Equalising Units Concept (Remainder)

Given: Before, One Side, Change Ratios, After Amount
Find: Change Amount

First
At first, the amount of money Abel had was 3/5 of the amount of money Simon had. They went for dinner together. Abel paid 1/4 of the bill and Simon paid for the rest. After paying for the dinner, Abel has $63 left and Simon has 1/4 of his money left. How much was the dinner?

(Before)
Abel : Simon
= 3 : 5
= 12 : 20

(Simon)
After : Change : Before
= 1 : 3 : 4
= 5 : 15 : 20

(Change)
Abel : Simon : Total
= 1 : 3 : 4
= 5 : 15 : 20

(After)
Abel = 12 u - 5 u = 7 u
7 u = $63

(Change)
Total = 20 u
20 u = $180

Equalising Units Concept (Unchanged Side)

First
There were 200 supporters for a school's soccer team during a match. 40% of them were girls. Later, more girls joined the group of supporters during half-time. The percentage of the girls increased to 70%. Find the number of girls who joined the group of supporters during half-time.

Equalising Unchanged Sides Method

(Before)
Girls : Boys
= 40% : 60%
= 2 : 3

(After)
Girls : Boys
= 70% : 30%
= 7 : 3

(Before)
Total = 2 u + 3 u = 5 u
5 u = 200

(Change)
Girls = 7 u - 2 u = 5 u = 200

Finding Unchanged Amount Method

(Before)
Girls = 40% x 200 = 80 girls
Boys = 60% x 200 = 120 boys

(After)
Boys (30%) = 120 boys
Girls (70%) = 120 / 30% x 70% = 280 girls

(Change)
Girls = 280 girls - 80 girls = 200 girls

Second
There were total of 720 pupils at a stadium. 30% of the pupils were girls. When more giris came into the stadium, the percentage of girls increased to 40%. How many more girls came into the stadium?

Equalising Units Method

(Before)
Girls : Boys
= 30% : 70%
= 3 : 7
= 9 : 21

(After)
Girls : Boys
= 40% : 60%
= 2 : 3
= 14 : 21

(Before)
Total = 9 u + 21 u = 30 u
30 u = 720

(Change)
Girls = 14 u - 9 u = 5 u
5 u = 720 / 6 = 120

Finding Amount Method

(Before)
Girls = 30% x 720 = 216 girls
Boys = 70% x 720 = 504 boys

(After)
Boys (60%) = 504 boys
Girls (40%) = 504 / 60% x 40% = 336 girls

(Change)
Girls = 336 - 216 = 120 girls

Third
Clifford had a total of 600 Malaysia and Singapore stamps. 48% of his stamps were Malaysia stamps. When he collected more Malaysia stamps, the percentage of Singapore stamps decreased to 25%. How many new Malaysia stamps had he collected?

Equalising Units Method

(Before)
Malaysia : Singapore
= 48% : 52%
= 12 : 13

(After)
Malaysia : Singapore
= 75% : 25%
= 3 : 1
= 39 : 13

(Before)
Total = 12 u + 13 u = 25 u
25 u = 600
1 u = 24

(Change)
Malaysia = 39 u - 12 u = 27 u
27 u = 648

Finding Amount Method

(Before)
Malaysia = 48% x 600 = 288
Singapore = 52% x 600 = 312

(After)
Singapore (25%) = 312
Malaysia (75%) = 312 x 3 = 936

(Change)
Malaysia = 936 - 288 = 648

Fourth
Matthew had a total of 450 Japanese and Korean comics. 36% of them were Japanese comics. When he bought more Japanese comics, the percentage of Korean comics decreased to 48%. How many new Japanese comics did Matthew buy?

Equalising Units Method

(Before)
Japanese : Korean
= 36% : 64%
= 9 : 16
= 27 : 48

(After)
Japanese : Korean
= 52% : 48%
= 13 : 12
= 52 : 48

(Before)
Total = 27 u + 48 u = 75 u
75 u  = 450 comics

(Change)
Japanese = 52 - 27 = 25 u
25 u = 450 / 3 = 150 comics

Gap Unit Concept (Headstart + Catchup)

First
Betty starts saving $30 per month from January. Dave starts saving $45 per month from March. In which month will Betty and Dave have an equal amount of savings?

