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Showing posts with label SAT Practice. Show all posts
Showing posts with label SAT Practice. Show all posts

Wednesday, September 23, 2026

SAT Math Probability: Conditional Probability, Two-Way Tables, Formulas and Practice

 

SAT Math Probability: Formulas, Conditional Probability, Two-Way Tables and Practice

Probability on the SAT Math section is often less about complicated calculations and more about identifying exactly what population or group the question is talking about.

A problem may give you a simple collection of objects, a two-way table, a survey, a repeated experiment, or a situation involving two events.

The calculation may be short.

The difficult part is deciding what belongs in the numerator and what belongs in the denominator.

The College Board's current SAT framework includes probability and conditional probability within the Problem-Solving and Data Analysis content area. The framework specifically includes one-way tables, two-way tables, tree diagrams, area models, relative frequency, probability, and conditional probability.

This makes probability an important skill to understand if you are preparing for the Digital SAT Math section in 2026.


SAT Probability Formulas and Main Ideas

Before working through examples, learn these basic relationships.

Basic probability

Probability = favorable outcomes ÷ total outcomes

Complement

P(not A) = 1 − P(A)

Probability of A or B

P(A or B) = P(A) + P(B) − P(A and B)

Probability of A and B for independent events

P(A and B) = P(A) × P(B)

Conditional probability

P(A | B) = P(A and B) ÷ P(B)

The notation P(A | B) means:

the probability of A given B.

The event after the vertical bar tells you which group you are working within.

That idea is especially important for SAT conditional probability questions.


What Is Probability?

Probability describes how likely an event is to occur.

For an event with equally likely outcomes:

Probability = favorable outcomes ÷ total possible outcomes

A probability can be written as a fraction, decimal, or percentage.

For example, suppose a container has:

  • 6 red balls

  • 4 blue balls

There are 10 balls altogether.

The probability of selecting a red ball is:

6/10 = 3/5 = 0.6 = 60%

A probability of 0 means the event cannot happen.

A probability of 1 means the event must happen.

Therefore:

0 ≤ P(A) ≤ 1

This simple range is also useful when checking an answer. A probability cannot be negative and cannot be greater than 1.


How to Solve Basic SAT Probability Questions

For a basic probability question, identify two things:

Numerator: the outcomes you want

Denominator: all possible outcomes in the relevant sample space

Example

A box contains 7 green pencils, 5 yellow pencils, and 8 red pencils.

What is the probability of randomly selecting a yellow pencil?

There are:

7 + 5 + 8 = 20 pencils

There are 5 favorable outcomes.

Therefore:

P(yellow) = 5/20

P(yellow) = 1/4

So the probability is:

1/4, 0.25, or 25%

The important step is not the division.

It is recognizing that the denominator is 20, because all 20 pencils are possible selections.


SAT Probability From a Table

Probability questions can also provide data in a table.

Suppose a school records whether 160 students participate in a sports program.

SportsNo SportsTotal
Grade 9423880
Grade 10463480
Total8872160

Question

What is the probability that a randomly selected student participates in sports?

The question does not restrict the student to a particular grade.

Therefore, use the entire group.

Favorable students:

88

Total students:

160

So:

P(sports) = 88/160

= 11/20

= 0.55

= 55%

This is an example of a probability calculated from a two-way table using the overall total.


How to Read a Two-Way Table on the SAT

A two-way table organizes observations according to two categories.

In the example above:

  • The rows represent grade level.

  • The columns represent sports participation.

  • The interior cells show the number of students in both categories.

  • The row totals show the number of students in each grade.

  • The column totals show the number of students in each sports category.

  • The bottom-right value is the total number of students.

When solving a SAT two-way table probability problem, do not automatically choose the bottom-right number as your denominator.

First read the wording.

The correct denominator depends on the population described by the question.


SAT Conditional Probability: The Most Important Idea

Conditional probability means that the question has restricted the group you are considering.

Look for wording such as:

  • given that

  • among

  • of the students who

  • if the student is

  • for students who

  • selected from those who

These phrases can indicate that you should work with a smaller group rather than the entire sample.

Using the previous table:

SportsNo SportsTotal
Grade 9423880
Grade 10463480
Total8872160

Question

A student is known to be in Grade 9.

What is the probability that the student participates in sports?

The condition tells us that we are considering Grade 9 students only.

There are 80 Grade 9 students.

Of those 80 students, 42 participate in sports.

Therefore:

P(Sports | Grade 9) = 42/80

= 21/40

= 0.525

= 52.5%

Notice what happened to the denominator.

We did not use 160.

The condition changed the relevant population from all students to Grade 9 students.

That is the central idea behind many SAT conditional probability problems.


How to Remember P(A | B)

Read the notation from left to right:

P(A | B)

means:

probability of A, given B

The condition is B.

So the denominator represents B.

For example:

P(plays soccer | Grade 10)

means:

Among Grade 10 students, what fraction plays soccer?

It does not mean:

Among soccer players, what fraction is in Grade 10?

Those are different questions.

In general:

P(A | B) ≠ P(B | A)

The order matters.


Joint Probability From a Two-Way Table

A joint event requires both conditions to be true.

For example:

Grade 9 and Sports

From the table:

42 students are both Grade 9 students and sports participants.

Out of 160 students:

P(Grade 9 and Sports) = 42/160

= 21/80

A joint probability from a table normally uses the relevant intersection cell as the numerator and the overall total as the denominator when no condition restricts the sample.


Marginal Probability vs. Conditional Probability

These two ideas can look similar but use different denominators.

Marginal probability

A marginal probability describes one category without restricting the sample.

For example:

What is the probability that a randomly selected student participates in sports?

Use:

88/160

Conditional probability

A conditional probability restricts the sample.

For example:

What is the probability that a Grade 9 student participates in sports?

Use:

42/80

The numerator may involve the same intersection cell.

The denominator changes because the question changes the population being considered.

This distinction is one of the most useful skills for SAT probability from two-way tables.


Independent Events on the SAT

Two events are independent when knowing that one event occurred does not change the probability of the other.

For independent events:

P(A and B) = P(A) × P(B)

A familiar example is flipping a fair coin twice.

The first flip does not change the probabilities for the second flip.

The probability of heads on each flip is:

1/2

Therefore:

P(two heads) = 1/2 × 1/2

= 1/4

The important word is independent.

Do not multiply probabilities automatically just because a question contains two events.

First determine whether the events affect one another.


Dependent Events and Sampling Without Replacement

Events are dependent when the outcome of one event changes the probabilities for another event.

A common example is selecting objects without replacement.

Suppose a bag contains:

  • 4 red marbles

  • 6 blue marbles

Two marbles are selected without replacement.

What is the probability that both are red?

First selection

There are 4 red marbles out of 10:

4/10

After one red marble is selected, there are:

3 red marbles

and:

9 marbles total

So the second probability is:

3/9

Therefore:

P(both red) = 4/10 × 3/9

= 12/90

= 2/15

The second denominator is 9 rather than 10 because the first marble was not returned.

This is why the phrase “without replacement” deserves immediate attention.


SAT Probability With Replacement

Now change the situation.

Suppose the first marble is selected and then returned to the bag before the second selection.

The total number of marbles remains 10.

The probability of red on each selection is:

4/10

Therefore:

P(two red) = 4/10 × 4/10

= 16/100

= 4/25

The difference between replacement and no replacement can completely change the answer.

When you see a repeated-selection probability question, ask:

Does the sample space change after the first selection?


Probability of A or B

The word “or” requires careful reading.

If two events cannot happen together, add their probabilities.

For example, on a standard six-sided die, a roll cannot be both 2 and 5.

Therefore:

P(2 or 5) = P(2) + P(5)

= 1/6 + 1/6

= 1/3

But some events overlap.

When A and B can happen at the same time:

P(A or B) = P(A) + P(B) − P(A and B)

The overlap is subtracted because it was counted twice.


SAT Example: King or Heart

Consider a standard 52-card deck.

What is the probability of selecting a king or a heart?

There are:

4 kings

13 hearts

But the king of hearts belongs to both groups.

Therefore, the overlap must be subtracted.

P(king or heart)

= 4/52 + 13/52 − 1/52

= 16/52

= 4/13

The key question is:

Can the two events happen at the same time?

If yes, check for overlap.


The Complement Rule

Sometimes the easiest way to calculate a probability is to find the probability of the opposite event.

The complement rule is:

P(not A) = 1 − P(A)

Example

Suppose the probability that a randomly selected machine passes a quality test is:

0.92

Then the probability that it does not pass is:

1 − 0.92 = 0.08

So:

P(not pass) = 0.08

The complement rule is particularly useful when the question asks for:

  • not

  • does not

  • neither

  • none

  • at least one

depending on how the problem is structured.


Probability of “At Least One”

“At least one” means one or more.

It can sometimes be easier to calculate the opposite:

none

and then subtract from 1.

Suppose a fair coin is flipped three times.

What is the probability of getting at least one head?

The opposite event is getting no heads.

That means all three flips are tails.

P(three tails) = 1/2 × 1/2 × 1/2

= 1/8

Therefore:

P(at least one head)

= 1 − 1/8

= 7/8

This approach is often shorter than listing every possible sequence containing one, two, or three heads.


Probability With Percentages

Probability and percentage are closely connected.

For example:

1/4 = 0.25 = 25%

3/5 = 0.6 = 60%

7/10 = 0.7 = 70%

If a SAT question gives you percentages instead of counts, pay attention to what the percentage represents.

A percentage could describe:

  • the entire population

  • a particular row

  • a particular column

  • a conditional group

Do not assume that every percentage in a table uses the same denominator.


Relative Frequency and SAT Probability

The College Board framework also includes relative frequency in its probability and conditional probability skill area.

Relative frequency is essentially the proportion of observations belonging to an outcome.

For example, suppose a machine produces 500 parts and 15 are defective.

The observed relative frequency of defective parts is:

15/500

= 0.03

= 3%

A question may use experimental or survey data rather than a theoretical sample space.

In that case, carefully identify whether the problem is asking you to calculate a proportion from the observed data.


