[October SAT] Top 5 HARDEST SAT Math Questions You MUST Understand

Top 5 HARDEST SAT Math Questions You MUST Understand

50+ SAT Math Questions:

John reviews 5 of the most arduous, challenging, formidable, complex, intricate, laborious, strenuous Digital SAT Math problems ever created. Knowing how to apply problem-solving skills as demonstrated in this video will help you on any SAT test you take: August SAT Math, October SAT Math, December SAT Math, March SAT Math, May SAT Math, or June SAT Math!

0:00 Level 1 - Circles (EASIEST)
5:45 Level 2 - Word Problem
10:17 Level 3 - Algebraic Manipulation
13:46 Level 4 - Circles & Squares
17:01 Level 5 - More Circles (IMPOSSIBLE)

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It is important to know some of the most important tips, tricks, and strategies for the new Digital SAT Math section. You will get practice with the Digital SAT Math questions by testing your skills on exponential functions, exponents, lines, special quadratics, and much more.

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[PrepworksEducation]

let us know which of the adjectives in the description was your favorite 😜

[rgamer]

honestly for #1 there is a way quicker way of doing it. Equilateral traingle has 60 60 60. If u split it in half it becomes 30 60 90 triangle so ur height is 8rt3. Then u just get the area of the triangle and subtract by the sector area of 1 circle x 3. which is pretty quick. Way faster method :)

[hki_komori]

in question 2 i made an equation s=20+2(2x) since they are equal
and let seniors be y, therefore we get an equation y+2x=80, i typed those equations in desmos and found the point of intersection

[shubhamgamer]

13:48 Well the thing i like the most about SAT questions that even if u didn't know about the depth of the chapter or anything about chapter from which question is asked still by keeping the basic in your mind by using simple logic you can able to tell answer the question

[mannatkukreja]

In question number 3, we can just do area of sector= 30/360 pi.r^2 and area of square =r^2, since r is the squares side. then when we divide the two we'll have 30/360pi=pi/12

[RaghavPant-xujt]

The fifth problem is actually the easiest, since the the ratio of areas is 1/25, then ratios of circumference will be 1/5. Now think of the larger circle's circumference 5 times bigger, then the smaller circle covered 2/5 of the total perimeter. Now just multiply 2/5 to 360 as in the total angle of bigger circle which gives x as 144.

[pastpaperprocrastinator]

The last one is way easier than shown. Once you get the distance traveled by the smaller circle (4pi), you can then figure out x by doing 4pi=5x to get the angle with the larger circle's circumference. You end up with 4pi/5, which converted into degrees is 144 degrees.

[Myeongji.]

Well, l got another aspects to figure out the 5# question by 3 steps. Due to the ratio of area is 1:25 which is for 2 dimensional, 1:125 of its sphere for 3 dimensional. The ratio of arcs going to be 1:5. And l just basically type 2:5 for small ball’s 2 complete rotations, then put 360/5*2 equal to 144 which exactly the degree is.

[Abhishek-nonl]

I think that everyone here got the last question wrong.
This is because of something called coin rotation paradox. The circle not only moves in the length of the circumference but it also goes around the circle. so the angle covered by it will actually be 2-2/6 *72=10/6*72=120 .
so the answer will be 120 degrees.

[shubhamgamer]

2:12 Well i want to point out there is a formal for funding area of a equilateral triangle

"√3/4*A^2"(A= side length of equilateral triangle)



I didn't know if there will any deduction in marks for using this formula because it is not given in SAT syllabus but if a MCQ question will come then you can use this from as cheat code to find the area really fast and getting rid of finding H by using Pythagoras

[diamondage]

for the circles question it took me 30 seconds since it makes sense, I just did 30/360 and got 1/12, and just placed pi instead of 1 and boom that's the answer

[dnnstuff]

For 4 I just made the equation (r^2*pi*(1/12))/r^2

Turns out that there is no need to find the radius since we are going to divide it out anyways.

[mia]

i’m just beginning my journey of preparation for the SAT and clicking on this video i was very scared, telling myself that this is the absolute worst that can happen. however having stopped the video before the solution for each question and doing it by myself i am now so relieved because these questions were so easy that i thought it was a mean joke lol. now i’m not worried at all 😂

[terrariariley]

I instantly knew what to do for the circle one and stuff but instead of wasting time on the right triangle i used herons formula and since you get a calculator this easy

[mujtabaalam]

Which test is number 5 from? Also, if the two circles were of equal size, would the answer be 360 degrees or 720 degrees?

[FaiajFida]

I would like to point out that the correct answer for problem V is not 144 degrees but 130.9 degrees due to the smaller circle travelling around a slightly bigger radius which is 5.5pi. Veritasium made a video about this problem you can check out if you want.

[vortexfitness]

somehow I found #1 to be the toughest because I misinterpreted the base and heigh to both be 16 lol
#2 was confusing but #5 I got 144 by literally using r=1 and r=5 for the two circles and finding the circumference of each by pi r^2. Then I set up a ratio of two times the circumference of circle A/ circumference of circle B= x/360 to find x=144. It was tedious but it's doable

[kaidarizz]

I think last one can be solved just a little faster. So as we know the ratio of areas A and B is 1/25, which mans that their circumference ratio is 1/5. The length of the arc(of the bigger circle) is (x/360)2pi r, and is equal to 2 circumferences of circle A( 2 x 2pi r"2"). However, as I previously mentioned the ratio is 1/5 meaning that (2 x 2pi r)/5 will be the circumference of A in terms of circumference B. Therefore, (x/360)2 pi r = (4pi r)/5, and from there we will find out that the answer is 144

[andrewcuber]

for question 4 theres honestly no reason to find the area, cuz it says to find the fraction. so the shaded area is (1/12)pi(r^2), cuz its 30/360 is 1/12 (so that shaded area is 1/12 of the area the circle is). the area of the square is r^2. so the fraction we are trying to find is ((1/12)pi(r^2))/(r^2). r^2 cancels on the top and bottom, so we are left with (1/12)pi

[SEWEYFOOTBAL]

1 is the hardest and everything else is pretty hard imo