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Chris Lele

10 Most Difficult SAT Questions

Here are the 10 toughest SAT questions from Magoosh SAT: see how many you get right by checking your answers at the end of the post!

(P.S. If you can get these right, then you should check out Harvard SAT scores and Yale SAT scores…)

 

Directions: Choose the words that best fit the blanks:

1. Cosmologist Martin Rees has cautioned that our present satisfaction with the big bang explanation for the creation of the universe may reflect the ——- of the data rather than the ——- of the theory.

  1. paucity . . validity
  2. genius . . accuracy
  3. relevance . . scope
  4. destruction . . core
  5. persuasiveness . . reality

Tip: Try to come up with your own word(s) for the blank. If you are unable to, it is okay, as a last resort, to plug the answer choices back in the blank. Sometimes meaning emerges this way and the sentence makes sense.

 

2. Apparently the groom was very nervous: one moment he would be ——-, rambling on to his best man about silly, meaningless things, and then abruptly he would turn ——- and could not be prompted to say anything

  1. garrulous . . reticent
  2. grandiose . . taciturn
  3. vociferous . . effusive
  4. melodious . . timorous
  5. munificent . . utilitarian

Tip: Match the clues with the blanks and then find a word that matches. Remember you only need to work one blank at a time, eliminating those answer choices that don’t work. Then, when you move on the other blank, you only have a few possible answers to deal with.

 

Directions: Choose the correct version of the sentence:

3. Regardless of the fact of the ridge-top condominiums’ aesthetics, every investor has enjoyed a high return on their investment.

  1. Regardless of the fact of the ridge-top condominiums’ aesthetics, every investor has enjoyed a high return.
  2. Regardless of the ridge-top condominium aesthetic, every investor has had a high return to enjoy.
  3. Regarding the aesthetics of the ridgetop condominiums, every investor has enjoyed a high return.
  4. Regardless of the fact of the ridge-top condominiums’ aesthetics, a high return by every investor has been enjoyed.
  5. Regardless of the aesthetics of the ridge-top condominiums, every investor has enjoyed a high return

Tip: Remember to retain the original meaning of the sentence – investors are enjoying an investment. If you remove investment than they are enjoying (having a good time) the high return (money). Which, while highly likely, changes the overall meaning of the sentence.

 

4. Included in the cost of many services and products sold in Great Britain, American tourists may not realize that they do not necessarily have to pay the value added tax (VAT).

  1. Included in the cost of many services and products sold in Great Britain, American tourists may not realize that they do not necessarily have to pay the value added tax (VAT).
  2. Included in the cost of many services and products which are sold in Great Britain, tourists from America may not realize that they do not necessarily have to pay the value added tax (VAT).
  3. American tourists may not realize that they do not necessarily have to pay the value added tax (VAT) that are included in the cost of many services and products sold in Great Britain.
  4. In addition to the cost of many services and products sold in Great Britain, American tourists may not realize that they do not necessarily have to pay the value added tax (VAT).
  5. American tourists may not realize that they do not necessarily have to pay the value added tax (VAT) that is included in the cost of many products and services sold in Great Britain.

Tip: Remember to make sure that the nouns in the sentence are being modified correctly. American tourists are not included in the cost of many services.

 

Select the answer that best answers the question:

5. The average (arithmetic mean) of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?

  1. 1
  2. 19
  3. 29
  4. 30
  5. 33

Tip: This is a logic question. Setting up an equation for average will only get you so far. Think in terms of what number could be the smallest possible value.

 

6. If square ABCD has area 25, and the area of the larger shaded square is 9 times the area of the smaller shaded square, what is the length of one side of the smaller shaded square?

Note: Figure not drawn to scale

  1.  3/4
  2. 5/4
  3. 6/5
  4. 4/3
  5. 5/3

 Tip: If you are not sure how to set up the question algebraically you can also solve using the given information. In this case you can assume the answer is (C). So if the side of the small square is 6/5 do we end up with 25 as the area of the big square? Remember the big square has an area that is twice as big as that of the small square (in this question the algebraic approach is better).

 

7. Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

  1. 250/3
  2. 500/3
  3. 400
  4. 480
  5. 600

Tip: You can solve this question by setting up an equation…or you can think of this problem as a weighted average.