Tree Diagrams and Probability

The SAT framework also identifies tree diagrams as one representation that can be used with probability and conditional probability.

A tree diagram is useful when a process happens in stages.

For example:

First selection → Second selection

Each branch represents a possible outcome.

When following a particular path, multiply the probabilities along that path.

If several different paths produce the result the question asks for, add the probabilities of those paths.

The important ideas remain the same:

Multiply along a path.

Add separate paths when appropriate.

Always check whether the events are independent or whether earlier outcomes change later probabilities.


How to Solve SAT Probability Word Problems

Probability word problems can contain a lot of information.

Do not calculate immediately.

Use this process.

Step 1: Identify the experiment

What is being selected, measured, rolled, flipped, or observed?

Step 2: Identify the event

What outcome does the question want?

Step 3: Find the relevant sample space

Ask:

What outcomes are still possible?

Step 4: Look for a condition

Words such as:

“given that”

“among”

“of those who”

“if the student is”

may restrict the sample.

Step 5: Check for multiple events

Does the question involve:

and

or

at least one

neither

Step 6: Check replacement

If objects are selected repeatedly, determine whether the first object is returned.

Step 7: Calculate

Only after identifying the correct sample space should you perform the arithmetic.

Step 8: Check the result

A probability must be between 0 and 1.


The Most Common SAT Probability Mistakes

Mistake 1: Using the grand total for every question

This is especially common with conditional probability.

If the question says:

“Among Grade 10 students…”

you are not considering the entire school anymore.

The Grade 10 total becomes the relevant denominator.


Mistake 2: Reversing conditional probability

These are different:

P(A | B)

and

P(B | A)

Read the wording carefully.

The condition determines the denominator.


Mistake 3: Ignoring overlap in an “or” question

If A and B can happen together, simply adding P(A) and P(B) counts the overlap twice.

Use:

P(A or B) = P(A) + P(B) − P(A and B)


Mistake 4: Forgetting that “without replacement” changes the sample

If an object is removed and not returned, the number of available objects changes.

Recalculate the next probability.


Mistake 5: Multiplying every time you see “and”

The multiplication rule depends on the structure of the events.

For independent events:

P(A and B) = P(A) × P(B)

For dependent events, the second probability must account for the first event.


Mistake 6: Reading the table before reading the question

A large table can contain much more information than you need.

Read the question first.

Then locate only the row, column, cell, or totals required to answer it.


SAT Probability Practice Questions

Question 1: Basic Probability

A container contains 12 white cards, 8 blue cards, and 10 green cards.

What is the probability of selecting a blue card at random?

Solution

Total cards:

12 + 8 + 10 = 30

Blue cards:

8

Therefore:

P(blue) = 8/30

= 4/15

Answer: 4/15


Question 2: Conditional Probability From a Two-Way Table

A school surveys 200 students.

Plays SoccerDoes Not Play SoccerTotal
Grade 9364480
Grade 105466120
Total90110200

A student is selected from the Grade 10 group.

What is the probability that the student plays soccer?

Solution

The condition is Grade 10.

Therefore, the denominator is:

120

The number of Grade 10 students who play soccer is:

54

So:

P(Soccer | Grade 10) = 54/120

= 9/20

= 0.45

Answer: 9/20


Question 3: Independent Events

A fair six-sided die is rolled twice.

What is the probability of rolling a 6 both times?

The probability of a 6 on one roll is:

1/6

The rolls are independent.

Therefore:

P(two 6s) = 1/6 × 1/6

= 1/36

Answer: 1/36


Question 4: Without Replacement

A bag contains 5 red balls and 7 blue balls.

Two balls are selected without replacement.

What is the probability that both balls are red?

First selection:

5/12

After one red ball is removed:

4 red balls remain out of 11 total balls.

Second selection:

4/11

Therefore:

P(both red) = 5/12 × 4/11

= 20/132

= 5/33

Answer: 5/33


Question 5: Complement

The probability that a randomly selected package arrives on time is 0.94.

What is the probability that it does not arrive on time?

Use the complement:

1 − 0.94 = 0.06

Answer: 0.06


Question 6: “Or” With Overlap

A standard 52-card deck contains 4 kings and 13 hearts.

What is the probability of selecting a king or a heart?

The king of hearts belongs to both groups.

Therefore:

P(king or heart)

= 4/52 + 13/52 − 1/52

= 16/52

= 4/13

Answer: 4/13


A Fast SAT Probability Checklist

When you see a probability problem, ask these questions:

1. What is the event?

What exactly am I trying to find?

2. What is the relevant population?

Am I considering everyone or only a particular group?

3. Is there a condition?

Look for “given,” “among,” or similar wording.

4. Is there an overlap?

If the question uses “or,” determine whether both events can happen together.

5. Are the events independent?

If one event happens first, does it change the next probability?

6. Is there replacement?

If not, the sample size may change.

7. Can the answer be checked?

The probability must be between 0 and 1.


How to Improve at SAT Probability Questions

The best way to improve is not to memorize dozens of isolated tricks.

Instead, practice identifying the structure of the problem.

When you miss a question, ask:

Did I choose the wrong numerator?

Did I choose the wrong denominator?

Did I miss a condition?

Did I overlook an overlap?

Did I treat dependent events as independent?

Did I forget that an item was removed without replacement?

These questions reveal whether the problem was a calculation error or a misunderstanding of the probability model.

For SAT Math probability practice, it is particularly useful to mix simple probability questions with two-way tables, conditional probability, complements, independent events, and dependent events.

That prevents you from solving every problem with the same formula.


SAT Probability: What You Actually Need to Remember

Probability questions become much easier when you stop looking for one formula that works for everything.

Instead, identify the situation.

Basic probability

Favorable outcomes ÷ relevant total outcomes

Conditional probability

Restrict the sample to the group named by the condition.

P(A | B) = P(A and B) ÷ P(B)

Independent events

One event does not change the probability of the other.

P(A and B) = P(A) × P(B)

Dependent events

The probability of a later event changes because of an earlier event.

“Or”

Check whether the events overlap.

P(A or B) = P(A) + P(B) − P(A and B)

Complement

Sometimes the opposite event is easier to calculate.

P(not A) = 1 − P(A)

Two-way tables

Read the question first.

Then determine whether the denominator should be:

  • the overall total

  • a row total

  • a column total

  • or another restricted group

The central skill behind many SAT probability and conditional probability questions is knowing which group the probability is being calculated within.

Once that becomes automatic, many probability problems reduce to a carefully chosen fraction rather than a complicated calculation.

Sunday, September 20, 2026

SAT Work Rate Problems: Complete Guide to Combined Work, Pipes, Machines & Formulas

 📘 SAT MATH: WORK RATE PROBLEMS

𝗛𝗼𝘄 𝘁𝗼 𝗦𝗼𝗹𝘃𝗲 𝗖𝗼𝗺𝗯𝗶𝗻𝗲𝗱 𝗪𝗼𝗿𝗸, 𝗣𝗶𝗽𝗲, 𝗠𝗮𝗰𝗵𝗶𝗻𝗲 𝗮𝗻𝗱 𝗥𝗮𝘁𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺𝘀

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Work rate problems can look like long word problems, but the mathematics underneath is usually built from one simple idea:

𝗥𝗮𝘁𝗲 × 𝗧𝗶𝗺𝗲 = 𝗪𝗼𝗿𝗸

The difficulty comes from translating the wording.

A person may finish a task in a certain number of hours.

A machine may produce a certain number of parts every minute.

A pipe may fill a tank while another pipe drains it.

Two workers may begin at different times.

One worker may leave before the job is finished.

These situations look different, but they can all be handled by thinking about how much work is completed during one unit of time.

This guide builds the method from the beginning and then moves into the harder forms that can appear in SAT-style rate and word problems.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

1️⃣ 𝗧𝗛𝗘 𝗠𝗢𝗦𝗧 𝗜𝗠𝗣𝗢𝗥𝗧𝗔𝗡𝗧 𝗜𝗗𝗘𝗔

Suppose Maya can complete a job in 8 hours.

If the entire job is called 1 job, then in one hour Maya completes:

1/8

of the job.

So her rate is:

1/8 job per hour

Now suppose Leo can complete the same job in 12 hours.

His rate is:

1/12 job per hour

If they work at the same time, the amount completed during one hour is:

1/8 + 1/12

= 3/24 + 2/24

= 5/24

So together they complete:

5/24 of the job per hour

The time needed for one complete job is therefore:

1 ÷ 5/24

= 24/5

= 4.8 hours

So:

Answer: 4.8 hours

This is the central idea behind almost every SAT work rate problem:

𝗔𝗱𝗱 𝗿𝗮𝘁𝗲𝘀, 𝗻𝗼𝘁 𝘁𝗶𝗺𝗲𝘀.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

2️⃣ 𝗪𝗛𝗔𝗧 𝗜𝗦 𝗔 𝗪𝗢𝗥𝗞 𝗥𝗔𝗧𝗘 𝗣𝗥𝗢𝗕𝗟𝗘𝗠?

A work rate problem tells you how quickly something gets completed.

The “something” could be:

• a construction job
• a repair
• a cleaning task
• a batch of products
• a number of pages
• a tank being filled
• a tank being emptied
• a computer process
• a production order

The people or machines are simply different ways of describing rates.

For example:

A worker finishes a task in 6 hours.

This means:

Rate = 1/6 job per hour

A machine produces 80 parts per hour.

This means:

Rate = 80 parts per hour

A pump removes 15 liters per minute.

This is a rate in the opposite direction:

−15 liters per minute

The numbers look different, but the basic concept is identical.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

3️⃣ 𝗧𝗛𝗘 𝗕𝗔𝗦𝗜𝗖 𝗙𝗢𝗥𝗠𝗨𝗟𝗔

The fundamental relationship is:

Work = Rate × Time

From this:

Rate = Work ÷ Time

and:

Time = Work ÷ Rate

When a problem describes one complete job, it is usually convenient to let:

Total work = 1

Then a worker who takes t hours has a rate of:

1/t

This reciprocal relationship is why fractions appear so often in SAT combined-work problems.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

4️⃣ 𝗧𝗪𝗢 𝗣𝗘𝗢𝗣𝗟𝗘 𝗪𝗢𝗥𝗞𝗜𝗡𝗚 𝗧𝗢𝗚𝗘𝗧𝗛𝗘𝗥

Suppose:

Worker A finishes a job in a hours.

