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.

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🔑 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.

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🧠 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”

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📊 3. Two-Way Tables Make Conditional Probability Easier

Many SAT questions present information in a table.

Consider this example:

	Uses Calculator	Does Not Use Calculator	Total
Group A	36	24	60
Group B	28	12	40
Total	64	36	100

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:

	Passed	Did Not Pass	Total
Studied	72	18	90
Did Not Study	24	36	60
Total	96	54	150

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%

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🧠 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%

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🔍 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.

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 Absolutely. This time I’m using actual Unicode mathematics throughout rather than LaTeX, and I’ve removed College Board and other website references from the article itself.

𝗦𝗔𝗧 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 & 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

𝗧𝗵𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝘁𝗲 𝗚𝘂𝗶𝗱𝗲 𝘁𝗼 𝗦𝗮𝗺𝗽𝗹𝗲 𝗦𝗽𝗮𝗰𝗲𝘀, 𝗧𝗮𝗯𝗹𝗲𝘀, 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 & 𝗧𝗿𝗶𝗰𝗸𝘆 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻𝘀

Probability questions can look much harder than they really are.

A question may talk about students, games, cards, machines, surveys, products or randomly selected objects. But underneath the story, the mathematics usually comes down to one central idea:

⭐ 𝗛𝗼𝘄 𝗺𝗮𝗻𝘆 𝗽𝗼𝘀𝘀𝗶𝗯𝗹𝗲 𝗼𝘂𝘁𝗰𝗼𝗺𝗲𝘀 𝗮𝗿𝗲 𝘁𝗵𝗲𝗿𝗲, 𝗮𝗻𝗱 𝗵𝗼𝘄 𝗺𝗮𝗻𝘆 𝗺𝗮𝘁𝗰𝗵 𝘄𝗵𝗮𝘁 𝗜 𝘄𝗮𝗻𝘁?

The biggest challenge is often not calculation.

It is identifying the correct group, especially when a question contains words such as:

𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁

𝗮𝗺𝗼𝗻𝗴

𝗼𝗳 𝘁𝗵𝗼𝘀𝗲 𝘄𝗵𝗼

𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲

𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁

Master those phrases and many SAT probability questions become much easier.

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① 𝗧𝗵𝗲 𝗕𝗮𝘀𝗶𝗰 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗙𝗼𝗿𝗺𝘂𝗹𝗮

When all outcomes are equally likely:

𝗣(𝗘) ＝ 𝗙𝗮𝘃𝗼𝗿𝗮𝗯𝗹𝗲 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀 ／ 𝗧𝗼𝘁𝗮𝗹 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A box contains 8 red balls and 4 blue balls.

One ball is selected randomly.

Total balls:

８ ＋ ４ ＝ １２

Favorable outcomes for blue:

４

Therefore:

𝗣(𝗯𝗹𝘂𝗲) ＝ ４／１２ ＝ １／３

⭐ 𝗦𝗔𝗧 𝗦𝗲𝗰𝗿𝗲𝘁

Before calculating anything, ask:

“𝗪𝗵𝗮𝘁 𝗶𝘀 𝗺𝘆 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿?”

The wrong denominator is one of the easiest ways to lose a probability question.

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② 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗖𝗮𝗻𝗻𝗼𝘁 𝗕𝗲 𝗟𝗲𝘀𝘀 𝗧𝗵𝗮𝗻 𝟬 𝗼𝗿 𝗚𝗿𝗲𝗮𝘁𝗲𝗿 𝗧𝗵𝗮𝗻 𝟭

Every probability satisfies:

０ ≤ 𝗣(𝗘) ≤ １

For example:

１／４ ＝ ０．２５ ＝ ２５％

A probability of:

０

means the event is impossible.

A probability of:

１

means the event is certain.

Therefore, an answer such as:

１．２

cannot be a probability.

---

③ 𝗧𝗵𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝗺𝗲𝗻𝘁 𝗧𝗿𝗶𝗰𝗸

The complement of an event means that the event does not happen.

𝗣(𝗻𝗼𝘁 𝗘) ＝ １ − 𝗣(𝗘)

This is particularly useful when you see:

𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲

𝗻𝗼𝗻𝗲

𝗻𝗼𝘁

𝗻𝗲𝘃𝗲𝗿

𝗱𝗼𝗲𝘀 𝗻𝗼𝘁

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A machine produces a defective item with probability ０．０８.

What is the probability that an item is not defective?

１ − ０．０８ ＝ ０．９２

Therefore:

𝗔𝗻𝘀𝘄𝗲𝗿 ＝ ９２％

---

④ 𝗧𝗵𝗲 “𝗔𝗡𝗗” 𝗥𝘂𝗹𝗲

When two independent events must both occur:

𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)

The symbol:

∩

means intersection, or the outcome where both events occur.

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A fair coin is tossed twice.

What is the probability of getting heads both times?

𝗣(𝗛) ＝ １／２

Therefore:

𝗣(𝗛 ∩ 𝗛) ＝ １／２ × １／２

＝ １／４

⭐ 𝗠𝗲𝗺𝗼𝗿𝘆 𝗧𝗿𝗶𝗰𝗸

𝗔𝗡𝗗 → 𝗧𝗵𝗶𝗻𝗸 𝗠𝗨𝗟𝗧𝗜𝗣𝗟𝗬

But remember: multiplication assumes the appropriate independence or conditional structure.

