 overlapping rings is 30 cm.

(b) Only the first ring is 9 cm while the other length of the rings is of the same length.

198 cm – 9 cm = 189 cm

189 cm ÷ 7 = 27

Number of rings -> 27 + 1 = 28

There are 28 rings.

Helen and Ivan had the same number of coins.

Helen had a number of 50-cent coins and 64 20-cent coins. Helen’s coins had a total mass of 1.134 kg. Ivan had a number of 50-cent coins and 104 20-cent coins.

(a) Who has more money in coins and by how much more?

(b) Given that each 50-cent coin is 2.7g heavier than a 20-cent coin, what is the total mass of Ivan’s coins in kilograms?

(a) Helen has more money because she has more 50-cent coins by the amount 40 x (50₵ -20₵) = 1200₵ = $12

Each 50₵ coin is 2.7g heavier than a 20₵ coin, Mass of Helen’s coins – Mass of Ivan’s coins = Difference in Total Mass 40 x 2.7 g = 108 g = 0.108 kg

Mass of Helen’s coins -> 1.134 kg Mass of Ivan’s coins -> 1.134 kg – 0.108 kg = 1.026 kg

(b) The total mass of Ivan’s coins is 1.026 kg.

Secondary 1

(a) Express 100 x 0.47 as a recurring decimal.

(b) Find the value of 100 x 0.47 - 0.47.

(c) Hence, express 0.47 as a fraction in its lowest terms.

(a) 100 x 0.47

= 100 x 0.474747 ...

= 47.474747 ...

= 47.47

Ans: **47.47**

(b) 100 x 0.47 - 0.47

= 47.474747 ... - 0.474747 ...

= 47

Ans: **47**

(c) 100 x 0.47 - 0.47 = 47

99 x 0.47 = 47

0.47 =

Secondary 2

Given that 𝓶 is inversely proportional to the square of 𝓷. It is known that 𝓶 = 8 for a particular value of 𝓷. When is increased by 200%, find

(a) the value of 𝓶.

(b) the percentage change in the value of 𝓶.

𝓶 is inversely proportional to the square root of 𝓷.

-> 𝓶 =

When 𝓶 = 8, 𝓷 = N

-> 8 =

(a) When 𝓷 increased by 200%, 𝓷 = 3N

𝓶 =

𝓶 =

=

=

(b) Percentage change in 𝓶

=

=

Secondary 3 E-Math

A point B is 8.8 km away and at a bearing of 065° from A. John cycles from B to a point C, where C is at a bearing of 282° from B and at a bearing of 341° from A. Calculate

(i) ∠ABC

(ii) ∠ACB

(iii) ∠AC,

(iv) the distance he has to cycle before he is closest to A.

John is taking a break at a point G which is equidistant from A and C. Find

(v) the bearing of G from A,

(vi) CG.

[ Hint: Make use of the 180° ]

(i)

[ Hint: Make use of the 360° ]

[ Hint: ∠ sum of 𝛥 ]

(ii) Using sine rule,

[ Hint: With 3 angles and 1 known side, we can use sine rule to find the unknown side AC ]

(iii) John is closest to 𝛢 when John is at the foot of the perpendicular from 𝛢 to 𝛣𝑪.

Let point D be the foot of the perpendicular from 𝛢 to 𝛣𝑪.

The distance he has to cycle before he is closest to A is** 7.03km**.

[ Hint: Shortest distance from A to BC is the perpendicular distance from A to BC. ]

(iv)

[ Hint: G is equidistant from A and C means G lies on the perpendicular bisector of A and C ]

[ Hint: 𝐴𝐶 = 𝐴𝐺 ]

(v)(vi) Let the midpoint of AC be E

[ Hint: Let the midpoint of AC be E so a right-angled triangle ACD is formed. Since we have ∠𝐸𝐶𝐺 and 𝐸𝐶 , we can use trigonometry to find CG. ]

Secondary 4 E-Math

A box contains 8 cards, numbered ‘1’ to ‘8’ respectively. A card is drawn with replacement from the box until an ‘8’ is obtained. Find the probability that

(i) the first draw was **not **an ‘8’,

(ii) it will take **exactly** *n* draws to obtain an ‘8’, giving your answer in terms of *n*,

(iii) it will take **at least** *n* draws to obtain an ‘8’, giving your answer in terms of *n*.