Worker B finishes the same job in b hours.

Their rates are:

1/a

and

1/b

Together:

1/a + 1/b

If T is the time they need together:

1/T = 1/a + 1/b

This can also be simplified to:

T = (a x b)/(a + b)

This is one of the most useful shortcuts for two-worker SAT problems.

Example

One technician completes a repair in 7 hours.

Another technician completes it in 21 hours.

Together:

T = (7 x 21)/(7 + 21)

T = 147/28

T = 21/4

T = 5.25 hours

So:

Answer: 5.25 hours

or:

5 hours 15 minutes

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

5️⃣ 𝗪𝗛𝗬 𝗗𝗢𝗡'𝗧 𝗪𝗘 𝗔𝗗𝗗 𝗧𝗛𝗘 𝗧𝗜𝗠𝗘𝗦?

This is one of the most common questions about work rate problems.

Suppose one person takes 10 hours and another takes 15 hours.

Adding the times gives:

10 + 15 = 25

But 25 hours has no meaning as their combined completion time.

The first person completes:

1/10

of the job each hour.

The second completes:

1/15

each hour.

Together:

1/10 + 1/15

= 3/30 + 2/30

= 5/30

= 1/6

So they complete the job at a rate of:

1/6 job per hour

Therefore:

Time = 6 hours

The reason is simple:

When two people work simultaneously, their work completed per hour is combined.

Their individual completion times are not.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

6️⃣ 𝗔 𝗙𝗢𝗨𝗥-𝗦𝗧𝗘𝗣 𝗠𝗘𝗧𝗛𝗢𝗗

When you see a SAT work rate question, use this sequence.

Step 1

Decide what the complete job represents.

Usually:

1 job

Step 2

Convert completion times into rates.

If someone takes 9 hours:

Rate = 1/9

Step 3

Combine the rates.

Workers helping:

Add

Drains or leaks:

Subtract

Step 4

Use:

Time = Work ÷ Rate

Then check the units.

This method works for two workers, three workers, machines, pipes and many harder variations.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

7️⃣ 𝗧𝗛𝗥𝗘𝗘 𝗪𝗢𝗥𝗞𝗘𝗥𝗦 𝗧𝗢𝗚𝗘𝗧𝗛𝗘𝗥

If three workers independently require:

a hours, b hours and c hours,

their combined rate is:

1/a + 1/b + 1/c

Therefore:

1/T = 1/a + 1/b + 1/c

Example

Three machines can finish an order individually in:

4 hours, 8 hours and 16 hours.

Their combined rate is:

1/4 + 1/8 + 1/16

= 4/16 + 2/16 + 1/16

= 7/16

Therefore:

T = 1 ÷ 7/16

= 16/7

hours.

Answer: 16/7 hours

or approximately:

2 hours 17 minutes

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8️⃣ 𝗠𝗔𝗖𝗛𝗜𝗡𝗘 𝗣𝗥𝗢𝗗𝗨𝗖𝗧𝗜𝗢𝗡 𝗣𝗥𝗢𝗕𝗟𝗘𝗠𝗦

Some SAT rate questions give the rate directly instead of giving the completion time.

Suppose:

Machine A produces 48 units per hour.

Machine B produces 32 units per hour.

Together:

48 + 32 = 80 units per hour.

If they need to produce 560 units:

Time = 560 ÷ 80

= 7 hours.

Answer: 7 hours

Notice that no reciprocal calculation was necessary.

The rates were already given.

That leads to an important rule:

𝗜𝗳 𝘁𝗵𝗲 𝗽𝗿𝗼𝗯𝗹𝗲𝗺 𝗴𝗶𝘃𝗲𝘀 𝗿𝗮𝘁𝗲𝘀 𝗱𝗶𝗿𝗲𝗰𝘁𝗹𝘆, 𝘂𝘀𝗲 𝘁𝗵𝗼𝘀𝗲 𝗿𝗮𝘁𝗲𝘀.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

9️⃣ 𝗣𝗔𝗥𝗧𝗜𝗔𝗟 𝗪𝗢𝗥𝗞

A worker may complete only part of a job before something changes.

The formula is:

Work completed = Rate x Time

Example

A machine completes 1/9 of an order every hour.

How much of the order will it complete in 5 hours?

Work:

5 x 1/9

= 5/9

So:

Answer: 5/9 of the order

The unfinished portion is:

1 − 5/9

= 4/9

This becomes especially important when another worker joins later.

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🔟 𝗪𝗛𝗘𝗡 𝗔 𝗦𝗘𝗖𝗢𝗡𝗗 𝗪𝗢𝗥𝗞𝗘𝗥 𝗝𝗢𝗜𝗡𝗦

These questions should be separated into time periods.

Example

A worker can finish a job in 12 hours.

A second worker can finish it in 8 hours.

The first worker works alone for 3 hours. Then the second worker joins.

How long do they work together after that?

First worker's rate:

1/12

Work completed in 3 hours:

3 x 1/12

= 1/4

Remaining:

1 − 1/4

= 3/4

Combined rate:

1/12 + 1/8

= 2/24 + 3/24

= 5/24

Time for the remaining 3/4:

(3/4) ÷ (5/24)

= 3/4 x 24/5

= 18/5

= 3.6 hours

Answer: 3.6 hours

If the question asks for total time from the beginning:

3 + 3.6 = 6.6 hours.

Always check exactly what the question asks.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

1️⃣1️⃣ 𝗪𝗛𝗘𝗡 𝗢𝗡𝗘 𝗪𝗢𝗥𝗞𝗘𝗥 𝗟𝗘𝗔𝗩𝗘𝗦

This is the reverse situation.

Example

Worker A can finish a task in 9 hours.

Worker B can finish it in 6 hours.

They work together for 2 hours. Then A leaves.

How much longer does B need?

Combined rate:

1/9 + 1/6

= 2/18 + 3/18

= 5/18

Work completed in 2 hours:

2 x 5/18

= 10/18

= 5/9

Remaining:

1 − 5/9

= 4/9

B's rate:

1/6

Time required:

(4/9) ÷ (1/6)

= 4/9 x 6

= 8/3

hours.

Answer: 8/3 hours

or:

2 hours 40 minutes

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1️⃣2️⃣ 𝗣𝗜𝗣𝗘 𝗔𝗡𝗗 𝗧𝗔𝗡𝗞 𝗣𝗥𝗢𝗕𝗟𝗘𝗠𝗦

A filling pipe is simply a positive rate.

A drain is a negative rate.

Suppose one pipe fills a tank in 10 hours.

Its rate is:

1/10

Another pipe empties the tank in 15 hours.

Its rate is:

−1/15

With both operating:

1/10 − 1/15

= 3/30 − 2/30

= 1/30

So the net rate is:

1/30 tank per hour

Therefore the tank takes:

30 hours

to fill.

This is exactly the same rate concept used for workers.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

1️⃣3️⃣ 𝗗𝗜𝗥𝗘𝗖𝗧 𝗣𝗜𝗣𝗘 𝗥𝗔𝗧𝗘𝗦

Sometimes the problem gives gallons per minute or liters per minute.

Suppose a pump adds:

36 liters/minute

and a leak removes:

9 liters/minute.

Net rate:

36 − 9 = 27 liters/minute.

For a 675-liter tank:

Time = 675 ÷ 27

= 25 minutes.

Answer: 25 minutes

No reciprocal formula is required because both rates were already given directly.

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1️⃣4️⃣ 𝗨𝗡𝗜𝗧𝗦 𝗠𝗨𝗦𝗧 𝗠𝗔𝗧𝗖𝗛

A very common mistake is combining rates with different time units.

Suppose:

Machine A produces 180 items per hour.

Machine B produces 2 items per minute.

You cannot immediately write:

180 + 2

because the time units are different.

Convert 180 items per hour:

180 ÷ 60 = 3 items per minute.

Now:

3 + 2 = 5 items per minute.

If 400 items are needed:

400 ÷ 5 = 80 minutes.

Answer: 80 minutes

Before adding rates, make sure their units agree.

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1️⃣5️⃣ 𝗪𝗢𝗥𝗞𝗘𝗥-𝗗𝗔𝗬 𝗣𝗥𝗢𝗕𝗟𝗘𝗠𝗦

Some problems assume that every worker has the same efficiency.

In that case:

Workers x Days = Constant

Example

5 workers complete a project in 18 days.

How many days would 15 workers require if all workers work at the same rate?

Total worker-days:

5 x 18 = 90

With 15 workers:

90 ÷ 15 = 6

Answer: 6 days

This shortcut should not automatically be used when workers have different efficiencies.

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1️⃣6️⃣ 𝗘𝗙𝗙𝗜𝗖𝗜𝗘𝗡𝗖𝗬 𝗥𝗔𝗧𝗜𝗢𝗦

Suppose A works twice as fast as B.

If B's rate is:

r

then A's rate is:

2r

Together:

r + 2r = 3r

Example

A is twice as efficient as B.

Together they complete a job in 9 hours.

Combined rate:

1/9

Let B's rate be r.

Then:

3r = 1/9

r = 1/27

A's rate:

2/27

Therefore A's individual time is:

27/2

= 13.5 hours.

Answer: 13.5 hours

B's individual time is:

27 hours.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

1️⃣7️⃣ 𝗙𝗜𝗡𝗗𝗜𝗡𝗚 𝗔 𝗠𝗜𝗦𝗦𝗜𝗡𝗚 𝗪𝗢𝗥𝗞𝗘𝗥

Suppose two workers together finish in 8 hours.

Worker A alone finishes in 12 hours.

Find Worker B's individual time.

Combined rate:

1/8

A's rate:

1/12

Therefore B's rate:

1/8 − 1/12

= 3/24 − 2/24

= 1/24

So B alone takes:

24 hours

This type of question is often easier than it first appears.

You already know the total rate.

Simply remove the known worker's rate.