---

⑤ 𝗧𝗵𝗲 “𝗢𝗥” 𝗥𝘂𝗹𝗲

The symbol:

∪

means union, or an outcome belonging to at least one of the events.

If two events cannot overlap:

𝗣(𝗔 ∪ 𝗕) ＝ 𝗣(𝗔) ＋ 𝗣(𝗕)

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A number from １ through １０ is selected.

What is the probability of selecting ２ or ９?

There are １０ possible numbers.

Favorable outcomes:

２

Therefore:

𝗣(２ 𝗼𝗿 ９) ＝ ２／１０ ＝ １／５

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⑥ 𝗧𝗵𝗲 𝗢𝘃𝗲𝗿𝗹𝗮𝗽 𝗧𝗿𝗮𝗽

Sometimes two events overlap.

Then simply adding their probabilities counts the shared outcomes twice.

The general rule is:

𝗣(𝗔 ∪ 𝗕) ＝ 𝗣(𝗔) ＋ 𝗣(𝗕) − 𝗣(𝗔 ∩ 𝗕)

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

In a group of students:

• ２８ play basketball
• ２２ play soccer
• ９ play both

How many play at least one of the two sports?

Start with:

２８ ＋ ２２ ＝ ５０

The ９ students who play both were counted twice.

Subtract them:

５０ − ９ ＝ ４１

So:

４１ students

play at least one sport.

⭐ 𝗥𝗲𝗺𝗲𝗺𝗯𝗲𝗿

𝗢𝗥 → 𝗔𝗗𝗗

If there is overlap:

𝗔𝗗𝗗 → 𝗦𝗨𝗕𝗧𝗥𝗔𝗖𝗧 𝗧𝗛𝗘 𝗢𝗩𝗘𝗥𝗟𝗔𝗣

---

⑦ 𝗜𝗻𝗱𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀

Two events are independent when the occurrence of one does not change the probability of the other.

For independent events:

𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A die is rolled and a coin is flipped.

The die result does not affect the coin result.

Therefore:

𝗣(６ 𝗮𝗻𝗱 𝗵𝗲𝗮𝗱𝘀)

＝ １／６ × １／２

＝ １／１２

---

⑧ 𝗗𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀

Sometimes the first event changes the probability of the second.

This commonly happens when objects are selected without replacement.

𝗘𝘅𝗮𝗺𝗽𝗹𝗲

A bag contains:

５ red balls

７ blue balls

Two balls are selected without replacement.

Probability of getting two red balls:

First red:

５／１２

After removing one red ball:

４／１１

Therefore:

５／１２ × ４／１１

＝ ２０／１３２

＝ ５／３３

⭐ 𝗖𝗿𝗶𝘁𝗶𝗰𝗮𝗹 𝗣𝗼𝗶𝗻𝘁

𝗪𝗶𝘁𝗵 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁 → 𝘁𝗵𝗲 𝗴𝗿𝗼𝘂𝗽 𝗿𝗲𝘁𝘂𝗿𝗻𝘀 𝘁𝗼 𝗶𝘁𝘀 𝗼𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝘀𝘁𝗮𝘁𝗲

𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁 → 𝘁𝗵𝗲 𝗴𝗿𝗼𝘂𝗽 𝗰𝗵𝗮𝗻𝗴𝗲𝘀

That difference can completely change the answer.

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⑨ 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

Conditional probability is the probability of one event after another condition has already been established.

It is written:

𝗣(𝗔｜𝗕)

Read it as:

“the probability of A given B.”

The fundamental formula is:

𝗣(𝗔｜𝗕) ＝ 𝗣(𝗔 ∩ 𝗕) ／ 𝗣(𝗕)

But for SAT questions, there is an even more useful way to remember it:

⭐ 𝗧𝗛𝗘 𝗖𝗢𝗡𝗗𝗜𝗧𝗜𝗢𝗡 𝗖𝗛𝗔𝗡𝗚𝗘𝗦 𝗧𝗛𝗘 𝗗𝗘𝗡𝗢𝗠𝗜𝗡𝗔𝗧𝗢𝗥.

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⑩ 𝗧𝗵𝗲 “𝗡𝗲𝘄 𝗧𝗼𝘁𝗮𝗹” 𝗧𝗿𝗶𝗰𝗸

Suppose a school has １００ students.

• ６０ are juniors.
• ２４ juniors play tennis.

A student is selected from the juniors.

What is the probability that the student plays tennis?

You are no longer choosing from all １００ students.

Your new group is:

６０ juniors

Of those:

２４ play tennis

Therefore:

𝗣(𝘁𝗲𝗻𝗻𝗶𝘀｜𝗷𝘂𝗻𝗶𝗼𝗿) ＝ ２４／６０

＝ ２／５

＝ ４０％

🚨 𝗧𝗵𝗲 𝘁𝗿𝗮𝗽

Do not calculate:

２４／１００

That answers a different question.

The condition “junior” has already reduced your sample space.

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⑪ 𝗧𝘄𝗼-𝗪𝗮𝘆 𝗧𝗮𝗯𝗹𝗲𝘀

Two-way tables are extremely useful for conditional probability.

Consider:

	𝗦𝗽𝗼𝗿𝘁𝘀	𝗡𝗼 𝗦𝗽𝗼𝗿𝘁𝘀	𝗧𝗼𝘁𝗮𝗹
𝗝𝘂𝗻𝗶𝗼𝗿𝘀	２４	３６	６０
𝗦𝗲𝗻𝗶𝗼𝗿𝘀	１０	３０	４０
𝗧𝗼𝘁𝗮𝗹	３４	６６	１００

Question:

Given that a student is a junior, what is the probability that the student plays sports?

