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 :)
That does speed up the process of finding the area of the triangle, yes, but, since equilateral triangles are so common and you’re already willing to memorize triangle formulas, knowing the formula for an equilateral triangle would be fastest. But if you really want the fast way for this problem, look at the answer choices. Only one had positive area :(. Would’ve been cooler if they made it one of the typed ones (though I doubt anyone would agree with me on that lmao)
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
Good use of desmos. It's funny that you just can use desmos to easily solve all algebraic problems (as long as you comprehend the subject at its bare minimum level)
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
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
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
There's no syllabus for the SAT, so I'm not sure what you're referring to. You also don't need to show work, so as long as your answer is right, you're 100% correct in the eyes of CB.
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.
Hey, we've gone over the coin rotation paradox in this comment section already. It doesn't apply here because a) we're not completing a full set of rotations around the entire larger circle and b) the radii are different lengths.
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.
I did the last one in less than 1 minute (mentally + calculator). Make and assumption of the areas as long as it keeps the ratio, in my case, 4(pi) and 100(pi) we know that A = (pi)r^2 so r1= 2 & r2= 10 so lets get the circumference 2(pi)r 1st circle = 4(pi) 2nd circle = 20(pi) the circle rolls twice so the arc is 8(pi). 20(pi) = 100% 8(pi) = 40% lets convert the degrees 360 = 100% and X degrees = 40% which is 144.
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 😂
You might want to make sure you also have a good grasp on quadratics. The Digital SAT test makers seem to really like these near the end of module 2 math
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
Honestly for #5 there is a simpler method, you know there can be 5 rotation and you know there is 360 degrees in a circle so divide 360 by 5 and you get 72. Now you can estimate there are 2 rotations from start to stop so multiply 72 by 2.
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
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.
Is this about the rolling coin paradox? The problem explicitly states the number of rotations thereby bypassing this paradox. It’s not testing whether you understand how rotations are counted…
@@PrepworksEducation I think he/she is right because if the circle completes two rotations over that arc, that means the length of the arc is not twice its circumference, but it's equal to it number of rotations = ratio of circumstances + 1 And I think that x should be Acute and equals 1/5 of 360 which is 72°
No, that only applies if it goes all the way around. The Veritasium video question asks how many revolutions the circle took, where this gives you (essentially) the number of revolutions. You can try it out for yourself at home.
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
Very true! We just wanted to give concrete examples instead of putting everything in variables. On the SAT itself, it might take more time to think about how you could do it with variables than just doing it with numbers (brute force).
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
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
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.
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.
For the second solution can seniors not also be equal to 50? Because then there are 15 juniors and 15 sophmores, and 20 freshman. A total of 100 students 20% are freshman. Does 50 seniors not meet all the rquirements of the problem?
Unfortunately, no. The number of seniors must be 20 more than TWICE the number of sophomores & juniors added together. 2(15 + 15) + 20 = 80, which can't be possible because then there would be no seniors at all.
@@PrepworksEducation I think the equation in your response above should've been 2(15 + 15) + 50 = 80, or 2(15 + 15) + 20 = 50. 50 does work, since 15 +15 = 30, and 20 more than 30 is 50. 15 soph, 15 jun, 50 seniors = 80, also = 80%. the remaining group is freshman, which has been given as 20, or 20%.
@@PrepworksEducation In your answer in the video, you said there are 60 seniors, but then your equation (20 + 2(x+x)) doesn't check out because then Sophomores have to equal to 20, and Juniors have to equal to 20 for sen=60.