(Headstart)
Gap Total = Betty = $30 per month x 2 months earlier = $60

(Catchup)
Gap Unit = Dave - Betty = $45 - $30 = $15 per month
Gap Number = $60 / $15 per month = 4 months

Second
Joseph reads 12 pages of a novel each day. His sister reads only 8 pages of the some novel each day but has started reading it 6 days earlier than Joseph. If Joseph began reading on 1st of February, on which date would his sister and he have read the same number of pages?

(Headstart)
Gap Total = Sister = 8 pages each day x 6 days earlier = 48 pages

(Catchup)
Gap Unit = Joseph - Sister = 12 - 8 = 4 pages each day
Gap Number = 48 pages / 4 pages each day = 12 days

1st of February + 12 days - 1 day = 12th of February (ordered range concept)

Third
Machine A prints 18 books per hour. Machine B prints 12 books per hour. How many hours must Machine A operate so that it will print the same number of books as Machine B given that Machine B has started operating 2 hours earlier than Machine A?

(Headstart)
Gap Total = Machine B = 12 books per hour x 2 hours earlier = 24 books

(Catchup)
Gap Unit = Machine A - Machine B = 18 - 12 = 6 books per hour
Gap Number = 24 books / 6 books per hour = 4 hours

Fourth
Patricia and James are selling funfair tickets. Every day, Patricia can sell 20 tickets while James can sell 14 tickets. If James started selling 9 days earlier than Patricia, how many days will Patricia take to sell the same number of tickets as James?

(Headstart)
Gap Unit Total = James = 14 tickets each day x 9 days earlier = 126 tickets

(Catchup)
Gap Unit Value = Patricia - James = 20 - 14 = 6 tickets each day
Gap Unit Number = 126 tickets / 6 tickets each day = 21 days

Area Concepts (Triangles with Common Bases or Heights)

Given: Common Base, Combined Height
Find: Combined Area

First
The figure below, not drawn to scale, shows two shaded triangles in a rectangle of length 20 cm and breadth 16 cm. What fraction of the whole figure is shaded?

(Express your answer in the simplest form.)

 

Method 1: Find Areas

Shaded Area
= Combined Area of 2 Triangles
= 1/2 × Common Base x Combined Height
= 1/2 × 12 cm × 16 cm
= 96 cm^2

Area of Rectangle
= length x breadth
= 20 cm x 16 cm
= 320 cm^2

Fraction (Shaded Area / Whole Area)
= 96 / 320 = 3 / 10

Method 2: Formulas in Ratios

Combined Height = Breadth

Shaded Area : Whole Area
= 1/2 × Common Base × Combined Height : Length x Breadth
= 1/2 × Common Base : Length
= 1/2 × 12 : 20
= 3 : 10
= 3 / 10

Units and Parts (Unchanged Amount and From Equality)

First
Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim’s sweets to chocolates became 1 : 7. The ratio of Ken’s sweets to chocolates became 1 : 4. How many sweets did Ken buy?

(After)
(Jim)
Sweets : Chocolates
= 1u : 7u

(Ken)
Sweets : Chocolates
= 1p : 4p

Reverse the Changes

(Before)
(Sweets)
Jim Sweets = Ken Sweets
1u + 12 sweets eaten = 1p
7u + 84 = 7p

(Chocolates)
Jim Chocolates = Ken Chocolates
7u = 4p + 18 chocolates eaten
7u - 18 = 4p

84 -(-18) = 7p - 4p
3p = 102

Ken Sweets (Before) = 1p = 34

Conversion (Units)

Time Conversion

Concepts: 
6 Times Tables

Big to Small

First (Single Unit)
Convert from hours to minutes

2 hours = 2 x 60 minutes = 120 minutes

5 hours = 5 x 60 minutes = 300 minutes


Convert from minutes to seconds

7 minutes = 7 x 60 seconds = 420 seconds

8 minutes = 8 x 60 seconds = 480 seconds


Second (Mixed Units)
Convert from hours and minutes to minutes

4 hours 5 minutes 
= 4 x 60 minutes + 5 minutes
= 240 minutes + 5 minutes
= 245 minutes

6 hours 24 minutes 
= 6 x 60 minutes + 24 minutes
= 360 minutes + 24 minutes
= 384 minutes