 

8. On a certain multiple-choice test, 9 points are awarded for each correct answer, and 7 points are deducted for each incorrect or unanswered question. Sally received a total score of 0 points on the test. If the test has fewer than 30 questions, how many questions are on the test?

  1. Cannot be determined
  2. 16
  3. 19
  4. 21
  5. 24

Tip: This is a question based more on logic. Do not try to set up an equation but think in terms of how many 7-point questions you need and how many 9-point questions you need for the two to cancel out.

 

9. A computer can perform c calculations in s seconds. How many minutes will it take the computer to perform k calculations?

  1. 60ks/c
  2. ks/c
  3. ks/60c
  4. 60c/ks
  5. k/60cs

Tip: Assign values to k, s, and c if you have difficulty thinking through this question algebraically.

 

10. If the circle with center O has area 9π, what is area of equilateral triangle ABC?

  1.  
  2. 18
  3. 24

Tip: Remember to think of the necessary steps to arrive at the answer. Once you’ve worked those steps at then apply the math. And don’t forget – the fundamental geometry formulas are always in the beginning of each math section.

For this last question, try it out in Magoosh SAT to see the answer and video explanation!

 

 

Answers:

1. A

2. A

3. E

4. E

5. E

6. B

7. E

8. B

9. C

10. C

About Chris Lele

For the last ten years, Chris has been helping students excel on the SAT and the GRE. In this time, he’s coached 5 students to a perfect SAT score. Some of his GRE students have raised their scores by nearly 400 points. He has taken many GMAT students from the doldrums of the 600s to the coveted land of the 700+. Rumor has it he does a secret happy dance when his students get a perfect score. You can read Chris's awesome blog posts on the Magoosh GRE blog and SAT blog.


78 Responses to “10 Most Difficult SAT Questions”

  1. Antaryirreree says:

    Hello! Just want to say thank you for this interesting article! =) Peace, Joy.

  2. mayank says:

    thanx dude got 9 of dem correct and i hav got 2300 in SAT

  3. julie says:

    Nice Dude It was pretty helpful but got just 6 so sad :(

    • Margarette Jung Margarette says:

      Hi, Julie! 6/10 certainly isn’t bad to start out with– I’m sure that after studying for a few weeks, you could score even higher :)

  4. Ankush says:

    sat exam math is so easy as compared to iits math,its almost like nothing

    • Chris Lele Chris says:

      Hi Ankush,

      The general SAT math portion is meant more to test one’s thinking/problem solving skills. It is definitely not too difficult, and sometimes getting a perfect score for us seasoned tutors, is making sure we don’t make a careless error. The SAT math subject tests, on the other hand, are much more difficult, and are targeted towards those who want to study in math/science fields.

    • Akshita Bhalla says:

      Are you Ankush Sharma from FIITJEE Chandigarh?????

  5. Dkgladr says:

    Hi there I’m 13 and in 8th grade in this test I got 4/10 ….I’m feeling bad

    • Chris Lele Chris Lele says:

      Dkgladr,

      Wow, that’s actually a great score, for someone who isn’t even required to take the PSAT. Some of the questions above are taken from GMAT and GRE tests, which are for those looking to go to graduate school (some of the smartest people around :)). So, I’d say you are doing pretty well :).

  6. Ned says:

    I got 9/10 correct and I have a 1960 SAT Score. These questions were helpful though

    • Chris Lele Chris Lele says:

      Hi Ned,

      These questions are very tough — so you did well. I’m guessing you do pretty well on the quant, maybe close to 700. Even then, on the SAT, those “easy” and “medium” questions can get you to go :).

      Good luck!

  7. Student says:

    Could you please explain how to do number 10? I keep getting the answer 27 times the square root of three. Thank you.

  8. Ash says:

    I’ve been studying for a few weeks and could only get 5 right (barely)…I am so gonna to bomb the SAT…

    • Chris Lele Chris Lele says:

      Hi Ash,

      Not at all — these are all difficult questions. Only about 25% of the questions on the SAT are this difficult. So if you are getting half of the difficult ones correct, you are probably do even better on the medium and easy questions.

      Good luck–and don’t get discouraged :)!

  9. mAdCap says:

    Well, tell me if I am wrong , but there is something off about question 8. Even I came up with 16(b) at first but then I considered this situation …

    Suppose Sally is as bad as a student as myself and managed to get 4/16 questions correct , earning her 4*9=36 marks .

    However she skipped or got wrong 12/16 questions , losing a total of 12*7=84 marks .