they are pretty easy except question 2 because i could not understand it properly (English is not my native language). Sadly, i was forced to take an ielts test not the SAT🙂
in qn 2, how tf did u suppose 40=20 + 2soph and also, seniors is actually = 20+2*(2 soph) because sen =20+2(soph+jun) and jun = soph in your video , you got soph =10 but when you plug that value in the eqn of senior, you will get seth like =20+2 *(10+10) =60 ? and you supposed senior =40 ? tf ? and after that you said senior =60, that's hella confusing and maybe not right
Hey, I did an in-between step you might've missed. The question states that the number of seniors is 20 more than twice the number of sophomores and juniors. We also know that the number of freshmen is 20 and 20%; thus, soph + jun + sen = 80 because each person is 1%. We know that seniors - 20 = 2(soph + juniors) from the problem. Both sides of this equation should be equal; since there are 2 sides to the equation, each side must be 40. I simply added the 20 to both sides in advance. Here, you can take the left half of the equation, which again we know to be equal to 40, and do seniors - 20 = 40 and add 20 to both sides to get 60. If you saw the check in the video, you'll know that this is the correct answer. If this was confusing, you may want to consider a simpler approach like this: 20 freshmen = 20%, so 80% = 80 seniors + juniors + fresh 80 = 2(x+x) + 20 + 2x 80 = 4x + 20 + 2x 80 = 6x + 20 60 = 6x x = 10 soph + junior = 10 + 10 = 20 80-20 = 60
Hey! Allow me to help you. 40 = 20 + 2 x soph, I suppose he resulted in such calculation because within the question it is stated that the number of juniors is equal to the number of sophomores. Considering that the numbers of seniors are 20 more than the sum of juniors and sophomores (which is the same, basically, because the numbers of juniors = the numbers of sophomores). I did the calculation like this: let x = the numbers of juniors y = the numbers of sophomores z = the numbes of seniors by the question, it is known that x = y. Thus, z = 20 + 2(x+y) = 20 + 2(x+x) = 20 + 2 (y+y) -- because x is y and y is x z = 20 + 2 (2x) = 20 + 4x. z = 20 + 4x. Now, given the fact that the number of freshman are 20, and 80% of the club members are constituted of seniors, sophomores, and juniors, we can understand that 20% of the club members are freshman, and that very 20% of the club is 20 (because the numbers of freshman are 20). Those calculations imply that if 20 freshman make up 20% of the club, the club members in total must be 100. So the rest of the members (100-20) is the total numbes of juniors, sophomores, and seniors. Given that we have written the equation, and we know that the total numbers of numbers of juniors, sophomores, and seniors are 80, that means: x + y +z = 80 x + x + 20 + 4x = 80 6x + 20 = 80 6x = 60 x = 10. Because x is 10, y must also be 10, which in total results that the number of sophomores and juniors altogether are 20. Those calculations boil down to the numbers of seniors, which basically is just 80 - (numbers of juniors and sophomores) = 80 - 20 = 60. We can also check it using our z equation: z = 20 + 4x z = 20 + 4(10) z = 60. So the numbers of seniors are 60 people. Hope this helps
We multiply 2pi by 2 because the small circle rolls around the larger one two times. If I were to mark a point on the edge of the small circle, we'd see that as we roll it, that point would travel 2pi twice (two circumferences of the circle). Does that help?
in the 2nd question, how did you just assume that the number of seniors is 40 and then solve the equation to get 60? Also the equation that you wrote which is "seniors = 20 + 2(soph)" should have been "seniors = 20 +2(2soph) = 20 + 4(soph)" (timestamp 8:57)
@@shivamkole2679 Here: “Hey, I did an in-between step you might've missed. The question states that the number of seniors is 20 more than twice the number of sophomores and juniors. We also know that the number of freshmen is 20 and 20%; thus, soph + jun + sen = 80 because each person is 1%. We know that seniors - 20 = 2(soph + juniors) from the problem. Both sides of this equation should be equal; since there are 2 sides to the equation, each side must be 40. I simply added the 20 to both sides in advance. Here, you can take the left half of the equation, which again we know to be equal to 40, and do seniors - 20 = 40 and add 20 to both sides to get 60. If you saw the check in the video, you'll know that this is the correct answer. If this was confusing, you may want to consider a simpler approach like this: 20 freshmen = 20%, so 80% = 80 seniors + juniors + fresh 80 = 2(x+x) + 20 + 2x 80 = 4x + 20 + 2x 80 = 6x + 20 60 = 6x x = 10 soph + junior = 10 + 10 = 20 80-20 = 60”
for question 5, i had no idea what to do going in but i think some sort of weird coincidence happened maybe where i accidentally solved it. i figured the scale factor from a to b was 1 to 5, and i also noticed that circle a went around twice which meant it spun a total of 720°. so then i completely guessed that this might've meant that 720 divided by the scale factor of 5 (144) would equal x° in circle B. it was a complete guess, but it yielded the right answer. this was just a coincidence though, right? or was there actually some direction to it? i'd appreciate any input!