Convert from minutes and seconds to seconds

12 minutes 54 seconds 
= 12 x 60 seconds + 54 seconds
= 720 seconds + 54 seconds
= 774 seconds

14 minutes 28 seconds 
= 14 x 60 seconds + 28 seconds
= 840 seconds + 28 seconds
= 868 seconds

21 minutes 41 seconds
= 21 x 60 seconds + 41 seconds
= 1260 + 41 seconds
= 1301 seconds


Small to Big

First (Single Unit: Whole Numbers)

Convert from minutes to hours

360 minutes = (360 / 60) hours = 6 hours



480 minutes = (480 / 60) hours = 8 hours



540 minutes = (540 / 60) hours = 9 hours


Convert from seconds to minutes


600 seconds = (600 / 60) minutes = 10 minutes


720 seconds = (720 / 60) minutes = 12 minutes


840 seconds = (840 / 60) minutes = 14 minutes

Second (Mixed Units, Mixed Numbers)

366 minutes = (366 / 60) hours = 6 hours 6 minutes
or
= 6 6/60 hours = 6 1/10 hours

492 minutes = (492 / 60) = 8 hours 12 minutes
= 8 12/60 hours = 8 1/5 hours

Patterns (Tokens)

Patterns (Rods)

Angles (Parallelogram and Triangle)

In the figure below, not drawn to scale, ABCD is a parallelogram, DE and AE are straight lines. Find the ∠AED.

Label the point with a 90 angle as F.

Method 1
CFE = 90 (adjacent angles on straight line BFC)
BCE = 65 (corresponding angles)
AED = 180 - 90- 65 = 25 (angle sum of triangle CFE)

Method 2
AFB = 90 (adjacent angles on straight line BFC)
ABC = 65 (diagonally opposite angles of parallelogram ABCD)
AED = BAF = 180 - 90 - 65 = 25 (angle sum of triangle BAF, alternate angles)

Angles (Isosceles Triangle)

The figure below is not drawn to scale. Given that TP = SP, find ∠RTS

TRS = 40 + 24 = 64 (exterior angle = interior opposite angles of triangle TPR)

TSP = (180 - 24) / 2 = 78 (base angles of isosceles triangle TSP)

RTS = 180 - 78 - 64 = 38 (angle sum of triangle RTS)

TRP = 180 - 40 - 24 = 116 (angle sum of triangle TRP)

TSP = (180 - 24) / 2 = 78 (base angles of isosceles triangle TSP)

RTS = 116 - 78 = 38 (exterior angle = interior opposite angles of triangle RTS)

Angles (2 Isosceles Triangles)

First

The figure below is not drawn to scale. ABC is an isosceles triangle where BA = BC. Given that D is the midpoint of BC and AC is 1/2 of BA, find ∠ADB.

Method 1

∠DCB = (180 - 20) / 2 = 80 (base angles of isosceles triangle ABC)

∠ADC = (180 - 80) / 2 = 50 (base angles of isosceles triangle ACD)

∠ADB = 180 - 50 = 130 (adjacent angles on straight line BDC)

Angles (Isosceles Triangle and Parallelogram)

First

The following figure is not drawn to scale. Given that ABCD is a parallelogram and ∠ADE is an isosceles triangle, find ∠y.

ADC = 180 - 118 = 62 (co-interior angles of parallelogram ABCD)

Y = 62 / 2 (exterior angle = interior opposite angles, base angles of isosceles triangle ADE)

DAB = 118 (diagonally opposite angles of parallelogram ABCD)

y = (180 - 118) / 2 = 31 (co-interior angles of trapezium ABCE)

Second

Study the figure below and find ∠FCB.

Method 1

FCD = EFA = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, corresponding angles)

DCB = 180 - 68 = 112 (co-interior angles of parallelogram ABCD)

FCB = 112 - 48 = 64

Method 2

CFB = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, vertically opposite angles)

FBC = 68 (diagonally opposite angles)

FCB = 180 - 48 - 68 = 64 (angle sum of triangle FCB)

Angles (Rhombus & Parallelogram)

In the figure below, ABCD is a rhombus and CDEF is a parallelogram, ∠ADE is 150° and ∠CFE is 115°. Find ∠x.