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1️⃣8️⃣ 𝗛𝗔𝗥𝗗𝗘𝗥 𝗪𝗢𝗥𝗞 𝗥𝗔𝗧𝗘 𝗣𝗥𝗢𝗕𝗟𝗘𝗠𝗦

Some questions hide the individual times behind a variable.

Example

Worker A takes 5 hours fewer than Worker B.

Together they complete a job in 6 hours.

Find the time Worker A takes alone.

Let A's time be:

t

Then B's time is:

t + 5

Their combined rate is:

1/6

Therefore:

1/t + 1/(t + 5) = 1/6

Multiply by:

6t(t + 5)

This gives:

6(t + 5) + 6t = t(t + 5)

6t + 30 + 6t = t² + 5t

12t + 30 = t² + 5t

t² − 7t − 30 = 0

Factor:

(t − 10)(t + 3) = 0

Therefore:

t = 10

or:

t = −3

A time cannot be negative.

So:

A takes 10 hours.

B takes:

10 + 5 = 15 hours.

Check:

1/10 + 1/15

= 3/30 + 2/30

= 5/30

= 1/6

Correct.

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1️⃣9️⃣ 𝗧𝗛𝗘 𝗟𝗖𝗠 𝗠𝗘𝗧𝗛𝗢𝗗: AVOIDING FRACTIONS

Sometimes fractions make a problem look harder than it is.

You can choose a convenient total amount of work.

Suppose:

Machine A completes a job in 5 hours.

Machine B completes it in 10 hours.

Choose:

Total work = 10 units

Then A produces:

10 ÷ 5 = 2 units/hour.

B produces:

10 ÷ 10 = 1 unit/hour.

Together:

2 + 1 = 3 units/hour.

Time:

10 ÷ 3

= 10/3 hours.

Answer: 3⅓ hours

This method is particularly useful when the individual completion times have a convenient common multiple.

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2️⃣0️⃣ 𝗦𝗔𝗡𝗜𝗧𝗬 𝗖𝗛𝗘𝗖𝗞𝗦

After calculating your answer, stop for a few seconds and ask whether it makes sense.

Two workers

The combined time should be less than the faster worker's individual time.

If one worker takes 5 hours and another takes 8 hours, their combined time must be less than 5 hours.

A worker joins

Once another worker joins, the rate should increase.

A drain opens

The net filling rate should decrease.

A worker leaves

The remaining rate should decrease.

More identical workers

More workers should mean less time, assuming the work conditions and efficiency remain the same.

These quick checks can catch many careless errors.


Trap 1: Adding times

Do not add completion times when workers operate simultaneously.

Use rates.

Do not add completion times when workers operate simultaneously.

Use rates.

⚠️ Trap 2: Forgetting the reciprocal

If a worker takes 8 hours, the rate is:

1/8

not 8.

⚠️ Trap 3: Forgetting to convert units

Minutes and hours must be made consistent.

⚠️ Trap 4: Treating a drain as positive

A drain removes water.

Subtract its rate.

⚠️ Trap 5: Answering total time when the question asks for additional time

Read the final sentence carefully.

⚠️ Trap 6: Using worker-days with unequal workers

The worker-days shortcut assumes equal efficiency.

⚠️ Trap 7: Rounding too early

Keep fractions until the final step whenever possible.

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2️⃣2️⃣ 𝗙𝗔𝗦𝗧 𝗦𝗛𝗢𝗥𝗧𝗖𝗨𝗧𝗦

Two workers

T = (a x b)/(a + b)

Three workers

1/T = 1/a + 1/b + 1/c

Direct production

Time = Total units ÷ Units per time

Partial job

Work = Rate x Time

Remaining job

Remaining = 1 − Completed

Equal-efficiency workers

Workers x Days = Constant

Filling and draining

Net rate = Fill rate − Drain rate

These formulas are useful, but the most important skill is knowing which formula fits the situation.

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2️⃣3️⃣ 𝗧𝗘𝗡 𝗙𝗥𝗘𝗦𝗛 𝗣𝗥𝗔𝗖𝗧𝗜𝗖𝗘 𝗤𝗨𝗘𝗦𝗧𝗜𝗢𝗡𝗦

1.

A technician can complete a calibration task in 14 hours. A second technician can complete it in 21 hours. How long will they need if they work together?

2.

A pipe fills a tank in 9 hours, while a drain empties the full tank in 18 hours. If both remain open, how long will the tank take to fill?

3.

A machine produces 72 components per hour. How many hours are needed to produce 540 components?

4.

12 workers can complete a project in 15 days. Assuming all workers have the same efficiency, how many days would 20 workers need?

5.

One worker completes a task in 16 hours and another completes it in 24 hours. They work together for 4 hours. What fraction of the job remains?

6.

Two workers together complete a job in 9 hours. One worker alone takes 15 hours. How long does the other worker take alone?

7.

A pump adds 64 liters per minute to a tank while a leak removes 16 liters per minute. How long will it take to add 960 liters?

8.

Three machines can individually complete an order in 6, 8 and 24 hours. How long will they take together?

9.

Worker A takes t hours to complete a task. Worker B takes t + 6 hours. Together they complete the task in 4 hours. Find t.

10.

A machine completes 3/5 of a production order in 2.4 hours at a constant rate. How long does it take to complete the entire order?

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2️⃣4️⃣ 𝗔𝗡𝗦𝗪𝗘𝗥𝗦

  1. 8.4 hours
  2. 18 hours
  3. 7.5 hours
  4. 9 days
  5. 1/2
  6. 22.5 hours
  7. 20 minutes
  8. 3 hours
  9. 6 hours
  10. 4 hours

Monday, August 31, 2026

SAT Inequalities Made Easy: The Sign Flip Trick + Hard Questions

 

╔══════════════════════════════════════════════╗
║ 𝐒𝐀𝐓 𝐌𝐀𝐓𝐇 ║
║ 𝐋𝐈𝐍𝐄𝐀𝐑 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 ║
║ 𝐓𝐡𝐞 𝐒𝐢𝐦𝐩𝐥𝐞 𝐌𝐞𝐭𝐡𝐨𝐝 𝐓𝐡𝐚𝐭 𝐏𝐫𝐞𝐯𝐞𝐧𝐭𝐬 𝐒𝐈𝐋𝐋𝐘 𝐌𝐢𝐬𝐭𝐚𝐤𝐞𝐬 ║
╚══════════════════════════════════════════════╝

𝐖𝐡𝐲 𝐝𝐨 𝐒𝐀𝐓 𝐢𝐧𝐞𝐪𝐮𝐚𝐥𝐢𝐭𝐲 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬 𝐜𝐚𝐭𝐜𝐡 𝐬𝐭𝐮𝐝𝐞𝐧𝐭𝐬?

Because they look almost exactly like equations.

You see:

𝟑𝐱 + 𝟒 = 𝟏𝟗

and you know what to do.

Then the SAT changes one symbol:

𝟑𝐱 + 𝟒 ≥ 𝟏𝟗

Now you are no longer looking for just one answer.

You are looking for a whole collection of values.

That is the central idea behind inequalities:

╭──────────────────────────────╮
│ 𝐄𝐐𝐔𝐀𝐓𝐈𝐎𝐍 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 │
│ 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 │
╰──────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏 — 𝐊𝐍𝐎𝐖 𝐓𝐇𝐄 𝐅𝐎𝐔𝐑 𝐒𝐘𝐌𝐁𝐎𝐋𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

There are four basic inequality signs:

𝐱 < 𝟕
→ x is less than 7

𝐱 > 𝟕
→ x is greater than 7

𝐱 ≤ 𝟕
→ x is less than or equal to 7

𝐱 ≥ 𝟕
→ x is greater than or equal to 7

The tiny horizontal line underneath the symbol is important.

It means:

𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐒 𝐀𝐋𝐋𝐎𝐖𝐄𝐃.

So:

𝐱 < 𝟓

does NOT include 5.

But:

𝐱 ≤ 𝟓

DOES include 5.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐 — 𝐓𝐇𝐄 𝐑𝐔𝐋𝐄 𝐘𝐎𝐔 𝐌𝐔𝐒𝐓 𝐍𝐎𝐓 𝐅𝐎𝐑𝐆𝐄𝐓
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Here is the most important rule in this entire guide:

╔══════════════════════════════════╗
║ 𝐌𝐔𝐋𝐓𝐈𝐏𝐋𝐘 𝐎𝐑 𝐃𝐈𝐕𝐈𝐃𝐄 𝐁𝐘 𝐀 ║
║ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐍𝐔𝐌𝐁𝐄𝐑? ║
║ ║
║ 𝐅𝐋𝐈𝐏 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍! ║
╚══════════════════════════════════╝

For example:

−𝟐𝐱 > 𝟏𝟎

Divide by −2.

Because −2 is negative:

𝐱 < −𝟓

Notice what happened:

became <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟑 — 𝐖𝐇𝐄𝐍 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐃𝐎𝐄𝐒 𝐍𝐎𝐓 𝐅𝐋𝐈𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Do not flip the sign every time you move something.

For example:

𝟓𝐱 − 𝟑 ≤ 𝟏𝟕

Add 3:

𝟓𝐱 ≤ 𝟐𝟎

Divide by +5:

𝐱 ≤ 𝟒

Nothing flips because 5 is positive.

A useful mental test is:

𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘

𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟒 — 𝐓𝐇𝐄 𝐒𝐀𝐓 𝐖𝐀𝐘 𝐓𝐎 𝐒𝐎𝐋𝐕𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Treat an inequality almost like an equation.

Example:

𝟒𝐱 + 𝟕 > 𝟐𝟑

Subtract 7:

𝟒𝐱 > 𝟏𝟔

Divide by 4:

𝐱 > 𝟒

That's it.

But always perform one final check:

𝐃𝐢𝐝 𝐈 𝐝𝐢𝐯𝐢𝐝𝐞 𝐛𝐲 𝐚 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞?