Sounds like this method would work! I'm going to translate into radians to show my logic since we can then work in a discussion of circumference: Circle spins 4pi times (i.e. 4pi*r of circumference). Divide this by the circumference of the larger object (5pi*r) -- scale factor of 5 is factored in as we would have 4pir/5pir, and the pi*r cancels out. So, yeah, that should work :)
@@PrepworksEducation Hey I also feel like I got the last question right coincidentally! First I found out that the circumference of the bigger circle was 5 times the circumfrence of the smaller circle which meant that the smaller circle completed the 360 degrees of the bigger circle by going through 5 equal angles or 360/5 which got me the value 72 degrees for each part. I also figured out the scale factor and realised that x was twice of 72 and got x=144 degrees.
How many problems are there in the sat and how much time do students have? Because these problems seem very easy for the "hardest ever" and unless you have like 1min per question all of the students should get very high results
Students receive about 1.5 minutes per question. Most of the math on the SAT is straightforward, you do not even need to have learned Algebra 2. However, it is critical to solve the last couple Module 2 questions which emulate the ones seen in the videos in order to score 750+
You’re fine! These typically would only show up at the end of Module 2. You could miss these and still end up with a 750-770+ on math granted you took care of business on the rest of the exam.
Thank you! And yes, it seems that a lot of international students find the SAT pretty easy. You can definitely tackle the math, but you may want to study up for the English section.
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@@PrepworksEducation the First question is an equilateral triangle why cant we use s^2rt 3/4?
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 :)
That is true! Using special triangles is a great hack.
You are amazing
Thanks!!
Why 8 rt 3?
That does speed up the process of finding the area of the triangle, yes, but, since equilateral triangles are so common and you’re already willing to memorize triangle formulas, knowing the formula for an equilateral triangle would be fastest.
But if you really want the fast way for this problem, look at the answer choices. Only one had positive area :(. Would’ve been cooler if they made it one of the typed ones (though I doubt anyone would agree with me on that lmao)
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
Great way to solve!
thats smart
Good use of desmos. It's funny that you just can use desmos to easily solve all algebraic problems (as long as you comprehend the subject at its bare minimum level)
Very true! Desmos makes everything quicker.
this video is great ive been looking for these types of actually hard problems
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
Exactly! SAT Math is also a test of your problem-solving abilities.
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
Yup! This works too! Sometimes, on the SAT, it's easier just to multiply stuff out (especially if it's like 3*3) so you don't get confused.
Thanks! I thought I was doing some unnecessary math when I kept seeing the same numbers
Yup, we get that. Unfortunately, the time constraints make it easy to second-guess yourself.
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
There's no syllabus for the SAT, so I'm not sure what you're referring to. You also don't need to show work, so as long as your answer is right, you're 100% correct in the eyes of CB.
👍😀@@PrepworksEducation
Of course!
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.
Hey, we've gone over the coin rotation paradox in this comment section already. It doesn't apply here because a) we're not completing a full set of rotations around the entire larger circle and b) the radii are different lengths.
Man I did the same after watching the paradox and I got 120 degrees but why is the answer different here?
@@PrepworksEducationCoin rotation paradox is applied for different radii too
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.
I did the last one in less than 1 minute (mentally + calculator). Make and assumption of the areas as long as it keeps the ratio, in my case, 4(pi) and 100(pi) we know that A = (pi)r^2 so r1= 2 & r2= 10 so lets get the circumference 2(pi)r 1st circle = 4(pi) 2nd circle = 20(pi) the circle rolls twice so the arc is 8(pi). 20(pi) = 100% 8(pi) = 40% lets convert the degrees 360 = 100% and X degrees = 40% which is 144.
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 😂
Haha, then you are perfectly fine for the SAT Math section!!
You might want to make sure you also have a good grasp on quadratics. The Digital SAT test makers seem to really like these near the end of module 2 math
@@Jake-cc7ie Very true! They've trended towards that lately.