Method 1
∠FCB = 150° (corresponding angles)
∠FCD = 180° - 115° = 65° (co-interior angles of a parallelogram)
∠DCB = 150° - 65° = 85°
∠DAB = 85° (diagonally opposite angles of a rhombus)
∠x = 85° / 2 = 42.5° (angle bisector of a rhombus)

Method 2
∠EDC = 115° (diagonally opposite angles of a parallelogram)
∠CDA = 360° - 150° - 115° = 95° (angles at a point)
∠DAB = 180° - 95° = 85° (co-interior angles of a parallelogram)
∠x = 85° / 2 = 42.5° (angle bisector of a rhombus)

Angles (Sum of Angles)

First

The figure below consists of 3 straight lines. What is the value of  ∠a + ∠b + ∠c?

Method 1

Using Vertically Opposite Angles, the angle in between ∠a & ∠c is equal to ∠b.

Using Adjacent Angles on a Straight Line, ∠a + ∠b + ∠c = 180° because ∠a, ∠b & ∠c lie on a straight line

Method 2

Using Vertically Opposite Angles, the other 3 angles are equal to ∠a + ∠b + ∠c.

Using Angles at a Point:

2 (∠a + ∠b + ∠c) = 360° or 2∠a + 2∠b + 2∠c = 360°

Therefore, ∠a + ∠b + ∠c = 360° / 2 = 180°

Second

The figure below is not drawn to scale. It is made up of 4 identical squares. find the sum of the 3 marked angles.

Label the topmost angle as a, the one directly beneath is as b and the one on its right as c.

b = 45 (angle bisector of a square)
a + c = 90 (double alternate angles)
a + b + c = 45 + 90 = 135


Third
The figure below is not drawn to scale. It shows two triangles inside a square ABCD. AE = CD. Find ∠BED.

Method 1 (Trans-Diagram)

Extend line AE upwards. Cut each of the two new exterior angles created into halves. Each of the exterior angle halves is equal to the interior opposite base angles of its own isosceles triangle.

(360 - 90) / 2 = 135 (angles at point A)

Method 2 (Algebraic)

AED = (180 - DAE) / 2

AEB = (180 - BAE) /2 

DAB =DAE + AEB = 90

AED + AEB 

= (180 - DAE) / 2 + (180 - BAE) /2 

= (360 - DAE - BAE) / 2

= (360 - DAB) / 2

= (360 - 90) /2

= 135

Circles (Terminology)

Circumference
= Line Around a Circle
= Perimeter

Circumference is slightly more than 3 times the diameter in a ratio given by Pi which is approximately 3.14 or 3 1/7

Centre of a Circle is a point that is of equal distance from every point on the circumference.

The Diameter is any line that starts from one point on a Circumference, passes through the Centre and ends on another point on the opposite side of a circumference.

The Radius is any line that starts from the Centre and ends on any point on the Circumference or vice versa. It is half the diameter.

A Chord is any line that starts from one point of a Circumference and ends at another point on the same Circumference. Diameter is a special type of Chord.

A Tangent is any line that starts and ends from outside the Circle but touches it only at one point.

A Secant is any line that starts and ends from outside the Circle and cuts the Circle at 2 points. It is like a Chord but with a line that extends betond the Circumference. A Tangent line is a special type of Secant.

A Semicircle is Half a Full Circle with a Central Angle of 180°. It's Area and Arc Length is also Half that of a Full Circle. It

A Quadrant is a Quarter of a Full Circle with a Central Angle of 90°. It's Area and Arc Length are also a Quarter of a Full Circle.

A Sector is a slice of a Full Circle. The Semicircle and the Quadrant are special types of Sectors.

An Arc is a portion of a Circumference.

Ordered Range Concept

The number of things between two ordinal numbers is equal to the difference and 1.

Ordered Range = Difference + 1

Taking a Difference eliminates the first object on an Ordered Range. Adding 1 restores that object back.

Derived from:

Elimination Concept
Difference Concept

Difference Concept

The Difference is gap between two things. The nearer they are, the smaller the difference. The further apart they are, the wider the difference.

Difference = Big Part - Small Part

The Small Part is separated from the Big Part by a Difference. To find something smaller from a bigger thing, we must Subtract.