If no, the sign remains unchanged.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟓 — 𝐓𝐇𝐄 𝐓𝐖𝐎-𝐒𝐈𝐃𝐄𝐃 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Sometimes the SAT gives you a sandwich:

𝟐 < 𝐱 + 𝟓 ≤ 𝟏𝟏

Subtract 5 from ALL THREE parts:

𝟐 − 𝟓 < 𝐱 ≤ 𝟏𝟏 − 𝟓

Therefore:

−𝟑 < 𝐱 ≤ 𝟔

The answer contains every number between −3 and 6, except −3 itself.

But 6 IS included.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟔 — 𝐓𝐇𝐄 𝐍𝐔𝐌𝐁𝐄𝐑 𝐋𝐈𝐍𝐄 𝐂𝐎𝐃𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A number-line question can often be solved almost instantly.

𝐎𝐏𝐄𝐍 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint NOT included

𝐂𝐋𝐎𝐒𝐄𝐃 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint included

So:

𝐱 > 𝟐

means:

○──────→
𝟐

while:

𝐱 ≥ 𝟐

means:

●──────→
𝟐

And direction matters:

←──────○
𝟐

means:

𝐱 < 𝟐

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟕 — 𝐓𝐇𝐄 𝐐𝐔𝐈𝐂𝐊 𝐆𝐑𝐀𝐏𝐇 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Whenever you see a number-line graph, ask two questions:

𝐐𝟏. 𝐈𝐬 𝐭𝐡𝐞 𝐞𝐧𝐝𝐩𝐨𝐢𝐧𝐭 𝐨𝐩𝐞𝐧 𝐨𝐫 𝐜𝐥𝐨𝐬𝐞𝐝?

𝐐𝟐. 𝐖𝐡𝐢𝐜𝐡 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐢𝐬 𝐬𝐡𝐚𝐝𝐞𝐝?

That gives you the inequality.

You do not need to guess.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟖 — 𝐓𝐇𝐄 𝐖𝐎𝐑𝐃𝐒 𝐇𝐈𝐃𝐈𝐍𝐆 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

SAT word problems often hide the inequality symbol inside ordinary English.

Memorize these translations:

𝐀𝐓 𝐋𝐄𝐀𝐒𝐓
→ ≥

𝐀𝐓 𝐌𝐎𝐒𝐓
→ ≤

𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ >

𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ <

𝐍𝐎 𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ ≤

𝐍𝐎 𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ ≥

𝐆𝐑𝐄𝐀𝐓𝐄𝐑 𝐓𝐇𝐀𝐍
→ >

𝐅𝐄𝐖𝐄𝐑 𝐓𝐇𝐀𝐍
→ <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟗 — 𝐓𝐇𝐄 “𝐀𝐓 𝐋𝐄𝐀𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose a problem says:

“A score of at least 80 is required.”

At least means 80 is acceptable.

Therefore:

𝐱 ≥ 𝟖𝟎

Not:

𝐱 > 𝟖𝟎

This tiny difference can decide the entire answer.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟎 — 𝐓𝐇𝐄 “𝐀𝐓 𝐌𝐎𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

“At most 25” means 25 is allowed.

Therefore:

𝐱 ≤ 𝟐𝟓

Compare:

“less than 25”

𝐱 < 𝟐𝟓

One word changes the mathematics.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟏 — 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 𝐈𝐍 𝐓𝐖𝐎 𝐕𝐀𝐑𝐈𝐀𝐁𝐋𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now the SAT can move from a number line to a coordinate plane.

Consider:

𝐲 > 𝟐𝐱 + 𝟏

First draw the boundary:

𝐲 = 𝟐𝐱 + 𝟏

Then determine which side belongs to the solution.

Because the inequality is:

𝐲 > ...

the solution is above the boundary.

Because equality is NOT included, the boundary is dashed.

So remember:

𝐲 > 𝐟(𝐱)
→ above + dashed

𝐲 < 𝐟(𝐱)
→ below + dashed

𝐲 ≥ 𝐟(𝐱)
→ above + solid

𝐲 ≤ 𝐟(𝐱)
→ below + solid

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟐 — 𝐓𝐇𝐄 𝐓𝐄𝐒𝐓-𝐀-𝐏𝐎𝐈𝐍𝐓 𝐌𝐄𝐓𝐇𝐎𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

If you are unsure which side of a boundary is correct, test a point.

Suppose:

𝐲 > 𝐱 + 𝟐

Try the point:

(𝟎,𝟎)

Substitute:

𝟎 > 𝟎 + 𝟐

That becomes:

𝟎 > 𝟐

False.

Therefore, the side containing (0,0) is NOT the solution.

This method is particularly useful when a graph is unfamiliar.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟑 — 𝐒𝐘𝐒𝐓𝐄𝐌𝐒 𝐎𝐅 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose:

𝐱 ≥ 𝟑

and

𝐱 < 𝟖

Both must be true.

Therefore:

𝟑 ≤ 𝐱 < 𝟖

Think of this as finding the common region.

𝐀𝐍𝐃 = 𝐎𝐕𝐄𝐑𝐋𝐀𝐏

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟒 — 𝐖𝐇𝐀𝐓 “𝐎𝐑” 𝐌𝐄𝐀𝐍𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now consider:

𝐱 < −𝟒

OR

𝐱 > 𝟑

These are two separate possibilities.

The solution is:

𝐱 < −𝟒 𝐎𝐑 𝐱 > 𝟑

Do not search for one continuous interval.

Remember:

𝐀𝐍𝐃 → intersection / overlap

𝐎𝐑 → either possibility

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟓 — 𝐀 𝐑𝐄𝐀𝐋 𝐖𝐎𝐑𝐋𝐃 𝐒𝐀𝐓 𝐌𝐎𝐃𝐄𝐋
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Imagine a student has $50.

A ticket costs $12 and each additional item costs $4.

If x represents the number of additional items, the total must not exceed $50.

Write:

𝟏𝟐 + 𝟒𝐱 ≤ 𝟓𝟎

Subtract 12:

𝟒𝐱 ≤ 𝟑𝟖

Divide:

𝐱 ≤ 𝟗.𝟓

But x represents a number of items.

You cannot buy half an item.

Therefore the greatest possible whole-number value is:

𝐱 = 𝟗

This illustrates an important SAT habit:

𝐀𝐋𝐆𝐄𝐁𝐑𝐀 𝐀𝐍𝐒𝐖𝐄𝐑 ≠ 𝐀𝐋𝐖𝐀𝐘𝐒 𝐅𝐈𝐍𝐀𝐋 𝐖𝐎𝐑𝐃

The context matters.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟔 — 𝐀𝐍𝐎𝐓𝐇𝐄𝐑 𝐖𝐎𝐑𝐃 𝐏𝐑𝐎𝐁𝐋𝐄𝐌
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A gym charges $20 to join and $8 per month.

A student can spend no more than $68.

How many months can the student afford?

Let x = number of months.

Write:

𝟐𝟎 + 𝟖𝐱 ≤ 𝟔𝟖

Subtract 20:

𝟖𝐱 ≤ 𝟒𝟖

Divide:

𝐱 ≤ 𝟔

Therefore:

𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐦𝐨𝐧𝐭𝐡𝐬 = 𝟔

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟕 — 𝐓𝐇𝐄 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐓𝐑𝐀𝐏 𝐑𝐄𝐕𝐈𝐒𝐈𝐓𝐄𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Solve:

𝟕 − 𝟑𝐱 ≥ 𝟏𝟔

Subtract 7:

−𝟑𝐱 ≥ 𝟗

Now divide by −3.

𝐒𝐓𝐎𝐏.

This is the danger point.

The sign must reverse:

𝐱 ≤ −𝟑

A useful habit:

Whenever the coefficient of x becomes negative immediately before division, mentally say:

“𝐅𝐋𝐈𝐏.”

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟖 — 𝐖𝐇𝐘 𝐃𝐎𝐄𝐒 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐅𝐋𝐈𝐏?
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

This is not an arbitrary SAT rule.

Take:

𝟐 < 𝟓

This is true.

Multiply both sides by −1:

−𝟐 > −𝟓

The order on the number line has reversed.

That is why:

< becomes >

and

becomes <

when multiplying or dividing by a negative number.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟗 — 𝐀 𝐅𝐀𝐒𝐓𝐄𝐑 𝐖𝐀𝐘 𝐓𝐎 𝐓𝐇𝐈𝐍𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Instead of memorizing dozens of separate rules, remember this chain:

╔═══════════════════════════════╗
║ 𝟏. 𝐓𝐑𝐀𝐍𝐒𝐋𝐀𝐓𝐄 ║
║ 𝟐. 𝐒𝐈𝐌𝐏𝐋𝐈𝐅𝐘 ║
║ 𝟑. 𝐒𝐎𝐋𝐕𝐄 ║
║ 𝟒. 𝐂𝐇𝐄𝐂𝐊 𝐅𝐎𝐑 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 ║
║ 𝟓. 𝐂𝐇𝐄𝐂𝐊 𝐓𝐇𝐄 𝐂𝐎𝐍𝐓𝐄𝐗𝐓 ║
╚═══════════════════════════════╝

This is much safer than trying to solve everything mentally.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟎 — 𝐓𝐇𝐄 𝐌𝐈𝐍𝐈 𝐒𝐀𝐓 𝐂𝐇𝐀𝐋𝐋𝐄𝐍𝐆𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

𝐐𝟏.

Solve:

𝟔𝐱 − 𝟓 > 𝟏𝟗

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

𝟔𝐱 > 𝟐𝟒

𝐱 > 𝟒

𝐐𝟐.

Solve:

−𝟓𝐱 + 𝟏𝟎 ≤ 𝟑𝟎

Subtract 10:

−𝟓𝐱 ≤ 𝟐𝟎

Divide by −5 and flip:

𝐱 ≥ −𝟒

𝐐𝟑.

Solve:

𝟑 ≤ 𝟐𝐱 + 𝟏 < 𝟏𝟏

Subtract 1:

𝟐 ≤ 𝟐𝐱 < 𝟏𝟎

Divide by 2:

𝟏 ≤ 𝐱 < 𝟓

𝐐𝟒.

A quantity must be no greater than 75.

Which inequality represents the statement?

𝐱 ≤ 𝟕𝟓

𝐐𝟓.

Which value satisfies:

𝐱 > −𝟐?