This comment shows how dumb i am lol.
You’re not dumb! Maybe you just need a little more practice :)
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.
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
Honestly for #5 there is a simpler method, you know there can be 5 rotation and you know there is 360 degrees in a circle so divide 360 by 5 and you get 72. Now you can estimate there are 2 rotations from start to stop so multiply 72 by 2.
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
Great method!
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.
Is this about the rolling coin paradox?
The problem explicitly states the number of rotations thereby bypassing this paradox. It’s not testing whether you understand how rotations are counted…
@@PrepworksEducation I think he/she is right because if the circle completes two rotations over that arc, that means the length of the arc is not twice its circumference, but it's equal to it
number of rotations = ratio of circumstances + 1
And I think that x should be Acute and equals 1/5 of 360 which is 72°
No, that only applies if it goes all the way around. The Veritasium video question asks how many revolutions the circle took, where this gives you (essentially) the number of revolutions. You can try it out for yourself at home.
Which test is number 5 from? Also, if the two circles were of equal size, would the answer be 360 degrees or 720 degrees?
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
Very true! We just wanted to give concrete examples instead of putting everything in variables. On the SAT itself, it might take more time to think about how you could do it with variables than just doing it with numbers (brute force).
@@PrepworksEducation thanks!
Of course!
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
Sure, you could solve it that way! Whatever works for you -- always remember that SAT questions have multiple ways to solve in almost every scenario.
how did u know circ ratio was 1/5?
@@s0phadays it was given that the area ratio is 1/25, which is the ratio of their radiuses(circumference) squared (this is area ratio rule/formula)
Yup, exactly!
for number Q number 2- 2(2x)+20+2x=80, where x is the number of sophomores or juniors.
this looks easier, why make it complicated
Time saving!
yep, just what i did
Sounds good! Remember, there are multiple ways to solve nearly all SAT problems.
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
That's the story of a lot of SAT math -- not enough time, dumb mistakes, busywork steps...
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.
yea, its the only one i got right
Question 2, how did you know they bith going to be 40 can you please explain
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.
That works!
As a indian I had no issues solving these
Great job!
Got them all right
Nice!!
For the second solution can seniors not also be equal to 50? Because then there are 15 juniors and 15 sophmores, and 20 freshman. A total of 100 students 20% are freshman. Does 50 seniors not meet all the rquirements of the problem?
Unfortunately, no. The number of seniors must be 20 more than TWICE the number of sophomores & juniors added together. 2(15 + 15) + 20 = 80, which can't be possible because then there would be no seniors at all.
@@PrepworksEducation oh okay, thanks for the clarification!
@@navkiransandhu3323 No problem! Be sure to check out our other SAT videos.
@@PrepworksEducation I think the equation in your response above should've been 2(15 + 15) + 50 = 80, or 2(15 + 15) + 20 = 50. 50 does work, since 15 +15 = 30, and 20 more than 30 is 50. 15 soph, 15 jun, 50 seniors = 80, also = 80%. the remaining group is freshman, which has been given as 20, or 20%.
@@PrepworksEducation In your answer in the video, you said there are 60 seniors, but then your equation (20 + 2(x+x)) doesn't check out because then Sophomores have to equal to 20, and Juniors have to equal to 20 for sen=60.
in q4 i dont think we even need to find radius. as it will cancel out in the end and we will get the answer without needing it.
Yup, as I explain in the video, it cancels out.
they are pretty easy except question 2 because i could not understand it properly (English is not my native language). Sadly, i was forced to take an ielts test not the SAT🙂
Makes sense. A lot of SAT Math questions are more reading comprehension than actual mathematics.
Bro I got 1080 on my practice test and I recently watch ur both vid can u give me some tips to improve my sat score before exam
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in qn 2, how tf did u suppose 40=20 + 2soph and also,
seniors is actually = 20+2*(2 soph) because sen =20+2(soph+jun) and jun = soph
in your video , you got soph =10 but
when you plug that value in the eqn of senior, you will get seth like
=20+2 *(10+10)
=60 ? and you supposed senior =40 ? tf ? and after that you said senior =60, that's hella confusing and maybe not right
Hey, I did an in-between step you might've missed.
