5 Most Difficult SAT Math Questions with Answers (2024)
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- Опубліковано 31 гру 2024
- Let's take a look at some of the trickiest math questions from SAT practice tests of 2024! If you can complete all these questions, you should be able to easily achieve a 770 or above!
For the k problem at 6:37 the solutions are -11/5 from the first factor, -sqrt(5k+10) from the second, and sqrt(5k+10) from the third. Then just use desmos regression to solve for k:
(-11/5) * (-sqrt(5k+10)) * (sqrt(5k+10)) ~ 154
PARAMETERS
k=12
Done
I didn’t get your point can you please explain it a bit more
could you make an in depht video on when we can apply regression?
@@user-vk3od9cp6n Rather than solving the equation by hand, you can use Desmos to solve the equation for you, i.e. once you get to the step where you have the product of the roots equal to 154, (-11/5)*(-sqrt(5k+10)(sqrt(5k+10) = 154, you can solve that in Desmos using regression. Or you can just type the equation (replace k with x) and then look for the vertical line at x= 12.
u could just set k as x and desmos will show ur x as 12 and x is k
Great work man, appreciate you hardwork. Please continue this series!!!
There is another easier way to get the answer of question 4:
Divide through by 3 to reduce the equation
18x4 + 73x2 + 35
This way when you equate x2 = A
it becomes: 18A2 + 73A + 35
Rather than using the formula method which seemed pretty much, you can factorize:
18A2 + 10A + 63A + 35
=> 2A(9A + 5) +7(9A + 5)
=> (2A + 7)(9A +5)
Is not the fifth question‘s answer 5/9?😶🌫You know 54(x^2+5/9)(x^2+7/2)
Thank you for your video! Some of the questions in the video actually came out in the real exam yesterday! I hope I did well. Thank you so much for making this video!
I think you can compare the numbers in the question no 2 like:
when you rearrange the first eqn y=7x/9 - 13/3 then just plug the value of the constant for eg in the first option 0 (3r+"0") then check for the same in y coordinate and you can see -13/3
for question two if you want to do less math you could just plug in the x and solve for the y, and if it matches up with the y in the solution you would have found your answer
In the first question, how do we know what is x? For example, when we're looking at x>k or x
x is referring to somewhere on our x-axis. For example, if we are saying x>k then we are looking at the part of the function when the values on the x-axis are greater than x=k
@@DrewWerbowski Ah, I see. Thank you!
@@DrewWerbowski but K is a y value. y are we comparing an x value to y value
@@Coming-Upppp k is a number corresponding to the y-value of the second POI. If we are looking at x>k, we are looking at values of x greater than when x=k. We can look at this, but it doesn’t make much sense for this question.
@@Coming-Upppp
You are making a valid argument. a, b are x coordinates, whereas j, k are y coordinates. As far as the x axis (or x values) is concerned, j and k are not defined or have no meaning. You can make the argument that choices A and B are incorrect for that reason. But you only need to worry if one of the answer choices is j
in the 3 question you could find the zeros just by the original form and made it equal to 154
I realized after recording this, thanks for pointing out
what do you mean the zeros? can u explain pls
@@maryamsowah2199 Zeros means the x-intercepts or roots of the function. Just another name for them!
@@maryamsowah2199 If the whole equation is equal to zero, then at least one of the 3 factors containing x must equal zero.
So you got 3 solutions:
5x + 11 = 0
x + √(5k + 10) = 0
x - √(5k + 10) = 0
@@Grizzly01-vr4pnman i am so behind i dont understand thiss
Bro I am gonna panic if any of these show up on the sat
No jokes fr
whens ur test? mines on this august lol
@ceoofbeingstoopid8490 same I wasn't scared before but I've been seeing vids saying it's harder than the practice test and now I'm scared🫠
Damn you guys are unlucky 😭 I’m a sophomore so im only doing october psat
Nah cause fr
Very helpful, thanks! Amazing job.
So useful, thabk you so much!!
thank you so much!!!!
where can i find more hard provlems like this
hi drew, studying for the sat and stumbled across your vid. it's been helpful but i wanted clarification on for the first question: how do you know x>k is wrong if you dont know k is?