    So, her total score is = ta ta ra ra … negative 48.
    You see my point?

    • Chris Lele Chris Lele says:

      Hi Madcap,

      It seems that we already have a given piece of information: the student scored zero points on the test. The only way this is possible, given the test has fewer than 30 questions, is if there are 16 questions on the test. The question is not saying, if the test had 16 questions, then how many points must Sally have gotten? She could have gotten a bunch of different points, including, as you mentioned, -48.

      Hopefully that makes sense :)

  10. Rose says:

    Im stuck on number 5 could you please explain? I was thinking that if there are no restrictions on what the other possible integers could be then couldn’t 1 be one of them and then the other two would add up to the rest? That was my thinking but I guess I was wrong. Could you please explain, thank you so much!!

    • Chris Lele Chris Lele says:

      Hi Rose,

      So we know that the sum of the average has to be 300 (75×4) = 300. If the largest number is 90, then the next largest numbers could be 89 and 88. Remember, we want to maximize these two numbers so that the remaining number can be as small possible. This gives us 90+89+88+x = 33 (E).

      Hope that makes sense :)

  11. Anzie says:

    Hi Chris,

    I don’t understand number 5 at all whatsoever. Could you please explain it to me? Also, regarding number 5, why is the answer not 1? The question doesn’t ask for the highest possible smallest integer that could result in the mean? It merely asks for the smallest, so 1 should be applicable, shouldn’t it?

    Thank you!

    • Chris Lele Chris Lele says:

      Hi Anzie,

      It does seem that way, that ‘1’ should be the answer. But this is a tricky one. If we actually add up the four integers, x + 88 + 89 + 90 = 300 we get 33 for x. x cannot equal ‘1’ because the two integers in the middle have to be less than 90, and the greatest possible numbers they can equal are 88 and 89 respectively.

      Hope that helps shed some light :).

      • Viraj says:

        I do not get number 9 at all because I tried to plug in 60 for all the values, and only choice B. worked using that method. The only way that I could see choice C. working is if there was a fraction bar between the ks and 60c like this:

        ks —–
        60c

        Is this how it was intended to be written? I tried to follow the choices given to me using order of operations.

        Thanks and Happy Holidays!

        • Viraj says:

          #9 was the only one I got wrong and I am in 7th grade. 😛

        • Chris Lele Chris Lele says:

          Hi Viraj,

          Good job on getting most of the questions correct!

          Yes, on question #9, it should be written as ks/60 (that mark is the same as the horizontal line dividing the numerator and the denominator).

          Hope that helps!

  12. Hey there,

    Thank you very much for these very helpful questions.

    I need to understand number 10 and 7, I am stuck in 7 for not knowing how many millimeters contain what percentage.

    in 10 it’s because I did 9^2÷π=25.78310078, and that’s close to the correct answer but I need to understand the idea!

    Thank you in advance, have a beautiful day 😀

    • Chris Lele Chris Lele says:

      On number 10, the key is to solve for the radius. Another important thing is on the SAT never change π to 3.14. Just keep it as π. If we know the circle has an area of 9π, then its radius must equal 3 (9π = πr^2, r = 3).

      Each length of the equi. triangle is therefore 6. Using the area for equi. formula (s^2√3)/4, where s corresponds to the side of the equi. triangle, we get 9√3.

      For number 7, think of the problem has a weighted average. If equal parts of each solution were put in a new container, then the resulting mixture would contain 20% alcohol. The resulting mixture, however, contains 25% alcohol. Therefore, more of solution Y has to be in the mix? How much more?

      Quick way to figure this is out is find the ratio between solution Y – resulting mixture (30 – 25) = 5, and resulting mixture and solution X (25 – 10) = 15. That gives us 1:3. Therefore there is three times as much solution Y as X: 3 x 200 = 600.

      Hope that makes sense!

      • Nikita Jivrajani says:

        Hey Chris,

        For number 10 the answer is 12 root 3 not 9 root 3. Yes, there is a 6 however that is the diameter of circle O. Once one has the diameter, he/she could simply use the 30-60-90 triangle as a reference and find the base of the triangle, segment CB. CB comes out to be 4 root 3. Using the formula A=(1/2)base times height : [(1/2)(4 root 3)(6)] the area comes out to be 12 root 3.

