Advanced SAT Type Questions Flashcards
This deck tests your core knowledge of all subjects reviewed in other decks with the advanced SAT type questions. The deck provides a recommended approach as well as key strategies for successful problem solving.
Congratulations! You’ve studied your way through the core concept decks and arrived here.
What is the purpose of this deck?
The purpose of this deck is to improve your critical thinking skills by showing you a logical solution path and guiding you through a detailed analysis of various SAT problems using the core concepts reviewed in other decks.
There are 4 to 7 problems per topic, in order of increasing difficulty.
Are the problems in this deck difficult?
Do they resemble the ones you would see on the test?
The problems in this deck range in difficulty from medium to advanced.
They definitely resemble the actual questions you might see on the test.
What do you think you might need to solve the problems in this deck?
Pen and paper?
A calculator?
The questions in this deck are not straightforward. Solving them may require you to work with pen and paper. Hey, Einstein wrote his ideas on napkins and the cuffs of his shirt. Why can’t you? Oh, you, guys don’t wear dress shirts…. LOL
If you can avoid using your calculator, please do so.
Does this deck include shortcuts and strategies that every student is looking for in order to beat the test?
Yes, this deck covers key strategies, time saving techniques and shortcuts to help you answer questions more efficiently and effectively.
The two most efficient and effective strategies are, however, to use your brain and practice, practice, practice.
It’s tempting to reveal the answer if the problem looks difficult or confusing. Resist the temptation.
What approach should you take to working through problems in this deck?
We suggest taking this approach to maximize the benefits of this deck:
- Give yourself sufficient time to solve each problem step-by-step
- Make sure you understand the explanation in the answer
- The first time you study this deck, rate all the cards “perfect” regardless of whether or not you solved the problem
- If many questions present difficulties for you, review the core concepts again
- In a week, come back to this deck and work through the questions again but now rate them fairly
You may want to take a diagnostic test. Depending on your score you could take different approaches to problems.
Are you an advanced student or an intermediate student?
If you are an advanced student; i.e scoring above 650 on the SAT Math section, please, feel free to ignore the steps and solve the problem any way you want. What matters is that you get right answers on a consistent basis.
Students with scores from 450 to 650, please do yourself a favor and stick to the recommended solution path.
In this card, we demonstrate the critical thinking process in action.
The cost of a coat is $200. At what retail price should this item be put out on the floor so that the store can offer a 25% discount and still make a 20% profit on the cost?
- (a) 290
- (b) 300
- (c) 320
- (d) 900
- (e) 150
* State the purpose. Find the percent increase/decrease
- Define the question. Find the retail price of the coat
- Extract key info. Cost of the coat is $200. The retail price has to allow for a 25% discount and a 20% profit
- Make logical conclusions. Profit is calculated from the cost while the discount is taken from the retail price
- Identify formulas and strategy.
Increased/Decreased Value =
(100% +/- % increase) x Original Value
- Apply all knowledge.
To make a 20% profit on $200, the store needs to get $240 from the customer.
200 * 1.2 = 240
To account for a 25% off sale, the retail price R (100%) minus 25% has to equal $240.
100% of R - 25% of R = 75% of R = 240.
- 75R = 240 ⇒ R = $320 - choice (c)
* Verify the answer against the question. Verify if it makes sense
The cost of a coat is $200. At what retail price should this item be put out on the floor so that the store can offer a 25% discount and still make a 20% profit on the cost?
- (a) 290
- (b) 300
- (c) 320
- (d) 900
- (e) 150
Is your answer (c)320? If not, you made a mistake somewhere in your thought process.
Where did your logic fail?
If you didn’t get the correct answer, it’s easy to predict your potential mistake:
- If your answer was (a) 290, you added 20% and 25% together and increased the cost by 45%
- If your answer was (b) 300, you increased $200 by 20%, then added 25% to that number
- (d) 900 and (e) 150 choices are way out of the ballpark
Obviously, both ways are incorrect because the problem clearly states that 20% should be added to the cost of the coat but the 25% discount should be taken off the retail price. Realize that:
cost + 20% = retail - 25%
To develop the ability to use many different pieces of information to come up with a solution process to a math problem is to think critically in math.