The question states that the number of seniors is 20 more than twice the number of sophomores and juniors. We also know that the number of freshmen is 20 and 20%; thus, soph + jun + sen = 80 because each person is 1%.
We know that seniors - 20 = 2(soph + juniors) from the problem. Both sides of this equation should be equal; since there are 2 sides to the equation, each side must be 40. I simply added the 20 to both sides in advance.
Here, you can take the left half of the equation, which again we know to be equal to 40, and do seniors - 20 = 40 and add 20 to both sides to get 60. If you saw the check in the video, you'll know that this is the correct answer.
If this was confusing, you may want to consider a simpler approach like this:
20 freshmen = 20%, so 80% = 80 seniors + juniors + fresh
80 = 2(x+x) + 20 + 2x
80 = 4x + 20 + 2x
80 = 6x + 20
60 = 6x
x = 10
soph + junior = 10 + 10 = 20
80-20 = 60
@@PrepworksEducation yea i myself did it this way but that senior =40 was a bit confusing. Anyways, great video
Hey! Allow me to help you. 40 = 20 + 2 x soph, I suppose he resulted in such calculation because within the question it is stated that the number of juniors is equal to the number of sophomores. Considering that the numbers of seniors are 20 more than the sum of juniors and sophomores (which is the same, basically, because the numbers of juniors = the numbers of sophomores).
I did the calculation like this:
let x = the numbers of juniors
y = the numbers of sophomores
z = the numbes of seniors
by the question, it is known that x = y. Thus,
z = 20 + 2(x+y) = 20 + 2(x+x) = 20 + 2 (y+y) -- because x is y and y is x
z = 20 + 2 (2x) = 20 + 4x.
z = 20 + 4x.
Now, given the fact that the number of freshman are 20, and 80% of the club members are constituted of seniors, sophomores, and juniors, we can understand that 20% of the club members are freshman, and that very 20% of the club is 20 (because the numbers of freshman are 20).
Those calculations imply that if 20 freshman make up 20% of the club, the club members in total must be 100. So the rest of the members (100-20) is the total numbes of juniors, sophomores, and seniors. Given that we have written the equation, and we know that the total numbers of numbers of juniors, sophomores, and seniors are 80, that means:
x + y +z = 80
x + x + 20 + 4x = 80
6x + 20 = 80
6x = 60
x = 10.
Because x is 10, y must also be 10, which in total results that the number of sophomores and juniors altogether are 20. Those calculations boil down to the numbers of seniors, which basically is just 80 - (numbers of juniors and sophomores) = 80 - 20 = 60.
We can also check it using our z equation:
z = 20 + 4x
z = 20 + 4(10)
z = 60.
So the numbers of seniors are 60 people.
Hope this helps
@@withIn40 Great explanation!
@@quillerity07 Thanks! 😁😁
For question 5 I didn’t get what multiplying 2pie by 2 does the question generally confused me could you explain it again? Please.
We multiply 2pi by 2 because the small circle rolls around the larger one two times. If I were to mark a point on the edge of the small circle, we'd see that as we roll it, that point would travel 2pi twice (two circumferences of the circle). Does that help?
1 is the hardest and everything else is pretty hard imo
It's all subjective!
in the 2nd question, how did you just assume that the number of seniors is 40 and then solve the equation to get 60? Also the equation that you wrote which is "seniors = 20 + 2(soph)" should have been "seniors = 20 +2(2soph) = 20 + 4(soph)" (timestamp 8:57)
Yeah, I took a couple of shortcuts. I explained my reasoning in another comment somewhere here… lmk if you can’t find it and I can repost.
@@PrepworksEducation I can't find the comment where you've explained the reasoning. Could you please repost it on this comment. Thanks
@@shivamkole2679 Here:
“Hey, I did an in-between step you might've missed.
The question states that the number of seniors is 20 more than twice the number of sophomores and juniors. We also know that the number of freshmen is 20 and 20%; thus, soph + jun + sen = 80 because each person is 1%.
We know that seniors - 20 = 2(soph + juniors) from the problem. Both sides of this equation should be equal; since there are 2 sides to the equation, each side must be 40. I simply added the 20 to both sides in advance.