        Hope this helped =)

        • Chris Lele Chris Lele says:

          Oops, it looks like I made a very basic error there. Clearly the diameter is not the same as the length of a leg of the equilateral triangle. Morale of the story: don’t rush :).

          • Hmm says:

            So .. trigonometry is in the syllabus of SAT ?

            • Chris Lele Chris Lele says:

              Actually, it’s not. You’ll just have questions that use basic geometry. For instance, you won’t have to know how to derive the length ratio of a 30:60:90 triangle. That’s given to you at the beginning of the section.

              The new SAT, the one set to debut in 2016, will have trigonometry, or so it’s rumored :).

  13. Danah says:

    I got 7/10 correct (yay), although, my SAT scores suck (except in math) ( ;__; )

    Anyway, I don’t understand question 1. The sentence doesn’t make sense to me with those answers.

    • Chris Lele Chris Lele says:

      Good job! Your SAT scores can’t suck too much if you did that well :).

      For #1, what the question is saying is that scientists are happy with the Big Bang theory because they simply don’t have enough data–data that could indicate the theory is not valid.

      Hope that helps!

  14. Shubham Mathur says:

    got a 2380 on the SAT, these questions were really helpful, thx for posting them

  15. Sepheroth says:

    Hey man i was wondering about the 5th question
    there are no restrictions about the other 3 intgers so couldnt the other 2 numbers also be 90 and 90 as 90 is the largest integer so in retrospect only one of the numbers need to be smaller than 90
    90+90+90+x=300
    x=30…..?

    • Lucas Fink Lucas says:

      You have great instincts! The SAT could definitely trap you with something like that. But in this case, the question does specify “four different integers,” so we know there can’t be three 90s. We instead have to use 90, 89, and 88 for the three large numbers, since those are the three largest different integers allowed. Notice how one word in the question can totally change the answer—this is a perfect example of the SAT being tricky and teaches an important lesson!

  16. DS says:

    Do you know of a book that has groups of questions that are ONLY the hard ones from or like the SAT or ACT? Thx.

  17. John says:

    I didn’t understand how to solve out 6,7, and 9. Also any general tips for the sat because I’m a really hard worker and still did poorly. Thanks!

    • Nick says:

      Problem 6. We know the total area of the square to be 25. Since that’s the case we can also infer that each side is 5. If we call a side of the small square x and of the larger square y then we have x + y = 5 since they span the length of the larger square. The areas of each square are x² and y². We are also given that the area of square y is 9 times the area of square x so y² = 9x². Now we have a system we can solve pretty easily.

      x + y = 5
      y² = 9x²

      Setting each equation equal to y in terms of x we get…

      y = 5 – x
      y = sqrt(9x²) or y = ±3x

      We’re only concerned about y = 3x since the length can only be positive and y = -3x will give a negative answer when setting -3x = 5 – x.

      So now we have

      y = 5 – x
      y = 3x

      Setting the right sides equal to one another we have

      5 – x = 3x
      5 = 3x + x
      5 = 4x
      5/4 = x

      Since x is the side length of the small square the answer is 5/4.

      Problem 7. We can also solve this one with a system of equations.
      We have a solution X that is 200 milliliters and we know that is 10% alcohol so 20 milliliters of that is alcohol. We want to get another solution that we’ll call Z that is 25% alcohol when we add the solutions X and Y together so we can model this by the following equation…

      .3y + .1(200) = .25z

      The 200 is the volume of solution X which we know from the given information so we can simplify this to….

      .3y + 20 = .25z

      We need to relate the totals now. The last equation models alcohol content. We know that solution X is 200 milliliters so this added to solution Y will give solution Z so we have the following equation…

      y + 200 = z so we can solve the following system

      .3y + 20 = .25z
      y + 200 = z

      We are looking for y so it would be quicker to put z in the first equation in terms of y and solve for y. Look at the second equation. We see that z = y + 200 so we can substitute y + 200 for z in the first equation to get

      .3y + 20 = .25(y + 200)
      .3y + 20 = .25y + 50
      .3y – .25y = 50 – 20
      .05y = 30
      y = 30/.05
      y = 600 so the answer is 600 milliliters. We can plug this into the system to check too…

      600 + 200 = 800 so z = 800
      .3(600) + 20 = .25(800)
      180 + 20 = 200
      200 = 200 so we’ve also proved our solution.

      Problem 9. A computer can perform c calculations in s seconds. How many minutes will it take the computer to perform k calculations?