A good critical thinker first picks apart the question at hand and uses the pieces of the information to devise a process to solve it.
How should you analyze an SAT problem in front of you?
Use this step-by-step analysis for any type of SAT problem. Train your brain to effectively extract key information, then use it to create a solution plan.
- State the purpose of the problem
- Define the question
- Extract key information
- Make logical conclusions from the information given
- Identify math formulas and strategies that may apply
- Apply all your knowledge to arrive at a logical answer
- Make sure your answer is the answer to the question
Understanding the purpose of a math problem is the first step in the critical thinking process.
What helps you determine the purpose of an SAT problem?
To determine the purpose of the problem, determine what type of problem it is.
Test makers love to test your ability to solve these types of arithmetic questions, commonly seen on the test:
- percents
- average
- ratios/proportions
- sets/sequences
- divisibility
- number properties
- probability
In the next few cards we will review some helpful strategies that may make the problem solving process easier and improve your timing on the test.
What are some strategies that we recommend using in order to get optimal results on the math section of the SAT?
Make these strategies part of your preparation routine for the SAT math section:
- Know core concepts and learn to apply them
- Pick Numbers
- Read actively
- Look for shortcuts
- Avoid traps
- Work back from the answers
- Guess if absolutely necessary
Nothing in the world will help you to score high on the SAT if you don’t have the proper core knowledge!
Knowledge of definitions, rules, and formulas should be cemented in your brain and subject to immediate recall. It will help you gain math confidence and solve problems faster.
How do you build a good “math” foundation?
This App is designed to help you study core concepts! Your goal is not to just memorize the facts in math but to understand them.
Study the decks repeatedly and answer practice questions before moving on to the advanced material deck.
Don’t skip tutorials and topic reviews in your other SAT materials.
The sum of two positive integers a and b is even. The sum of 2a, b and c is odd. Which statement below must be true?
- (a) a is even
- (b) a is odd
- (c) the product of a and b is even
- (d) if b is even, then c is odd
- (e) if c is even, then the product of a and b is even
If a, b, and c are too abstract, maybe substituting them with numbers would help?
As a matter of fact, this problem lends itself perfectly to picking easy, small numbers to stand for variables. Make sure they abide by the restrictions set in the problem and are positive integers.
a + b = even and 2a + b + c = odd
Realize that an even sum could be the result of either adding two even numbers or adding two odd numbers. So, test odd/odd and even/even pairs.
- a* = 4, b = 2 ⇒ 4 + 2 = 6
- a* = 3, b = 3 ⇒ 3 + 3 = 6
- a* and b could be either odd or even, but 2a (8 or 6) will always be even.
even 2a + b + c = odd
6 + 3 + 2 = 11 or 8 + 2 + 1 = 11
⇒ b + c must be odd
So, either b is even when c is odd or vice versa. Therefore, choice (d) must be true.
How do you know what problems lend themselves to using the “pick numbers” strategy?
It would be helpful to use the “pick numbers” strategy when you see:
- variables in the body of the question or in the answer choices
- problems pertaining to number properties asking for what “must be”, “could be” or “cannot be”
- problems asking to find a fraction or a percent of an unknown whole
When you decide to use numbers instead of variables, how do you know which numbers to pick?
- Numbers have to be small and easy to work with
Don’t pick 368 or 5, 246. Pick 1 or 2 or 3. Or 10 or 100.
- Numbers have to fit the problem. Abide by the restrictions
If the problem calls for a positive odd number, don’t pick 4 or a negative number. If it says that a > b, picking b = 5 and a = 3 is wrong!
*** Note: once you have picked your numbers in a word problem, re-read it with numbers instead of variables. It will be easier to process the information.
The “pick numbers” strategy sometimes involves testing various options. When should you do that?
Make sure you test all possible choices when not much is specified in a problem.
- Example:* If only “positive” is specified, make sure to pick both a whole number and a fraction to test.
- Example*: If a problems calls for an integer, make sure you test both positive and negative numbers and zero.
The Blue Sky limousine service charges $3 for the first mile plus $1.50 for each additional mile traveled. Yellow Cab charges an initial fee of $5 dollars plus $0.50 for each half-mile traveled. Suppose you take a ride for 10 miles with Blue Sky Limo. How many miles can you travel with Yellow Cab if you spend the same amount of money as you did with Blue Sky?