The condition is:

𝗝𝘂𝗻𝗶𝗼𝗿

Therefore, use the junior total:

６０

Favorable students:

２４

So:

𝗣(𝘀𝗽𝗼𝗿𝘁𝘀｜𝗷𝘂𝗻𝗶𝗼𝗿) ＝ ２４／６０

＝ ２／５

＝ ４０％

🔑 𝗧𝗵𝗲 𝟯-𝗦𝗲𝗰𝗼𝗻𝗱 𝗧𝗿𝗶𝗰𝗸

When you see:

“𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”

immediately ask:

“𝗪𝗵𝗮𝘁 𝗴𝗿𝗼𝘂𝗽 𝗮𝗺 𝗜 𝗻𝗼𝘄 𝗹𝗼𝗼𝗸𝗶𝗻𝗴 𝗮𝘁?”

That group usually supplies the denominator.

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⑫ 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗪𝗶𝘁𝗵 𝗣𝗲𝗿𝗰𝗲𝗻𝘁𝗮𝗴𝗲𝘀

Suppose:

７０％ of students own a laptop.

Among laptop owners:

４０％ also own a tablet.

What percentage of all students own both?

Translate:

𝗣(𝗟) ＝ ０．７０

and:

𝗣(𝗧｜𝗟) ＝ ０．４０

Therefore:

𝗣(𝗟 ∩ 𝗧) ＝ ０．７０ × ０．４０

＝ ０．２８

Therefore:

𝗔𝗻𝘀𝘄𝗲𝗿 ＝ ２８％

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⑬ 𝗧𝗵𝗲 “𝗔𝘁 𝗟𝗲𝗮𝘀𝘁 𝗢𝗻𝗲” 𝗦𝗵𝗼𝗿𝘁𝗰𝘂𝘁

“At least one” means:

one or more.

Instead of calculating every possibility separately, calculate the probability of none.

Then subtract from １.

Suppose an event has probability:

０．２０

on each independent trial.

Over two trials:

Probability of failure each time:

１ − ０．２０ ＝ ０．８０

Probability of no success twice:

０．８０ × ０．８０ ＝ ０．６４

Therefore:

𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲) ＝ １ − ０．６４

＝ ０．３６

＝ ３６％

⭐ 𝗠𝗲𝗺𝗼𝗿𝘆 𝗧𝗿𝗶𝗰𝗸

𝗔𝗧 𝗟𝗘𝗔𝗦𝗧 𝗢𝗡𝗘 ＝ １ − 𝗡𝗢𝗡𝗘

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⑭ 𝗧𝗿𝗲𝗲 𝗗𝗶𝗮𝗴𝗿𝗮𝗺 𝗧𝗵𝗶𝗻𝗸𝗶𝗻𝗴

Tree diagrams are useful when a probability problem happens in stages.

A box contains:

３ green balls

２ yellow balls

Two balls are selected without replacement.

First green:

３／５

After one green is removed:

２ green + ２ yellow = ４ balls

Second green:

２／４ ＝ １／２

Therefore:

𝗣(𝗴𝗿𝗲𝗲𝗻 𝗮𝗻𝗱 𝗴𝗿𝗲𝗲𝗻)

＝ ３／５ × １／２

＝ ３／１０

🧠 𝗧𝗿𝗲𝗲 𝗗𝗶𝗮𝗴𝗿𝗮𝗺 𝗥𝘂𝗹𝗲

𝗠𝘂𝗹𝘁𝗶𝗽𝗹𝘆 𝗮𝗹𝗼𝗻𝗴 𝗮 𝗯𝗿𝗮𝗻𝗰𝗵.

𝗔𝗱𝗱 𝘀𝗲𝗽𝗮𝗿𝗮𝘁𝗲 𝗯𝗿𝗮𝗻𝗰𝗵𝗲𝘀 𝘁𝗵𝗮𝘁 𝗽𝗿𝗼𝗱𝘂𝗰𝗲 𝘁𝗵𝗲 𝗱𝗲𝘀𝗶𝗿𝗲𝗱 𝗿𝗲𝘀𝘂𝗹𝘁.

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⑮ 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 𝗣𝗼𝘀𝘀𝗶𝗯𝗹𝗲 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀

Sometimes the fastest probability method is simply counting.

Suppose a code contains:

• one digit from ０–９
• one letter from Ａ–Ｚ

There are:

１０ × ２６ ＝ ２６０

possible codes.

If one particular code is selected:

𝗣(𝘁𝗵𝗮𝘁 𝗰𝗼𝗱𝗲) ＝ １／２６０

⭐ 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 𝗣𝗿𝗶𝗻𝗰𝗶𝗽𝗹𝗲

If one stage has m possibilities and another has n possibilities:

𝗧𝗼𝘁𝗮𝗹 𝗽𝗼𝘀𝘀𝗶𝗯𝗶𝗹𝗶𝘁𝗶𝗲𝘀 ＝ m × n

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⑯ 𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

Probability can also be estimated from actual results.

Suppose a spinner is used:

２００ times

It lands on blue:

５８ times

Experimental probability:

５８／２００ ＝ ０．２９

Therefore:

𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 ＝ ２９％

𝗗𝗼𝗻'𝘁 𝗖𝗼𝗻𝗳𝘂𝘀𝗲

𝗧𝗵𝗲𝗼𝗿𝗲𝘁𝗶𝗰𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

comes from the mathematical structure.

𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

comes from observed results.

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⑰ 𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝗩𝗮𝗹𝘂𝗲

Expected value describes the long-run average outcome.

Suppose a game gives:

• ₹１００ with probability ０．２０
• ₹０ with probability ０．８０

Then:

𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝗩𝗮𝗹𝘂𝗲

＝ １００ × ０．２０ ＋ ０ × ０．８０

＝ ２０

Therefore:

𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝘃𝗮𝗹𝘂𝗲 ＝ ₹２０

This does not mean every player receives ₹２０.

It means that over many plays, the average outcome approaches ₹２０ per play.

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⑱ 𝗔 𝗛𝗮𝗿𝗱𝗲𝗿 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗣𝗿𝗼𝗯𝗹𝗲𝗺

A survey contains ５００ people.

• ３００ use App A
• ２５０ use App B
• １５０ use both

A person is selected from the people who use App B.

What is the probability that the person also uses App A?

The phrase:

“from the people who use App B”

changes the sample space.

The new total is:

２５０

The favorable group is:

１５０

Therefore:

𝗣(𝗔｜𝗕) ＝ １５０／２５０

＝ ３／５

＝ ６０％

❌ 𝗧𝗵𝗲 𝗧𝗿𝗮𝗽

You might calculate:

１５０／５００ ＝ ３０％

But that is the probability of selecting someone who uses both apps from the entire population.

The question does not ask that.

It asks for the probability among App B users.

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⑲ 𝗦𝗔𝗧 𝗪𝗼𝗿𝗱𝘀 𝗧𝗵𝗮𝘁 𝗖𝗵𝗮𝗻𝗴𝗲 𝘁𝗵𝗲 𝗠𝗮𝘁𝗵

“𝗚𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”

Think:

𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

---

“𝗔𝗻𝗱”

Think:

𝗕𝗼𝘁𝗵 𝗺𝘂𝘀𝘁 𝗵𝗮𝗽𝗽𝗲𝗻

Often:

𝗠𝘂𝗹𝘁𝗶𝗽𝗹𝘆

---

“𝗢𝗿”

Think:

𝗘𝗶𝘁𝗵𝗲𝗿 𝗰𝗮𝗻 𝘀𝗮𝘁𝗶𝘀𝗳𝘆 𝘁𝗵𝗲 𝗰𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻

Usually:

𝗔𝗱𝗱

But check for overlap.

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“𝗔𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲”

Think:

１ − 𝗻𝗼𝗻𝗲

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“𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁”

Think:

𝗗𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝗶𝗲𝘀

---

“𝗔𝗺𝗼𝗻𝗴…”

Think:

𝗡𝗲𝘄 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿

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⑳ 𝗧𝗵𝗲 𝗕𝗶𝗴𝗴𝗲𝘀𝘁 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗧𝗿𝗮𝗽𝘀

❌ 𝗧𝗿𝗮𝗽 𝟭 — 𝗪𝗿𝗼𝗻𝗴 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿

If the question gives a condition, the denominator may no longer be the grand total.

❌ 𝗧𝗿𝗮𝗽 𝟮 — 𝗙𝗼𝗿𝗴𝗲𝘁𝘁𝗶𝗻𝗴 𝗼𝘃𝗲𝗿𝗹𝗮𝗽

If two groups share members:

𝗔 ＋ 𝗕

may count some people twice.

❌ 𝗧𝗿𝗮𝗽 𝟯 — 𝗜𝗴𝗻𝗼𝗿𝗶𝗻𝗴 “𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁”

The second probability may be different.

❌ 𝗧𝗿𝗮𝗽 𝟰 — 𝗖𝗼𝗻𝗳𝘂𝘀𝗶𝗻𝗴 “𝗮𝘁 𝗹𝗲𝗮𝘀𝘁” 𝘄𝗶𝘁𝗵 “𝗲𝘅𝗮𝗰𝘁𝗹𝘆”

𝗔𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲

means one or more.

𝗘𝘅𝗮𝗰𝘁𝗹𝘆 𝗼𝗻𝗲

means one and only one.

❌ 𝗧𝗿𝗮𝗽 𝟱 — 𝗟𝗲𝘁𝘁𝗶𝗻𝗴 𝘁𝗵𝗲 𝗰𝗮𝗹𝗰𝘂𝗹𝗮𝘁𝗼𝗿 𝗱𝗲𝗰𝗶𝗱𝗲 𝘁𝗵𝗲 𝗺𝗮𝘁𝗵

A calculator can calculate the wrong expression perfectly.

𝗥𝗲𝗮𝘀𝗼𝗻 𝗳𝗶𝗿𝘀𝘁.

𝗖𝗮𝗹𝗰𝘂𝗹𝗮𝘁𝗲 𝘀𝗲𝗰𝗼𝗻𝗱.

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㉑ 𝗢𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗣𝗿𝗮𝗰𝘁𝗶𝗰𝗲 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻𝘀

🟢 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟭

A box contains ９ white counters and ６ black counters.

One counter is selected randomly.

What is the probability of selecting a black counter?

Total:

９ ＋ ６ ＝ １５

Therefore:

𝗣(𝗯𝗹𝗮𝗰𝗸) ＝ ６／１５

＝ ２／５

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ２／５

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🟢 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟮

A fair die is rolled twice.

What is the probability that both results are even?