Small Part = Big Part - Difference

The Big Part is separated from the Small Part by a Difference. To find something bigger from a smaller thing, we must Add.

Big Part = Small Part + Difference

Derived From:
Elimination Concept
Small to Big Concept
Big to Small Concept

Application:

1) Algebra

Total Concept

The Total is a Sum of its Parts. It is bigger than every Part within it. To find something bigger from smaller quantities, we have to Add.

First Part + Second Part = Total

Each Part is smaller than the Total. To find something smaller from a bigger quantity, we have to Subtract.

First Part = Total - Second Part
Second Part = Total - First Part

Application:

1) Geometry
- Angle Sum of Triangle
- Adjacent Angles on a Straight Line
- Angles about a Point
- Pythagoras' Theorem
2) Algebra
3) Arithmetic
4) Trigonometry

Etc

This concept is derived from the Small to Big Concept and the Big to Small Concept

Trigonometry (Pythagorean Identity)

(sin A)^2 + (cos A)^2 = 1
Square of Height + Square of Base = Square of Slope

Using a Unit Circle with Radius = 1, with a Centre about the Origin (0,0) on the Cartesian Plane, a perpendicular line drawn from any point on the circle's circumference to the X-Axis produces a right angle.

That perpendicular line (height of the triangle) is Opposite to the Angle between the X-Axis and the hypotenuse. It represents the values about the Y-Axis. Since Hypotenuse = 1, sine of angle equals the y values.

The base of the triangle is along the X-Axis and thus represents the values about the X-Axis. It is Adjacent to the Angle. Since Hypotenuse = 1, cosine of angle equals the x values

The slope of the triangle which has a value of 1 which is also the radius of the Unit Circle, is the Hypotenuse. It is opposite to the right angle of the triangle.

Using Pythagoras' Theorem,

Y^2 + X^2  = 1^2
(sin A)^2 + (cos A)^2 = 1

Circles Concepts (Difference in Area)

The diagram below shows a circle with center O. AB and XY are diameters of the circle. Find the difference between the shaded and unshaded areas 

Rate Concept (Candle)

First
Two candles of the same height are lit at the same time. The first candle takes 5h to burn completely. The second candle takes 4h to burn completely. If each candle burns at a constant rate, how long does it take, in hours, for the height of the first candle to be four times that of the second candle?

Part I
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Method 1: (Common Total Concept)
1st Candle (Rate) = 1/5 candle/hour = 4/20 candle/hour
2nd Candle (Rate) = 1/4 candle/hour = 5/20 candle/hour

(Amount Burnt)1st Candle : 2nd Candle
= 4u : 5u

Method 2: (Inverse Proportionality + Proportionality Concept)

(Time)
1st Candle : 2nd Candle
= 5 : 4

(Rate)
1st Candle : 2nd Candle
= 4 : 5

(Amount Burnt)
1st Candle : 2nd Candle
= 4u : 5u

Part II
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(Remaining Amount)
1st Candle : 2nd Candle
= 4p : 1p

Method 1: Elimination  and Substitution Concept

(Full Candle): 4u + 4p = 5u + 1p

(Elimination): 3p = 1u  or 1u = 3p

(Expansion): 4u = 12p

(Substitution) 
Total Candle = 4u + 4p = 12p + 4p = 16p

(Amount Burnt)1st Candle = 4u = 12p

12/16 x 5 hours = 3 3/4 hours = 3.75 hours = 3 hours 45 minutes

Method 2: Bar Model Drawing


1st Candle: 
 U   U   U   U PPP P
 U   U   U   U   U  P
2nd Candle:

1st Candle: 
PPP PPP PPP PPP PPP P
PPP PPP PPP PPP PPP P
2nd Candle:

12/16 x 5 hours = 3 hours 45 minutes

Rate Concept (Taps and Tanks)

First
Tap X flows at a rate of 2100 ml/min while Tap Y flows at a rate of 2500ml/min. Both taps were turned on at the same time to fill a tank with dimensions 50 cm x 40 cm x 30 cm. After 5 minutes, the plug at the bottom of the tank is removed, with the two taps still running. If the water is drained at a rate of 600 ml/min, what is the water level 2 minutes after the plug is removed?
(Ans: 15.5cm)

