Word problems
1 Algebraic translations
Almost any word problem can be broken down into four steps:
1. Identify what value the question is asking for. We’ll this the desired value
2. Identify unknown values and label them with variables.
3. Identify relationships and translate them into equations.
4. Use the equations to solve for the desired value.
You need to turn a word problem into a system of equations, and use those equations to
solve for the desired value.
Pay attention to units
Every equation that correctly represents a relationship has units that make sense. Most
relationships are either additive ore multiplicative. The additive relationships make sense
because the units of every term are the same. Also adding terms whit he same units does
not change the unit.
For multiplicative relationships, treat units like numerators and denominators. Units that are
multiplied together DO change.
Common relationships
You must know the following relationships:
Total cost = unit price (€/unit) x quantity purchased (units)
Profit = revenue – cost
Total earnings = wage rate (€/hour) x hours worked
Miles = miles per hour x hours
Miles = miles per gallon x gallon
When values with units are multiplied or divided, the units change. This property is the basis
of using conversion factors to convert units (omrekeningsfactoren om eenheden te
converteren.). a conversion factor is a fraction whose numerator and denominator have
different units but the same value. 60 seconds / 1 minute is a conversion factor.
Integer constraint (beperking)
If Kelly received 1/3 more votes than Mike, which of the following could have been the total
number of votes cast for the two candidates?
A 54 B 55 C 56 D 57 E 58
The total number must be a whole number. Kelly received 4/3 the number of votes. What
times 4/3 will equal an integer? Only multiples of 3 will cancel out the 3.
Mike Kelly Total
3 4 7
6 8 14
9 12 21
By mike it is a multiply of 3, and by Kelly a multiply of 4. Taken together, the total number of
votes cast must then be a multiple of 7. The answer is C. following equations
2 Rates & work
Rate problems all are marked by three primary components: rate, time and distance or work.
These elements are related by the following equations:
Rate x time = distance RT = D
OR
Rate x time = work RT = W
, Basic motion: the RTD chart
All basic motion problems involve three elements: rate, time and distance.
Rate is expressed as a ratio of distance and time, with two corresponding units. VB:
30 miles per hour.
Time is expressed using a unit of time. VB: 6 hours, 23 seconds
Distance is expressed using a unit of distance. VB: 18 miles, 20 meters.
You can make an RTD chart to solve a basic motion problem. VB: if a car is traveling at 30
miles per hour, how long does it take to travel 75 miles?
Rate X Time = Distance
Car 30 mi/hr X - 75 mi
30t = 75 or t = 2,5 hours
Matching units in the RTD chart
All the units in the RTD chart must match up with one another. Always express rates as
distance over time, not as time over distance. VB: 1 floor/ 4 seconds = 0,25 floor/second and
not 4 seconds per floor.
Multiple rates
Some rate questions will involve more than one trip or traveller. To deal with this, you need to
deal with multiple RT = D relationships. Example:
Harvey runs a 30 mile course at a constant rate of 4 miles per hour. If clyde runs the same
track at a constant rate and completes the course in 90 fewer minutes, how fast did clyde
run?
R X T (hours) = D
Harvey 4 T 30
Clyde T – 1,5 30
4t = 30 = t = 7,5
7,5 – 1,5 = 6 for clyde
R x 6 = 30 = r = 5
Relative rates
This are subset of multiple rate problems. The defining aspect of relative rate problems is
that two bodies are traveling at the same time. Three possible scenarios:
1. The bodies move towards each other
2. The bodies move away from each other
3. The bodies move in the same direction on the same path.
You can save valuable time and energy by creating a third RT = D equation for the rate at
which the distance between the bodies changes.
Als twee objecten naar elkaar toekomen, dus de afstand neemt af tussen de twee
objecten, mag je dit bij elkaar optellen. O1 5 mph 6 mph O2 = 5 + 6 = 11 mph
Als twee objecten van elkaar weggaan, dus de afstand neemt toe tussen de twee
objecten, mag je dit bij elkaar optellen. O1 30 mph 45 mph O2 = 30 +45 = 75
mph.
Als twee objecten de zelfde kant op gaan, mag je de afstanden tot het eindpunt van
elkaar aftrekken. O1 8 mph O2 5 mph = 8 – 5 = 3 mph
Two people are 14 miles apart an begin walking towards each other. Person A walks 3 miles
per hour, and person B walks 4 miles per hour. How long will it take them to reach each
other?