Even results:

２，４，６

Therefore:

𝗣(𝗲𝘃𝗲𝗻) ＝ ３／６ ＝ １／２

Both rolls even:

１／２ × １／２

＝ １／４

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: １／４

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🟡 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟯

A club has ４０ members.

２５ study mathematics.

１０ of those mathematics students also study physics.

If a mathematics student is selected randomly, what is the probability that the student studies physics?

The condition says:

𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝘀 𝘀𝘁𝘂𝗱𝗲𝗻𝘁

Therefore:

𝗗𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿 ＝ ２５

Favorable:

１０

Therefore:

𝗣(𝗽𝗵𝘆𝘀𝗶𝗰𝘀｜𝗺𝗮𝘁𝗵) ＝ １０／２５

＝ ２／５

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ２／５

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🟡 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟰

A box contains ５ red counters and ７ blue counters.

Two counters are selected without replacement.

What is the probability that both are red?

First red:

５／１２

Second red:

４／１１

Therefore:

５／１２ × ４／１１

＝ ２０／１３２

＝ ５／３３

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ５／３３

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🟠 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟱

A survey of １２０ students shows:

• ７０ take mathematics
• ５５ take physics
• ３５ take both

A mathematics student is selected randomly.

What is the probability that the student also takes physics?

The condition is:

𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝘀

So:

𝗗𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿 ＝ ７０

Both:

３５

Therefore:

𝗣(𝗽𝗵𝘆𝘀𝗶𝗰𝘀｜𝗺𝗮𝘁𝗵) ＝ ３５／７０

＝ １／２

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: １／２

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🔴 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟲

An event has a probability of ０．１５ on each independent trial.

What is the probability that it occurs at least once in three trials?

Probability it does not occur:

１ − ０．１５ ＝ ０．８５

Probability it never occurs:

０．８５ × ０．８５ × ０．８５

＝ ０．６１４１２５

Therefore:

𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲)

＝ １ − ０．６１４１２５

＝ ０．３８５８７５

Therefore:

✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ３８．５８７５％

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㉒ 𝗧𝗵𝗲 𝗨𝗹𝘁𝗶𝗺𝗮𝘁𝗲 𝗦𝗔𝗧 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗖𝗵𝗲𝗮𝘁 𝗦𝗵𝗲𝗲𝘁

𝗕𝗮𝘀𝗶𝗰 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

𝗣(𝗘) ＝ 𝗙𝗮𝘃𝗼𝗿𝗮𝗯𝗹𝗲 ／ 𝗧𝗼𝘁𝗮𝗹

𝗖𝗼𝗺𝗽𝗹𝗲𝗺𝗲𝗻𝘁

𝗣(𝗻𝗼𝘁 𝗘) ＝ １ − 𝗣(𝗘)

𝗜𝗻𝘁𝗲𝗿𝘀𝗲𝗰𝘁𝗶𝗼𝗻

𝗣(𝗔 ∩ 𝗕)

means A and B.

𝗨𝗻𝗶𝗼𝗻

𝗣(𝗔 ∪ 𝗕)

means A or B.

𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆

𝗣(𝗔｜𝗕) ＝ 𝗣(𝗔 ∩ 𝗕) ／ 𝗣(𝗕)

𝗜𝗻𝗱𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀

𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)

𝗔𝘁 𝗟𝗲𝗮𝘀𝘁 𝗢𝗻𝗲

𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲) ＝ １ − 𝗣(𝗻𝗼𝗻𝗲)

𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗥𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁

𝗧𝗵𝗲 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗺𝗮𝘆 𝗰𝗵𝗮𝗻𝗴𝗲.

𝗪𝗶𝘁𝗵 𝗥𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁

𝗧𝗵𝗲 𝗼𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗰𝗼𝗺𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝗶𝘀 𝗿𝗲𝘀𝘁𝗼𝗿𝗲𝗱.

---

🏆 𝗧𝗵𝗲 𝗢𝗻𝗲 𝗜𝗱𝗲𝗮 𝗬𝗼𝘂 𝗠𝘂𝘀𝘁 𝗥𝗲𝗺𝗲𝗺𝗯𝗲𝗿

The most important probability skill is not memorizing a dozen formulas.

It is identifying the correct sample space.

If the question asks about everyone:

𝗨𝘀𝗲 𝘁𝗵𝗲 𝗳𝘂𝗹𝗹 𝗴𝗿𝗼𝘂𝗽.

If it asks about students who belong to a particular category:

𝗨𝘀𝗲 𝘁𝗵𝗮𝘁 𝗴𝗿𝗼𝘂𝗽 𝗮𝘀 𝘆𝗼𝘂𝗿 𝗻𝗲𝘄 𝘀𝗮𝗺𝗽𝗹𝗲 𝘀𝗽𝗮𝗰𝗲.

If you see:

“𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”

stop and identify the condition.

Then think:

𝗖𝗢𝗡𝗗𝗜𝗧𝗜𝗢𝗡 → 𝗡𝗘𝗪 𝗚𝗥𝗢𝗨𝗣 → 𝗡𝗘𝗪 𝗗𝗘𝗡𝗢𝗠𝗜𝗡𝗔𝗧𝗢𝗥

That single habit can turn a confusing probability problem into a one-line calculation.

⭐ 𝗙𝗜𝗡𝗔𝗟 𝗠𝗘𝗠𝗢𝗥𝗬 𝗧𝗥𝗜𝗖𝗞

𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗶𝘀 𝗻𝗼𝘁 𝗮𝗯𝗼𝘂𝘁 𝗴𝘂𝗲𝘀𝘀𝗶𝗻𝗴.