(iv) Krystal said that since she has not drawn an ‘8’ for the last 15 draws, her next draw will likely be an ‘8’. Comment on the validity of her statement.

(i) P(first draw not an ‘8’) =

[ Hint: There are 7 cards that are not ‘8’ ]

(ii) it will take **exactly** *n* draws to obtain an ‘8’, giving your answer in terms of *n*,

P(it will take exactly *n* draws) =

[ Hint: The first n-1 draws are not ‘8’ and for each of the first n-1 draws, the probability is 7/8. The nth draw is an ‘8’ and its probability is 1/8. ]

(iii) it will take **at least** *n* draws to obtain an ‘8’, giving your answer in terms of *n*.

P(it will take at least *n* draws) =

[ Hint: To obtain an ‘8’ with at least n draws is the same as not obtaining an ‘8’ with n-1 draws. ]

(iv) Statement is not valid. As the balls were drawn with replacement, her chances of getting an ‘8’ in the draw is independent of her previous draws.

[ Hint: Since a card is drawn with replacement from the box, each draw is independent and there are always cards that are not 8 in each draw. ]

Secondary 3 A-Math

A wireless access point is installed at point 𝑀. The coverage of the wireless access point can be modelled by a circle 𝐶_{1 } whose equation is 𝑥_{2} + 𝑦_{2} - 80𝑥 - 160𝑦 - 2000 = 0 with centre 𝑀 and radius 𝑟.

(a) Find the coordinates of 𝑀 and the value of 𝑟.

(b) Circle 𝐶_{1} intersects the 𝑥-axis at points 𝐽 and 𝑆. A second circle, 𝐶_{2} , comes from an identical wireless access point installed at point 𝑆. The 𝑥-coordinate of 𝑆 is positive. Explain why the point (0,20) lies within only one circles 𝐶_{1 } and 𝐶_{2}.

(c) he wireless access point at 𝑀 is replaced with a new model which can cover 4 times the area. Find the equation of the new circle 𝐶_{3}.

(a) 𝑥_{2} + 𝑦_{2} - 80𝑥 - 160𝑦 - 2000 = 0

Using completing square to form the standard form of equations of circle.

(𝑥 - 40)^{2} - (𝑦 - 80)^{2} - 2000 - 1600 - 6400 = 0

(𝑥 - 40)^{2} - (𝑦 - 80)^{2} = 100^{2}Therefore the coordinate of **𝑀**** **is **(40, 80)** and the **radius** is **100 units.**(b)

Sub y = 0 into (𝑥 - 40)

(𝑥 - 40)

(𝑥 - 100)(𝑥 + 20) = 0

𝑥 = 100 or 𝑥 = -20

S(100,0) , J(-20,0)

Length of M to (0,20)

Since

** Therefore (0,20) lies in the circle.**

(c) Old radius = 100, old diameter = 200.

4 x 𝜋(100)^{2} = 40000𝜋

𝑟^{2} = 40000^{2} ---> 𝑟 = 200 units

Therefore equation of circle is (𝑥 - 40)^{2} + (𝑦 - 80)^{2} = 40000

Secondary 4 A-Math

A particle is moving in a straight line passes a fixed point A with a velocity of -8𝑐𝑚𝑠^{-1}. The acceleration, 𝑎 𝑐𝑚𝑠^{-2} of the particle, 𝑡 seconds after passing A is given by 𝑎 = 10 - 𝑘𝑡 where 𝑘 is a constant. The particle first come to instantaneous rest at 𝑡 = 1 and reaches maximum speed at 𝑇 seconds. (the particle does not come instantaneous to rest at 1 < 𝑡 < 𝑇 ).

(a) Find the value of 𝑘.

(b) Find the total distance travelled by the particle when 𝑡 = 𝑇.

Since at point A, 𝑡 = 0, 𝑣 = -8

Therefore

Instantaneous rest at 𝑡 = 1, 𝑣 = 0

(a) Maximum speed = Maximum velocity ➔ Acceleration = 0

Sub 𝑎 = 0 into 𝑎 = 10 - 4𝑡

0 = 10 - 4𝑡

𝑇 = 2.5𝑠

At fixed point A, Sub 𝑡 = 0, 𝑠 = 0 into 𝑠

At first instantaneous rest, sub 𝑡 = 1 into 𝑠

At T = 2.5s,

Distance travelled