1 Algebraic translations
Almost any word problem can be broken down into four steps:
1. Identify what value the question is asking for. We’ll this the desired value
2. Identify unknown values and label them with variables.
3. Identify relationships and translate them into equations.
4. Use the equations to solve for the desired value.
You need to turn a word problem into a system of equations, and use those equations to
solve for the desired value.
Pay attention to units
Every equation that correctly represents a relationship has units that make sense. Most
relationships are either additive ore multiplicative. The additive relationships make sense
because the units of every term are the same. Also adding terms whit he same units does
not change the unit.
For multiplicative relationships, treat units like numerators and denominators. Units that are
multiplied together DO change.
Common relationships
You must know the following relationships:
Total cost = unit price (€/unit) x quantity purchased (units)
Profit = revenue – cost
Total earnings = wage rate (€/hour) x hours worked
Miles = miles per hour x hours
Miles = miles per gallon x gallon
When values with units are multiplied or divided, the units change. This property is the basis
of using conversion factors to convert units (omrekeningsfactoren om eenheden te
converteren.). a conversion factor is a fraction whose numerator and denominator have
different units but the same value. 60 seconds / 1 minute is a conversion factor.
Integer constraint (beperking)
If Kelly received 1/3 more votes than Mike, which of the following could have been the total
number of votes cast for the two candidates?
A 54 B 55 C 56 D 57 E 58
The total number must be a whole number. Kelly received 4/3 the number of votes. What
times 4/3 will equal an integer? Only multiples of 3 will cancel out the 3.
Mike Kelly Total
3 4 7
6 8 14
9 12 21
By mike it is a multiply of 3, and by Kelly a multiply of 4. Taken together, the total number of
votes cast must then be a multiple of 7. The answer is C. following equations
2 Rates & work
Rate problems all are marked by three primary components: rate, time and distance or work.
These elements are related by the following equations:
Rate x time = distance RT = D
OR
Rate x time = work RT = W
, Basic motion: the RTD chart
All basic motion problems involve three elements: rate, time and distance.
Rate is expressed as a ratio of distance and time, with two corresponding units. VB:
30 miles per hour.
Time is expressed using a unit of time. VB: 6 hours, 23 seconds
Distance is expressed using a unit of distance. VB: 18 miles, 20 meters.
You can make an RTD chart to solve a basic motion problem. VB: if a car is traveling at 30
miles per hour, how long does it take to travel 75 miles?
Rate X Time = Distance
Car 30 mi/hr X - 75 mi
30t = 75 or t = 2,5 hours
Matching units in the RTD chart
All the units in the RTD chart must match up with one another. Always express rates as
distance over time, not as time over distance. VB: 1 floor/ 4 seconds = 0,25 floor/second and
not 4 seconds per floor.
Multiple rates
Some rate questions will involve more than one trip or traveller. To deal with this, you need to
deal with multiple RT = D relationships. Example:
Harvey runs a 30 mile course at a constant rate of 4 miles per hour. If clyde runs the same
track at a constant rate and completes the course in 90 fewer minutes, how fast did clyde
run?
R X T (hours) = D
Harvey 4 T 30
Clyde T – 1,5 30
4t = 30 = t = 7,5
7,5 – 1,5 = 6 for clyde
R x 6 = 30 = r = 5
Relative rates
This are subset of multiple rate problems. The defining aspect of relative rate problems is
that two bodies are traveling at the same time. Three possible scenarios:
1. The bodies move towards each other
2. The bodies move away from each other
3. The bodies move in the same direction on the same path.
You can save valuable time and energy by creating a third RT = D equation for the rate at
which the distance between the bodies changes.
Als twee objecten naar elkaar toekomen, dus de afstand neemt af tussen de twee
objecten, mag je dit bij elkaar optellen. O1 5 mph 6 mph O2 = 5 + 6 = 11 mph
Als twee objecten van elkaar weggaan, dus de afstand neemt toe tussen de twee
objecten, mag je dit bij elkaar optellen. O1 30 mph 45 mph O2 = 30 +45 = 75
mph.
Als twee objecten de zelfde kant op gaan, mag je de afstanden tot het eindpunt van
elkaar aftrekken. O1 8 mph O2 5 mph = 8 – 5 = 3 mph
Two people are 14 miles apart an begin walking towards each other. Person A walks 3 miles
per hour, and person B walks 4 miles per hour. How long will it take them to reach each
other?