Too many words…. What can help you to process the information better?
For any word problem, you should use active reading skills to sort out the information.
Active Reading involves reading aloud, re-telling, highlighting and drawing diagrams of the problem. Chances are that if you do all of it or some of it, you won’t say “Ooohh, i didn’t notice that” or “That’s what it is??”.
The key here is to realize that the 1st mile with Blue Sky Limo is already accounted for. With Yellow Cab, notice that their charges are per half-mile. If you miss those, you’ll fall into traps that test makers prepared for you.
3 + 1.50 * 9 = 5 + 0.5 * 2x ⇒ x = 11.5
While it’s a good habit to always read actively, it’s simply necessary to do when solving word problems on the SAT.
What does “read actively” mean?
Active reading involves keeping your mind working at all times while trying to anticipate where the information is leading as you read it. Don’t let the words just wash over you. Solving word problems requires strong reading comprehension skills.
Ok, let’s learn a few helpful techniques.
What are some simple techniques you should practice to improve your active reading skills?
- Read the problem aloud when you practice. Accentuate key information with your voice
- Every word problem is a story. Practice re-telling it after reading it once or twice
- Most important: highlight and/or underline key information
- Create a simple diagram with the information given
The product of three consecutive integers is given by the formula: x3 + 3x2 + 2x, where x is the first integer. Find the product of 372, 373, and 374.
On the test, you will have five answer choices to pick from.
Let’s try to anticipate what you should look for in the correct answer.
Certainly don’t plug in 372 into the formula. Also, don’t multiply 372 x 373 x 374 even if you have a calculator.
You look for the asnwer choice ending in 4. Why? Multiplying just the unit digits of three numbers results in 24.
What if there is more than one answer that ends in 4? Pick the one that is divisible by 6. Why? The product must be divisible by 3 since it’s the product of three consecutive integers. The two factors are even, so the product is divisible by 2.
Look for shortcuts when you see questions that look like they can take a long time to solve.
What helps you find shortcuts?
- It’s hard to find a shortcut unless you know how to solve the problem the “long” way.
- Shortcuts don’t always exist so don’t look for a shortcut in every problem.
- Your ability to find a shortcut is a result of being comfortable with math problems which is achieved by hours of practice.
What problems and/or math operations make it necessary to look for shortcuts?
Look for shortcuts when you see:
- Large or confusing fractions or fractional expressions. Always try to reduce first
- Large numbers. Try to factor them into simple, small numbers for easy calculations
- Problems that require long arithmetic calculations
a + b + 2c = 200, where a, b, and c are positive integers.
If 2a = 5b and 3a = 10c, find the value of c?
(a) 10
(b) 15
(c) 20
(d) 25
(e) 30
You can solve this problem algebraically by expressing b and c in terms of a, solving for a, then finding c. A faster way for some of you may be to plug the answer choices back into the question.
Let’s pick (c) 20 and plug into the equation 3a = 10c. If c = 20, then 3a = 200. Clearly, this choice doesn’t work since a, in this case, is not an integer.
Since a has to be an integer, 10c must be divisible by 3.
Choice (b) 15 and choice (e) 30 meet this requirement. 15 is too small so the correct answer is (e) 30.
When does it make sense to work back from the answer choices?
Backsolve when you have no idea how to solve the problem mathematically or it looks like solving it might take a long time.
Use this strategy when the answer choices are simple, small numbers.
While for some of you plugging in answer choices saves time, we recommend to use backsolving as a “fall back” strategy.
If you decide to backsolve, which answer choice do you plug in first?
Start with the middle answer choice “C”.
In some cases work backwards from choice “E”. For example, when you are asked to find the largest possible value.
The makers of the SAT like to take advantage of your uncertainty in math.
They can anticipate where and what kind of mistakes you might make so they intentionally put the most common wrong answers as answer choices, creating so called “traps”.
Is there a way to recognize and avoid “traps”?
To avoid “traps”:
- improve your core knowledge
- read slowly
- solve problems to the end
Don’t stop half-way and try not to semi-guess. Arrive at your own answer, then compare it to the choices available.
k, 3k, 6k, 9k, 12k….. where k is not zero. Which of these five numbers is the smallest?