A. −𝟓
B. −𝟑
C. −𝟐
D. 𝟎

Answer:

𝐃. 𝟎

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟏 — 𝐓𝐇𝐄 𝐅𝐈𝐕𝐄-𝐒𝐄𝐂𝐎𝐍𝐃 𝐅𝐈𝐍𝐀𝐋 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Before submitting an inequality answer, run this mental checklist:

✓ Did I translate the words correctly?

✓ Did I distribute brackets correctly?

✓ Did I isolate x?

✓ Did I multiply or divide by a negative?

✓ If yes, did I reverse the sign?

✓ Is the endpoint included?

✓ Does the answer make sense in the real-world situation?

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟐 — 𝐓𝐇𝐄 𝐔𝐋𝐓𝐈𝐌𝐀𝐓𝐄 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐌𝐄𝐌𝐎𝐑𝐘 𝐂𝐀𝐑𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

╭────────────────────────────────╮
│ < → 𝐋𝐄𝐒𝐒 │
│ > → 𝐌𝐎𝐑𝐄 │
│ ≤ → 𝐋𝐄𝐒𝐒 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ ≥ → 𝐌𝐎𝐑𝐄 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ │
│ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏 │
│ 𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘 │
│ │
│ 𝐎𝐏𝐄𝐍 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐄𝐗𝐂𝐋𝐔𝐃𝐄𝐃 │
│ 𝐂𝐋𝐎𝐒𝐄𝐃 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐍𝐂𝐋𝐔𝐃𝐄𝐃│
│ │
│ 𝐀𝐍𝐃 → 𝐎𝐕𝐄𝐑𝐋𝐀𝐏 │
│ 𝐎𝐑 → 𝐄𝐈𝐓𝐇𝐄𝐑 𝐏𝐎𝐒𝐒𝐈𝐁𝐈𝐋𝐈𝐓𝐘 │
╰────────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐅𝐈𝐍𝐀𝐋 𝐒𝐀𝐓 𝐓𝐀𝐊𝐄𝐀𝐖𝐀𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Linear inequalities are not really about memorizing a large collection of formulas.

They are about controlling one idea:

𝐖𝐇𝐈𝐂𝐇 𝐕𝐀𝐋𝐔𝐄𝐒 𝐀𝐑𝐄 𝐀𝐋𝐋𝐎𝐖𝐄𝐃?

Once you see the question that way, the symbols become easier.

If the SAT says:

“at least”

think:

≥

If it says:

“at most”

think:

≤

If you divide by a negative:

𝐅𝐋𝐈𝐏 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍.

If a graph is involved:

𝐎𝐏𝐄𝐍 = 𝐍𝐎 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘

𝐂𝐋𝐎𝐒𝐄𝐃 = 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐍𝐂𝐋𝐔𝐃𝐄𝐃

And when a word problem produces a mathematical answer, always return to the original situation.

That final step is where many avoidable SAT mistakes disappear.

𝐓𝐡𝐞 𝐛𝐞𝐬𝐭 𝐢𝐧𝐞𝐪𝐮𝐚𝐥𝐢𝐭𝐲 𝐬𝐭𝐫𝐚𝐭𝐞𝐠𝐲 𝐢𝐬:

╔════════════════════════════════════╗
║ 𝐓𝐑𝐀𝐍𝐒𝐋𝐀𝐓𝐄 → 𝐒𝐎𝐋𝐕𝐄 → 𝐅𝐋𝐈𝐏 ║
║ → 𝐂𝐇𝐄𝐂𝐊 → 𝐈𝐍𝐓𝐄𝐑𝐏𝐑𝐄𝐓 ║
╚════════════════════════════════════╝

Master that sequence and a large class of SAT inequality questions becomes much more predictable.

other pages to explore

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 


ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]


SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]


SYSTEM OF EQUATIONS  [ PART  I ]


SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2


 PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS


QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 


QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 


Linear Inequality

linear inequalities


PERCENTAGES


PERCENTAGES [introduction]

PERCENTAGES [part 1]


PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]


SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems


SAT Reverse Percentage Problems




GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions





SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster



The Big Word-Problem Families:  SAT Math Guide


How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier



SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions





SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions




SAT Conditional Probability

Thursday, August 27, 2026

SAT Conditional Probability Trick: Master “Given That” Questions Fast

 

🎯 SAT Conditional Probability: The Hidden “Given That” Trick

Conditional probability looks complicated on the SAT because the question often hides the mathematics inside a table, survey, experiment, or real-life situation.

But the core idea is surprisingly simple:

When the question says “given that,” your universe becomes smaller.

That one idea can turn a difficult-looking SAT probability question into a short calculation.


🔑 1. The Basic Conditional Probability Formula

If the question asks for the probability of A given B, write:

P(A | B) = P(A and B) ÷ P(B)

The vertical bar | means:

“given that”

So:

P(A | B)

means:

“the probability of A, given that B has already happened.”

The most important part is the denominator:

⭐ The denominator is the condition.

If you see:

P(A | B)

start by asking:

“How many outcomes satisfy B?”

That becomes your new total.


🧠 2. The SAT Shortcut

Suppose a survey contains 200 students.

• 120 study mathematics
• 80 study physics
• 50 study both mathematics and physics

What is the probability that a randomly selected student studies mathematics given that the student studies physics?

The phrase “given that the student studies physics” changes the problem.

You are no longer choosing from all 200 students.

You are choosing only from the 80 students who study physics.

Among those 80 students, 50 also study mathematics.

Therefore:

P(Math | Physics) = 50 ÷ 80

= 5 ÷ 8

= 0.625

= 62.5%

🚨 SAT trap:

A common mistake is:

50 ÷ 200

That would answer a different question:

What percentage of ALL students study both subjects?

The SAT is testing whether you notice the words:

“given that”


📊 3. Two-Way Tables Make Conditional Probability Easier

Many SAT questions present information in a table.

Consider this example:

Uses CalculatorDoes Not Use CalculatorTotal
Group A362460
Group B281240
Total6436100

Suppose a student is selected from those who use a calculator.

What is the probability that the student belongs to Group A?

The condition is:

Uses Calculator

So the denominator is:

64

The favorable outcomes are Group A students who use a calculator:

36

Therefore:

P(Group A | Uses Calculator)

= 36 ÷ 64

= 9 ÷ 16

= 0.5625

So the answer is:

56.25%


⚠️ 4. The Denominator Test

Whenever you see a conditional probability problem, perform this three-second test:

Step ① Find the words after “given that.”

Step ② Find the total number belonging to that condition.

Step ③ Divide the desired intersection by that total.

For example:

P(A | B)

means:

Desired = A and B

Total = B

Therefore:

P(A | B) = (A and B) ÷ B

This is one of the most useful patterns to recognize on the SAT.


🔥 5. “And” vs “Given”

These two expressions look similar but mean very different things.

P(A and B)

asks for the probability that both events happen.

P(A | B)

asks for the probability that A happens among the cases where B is already known to happen.

For example, suppose 100 students are surveyed.

• 40 play basketball
• 30 play soccer
• 15 play both

Then:

P(Basketball and Soccer)

= 15 ÷ 100

= 15%

But:

P(Basketball | Soccer)

= 15 ÷ 30

= 50%

Same intersection.

Different denominator.

That is the entire trick.


🎯 6. A SAT-Style Example

A school surveys 300 students about whether they participate in music or sports.

The results are:

• 180 participate in sports
• 120 participate in music
• 75 participate in both

A student who participates in music is selected at random.

What is the probability that the student also participates in sports?

The phrase:

“A student who participates in music is selected”

creates the condition.

Therefore, the total possible students are:

120

The students satisfying both conditions are:

75

So:

P(Sports | Music)

= 75 ÷ 120

= 5 ÷ 8

= 0.625

Answer:

62.5%


🧩 7. Watch for “Among”

The SAT may avoid the words “given that” and use another phrase.

Watch for:

among

of those who

for students who

from the group that

if the selected student is known to

All of these can signal a restricted sample.

For example:

Among students who own a bicycle, 18 out of 30 ride to school.

The probability that a randomly selected bicycle owner rides to school is:

18 ÷ 30 = 60%

The denominator is 30, not the total number of students in the school.


📐 8. Conditional Probability From a Formula

Sometimes the SAT gives probabilities instead of counts.

Suppose:

P(A) = 0.40

P(B) = 0.50

and

P(A and B) = 0.20

Find:

P(A | B)

Use:

P(A | B) = P(A and B) ÷ P(B)

Therefore:

P(A | B) = 0.20 ÷ 0.50

= 0.40

Answer:

40%

Notice that you don't need to create a table.

The same idea works with probabilities, percentages, fractions, or counts.


💡 9. The Reverse Conditional Probability Trap

This is an especially important SAT idea.

In general:

P(A | B) ≠ P(B | A)

For example:

Suppose:

• 60 students play soccer
• 40 students play basketball
• 20 play both

Then:

P(Soccer | Basketball)

= 20 ÷ 40

= 50%

But:

P(Basketball | Soccer)

= 20 ÷ 60

= 33⅓%

The numerator is the same.

The denominator changes.

⭐ Remember:

The condition controls the denominator.


🧮 10. Turning Percentages Into Counts

SAT questions sometimes give percentages rather than actual numbers.

Suppose 40% of students own a tablet.

Among students who own a tablet, 75% also own a laptop.

What percentage of ALL students own both?

Imagine there are 100 students.

Tablet owners:

40

Of those 40, 75% own a laptop:

0.75 × 40 = 30

Therefore:

30%

of all students own both.

This gives a useful relationship:

P(A and B) = P(B) × P(A | B)

So:

P(A and B) = 0.40 × 0.75

= 0.30

= 30%


🚨 11. A Classic SAT Mistake

Suppose:

P(A) = 60%

and

P(B | A) = 25%

A student might incorrectly say:

P(A and B) = 60% + 25%

That is wrong.

The 25% applies only to the group A.

Use:

P(A and B) = P(A) × P(B | A)

Therefore:

= 0.60 × 0.25

= 0.15

So:

15%

of the entire population belongs to both groups.


📈 12. Conditional Probability and Tables

A table can often be converted directly into a probability.