𝗜𝘁 𝗶𝘀 𝗮𝗯𝗼𝘂𝘁 𝗱𝗲𝗳𝗶𝗻𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗴𝗿𝗼𝘂𝗽, 𝗳𝗶𝗻𝗱𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗼𝘂𝘁𝗰𝗼𝗺𝗲𝘀, 𝗮𝗻𝗱 𝗳𝗼𝗿𝗺𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗿𝗮𝘁𝗶𝗼.

𝗪𝗵𝗲𝗻 𝘆𝗼𝘂 𝗰𝗮𝗻 𝗳𝗶𝗻𝗱 𝘁𝗵𝗲 𝗰𝗼𝗿𝗿𝗲𝗰𝘁 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿, 𝘆𝗼𝘂 𝗵𝗮𝘃𝗲 𝗮𝗹𝗿𝗲𝗮𝗱𝘆 𝘀𝗼𝗹𝘃𝗲𝗱 𝗺𝘂𝗰𝗵 𝗼𝗳 𝘁𝗵𝗲 𝗽𝗿𝗼𝗯𝗹𝗲𝗺.

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 ]

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

to find the parameter, unique solution, no solution infinite number of solutions

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

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

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

geometry notes  [ part 1 ] 

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

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

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

How SAT word problems hide equations

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

SAT Word Problems Made Easy

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

SAT function transformations tips and tricks

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

SAT probability guide

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.

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🔹 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.

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🔹 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.

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🔹 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.

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🔥 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

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🔹 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

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🔹 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

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🔹 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.

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🔹 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

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🔹 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

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🔥 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

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🔹 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

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🔹 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

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💡 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.

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🔹 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.

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🔥 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

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🔹 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

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🔹 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

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🔥 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.

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🔥 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)

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🔥 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?”

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🔹 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

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🔹 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%

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🔹 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.

Explore other 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 ]

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

to find the parameter, unique solution, no solution infinite number of solutions

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

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

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

geometry notes  [ part 1 ] 

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

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

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

How SAT word problems hide equations

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

SAT Word Problems Made Easy

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

SAT function transformations tips and tricks

SAT Function Transformations: The Tricks That Make Hard Questions Easy

 

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

Function transformation questions on the SAT can look intimidating because a familiar function may suddenly appear with numbers, negative signs, fractions, and parentheses in unfamiliar places.

But there is a simple way to read them.

Instead of trying to memorize every possible graph, learn to identify what happened to the input and what happened to the output.

Once you see that pattern, many SAT transformation questions become quick recognition problems.

This guide develops that idea from the ground up with original examples and SAT-style practice.

---

1. What Does “Transformation” Mean?

Start with a function:

f(x)

A transformation creates a new function whose graph has been moved, flipped, stretched, or compressed.

For example:

g(x) = f(x) + 4

does not change the x-values. It changes the y-values.

On the other hand:

g(x) = f(x − 4)

changes where the same output occurs.

That distinction is the foundation of function transformations.

A useful first question is:

Did the change happen outside f(x), or inside its input?

---

2. The Two Big Categories

Most SAT transformations can be organized into two groups.

Changes outside the function

Examples:

f(x) + 5

f(x) − 2

3f(x)

−f(x)

These primarily affect the output, or y-coordinate.

Changes inside the function

Examples:

f(x − 5)

f(x + 2)

f(−x)

f(3x)

These primarily affect the input, or x-coordinate.

This “inside versus outside” idea is more useful than trying to memorize a huge collection of unrelated rules.

---

3. Moving a Graph Up or Down

Consider:

g(x) = f(x) + 6

Every output increases by 6.

So the graph moves:

6 units upward

If a point on the original graph is:

(3, 4)

the new point is:

(3, 10)

The x-coordinate stays exactly where it was.

Downward movement

Now consider:

g(x) = f(x) − 6

Every y-value decreases by 6.

Therefore, the graph moves:

6 units downward

The basic rule is:

f(x) + k → up k units

f(x) − k → down k units

---

4. The Horizontal Shift Trap

Horizontal transformations are where SAT questions often catch students.

Consider:

g(x) = f(x − 5)

The graph moves:

5 units to the right

Not left.

Now consider:

g(x) = f(x + 5)

The graph moves:

5 units to the left

So:

f(x − h) → right h

f(x + h) → left h

Why does the sign seem reversed?

Suppose the original function produces a particular output when:

x = 2

For:

g(x) = f(x − 5)

we want the new input to produce that same output.

Set:

x − 5 = 2

Therefore:

x = 7

The point has moved from x = 2 to x = 7.

So the graph shifted right by 5.

This is why the horizontal sign behaves differently from the vertical sign.

---

5. A Quick Memory Trick

When the number is outside the function:

f(x) + 4

think:

“The output goes up.”

When the number is inside the function:

f(x − 4)

think:

“The input needs 4 more.”

That produces a movement to the right.

A short version worth remembering is:

Outside = vertical. Inside = horizontal.

---

6. Reflections: Which Axis?

Negative signs can also flip a graph.

There are two important forms.

Reflection across the x-axis

g(x) = −f(x)

Every y-coordinate changes sign:

(x, y) → (x, −y)

For example:

(4, 7) → (4, −7)

The graph is reflected across the x-axis.

---

Reflection across the y-axis

g(x) = f(−x)

Every x-coordinate changes sign:

(x, y) → (−x, y)

For example:

(4, 7) → (−4, 7)

The graph is reflected across the y-axis.