We present this problem to illustrate what common traps you should learn to avoid on the SAT math section.
You cannot determine which of the five numbers in the problem is the smallest with the information given.
If k is a positive integer, then the smallest number is k. But not all numbers are positive integers!
k can be negative. In this case, k is the greatest number, not the smallest.
What should you be careful about when you pick numbers to substitute for variables?
When you pick numbers to substitute for variables, make sure to test all possible choices.
Example:
If the problem states that n > 0, n can be a whole number as well as a fraction.
If it says that n is an integer, don’t forget to test negative numbers.
Consecutively numbered raffle tickets were given to all the kids in your school. The first ticket was number 29. What was the number of the last ticket if 254 students (one ticket per student) received raffle tickets?
of tickets distributed =
Last ticket # - First ticket # + 1
- x* - 29 + 1 = 254 ⇒ x = 254 + 29 -1 ⇒
- x* = 282
If your answer was 283, you added 29 and 254 but didn’t subtract 1. You’ve fallen into a trap the test makers prepared for you.
*** Change the numbers to small numbers (like 2 and 10) and count.
What is one common mistake students make when they count consecutive integers?
*** Note: that mistake may be set as a trap answer by the test makers.
To count consecutive integers in a range, subtract the smaller integer from the larger and then add 1 to the difference.
Example:
How many consecutive integers are between 5 and 10 inclusively? Let’s count. 5, 6, 7, 8, 9, and 10. There are six integers between 5 and 10.
10 - 5 + 1 = 6
Suppose you solved a problem correctly but still got a quarter of a point deduction. How could this happen?
You probably fell into a trap….
Make sure you answer the question that is being asked.
Yes, you could have solved the problem correctly but didn’t read the instructions carefully.
The last step in the successful solution process is to evaluate the answer.
The ratio of wins to losses of games played this year by your favorite team is 7/13.
- Does it mean that the team played 20 games?
- If you subtract from or add 5 to both parts of the ratio above, will it be the same ratio?
- If you multiply or divide both parts of the ratio by 5, will it be the same ratio?
- The team didnt necessarily play 20 games this year. It may have played 20, 40, … or any multiple of 20 games this year.
- Subtracting from or adding numbers to the ratio will change the ratio. A 7 to 13 ratio doesn’t equal a 2 to 8 ratio.
- Multiplying or dividing both parts of the ratio by the same number will not change the ratio. A 7 to 13 ratio equals a 35 to 65 ratio.
The ratio of blondes to red heads in your school is 9 : 2, and the ratio of blondes to brunettes is 3 : 5. What is the ratio of brunettes to red heads?
- (a) 5 : 2
- (b) 2 : 5
- (c) 8 : 11
- (d) 15 : 2
- (e) 2 : 15
Hint: beware of traps…
The common part (blonde students) of two ratios is not the same number. The numbers in these ratios are not proportional because the ratios don’t refer to the same whole.
- Blondes* : Reds = 9 : 2
- Blondes* : Brunettes = 3 : 5
Re-state the ratios so that the common part is represented by the same number. The least common multiple of 3 and 9 is 9. Re-state 2nd ratio.
Blondes : Brunettes = 9 : 15.
Only now you can compare the ratios. But… make sure you are answering the question! You are asked for the ratio of brunettes to red heads, so it’s 15 : 2.
What are the most common misconceptions about ratios that could be set as trap answers on the SAT?
Students are often confused about these simple concepts.
- Ratios may not represent actual quantities.
- You cannot add or subtract from a ratio or part of the ratio.
- When common parts of two ratios don’t refer to the same whole, the numbers in the ratios are not proportional.
*** Remember, ratios are a lot like fractions. The same rules apply.
You read n pages of a book at an average rate of 10 pages a day. On vacation, there was more time to read so you read the remaining n pages at an average rate of 30 pages a day. What was your average rate of reading for the entire book?
- (a) 14
- (b) 15
- (c) 17
- (d) 19
- (e) 20
Hint: try to avoid the trap.