Suppose:

PassedDid Not PassTotal
Studied721890
Did Not Study243660
Total9654150

Question:

What is the probability that a student studied, given that the student passed?

The condition is:

Passed

So use the Passed column.

Total who passed:

96

Passed and studied:

72

Therefore:

P(Studied | Passed)

= 72 ÷ 96

= 3 ÷ 4

= 75%


🧠 13. The “Shrink the Universe” Method

Here is a powerful way to think about every conditional probability problem.

Imagine that the entire group is a large circle.

When the question says:

“given that B”

you throw away everything outside B.

Now your entire universe is:

B

Then ask:

How much of B is also A?

That gives:

A ∩ B ÷ B

or:

P(A | B) = P(A ∩ B) ÷ P(B)

This mental picture is often easier than memorizing a formula.


🎯 14. SAT Challenge Question

A survey of 400 students found:

• 240 students use a particular study app.
• 160 students use a particular online course.
• 100 students use both.

If a student who uses the online course is selected at random, what is the probability that the student also uses the study app?

Step 1: Identify the condition.

The student uses the online course.

So the denominator is:

160

Step 2: Find the intersection.

Both:

100

Step 3: Divide.

100 ÷ 160

= 5 ÷ 8

= 0.625

Answer:

62.5%


🔍 15. The Fastest Way to Solve These Questions

When you see a conditional probability question, don't immediately calculate.

First write:

Condition = ______

Then:

Total in condition = ______

Then:

Favorable within condition = ______

Finally:

Probability = favorable ÷ condition

For:

P(A | B)

write:

Condition → B

Total → B

Favorable → A ∩ B

Answer → (A ∩ B) ÷ B

This prevents one of the most common denominator errors.


🏆 16. What to Remember on Test Day

You do not need a complicated strategy.

Remember these five rules:

① “Given that” means the sample space changes.

② The condition becomes the denominator.

③ “And” usually points toward the intersection.

④ P(A | B) and P(B | A) are usually different.

⑤ For a two-way table, locate the condition's row or column first.


⚡ The 10-Second SAT Method

When the clock is running:

GIVEN → DENOMINATOR

AND → INTERSECTION

CONDITIONAL → SHRINK THE GROUP

PROBABILITY → FAVORABLE ÷ TOTAL

So if you see:

P(A | B)

think immediately:

“Within B, how many are also A?”

That single question can solve many conditional probability problems quickly.


📌 Final Takeaway

Conditional probability is not really about complicated probability.

It is about choosing the correct universe.

The SAT may disguise the condition inside:

• a survey
• a two-way table
• percentages
• a real-world situation
• a probability statement
• or a sentence containing “among”

But the underlying structure remains:

P(A | B) = P(A and B) ÷ P(B)

And the most important memory trick is:

⭐ GIVEN = DENOMINATOR ⭐

Once that becomes automatic, many SAT probability questions become much shorter.

explore more pages

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 


ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]


SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]


SYSTEM OF EQUATIONS  [ PART  I ]


SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2


 PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS


QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 


QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 


PERCENTAGES


PERCENTAGES [introduction]

PERCENTAGES [part 1]


PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]


SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems


SAT Reverse Percentage Problems




GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions





SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster



The Big Word-Problem Families:  SAT Math Guide


How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier



SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions





SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions




SAT Conditional Probability

Sunday, August 23, 2026

SAT Word Problems Made Easy: The Tricks for Speed, Work, Age & Profit

 

SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

🎯 Why SAT Word Problems Feel Harder Than They Are

A long SAT word problem can look intimidating because it contains a story, several numbers, units, percentages, and extra information.

But underneath the wording, there is usually one simple mathematical relationship.

The real SAT skill is not just calculation.

It is translation.

You need to turn:

Words → Quantities → Relationship → Equation → Answer

Once you learn to recognize the hidden structure, many word problems become much quicker.


🔹 1. The Golden Rule: Find the Relationship Before Calculating

Suppose the SAT says:

A cyclist travels 84 miles in 4 hours. What is the cyclist's average speed?

Do not immediately start calculating.

Identify:

Distance = 84 miles

Time = 4 hours

Unknown = speed

The relationship is:

Speed = Distance ÷ Time

Therefore:

r = d/t

r = 84/4

r = 21 miles per hour

✅ Answer: 21 miles per hour

The important step was not the division.

It was recognizing the relationship.


🔹 2. The SAT Distance Formula You Should Know Instantly

For motion problems:

d = rt

where:

d = distance

r = rate

t = time

From this:

r = d/t

and:

t = d/r

Think of the three quantities as connected:

Distance = Rate × Time

If the SAT gives you any two, you can find the third.


🔹 3. Units Can Quietly Destroy a Correct Solution

Suppose a car travels at:

60 miles per hour

for:

30 minutes

A common mistake is:

60 × 30

That would be wrong because the rate is measured in hours, while the time is given in minutes.

Convert:

30 minutes = 30/60 hour

= 1/2 hour

Now:

d = rt

d = 60 × 1/2

d = 30 miles

🧠 SAT habit

Before calculating, ask:

“Are my units speaking the same language?”

If the rate is in miles per hour, the time should be in hours.


🔹 4. Useful Time Conversions

Memorize these:

15 min = 1/4 hr

20 min = 1/3 hr

30 min = 1/2 hr

40 min = 2/3 hr

45 min = 3/4 hr

60 min = 1 hr

90 min = 3/2 hr

Fractions can often make SAT calculations easier than decimals.


🔹 5. When a Trip Has More Than One Speed

Suppose a driver travels:

60 miles at 30 mph

and then:

90 miles at 45 mph

To find the total travel time, handle each section separately.

First section:

t₁ = 60/30

t₁ = 2 hours

Second section:

t₂ = 90/45

t₂ = 2 hours

Therefore:

Total time = 2 + 2

= 4 hours

⭐ Key idea

When a journey has different speeds, break it into sections.

Do not try to force the entire trip into one speed equation.


🔥 6. The Average-Speed Trap

This is one of the most useful SAT traps to understand.

Suppose a car travels:

100 miles at 50 mph

and then:

100 miles at 100 mph

A tempting answer is:

(50 + 100)/2 = 75 mph

❌ Not correct.

Average speed means:

Total Distance ÷ Total Time

First journey:

100/50 = 2 hours

Second journey:

100/100 = 1 hour

Total distance:

200 miles

Total time:

3 hours

Therefore:

Average speed = 200/3

≈ 66.67 mph

🚨 Remember

Average speed ≠ average of speeds

Instead:

Average speed = Total Distance ÷ Total Time


🔹 7. Catch-Up Problems: Think “Gap”

Suppose Runner A is:

20 meters ahead

Runner A runs at:

5 m/s

Runner B runs at:

7 m/s

How long does B take to catch A?

The important quantity is not either speed by itself.

It is the speed at which B closes the gap.

Relative speed = 7 − 5

= 2 m/s

Initial gap:

20 m

Therefore:

Time = Gap ÷ Relative Speed

t = 20/2

t = 10 seconds

🧠 Shortcut

For objects moving in the same direction:

Relative speed = faster speed − slower speed


🔹 8. Objects Moving Toward Each Other

Suppose two cyclists are:

120 miles apart

One travels at:

30 mph

The other travels at:

50 mph

They move toward each other.

Their separation decreases at:

30 + 50 = 80 mph

Therefore:

t = 120/80

= 1.5 hours

Remember

Same direction:

Subtract speeds

Opposite directions:

Add speeds


🔹 9. Work Problems Are Really Rate Problems

Work questions sometimes look completely different from distance questions.

They aren't.

They use the same basic idea:

Amount completed = Rate × Time

If a worker completes an entire job in 6 hours, the worker completes:

1/6 of the job per hour

So:

Work rate = 1/6

If another worker completes the same job in 3 hours:

Work rate = 1/3

Together:

1/6 + 1/3

= 1/6 + 2/6

= 1/2

Together they complete:

1/2 of the job per hour

Therefore:

Time = 1 ÷ 1/2

= 2 hours

⭐ The key idea

In work problems:

Add rates, not times.


🔹 10. A Work-Rate Template Worth Memorizing

If someone completes a job in T hours:

Rate = 1/T

For multiple workers:

Combined Rate = Rate₁ + Rate₂ + Rate₃ + ...

Then:

Time = Total Work ÷ Combined Rate

If the entire job is represented by 1:

Time = 1 ÷ Combined Rate


🔹 11. Pipes and Tanks Use the Same Trick

A pipe fills a tank in:

4 hours

Its rate is:

1/4 tank per hour

Another pipe fills it in:

6 hours

Its rate is:

1/6 tank per hour

Together:

1/4 + 1/6

= 3/12 + 2/12

= 5/12

So they fill:

5/12 of the tank per hour

Therefore:

Time = 1 ÷ 5/12

= 12/5 hours

= 2.4 hours


🔥 12. What If There Is a Drain?

A drain removes water, so its rate is subtracted.

Suppose:

Pipe A = 1/4 tank/hour

Pipe B = 1/6 tank/hour

Drain = 1/12 tank/hour

Net rate:

1/4 + 1/6 − 1/12

Convert to twelfths:

3/12 + 2/12 − 1/12

= 4/12

= 1/3

Therefore the tank fills at:

1/3 tank per hour

and takes:

3 hours


🔹 13. Age Problems: Define the Present Age

Age questions become much easier when you choose one person's age as x.

Suppose Maya is:

4 years older than Leo

Let Leo's age be:

x

Then Maya's age is:

x + 4

If their ages add to 30:

x + (x + 4) = 30

2x + 4 = 30

2x = 26

x = 13

Therefore:

Leo = 13

Maya = 17


🔹 14. Future-Age Problems

Suppose a father is currently three times as old as his son.

Let the son's age be:

x

Father's age:

3x

Five years later:

Son:

x + 5

Father:

3x + 5

If the father will then be twice the son's age:

3x + 5 = 2(x + 5)

Expand:

3x + 5 = 2x + 10

Therefore:

x = 5

So the son is currently:

5 years old

and the father is:

15 years old


💡 15. The Age Difference Never Changes

This is one of the most useful observations in age problems.