Filling Tap X (Rate) = 2100 ml/min
Filling Tap Y (Rate) = 2500 ml/min
Draining Plug (Rate) = 600 ml/min

(First 5 minutes)
Tap X + Tap Y (Rate) = 2100 + 2500 = 4600 ml/min
Volume of Water = 4600 ml/min x 5 min = 23 000 ml

(2 minutes after first 5 minutes)
Tap X + Tap Y - Plug (Rate) = 4600 - 600 = 4000 ml/min
Volume of Water = 4000 ml/min x 2 minutes = 8000 ml 

Total Volume of Water = 23 000 ml + 8000 ml = 31 000 ml
Water Level (Height) = Volume / Length / Breadth 

= 31 000 / 50 / 40 = 15.5cm


Second
At 9am, Mr Fernandez used 2 taps to fill up a tank. The first tap could fill the tank in 4 hours. The second tap could fill the tank in 3 hours. An hour after both taps were turned on, the second tap were faulty and stopped working. Mr Fernandez accidentally opened the 3rd tap which could drain a full tank completely in 2 hours. Instead of being filled, the tank was being emptied. How long did it take for the tank to be completely empty ?


1st Filling Tap (Rate) = 1/4 tank/hour
2nd Filling Tap (Rate) = 1/3 tank/hour
3rd Draining Tap (Rate) = 1/2 tank/hour

(1st Hour: 9am - 10am) (1st and + 2nd Filling Tap)
1st + 2nd Filling Tap (Rate) = 1/4 + 1/3 = 7/12 tank/hour
Amount of Tank = 7/12 tank/hour x 1 hour = 7/12 tank


(After 1st Hour) (1st Filling Tap - 3rd Draining Tap)
3rd - 1st Tap (Rate) = 1/2 - 1/4 = 1/4 tank/hour
Remaining Amount of Tank = 1 tank - 7/12 tank = 5/12 tank

Drainage (Time) = (5/12 tank) / (1/4 tank/hour) = 1 2/3 hour = 1 hour 40 minutes

Rate Concept (Labour)

Additive Rate
James takes 6 hours to paint a house. If Diana helps him, they would take 4 hours. How long will Diana take to paint the house herself?


James (Rate) = 1 house / 6 hours = 1/6 house/hour
Diana (Rate) = 1/4 - 1/6 = 1/12 house/hour (Total Concept)
James + Diana (Rate) = 1 house / 4 hours = 1/4 house/hour
Diana (Time) = 1 house / (1/12 house/hour) = 12 hours
Algebra Method:
1 J = 1/6 house
1/12 house = 1 D
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Single Combo + Gap Concept
Desmond (Rate) = 1/9 kitchen/day
If ALL 13 days are done by Philip:
Gap Amount = 1 kitchen - 13/21 kitchen = 8/21 kitchen
Philip (Time) = (8/21 kitchen) / (4/63 kitchen/day)
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Double Combos + Elimination
Double Combos + Elimination Method
4 C + 2 E = 44
4 E = 84 - 44 = 40 computers
3 E = 30 computers
2 C = 42 - 30 = 12 computers
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Given: Paired Rate
Find :

First
Ahmad and Halim together took 5 days to paint their house. If Ahmad and Halim work together for 2 days, followed by Ahmad working alone for 8 days, Halim will take 1 more day to complete the remaining work. How long will Ahmad take to paint the house all by himself?

Ahmad + Halim (Rate) = 1/5 house /day
(First 2 days) (Together)
2 A + 2 H = 2/5 house

Remaining Amount of House = 3/5 house
8 A + 1 H = 3/5 house
1 A + 1 H = 1/5 house
7 A = 2/5


1 house = 7 A /2 x 5 = 17.5 days


Second
Amin and Haiqing together took 5 days to paint their house. If Amin and Haiqing work together for 2 days, followed by Amin working alone for 8 days, Haiqing would take 1 more day to complete the remaining work. How long will Amin take to paint the house all by himself?