      If the computer does c calculations in s seconds then it does c/s calculations in 1 second and hence does 60c/s calculations in one minute so if a minute is m then k/m is the amount of calculations done per minute which is equivalent to 60c/s so we can make the equation.

      K/m = 60c/s
      Multiplying both sides by m we get

      K = (60c/s)m

      Now we can multiply both sides by s now to get

      Ks = 60cm

      And to finally isolate m we divide both sides by 60c to get.

      Ks/60c = m

      So the answer is m = ks/60c which is answer choice C.

      There won’t be too many of these problems on the SAT that are this difficult nor should they take as long as they look. I just went through them step by step so you could see how they are done.

  18. Ayush M Agrawal says:

    Got 8 of them correct. Thank you for the questions Chris. And i will give my SAT exam this June. :)

  19. Chirag Bansal says:

    Thanks and i got 7 out of them all from maths and 1 from English the first question only and i think I need major improvement in English section. I am giving my SAT exam in June this year

  20. Omar Yasser says:

    I got 7 correct, 2 wrong & left one. I mistook in questions 1 & 4, left 3.

    I got all others correct. I’m a 10th grader & going to have my first test next June.

    I hope I can get a good score at critical & writing sections, & a full mark in math’s :-)

    • Chris Lele Chris Lele says:

      Hi Omar,

      That’s a great start! With a little more prep you could do very well in all three sections.

      Keep up the good work :)

  21. yash says:

    Got 10 out of 10… Giving SAT this Saturday I hope a get a decent score :-)

  22. Lauren says:

    OMG! Chris you are so hansome. I think im in love lol

  23. JOUD says:

    Hi, please can you tell me how the question 6 algebraically solve
    thanks

    • Chris Lele Chris Lele says:

      Hi Joud,

      This is part of Nick’s comment below. Hopefully that helps :)

      Problem 6. We know the total area of the square to be 25. Since that’s the case we can also infer that each side is 5. If we call a side of the small square x and of the larger square y then we have x + y = 5 since they span the length of the larger square. The areas of each square are x² and y². We are also given that the area of square y is 9 times the area of square x so y² = 9x². Now we have a system we can solve pretty easily.
      x + y = 5
      y² = 9x²
      Setting each equation equal to y in terms of x we get…
      y = 5 – x
      y = sqrt(9x²) or y = ±3x
      We’re only concerned about y = 3x since the length can only be positive and y = -3x will give a negative answer when setting -3x = 5 – x.
      So now we have
      y = 5 – x
      y = 3x
      Setting the right sides equal to one another we have
      5 – x = 3x
      5 = 3x + x
      5 = 4x
      5/4 = x
      Since x is the side length of the small square the answer is 5/4.
      – See more at: http://magoosh.staging.wpengine.com/sat/2011/10-most-difficult-sat-questions/#comment-200951

  24. Nishit Bade says:

    I only got the 9th one wrong, but my biggest difficulty in the SAT is the time. Are there any useful tips to do the questions faster so that I don’t run out of time.

    • Chris Lele Chris Lele says:

      That’s a great question, and one that is not easy to answer. For one, not sure how you approached these questions. Secondly, I don’t want to hurt your accuracy by suggesting a strategy–esp. if you are doing okay time-wise on an Official SAT test.

      But I don’t want to leave you hanging, so here are a few tips:

      1) Assuming that you do math questions the long way (write every step out), get used to doing more math in your head (again, make sure that accuracy doesn’t suffer). Next, revisit the problem and see if there is a shorter way to get the solution.

      2) Know your grammatical errors. That way, you are scanning the sentences intelligently, looking for the mistake, instead of hopping that one just kind of “pops out” at you.

      Hopefully, that helps somewhat. If not, I’d be happy to give you some more tips (just give me some more specifics on you are currently approaching problems).

  25. Shelly says:

    Hello! This was very good practice for me! This is a really helpful website! The only question I didn’t understand was number 8; could you please help me with this?

    • Chris Lele Chris Lele says:

      Hi Shelly,

      We know that we have to find the lowest common multiple the two numbers have. That way, we can “balance” out the 9 point and 7 point questions to get ‘0’ total points. 63 is the lowest common denominator. That means 7 questions have to be 9 points (7×9 = 63) and 9 questions have to be 7 points (9×7=63). That is a total of 16 questions.