Remember:

−f(x) → x-axis reflection

f(−x) → y-axis reflection

The location of the negative sign tells you which coordinate changes.

---

7. Vertical Stretch and Compression

Suppose:

g(x) = 3f(x)

Every output is multiplied by 3.

So if:

f(2) = 4

then:

g(2) = 12

The graph becomes taller vertically.

This is called a vertical stretch by a factor of 3.

---

What if the multiplier is between 0 and 1?

Consider:

g(x) = ½f(x)

Every y-value is cut in half.

For example:

(2, 8) → (2, 4)

The graph becomes shorter vertically.

This is a vertical compression.

So:

|a| > 1 → vertical stretch

0 < |a| < 1 → vertical compression

for:

g(x) = af(x)

---

8. Horizontal Stretch and Compression

Horizontal scaling requires extra attention.

Suppose:

g(x) = f(2x)

It is tempting to say “2 means stretch by 2.”

That is incorrect.

The horizontal scale factor is the reciprocal.

Therefore:

f(2x)

produces a horizontal compression by:

½

See it through a point

Suppose:

f(6) = 9

For the transformed function:

g(x) = f(2x)

we need:

2x = 6

so:

x = 3

The point has moved from:

(6, 9)

to:

(3, 9)

The graph has been compressed horizontally.

---

9. Horizontal Stretch Example

Consider:

g(x) = f(x/4)

The input is multiplied by:

¼

Therefore, the graph is stretched horizontally by:

4

If the original graph contains:

(3, 8)

the corresponding point becomes:

(12, 8)

The output stays 8 while the x-coordinate becomes four times as large.

---

10. The Transformation Formula

Many complicated-looking SAT equations can be organized using:

g(x) = af(b(x − h)) + k

Each component tells you something.

a

Controls vertical scaling and may produce a reflection.

b

Controls horizontal scaling and may produce a horizontal reflection.

h

Controls horizontal movement.

k

Controls vertical movement.

For example:

g(x) = 2f(3(x − 4)) + 7

contains several transformations at once.

You do not need to panic when you see an expression like this.

Break it into pieces.

---

11. A Point Transformation Formula

Suppose:

(x₀, y₀)

is a point on f.

For:

g(x) = af(b(x − h)) + k

the corresponding horizontal coordinate satisfies:

b(x − h) = x₀

Therefore:

x = x₀/b + h

The new y-coordinate is:

y = ay₀ + k

So the point becomes:

(x₀/b + h, ay₀ + k)

This can be an extremely efficient technique when the SAT gives you a point on the original function.

---

12. SAT Example: Transforming a Point

Suppose:

f(−1) = 6

and:

g(x) = 2f(x − 3) − 5

Find the corresponding value of g.

The original point is:

(−1, 6)

First deal with the horizontal transformation:

x − 3 = −1

Therefore:

x = 2

Now transform the output:

2(6) − 5 = 12 − 5 = 7

Therefore:

g(2) = 7

No complete formula for f(x) was necessary.

---

13. Why Point Questions Can Be Faster

Suppose a question gives you:

f(5) = −3

and asks about:

g(x) = 4f(x − 2) + 1

You do not need to discover the equation for f.

You already know one point:

(5, −3)

Find the new x-coordinate:

x − 2 = 5

so:

x = 7

Then find the new y-coordinate:

4(−3) + 1 = −11

Therefore:

g(7) = −11

This type of question rewards recognizing the transformation rather than doing lengthy algebra.

---

14. The Difference Between Similar Expressions

Pay close attention to these:

Expression A

f(x − 4)

Graph moves:

4 units right

Expression B

f(x) − 4

Graph moves:

4 units down

They look almost identical, but their effects are completely different.

This is one of the most common transformation traps.

---

15. Another Important Difference

Compare:

−f(x)

and

f(−x)

The first changes the y-coordinate:

(x, y) → (x, −y)

The second changes the x-coordinate:

(x, y) → (−x, y)

Therefore:

−f(x) → reflection across x-axis

f(−x) → reflection across y-axis

If you keep track of which coordinate changes, the distinction becomes much easier.

---

16. Quadratic Transformations

Quadratic functions are particularly friendly because their transformations can often be seen directly from vertex form.

The standard form is:

y = a(x − h)² + k

The vertex is:

(h, k)

Consider:

y = 3(x − 2)² − 4

The vertex is:

(2, −4)

Compared with:

y = x²

the graph has:

• shifted 2 units right
• shifted 4 units down
• undergone a vertical stretch by a factor of 3

Because the coefficient is positive, the parabola still opens upward.

---

17. What If the Quadratic Coefficient Is Negative?

Consider:

y = −2(x + 3)² + 1

The vertex is:

(−3, 1)

The graph:

• moves 3 units left
• moves 1 unit up
• is vertically stretched by 2
• reflects across the x-axis

The negative coefficient means the parabola opens downward.

---

18. Function Transformations and Zeros

Transformations can move zeros from one location to another.

Suppose:

f(4) = 0

Therefore, f has a zero at:

x = 4

Now define:

g(x) = f(x − 6)

For g to equal zero, the input to f must be 4:

x − 6 = 4

Therefore:

x = 10

The zero moved from:

4 → 10

because the graph shifted 6 units right.

---

19. Transformations Seen in Tables

You can recognize transformations without seeing a graph.

Suppose:

x	f(x)
0	2
1	5
2	9

Now another function has:

x	g(x)
3	2
4	5
5	9

The output sequence has not changed.