5 A + 5 H = 5/5 house
1 A + 1 H = 1/5 house
2 A + 2 H = 2/5 house

Remainder = 1 - 2/5 = 3/5 house
8 A + 1 H = 3/5 house
1 A + 1 H = 1/5 house
7 A = 2/5 house

1 house = 5/5 house = 7 A / 2 x 5 = 17.5 A

Amin will take  10.5 days to paint the house by himself


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If Allan and Bob work together, they can complete a job in 12 days. If they work together for 3 days, followed by Bob who worked alone for the next 5 days, they can finish 5/12 of the job. How many days will each of them need to complete the job if they were to work alone?
(Ans: Bob 30 days, Alan 20 days)
12 A + 12 B = 1 job
3 A + 3 B + 5 B = 5/12 job
32 B - 12 B = 5/3 job - 1 job
Bob takes 30 days to finish the job alone
30 B = 1 job
12 A = 1 job - 2/5 job = 3/5 job
Ahmad takes 20 days to complete the job alone 
Triple Combos
John and Rauf take 4 days to build a model train. Rauf and Sean take 6 days to build the same train while John and Sean take only 3 days. How long would the three of them take to build the same model train?
1 J + 1 R = 1/4 train
1 J + 1 R + 1 R + 1 S + 1 J + 1 S = 1/4 + 1/6 + 1/3
1 / 3/8 = 8/3 = 2 2/3 days
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1 A + 1 B = 1/4 aircraft






1 J + 1 D = 1/4 house
1 D = 1/4 - 1/6 = 1/12 house

1 house = 12 D


Desmond takes 9 days to renovate 1 kitchen while Philip takes 21 days to renovate the same kitchen. If Philip starts renovating the kitchen first and leaves the rest to be completed by Desmond, they will take 13 days in total to complete renovating the kitchen. How many days does Desmond spend on the job?

Philip (Rate) = 1/21 kitchen/day
Gap Rate = 1/9 - 1/21 = 4/63 kitchen/day

Amount of Kitchen by Philip
= (1/21 kitchen/day) x 13 days
= 13/21 kitchen = 13/21 kitchen


= 6 days


When Caleb spent 2 days and Ethan spent 3 days assembling computers, they produced 42 sets of PCs. When Caleb spent 4 days and Ethan spent 2 days assembling computers, they produced 44 sets of PCs. How long will each boy require to assemble 30 sets of computers?

2 C + 3 E = 42 computers
4 C + 6 E = 84 computers


1 E = 10 computers
Ethan can assemble 10 computers in 1 day

Ethan can assemble 30 computers in 3 day

1 C = 6 computers
Caleb can assemble 6 computers in 1 day
5 C = 30 computers
Caleb can assemble 30 computers in 5 days




Amount of House = 1/5 x 2 = 2/5 house


16 A + 2 H = 6/5 house








3 A + 8 B = 5/12 job
12 A + 32 B = 5/3 job

20 B = 2/3 job
1 job = 30 B


3 B = 1/10 job
12 B = 4/10 = 2/5 job

1 job = 20 A


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1 R + 1 S = 1/6 train
1 J + 1 S = 1/3 train

2 J + 2 R + 2 S = 3/4 train
1 J + 1 R + 1 S = 3/8 train

Abel and Ben can build a model aircraft together in 4 days. Ben and Calvin can do the same job together in 1 1/2 times as many days. Abel and Calvin can complete the same job together in twice the number of days Ben and Calvin can do together. If the three boys decide to work together, how long will they take to complete their job?

4 days x 1 1/2 = 6 days
1 B + 1 C = 1/6 aircraft
6 days x 2 = 12 days
1 A + 1 C = 1/12 aircraft

2 A + 2 B + 2 C = 1/4 + 1/6 + 1/12 = 1/2 aircraft
4 A + 4 B + 4 C = 1 aircraft

They would take 4 days

Circles Concept (Complex Figures)

Shaded Area = (Square - Circle) / 2

Perimeter A = Circumference / 2

Gap Units Concept (Rift Unit)

First
Darren bought some tarts for a party. 40% of them were strawberry tarts and the rest were chocolate tarts. Each strawberry tart cost $2. The price of each chocolate tart was 90% of the price of each strawberry tart. Darren spent $30.80 more on chocolate tarts than he did on strawberry tarts. How many tarts did he buy altogether? 


