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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SOLVING LINEAR EQUATIONS [PART II]
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SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks
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