Avg Rate = Total # of pages / Total time
We know that the book has 2n pages. We don’t know the time it took you to read each part but we can write equations to find it:
time 1 = n/10 (n pages at 10ppd rate)
time 2 = n/30 (n pages at 30 ppd rate)
To make it easier, pick a number for n that is a multiple of both 10 and 30. Or solve it algebraically.
Total time = time 1 + time 2 = n/10 + n/30 = <em>3n + n</em>/30 = <em>4n</em>/30
Avg Rate = 2n ÷ 4n/30 = 2n * 30/4n = 15
The correct answer is choice (b) 15.
You read n pages of a book at an average rate of 10 pages/day. On vacation, there was more time to read so you read the remaining n pages at an average rate of 30 pages/day. What was your average rate of reading for the entire book?
- (a) 14
- (b) 15
- (c) 17
- (d) 19
- (e) 20
Have you picked (e) 20 answer choice? Clearly, it’s wrong. Why?
Taking the average of 10 and 30 to find the average rate of reading is wrong! It doesn’t take into account that it took you more time to read the first part of the book because your average rate of reading was slower.
You can pick numbers to solve this problem instead of writing equations.
The LCM of 10 and 30 is 30. Let’s set n as 30 pages. There are 60 pages in the entire book. It took you 3 days (30 ÷ 10) to read the 1st part and 1 day (30 ÷ 30) to read the 2nd part. So, you read 60 pages in 4 days. Your average rate of reading for the entire book is 15 pages/day.
What is the most common misconception about finding the average rate that could be set as a trap answer on the SAT?
Many students forget that the average rate of A per B is the total of A divided by the total of B.
For example, the average rate of speed of the whole trip is not the average of the rates of speed on different legs of the trip.
Avg Rate of Speed = Total Distance/Total Time
The population of town A increased by 10% in 5 years. In the next 5 years, the population increased again by 20%. What is the percent increase in the population in 10 years compared to the original population?
- (a) 30%
- (b) 32%
- (c) 38%
- (d) 40%
- (e) 42%
You can solve the problem algebraically but let’s practice the “pick numbers” strategy.
Suppose the original population of town A was 100. In 5 years, it became 110 (100 * 1.1). Over the next 5 years it increased upto 132 (110 * 1.2). The change in population is 32, the original population is 100. Therefore, the population increased by 32%, choice (b).
If you picked choice (a) 30%, you’ve added 20% to 10% and got caught in a trap.
You invested $1,000 in stocks. After the first year the value of your portfolio increased by 25%. After the second year, it shrunk by 25%.
- Does the value of the increase equal the value of the decrease?
- Is the value of your portfolio after two years the same as the original value? Less? More?
- Even without calculating exact amounts you can determine that the increase is smaller than the decrease. The reason is that the 25% increase is calculated off of 1,000 while the 25% decrease is taken off of a greater amount.
- The value of your portfolio after two years is smaller than the original value.
1st year: 1,000 x 1.25 = 1,250
2nd year: 1,250 x 0.75 = 937.50
On the SAT, this problem would be presented with multiple choice answers where some may be designed to be traps.
You invested $1,000 in stocks. After the first year the value of your portfolio increased by 25%. After the second year, it shrunk by 25%. What percent of the original value is the value of your portfolio after two years?
- (a) 150%
- (b) 125.6%
- (c) 100%
- (d) 93.8%
- (e) 75%
1st year: 1,000 x 1.25 = 1,250
2nd year: 1,250 x 0.75 = 937.5
937.5/1000 * 100% = 93.75%, choice (e).
Answer choices (a) and (c) are traps.
If you picked (a), you didn’t read the problem carefully and thought that both years the value increased by 25%. Then, you added up percent increases.
If you picked (c), you mistakenly assumed that the amount of the increase equals the amount of the decrease.
What are some common misconceptions about figuring out percent changes that may be set as trap answers on the SAT?
- Multiple percent changes are not additive. A 20% decrease and an additional 60% decrease off the original price do not equate to an 80% decrease. Percentages can be added or subtracted only when they are calculated from the same amount.
- The same percent increases or decreases don’t necessarily equal the same amounts of decrease of increase.
You are having a mental block and can’t solve a problem.
When should you guess?
Guessing is smart as long as you are using “educated” guessing.