(Big Unit) 
1 Strawberry = $2

(Small Unit)
1 Chocolate = 90% × $2 = $1.80

(Big Quantity : Small Quantity)
Strawberry : Chocolate
= 40 : 60 
= 2 : 3

(Rift Unit)
 3 Chocolate - 2 Strawberry
=  3 × $1.80 - 2 × $2
= $1.40

(Rift Quantity)
$30.80 ÷ $1.40 = 22

(Group Unit) 

3 + 2 = 5

Total Quantity = 22 × 5 = 110

Replacement

Single Equation + Substitution

First
Kelly spent 1/3 of her money on 5 pens and 11 erasers. The cost of each pen is 3 times the cost of each eraser. She bought some more pens with 3/4 of her remaining money. How many pens did she buy altogether?

(Single Equation)
5 P + 11 E = 1/3 money

(Cost)
Pen : Eraser
= 3u : 1u

(Equivalent Quantity)
3u = 1 Pen = 3 Erasers

(Replacement Quantity)
11 Erasers = 1 Pen / 3 Eraser x 11 Erasers = 11/3 Pens = 3 2/3 Pens

(Replacement)
5 P + 3 2/3 P = 1/3 money
8 2/3 P = 1/3 money
1 money = 8 2/3 x 3 = 26 P

3/4 of remaining money = 3/4 x 2/3 = 1/2 money
1/2 money = 13 P

5 P + 13 P = 18 P altogether

Second
Danny spent 3/8 of his money on 10 identical markers and 10 identical files. He then spent 1/10 of the remainder on 6 pens. 
Each file cost 5 times as much as each marker and each pen cost $2.50 more than a marker.How much did each marker cost?

(Single Equation)
10 M + 10 F = 3/8 money

(Cost)
File : Marker
= 5u : 1u
5u = 1 F = 5 M
10 F = 50 M

10 M + 50 M = 3/8 money
60 M = 3/8 money
6 M = 

1/10 of remainder = 1/10 x 5/8 = 1/16  money
6 P = 1/16 money








3 shirts and 4 dresses cost $180. A dress cost 1 1/2  times as much as a shirt. What is the cost of a dress? 

Units Method

1 1/2 = 3/2

1 Dress = 3u

4 Dresses = 12u

1 Shirt = 2u

3 Shirts = 6u

12u + 6u = 18u

18u = $180

1 Dress = 3u = $30

Inverse Proportion Method

1 Dress : 1 Skirt

= 3u : 2u

6u = 2 Dresses = 3 Skirts

2 Dresses + 4 Dresses = $180

6 Dresses = $180

1 Dress = $30

Madam Halimah bought 7kg of sugar and 10kg of flour for $26.25.
If 3/5kg of sugar costs as much as 3/4kg as flour, find the cost of 1kg of flour.

A) Substitution and Elimination

1) John and Peter have a total mass of 66.6kg. Peter and Yenni have a total mass of 48.6kg. John is thrice as heavy as Yenni.  What is John's mass?

Working Backwards Concept

Step-by-Step Changes

1) A train left Station A with some passengers. At Station B, 1/3 of the passengers alighted and 20 passengers boarded the train. At Station C, 1/4 of the passengers alighted. In the end, there were 36 passengers left on the station. How many passengers were there on the train when it first left Station A?

2) A train with some passengers left Station A. At Station B, 1/5 of the passengers alighted and 26 passengers boarded the train. At Station C, 1/3 of the passengers alighted and 12 passengers boarded the train. If there were 328 passengers on the train then, how many passengers were there on the train when it left Station A?

3) A bus with some passengers left Bus Stop A for Bus Stop B. At Bus Stop B, 1/7 of the passengers alighted and 6 more boarded the bus. When the bus moved to Bus Stop C, 1/6 of the passengers alighted and there were 75 passengers that remained. How many passengers were there in the bus when it first left Bus Stop A?

4) When an MRT train arrived at City Hall Station, 2/7 of its passengers alighted and 150 passengers boarded the train. When the train arrived at Bugis Station, 1/5 of the passengers on the train alighted and 110 passengers boarded it. After this, there were 790 passengers on the train. How many passengers were there on the train at first?

5) There were some passengers on a bus.
At the first bus stop, 0.5 of the passengers alighted and 14 passengers came on board.
At the next bus stop, 0.25 of the passengers alighted and 10 passengers came on board.
If there 46 passengers now, how many passengers were on the bus at first?