Here, you can take the left half of the equation, which again we know to be equal to 40, and do seniors - 20 = 40 and add 20 to both sides to get 60. If you saw the check in the video, you'll know that this is the correct answer.
If this was confusing, you may want to consider a simpler approach like this:
20 freshmen = 20%, so 80% = 80 seniors + juniors + fresh
80 = 2(x+x) + 20 + 2x
80 = 4x + 20 + 2x
80 = 6x + 20
60 = 6x
x = 10
soph + junior = 10 + 10 = 20
80-20 = 60”
thanks for the confidence boost but these are def not the hardest sat math problems lol
Best of luck on your SAT! :)
Ok questions were easier than I thought
Awesome! You will be set for the SAT Math section!
for question 5, i had no idea what to do going in but i think some sort of weird coincidence happened maybe where i accidentally solved it. i figured the scale factor from a to b was 1 to 5, and i also noticed that circle a went around twice which meant it spun a total of 720°. so then i completely guessed that this might've meant that 720 divided by the scale factor of 5 (144) would equal x° in circle B. it was a complete guess, but it yielded the right answer. this was just a coincidence though, right? or was there actually some direction to it? i'd appreciate any input!
Sounds like this method would work! I'm going to translate into radians to show my logic since we can then work in a discussion of circumference:
Circle spins 4pi times (i.e. 4pi*r of circumference). Divide this by the circumference of the larger object (5pi*r) -- scale factor of 5 is factored in as we would have 4pir/5pir, and the pi*r cancels out. So, yeah, that should work :)
@@PrepworksEducation Hey I also feel like I got the last question right coincidentally! First I found out that the circumference of the bigger circle was 5 times the circumfrence of the smaller circle which meant that the smaller circle completed the 360 degrees of the bigger circle by going through 5 equal angles or 360/5 which got me the value 72 degrees for each part. I also figured out the scale factor and realised that x was twice of 72 and got x=144 degrees.
Nice video
Appreciate it!! Hope this video helped you prep for the SAT!
@@PrepworksEducation Thanks!
Good luck!
All solved without under 5min
Wonderful work!
Thank you : )
Of course!
How many problems are there in the sat and how much time do students have? Because these problems seem very easy for the "hardest ever" and unless you have like 1min per question all of the students should get very high results
Especially if desmos is allowed which makes half of the questions just plugging stuff in there and righting down the answer
Students receive about 1.5 minutes per question. Most of the math on the SAT is straightforward, you do not even need to have learned Algebra 2. However, it is critical to solve the last couple Module 2 questions which emulate the ones seen in the videos in order to score 750+
Are there 22 questions in each maths module or 27??
As for english is it also 66 in total or less?
Hey, there are 27 Qs for each English module. 22 Qs for each math module.
@@PrepworksEducation ok many tnxxx 👍
Are these questions from pt5, 6
No!
Although I am from different country, but these were pretty basic qsns.. I wish I could give this exam 😢😂
Haha, there are very stark differences between education systems… 🤷♂️🤗
I bet class 10 CBSE students can easily solve first three and any class 11 JEE aspirant can easily solve ANY of these questions
There are differences in mathematical abilities across regions!
I think I'm cooked
You’re fine! These typically would only show up at the end of Module 2. You could miss these and still end up with a 750-770+ on math granted you took care of business on the rest of the exam.
uhmmm i can solve these with like 13 year old me's math level from india(all under 30 secs....)
(loved the video) honestly, after watching your video i might consider giving the sats in 2-3 years when i'm of age, they seem pretty easy
Thank you! And yes, it seems that a lot of international students find the SAT pretty easy. You can definitely tackle the math, but you may want to study up for the English section.
This is hard? Is this a joke or.
If this was easy for you, the SAT Math section should be a breeze! Check out our Full English Review guide.
@@PrepworksEducation I got 1580 in my SAT but thanks
Awesome!
All of em were easy asf bruh
Great to hear! You will be ready for the upcoming SATs!! :D
@@PrepworksEducation im not giving SAT, way too expensive in my country
@@jxtk3154 Indian?
@@Yeah_its_desi half
How much does it cost?