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
Friday, 26 August 2016
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
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
Thursday, 25 August 2016
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
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
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
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?
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.
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?
Sunday, 21 August 2016
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?
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