Every x-value has increased by 3.

Therefore:

g(x) = f(x − 3)

The function has shifted right by 3.

---

20. Recognizing Transformations From Graphs

The SAT may give you two graphs instead of equations.

Start by looking for a feature that is easy to identify.

Depending on the graph, this could be:

• a vertex
• an intercept
• a maximum
• a minimum
• an endpoint
• a distinctive curve feature
• an asymptote

Example

Suppose a parabola's vertex changes from:

(−2, 3)

to:

(4, −1)

The horizontal change is:

+6

The vertical change is:

−4

So the graph moved:

6 units right and 4 units down

A corresponding transformation is:

g(x) = f(x − 6) − 4

---

21. A Reliable SAT Method

When faced with a transformation question, use this checklist.

Step 1: Find f(x)

Identify the original function.

Step 2: Examine the inside

Look for changes such as:

x − h

x + h

−x

bx

These affect the horizontal behavior.

Step 3: Examine the outside

Look for:

+ k

− k

af(x)

−f(x)

These affect the vertical behavior.

Step 4: Look for multiple transformations

A single function may be shifted, stretched, and reflected at the same time.

Step 5: Use a known point when possible

If the question gives a point or function value, transform that point instead of rebuilding the whole function.

---

22. Practice Question 1

A function f contains the point:

(−2, 7)

A new function is:

g(x) = f(x − 5) + 2

Which point must lie on g?

A) (−7, 9)

B) (−2, 9)

C) (3, 9)

D) (3, 5)

Solution

The expression:

f(x − 5)

moves the graph 5 units right.

So:

−2 → 3

Then:

+2

moves the graph upward:

7 → 9

Therefore:

(−2, 7) → (3, 9)

Answer: C

---

23. Practice Question 2

Suppose:

f(8) = 11

and:

g(x) = f(2x)

For what value of x is:

g(x) = 11?

We need:

2x = 8

Therefore:

x = 4

Answer: 4

---

24. Practice Question 3

The function:

f(x) = x²

is transformed into:

g(x) = −(x + 4)² + 2

Describe the transformation.

The expression:

x + 4

means the graph moves:

4 units left

The negative sign outside the squared expression means:

reflection across the x-axis

The +2 means:

2 units up

Therefore, the graph is shifted 4 units left, reflected across the x-axis, and shifted 2 units upward.

Its vertex is:

(−4, 2)

---

25. Practice Question 4

A function f has a zero at:

x = 9

The transformed function is:

g(x) = f(x + 4)

Where is the corresponding zero of g?

Set the input equal to 9:

x + 4 = 9

Therefore:

x = 5

Answer: 5

---

26. Practice Question 5

The point:

(−3, 4)

lies on f.

A new function is:

g(x) = 3f(−x) − 2

Find the corresponding point on g.

First, the input is reflected:

−3 → 3

Then multiply the output by 3:

4 → 12

Finally subtract 2:

12 → 10

Therefore:

(−3, 4) → (3, 10)

Answer: (3, 10)

---

27. Practice Question 6

Suppose:

f(2) = −5

and:

g(x) = −2f(x − 6) + 4

Find g(8).

Check the input:

8 − 6 = 2

Therefore:

f(8 − 6) = f(2) = −5

Now substitute:

g(8) = −2(−5) + 4

g(8) = 10 + 4

g(8) = 14

Answer: 14

---

28. Practice Question 7

A graph of f has a point at:

(6, −2)

The transformed function is:

g(x) = ½f(x + 3) − 1

Find the corresponding point.

For the horizontal coordinate:

x + 3 = 6

so:

x = 3

For the vertical coordinate:

½(−2) − 1

= −1 − 1

= −2

Therefore the new point is:

(3, −2)

Answer: (3, −2)

---

29. The Transformation Rules Worth Memorizing

If you are short on study time, concentrate on these.

Vertical shifts

f(x) + k → up k

f(x) − k → down k

Horizontal shifts

f(x − h) → right h

f(x + h) → left h

Reflections

−f(x) → reflection across x-axis

f(−x) → reflection across y-axis

Vertical scaling

af(x) → y-values multiplied by a

Horizontal scaling

f(bx) → x-values scaled by 1/|b|

Quadratic vertex form

a(x − h)² + k

has vertex:

(h, k)

---

30. A Final SAT Strategy

Do not treat a transformation equation as one giant expression.

Break it apart.

For example:

g(x) = −3f(2(x − 4)) + 7

Instead of staring at the entire equation, identify:

x − 4

→ horizontal shift

2

→ horizontal compression

−3

→ vertical stretch and reflection

+7

→ vertical shift

The equation becomes much easier when each component is assigned a job.

---

31. Final Takeaway

Function transformations are really about tracking coordinates.

If the change affects the output, think vertically.

If the change affects the input, think horizontally.

The most important patterns are:

f(x) + k → up k

f(x) − k → down k

f(x − h) → right h

f(x + h) → left h

−f(x) → x-axis reflection

f(−x) → y-axis reflection

af(x) → vertical scaling

f(bx) → horizontal scaling by 1/|b|

And when a question gives you a known point, use it.

You often do not need to find the original function at all.

For SAT Math, that is one of the biggest advantages of understanding transformations: you can often solve the problem by tracking what happened to one point instead of doing a page of algebra.

Explore the following

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 ]

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

to find the parameter, unique solution, no solution infinite number of solutions

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

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

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

geometry notes  [ part 1 ] 

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

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

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

How SAT word problems hide equations