If two people have an age difference of:

12 years

today, their difference will still be:

12 years

in:

5 years

10 years

20 years

The numbers change.

The difference does not.

This can sometimes eliminate the need for a long equation.


🔹 16. Mixture Problems: Find the Amount of Pure Substance

Mixture questions often involve concentration.

The fundamental relationship is:

Amount of pure substance = Total amount × Concentration

Suppose you have:

50 liters

of a:

20% salt solution

Then:

0.20 × 50 = 10 liters

of the mixture is salt.

The remaining:

50 − 10 = 40 liters

is the other component.


🔥 17. Mixing Two Different Concentrations

Suppose:

20 liters of a 30% solution

are mixed with:

x liters of a 50% solution

to create a:

40% solution

First solution contributes:

0.30(20) = 6

Second solution contributes:

0.50x

Total amount:

20 + x

Final amount of pure substance:

0.40(20 + x)

Therefore:

6 + 0.50x = 0.40(20 + x)

Expand:

6 + 0.50x = 8 + 0.40x

0.10x = 2

x = 20

✅ Answer:

20 liters


🔹 18. The Universal Mixture Equation

When appropriate, use:

Amount₁ × Concentration₁ + Amount₂ × Concentration₂

= Total Amount × Final Concentration

This single pattern can solve many mixture questions.

And always convert:

25% → 0.25

40% → 0.40

7% → 0.07


🔹 19. Profit: Revenue Minus Cost

Profit questions are usually straightforward once you identify the two quantities.

Profit = Revenue − Cost

Suppose a store buys an item for:

$40

and sells it for:

$55

Then:

Profit = 55 − 40

= $15


🔥 20. Profit Percentage

Profit percentage is calculated relative to the cost.

Formula:

Profit % = Profit/Cost × 100

Using the previous example:

Profit % = 15/40 × 100

= 37.5%

🚨 SAT warning

Do not divide the profit by the selling price unless the question specifically asks for a percentage based on the selling price.

For ordinary profit percentage:

Cost is the reference value.


🔹 21. Markups and Discounts

Suppose an item costs:

$80

The store marks it up by:

25%

Markup:

0.25 × 80 = 20

New price:

80 + 20 = $100

Now suppose the store gives a:

20% discount

Discount:

0.20 × 100 = 20

Final price:

100 − 20 = $80

This example illustrates an important SAT idea:

Percentages are applied to the current value.


🔥 22. Successive Percentage Changes

Suppose a price increases by:

20%

and then decreases by:

20%

Start with:

100

After the increase:

100 × 1.20 = 120

After the decrease:

120 × 0.80 = 96

Final value:

96

So the overall change is:

−4%

⭐ The multiplier method

Increase by 20%:

× 1.20

Decrease by 20%:

× 0.80

Combined:

1.20 × 0.80 = 0.96

Therefore:

96% of the original value remains.


🔹 23. Commission Problems

Suppose a salesperson earns:

6% commission

on sales.

If the salesperson sells:

$4,000

then:

Commission = 0.06 × 4000

= $240

If the salesperson also has a fixed salary:

Total earnings = Salary + Commission

The SAT may hide this simple structure inside a longer story.


🔹 24. Tax Problems

An item costs:

$500

and the tax rate is:

8%

Tax:

0.08 × 500 = 40

Total:

500 + 40 = $540

Or use the multiplier:

500 × 1.08 = 540

🧠 Quick rule

Tax added:

Original × (1 + tax rate)

Discount applied:

Original × (1 − discount rate)


🔥 25. Combined Word Problems

The SAT may combine several ideas in one question.

Example:

A cyclist travels the first:

30 miles at 15 mph

and the remaining:

45 miles at 30 mph

What is the average speed?

First section:

30/15 = 2 hours

Second section:

45/30 = 1.5 hours

Total distance:

30 + 45 = 75 miles

Total time:

2 + 1.5 = 3.5 hours

Average speed:

75/3.5 ≈ 21.43 mph

Notice what happened.

The problem looked complicated.

But it was simply:

Distance ÷ Rate → Time

followed by:

Total Distance ÷ Total Time


🔹 26. Decode SAT Wording

Certain words provide mathematical clues.

“Per”

Usually indicates a rate.

240 miles per 4 hours

means:

240/4

“Each”

Often indicates multiplication or a unit rate.

“Of”

Usually indicates multiplication.

30% of 80

means:

0.30 × 80

“At least”

means:

≥

“At most”

means:

≤

“More than”

means:

>

“Less than”

means:

<


🔹 27. “Difference” Questions

The difference between two quantities is generally represented by:

|A − B|

For example:

A = 19

B = 12

Difference:

|19 − 12| = 7

The absolute value makes the result nonnegative.


🔥 28. “How Much Greater?” vs. “How Much Greater Percent?”

These questions are not the same.

Suppose:

A = 30

B = 20

“How much greater is A than B?”

30 − 20 = 10

But:

“What percent greater is A than B?”

Use B as the reference:

(30 − 20)/20 × 100

= 50%

🚨 Always ask:

“Percent relative to what?”


🔹 29. The Five-Step SAT Translation System

When a word problem looks enormous, use this system.

STEP 1 — Identify the quantities

What numbers and units are given?

STEP 2 — Define the unknown

Write:

x = ...

STEP 3 — Identify the relationship

Is it:

distance?

rate?

work?

percentage?

mixture?

age?

profit?

STEP 4 — Build the equation

Translate the words into mathematics.

STEP 5 — Check the result

Ask:

Does this answer make sense?

This final question is surprisingly powerful.


🔥 30. Six SAT Word-Problem Traps to Avoid

❌ Trap 1: Averaging speeds directly

Do not automatically calculate:

(r₁ + r₂)/2

Use:

Total Distance ÷ Total Time

when appropriate.

❌ Trap 2: Ignoring units

Minutes, hours, seconds, miles, kilometers, and meters must be handled consistently.

❌ Trap 3: Adding work times

Workers' rates are added.

❌ Trap 4: Treating percentage changes as ordinary addition

A 20% increase followed by a 20% decrease does not equal 0%.

❌ Trap 5: Using the wrong reference value

Profit percentage normally uses:

Cost

❌ Trap 6: Solving for the wrong thing

You may find x correctly and still choose the wrong answer if the question asks for another quantity.


🔹 31. Practice Challenge #1

A train travels:

180 miles

at:

60 miles per hour

How long does the trip take?

A) 2 hours
B) 3 hours
C) 4 hours
D) 6 hours

Use:

t = d/r

t = 180/60

t = 3

✅ Answer: B


🔹 32. Practice Challenge #2

One worker can complete a task in:

10 hours

Another can complete it in:

15 hours

How long will they take together?

A) 5 hours
B) 6 hours
C) 7.5 hours
D) 25 hours

Rates:

1/10

and:

1/15

Combined:

1/10 + 1/15

= 3/30 + 2/30

= 5/30

= 1/6

Therefore:

Time = 6 hours

✅ Answer: B


🔹 33. Practice Challenge #3

A mother is:

24 years older

than her daughter.

In 4 years, the mother will be twice the daughter's age.

How old is the daughter now?

Let:

Daughter = x

Mother:

x + 24

Four years later:

Daughter:

x + 4

Mother:

x + 28

Equation:

x + 28 = 2(x + 4)

x + 28 = 2x + 8

x = 20

✅ Answer: 20 years old


🔹 34. Practice Challenge #4

A 20% solution is mixed with a 50% solution to produce:

30 liters of a 40% solution

How many liters of the 50% solution are needed?

Let:

x = liters of 50% solution

Then:

30 − x = liters of 20% solution

Equation:

0.20(30 − x) + 0.50x = 0.40(30)

Expand:

6 − 0.20x + 0.50x = 12

0.30x = 6

x = 20

✅ Answer: 20 liters


🔹 35. Practice Challenge #5

A store buys an item for:

$60

and sells it for:

$75

What is the profit percentage?

Profit:

75 − 60 = 15

Profit percentage:

15/60 × 100

= 25%

✅ Answer: 25%


🔹 36. Practice Challenge #6

A car travels:

120 miles at 40 mph

and then:

180 miles at 60 mph

What is its average speed?

First section:

120/40 = 3 hours

Second section:

180/60 = 3 hours

Total distance:

300 miles

Total time:

6 hours

Average speed:

300/6 = 50 mph

✅ Answer: 50 miles per hour


🧠 37. The SAT Word-Problem Formula Bank

🚗 Motion

d = rt

r = d/t

t = d/r

⚡ Relative motion

Same direction:

difference of rates

Opposite directions:

sum of rates

🛠 Work

Rate = 1/time

Combined rate = sum of individual rates

🧪 Mixtures

Amount × Concentration

💰 Profit

Profit = Revenue − Cost

📈 Profit percentage

Profit/Cost × 100

📊 Percentage increase

Increase/Original × 100

📉 Percentage decrease

Decrease/Original × 100

👨‍👩‍👧 Ages

Future age = Current age + Years Passed

🔢 Inequalities

At least → ≥

At most → ≤

More than → >

Less than → <


🚀 38. The 10-Second SAT Word-Problem Scan

When you see a long word problem, mentally ask:

① What quantities are given?

② What is unknown?

③ What are the units?

④ What relationship connects the quantities?

⑤ What exactly does the question ask me to find?

Then solve.

Do not let the story control your thinking.

You control the story by translating it into mathematics.


🏆 Final SAT Takeaway

The SAT can wrap a simple equation inside a paragraph designed to make you hesitate.

A distance question may hide:

d = rt

A work question may hide:

rate = 1/time

A mixture question may hide:

amount × concentration

A profit question may hide:

revenue − cost

An age question may hide:

x + years

A percentage question may hide:

original × multiplier

The secret is not memorizing dozens of special tricks.

It is learning to recognize the mathematical structure hiding inside the language.

Remember this sequence:

READ → IDENTIFY → TRANSLATE → EQUATE → SOLVE → CHECK

When you can consistently turn an SAT word problem into a clean equation, the paragraph stops being the problem.

The equation is the problem.

And equations are much easier to solve.


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