- Educated guesses are guesses that are made based on a familiarity with the math concepts in the question being asked
- Advanced students should make educated guesses whenever they can eliminate one answer choice
- Intermediate students should make educated guesses only when they can eliminate two answer choices
- Always guess on grid-ins because points are not lost for wrong answers!
How should you manage the limited amount of time you have to get through all questions in a section?
Math questions are arranged in order of difficulty - easy to hard. On the average, you can spend 60 to 90 seconds per question. But, with practice, you will be able to solve easier questions in the beginning of each section faster, leaving more time for harder ones.
- In a section that gives you 25 minutes for 20 questions, try to get those first 10 easy questions solved in about 7 minutes.
- Try to have 2 mins per question towards the end of the section, on the last 3 or 4 problems.
What are some of the things you can do to make sure you don’t run out of time?
It’s very important that you manage your time wisely during the test.
- Don’t get stuck on a question. If you see that it’s taking you too long, move on. If time allows, you can come back to it
- All questions are worth the same points. Make sure you answer all easy questions (usually 1 through 10) in each section
- Keep track of time. To get the feel of how much time you are spending per question, take a lot of practice tests as if they were the real tests
Remainder problems are very common on the SAT. Most of them are medium difficulty.
What is our strategic advice on how to approach remainder problems on the SAT?
To solve remainder problems, you can use this formula to figure out any missing element.
N = Q x D + R
- N - Number in question (or Dividend)
- Q - quotient
- D - divisor
- R - remainder
Example:
When N is divided by 5, the remainder is 4. What are the first three values of N?
N = Q x 5 + 4
The first three values of N are 9 (when Q = 1), 14 (when Q = 2), and 19 (when Q = 3).
When a number is divided by 5, the remainder is 3. What is the remainder when double that number is divided by 5?
How do you solve this problem using the recommended step-by-step approach and our strategic advice?
- The purpose of the problem is to divide certain numbers to find a remainder
- The question is to find a remainder of a certain number when divided by 5
- Extract key info: when a divisor of a certain number is 5, the remainder equals 3
- Identify formula/strategy:
N = Q x D + R
(N - number in question, Q - quotient, D - divisor, R - remainder)
- Apply all knowledge to solve:
If N = Q x 5 + 3 ⇒
2N = 2 (5Q + 3) = 10Q + 6
10Q is divisible by 5. When 6 is divided by 5, the remainder is 1.
- Make sure your answer makes sense and is the answer to the question.
When a certain number N is divided by 6, the remainder is 4. Which of the following will have no remainder when divided by 6?
- (a) N + 1
- (b) N + 8
- (c) N + 3
- (d) N + 5
- (e) N + 4
Go through all the steps of the critical thinking process on your own. The back of the card gives you a condensed version.
Once you’ve analyzed the information, set up the formula N = Q x D + R and plug in what you know.
(N - number in question, Q - quotient, D - divisor, R - remainder).
N = Q x 6 + 4. Possible choices for N are 10 (Q = 1), 16 (Q = 2), 22 (Q = 3), etc. Examining answer choices, you will see that adding 8 to those numbers will leave no remainder when divided by 6. The correct answer is (b) N + 8.
Another way to solve it is to use logic. A zero remainder when divided by 6 means that we are looking for multiples of 6; i.e. 6, 12, 18, … What number should you add to 4 to make it one of those multiples? 2, 8, 16, … N + 8 is choice (b).
When each student gets 6 pens, there are 4 pens left in the box. If two of the students are absent, then each student in the class can be given 8 pens. How many pens were in the box before distribution?
The back of the card gives you a short version of the process, but if you are an intermediate student, you shouldn’t skip any steps of critical thinking process.
The essence of this divisibility problem is that if a certain number is divided by 6, the remainder is 4. When that number is reduced by 2 and divided by 8, there is no remainder.
Let’s set up the remainder formula.
N = Q x D + R
Realize that Q is, in this case, the number of students, D is the number of pens given to each.
- N* (# of pens) = S (students) x 6 + 4
- N* (# of pens) = S - 2 (students) x 8
6S + 4 = (S - 2) x 8 ⇒ 6S + 4 = 8S - 16
Solving for S, you get that there are 10 students in the class originally. When 2 students are absent, there are 8 students left in the class. 8 x 8 = 64 - # of pens before distribution.