      Hope that helps :)

  26. Ny says:

    Im actually studying for the shsat and i got most of these wrong. I especially want to know how to solve number ten. Thx…

    • Chris Lele Chris Lele says:

      Hi Ny,

      Solve for the radius of the triangle to get 3. Double that to get the diameter. Notice how the diameter is the same as altitude of the equilateral triangle. Use 30:60:90 properties to reason that the small side of the triangle is 6/√3. Use (basexheight)/2 to arrive at (3×6√3)2 = (C).

      Hope that helps!

      • Sergio says:

        How can we exactly assume that the triangle is 30:60:90? I got to 3 and the altitude.. but then what property am I forgetting to know that the triangle is one of those… thanks btw

        • swagoo says:

          When you split an equilateral triangle like that in half it becomes 30-60-90.

          This is because all the angles in an equilateral triangle are 60. The line that splits it in half bisects the angle into two 30 degree angles. The line is also perpendicular to one side so there are two 90 degree angles. The other two angles are unaffected and remain 60 degrees. You thus get two triangles with degree measures of 30, 60, and 90.

  27. lina says:

    The post is soo late…
    I tried to solve #5. and if the 3 other remaining integers add up to 210 (300-90), the least possible is 1 in my calculations: 209 which is the sum of remaining 2 middle integers is impossible because one of them would have to be greater than 90 to make 209 possible.

    • Chris Lele Chris Lele says:

      So, you have to be careful with ‘1’, because, as you noted, that would give you a number larger than ’90’, and a sum larger than 300. The best way to attack is to maximize the other two integers (giving you 89 and 88), which leaves you with 33: 33 + 88 + 89 + 90 = 300.

      Hope that helps resolve any confusion :)

  28. Nikita says:

    I got 9/10 right. I just CANNOT understand problem no. 7. These mixture sums always get me :(

    • Chris Lele Chris Lele says:

      Yes, they can be quite frustrating :)

      Here’s an interesting way to work a problem: just “backsolve” using the answer choices. First off, you can figure out that the number of ml has to be greater than 200, because if you added 200 ml of Y, then that would give you a 20% solution. Since, we need a 25% solution, we know we have to have more of solution Y.

      Answer (C) 400 is a good place to start backsolving. Doing so give us:

      [(400)30 + (200)10]/ 600 = 12,000 + 2,000 = 14,000/600 = 23 something. Therefore, we have to choose a large number.

      By the way, I get 600 by adding the total ml of both solutions (400 + 200).

      Backsolving using (E), since it is an easier number to deal with than 480, I get:

      [(600)30 + (200)10]/ 800 = (18,000 + 2,000)/ 800 = 25%, the answer.

      This might seem long, but I wrote out all the calculation steps, something that only takes a few seconds on the calculator.

      Hope that helps provide a helpful approach to a mixture problem :)

    • Qdvoice says:

      If 200 ml of solution x is used that means it has 20 ml of alcohol in it because 10% of 200 ml (0.1*200). Let’s say we will use y ml for solution y then 0.3*y of alcohol will be used. Since the total volume of mixture is 200+y, then (20 + 0.3*y)/(200+y) = 0.25. Solving that equation, you will get 600 m for solution y.

  29. Someone says:

    Do you need to memorize random words and their meaning in order to pass the reading portion? cause that how i see it. P.S. not a native speaker

  30. feyikemi says:

    i got 8 i really don’t get 1 and 3

  31. Solomon says:

    your questions are very easy,,,,,,,,,,,,its just a speed test for me (!_!)

  32. James says:

    I took the SAT, and got a 1680 on it,but i got 9/10. The only one i got wrong was number 9 if you could explain it would help me a lot. Also if you know how to improve my writing section score it will help a lot. Thanks

    • Matt says:

      Just use dimensional analysis. If you start with K calculations, set it up so the units cancel.

      K calcs * (S sec/C calcs)

      Here, the calcs canceled so you’re left with KS/C seconds. Next, convert seconds to minutes.

      K calcs * (S sec/C calcs) * (1 min/60 sec)

      Now, when you multiply that you, you are left with KS/60C

  33. toffy says:

    to be honest i didn’t read through just wanted to give a comment. My SAT is 4 days away and i thought i was ready now it seems like my it on head is empty. sorry i wrote this here i did because i know the wont even put on the site. you did a great work with the questions though , will try to go through it.

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