We know if there are 2 POI’s then the graphs have to look something like the sketch we drew. The exponential function will eventually grow faster than the linear one after the second POI, so when x>k we have g(x) < f(x) instead
Graph it. You can see it. You can graph equations on Desmos - it's provided on Bluebook.
for 12:13, how did you know to assume that triangle BAC was a right triangle? was it just a visual assumption or is there a ratio/rule that you used to determine that?
It's not an assumption. Although he didn't mention it, he invoked Thales's theorem, which is a special case of the inscribed angle theorem: 'If X, Y, and Z are distinct points on a circle where the line XZ is a diameter, the angle ∠XYZ is a right angle.'
Because its a semicircle, its angle at the circumference is 90 degrees
3rd question was the hardest on here imo
9:47
Can I consider x^2-5k-10 as a qudratic function and find its roots instead of equating x^2 to zero? I've done it this way but cannot get the correct answer for some reason, is there any explanation why my method does not work? Thank you for a great video!
no, for x^2 -5k -10 to be considered a quadratic function 5k should act as a coefficient to x. it would have to be x^2 - 5kx -10. think of (-5k -10) as c in the standard quadratic equation, as it has no variable and, therefore, is a constant.
btw these questions are from preppros 150 hardest dsat questions book i think you should mention him just in case
Thank you PrepPros! I pulled these from Reddit.
thanks man
where did the 2A+7 and 9A +5 come for the last question?
Quadratic equation gives us A=-7/2. Rearrange to get 2A+7=0 for our root in factored form. Same for the other root.
Question for the last one: are the roots of the equation in the quadratic not effected by K? meaning when you did the quadratic formula, you disregarded K as if it wouldnt affect the entire equations factored form. isnt this a problem?
The roots of the function will remain the same regardless of the value of k. If you wanted to find k for this question, you could factor out 3 from the original function, then find the roots. You can verify that the roots will remain the same.
@@DrewWerbowski
It is a good habit to factor out k before applying the quadratic formula. You will be working with smaller numbers. And you may be able just to factor by hand.
@@OverclockingCowboyThis is a great point!
Hi! for the last one, why are both factors ax^2 +b? because one of the other ones is cx^2+d. Is it because of the uncertainty of how we should order the factors?
This was very useful btw, I gave myself a pat on the back when I knew I could solve the first three correctly without any problems, your idea with using theta for the circle problem actually gave me a lot less confusion from the way I'd normally solve it (by marking which sides looked equal lmao) thank you!
because both factors consist of (coefficient)*x^2 + (constant), it’s not clear which factor, either (2x^2+7) or (9x^2+5), represents (ax^2+b). because of this, we have to multiply the coefficient and constant from each factor to see which is smallest, and therefore which product could be the smallest possible value for ab. sorry if that is hard to understand or explained poorly, i’m not to good at articulating these kind of things 😂
@@LukieReal why are we looking for the smallest possible value for ab?
@@LukieReal why are we looking for the smallest possible value for ab?
@@noname-ed2un
That is what the question is asking for.
@@noname-ed2unthat is what is being asked. we know that one of the factors is (ax^2 + b) but either factor could be in this form because it is not stated. so if we try multiplying each factor’s coefficient (a) and constant (b) and seeing which is the smallest, that will represent the smallest value that ab could be.
for the triangle question how did you find the adjacent and hypotonuse especially since it isnt really clear for the entire triangle ABC. Also thanks Drew for the walkthrough for the questions
We see a right angle triangle with one angle as theta, so we can identify the adjacent, opposite, and hypotenuse.
How much would i get if i got 3 difficult questions wrong.
In the third question, why did you reciprocate 11/5 to 5/11? thanks for the video btw.
Multiplying both sides by 5 then dividing both sides by 11. We are trying to isolate for k
question for the first one, where did 0,1 and 1,3 came from?
(x,y) = (x, 3^x). At x=0, 3^0=1 and at x=1, 3^1=3
at 20:10 why are you allowed to just add a k wouldn't it alter the value of the equation to get the K wouldn't you need to factor out something from the two solutions
My notation was a bit sloppy there. There’s no need to add k. Just showing how by finding the roots we can express the function in the factored form shown in the question
I dont understand the first question solution. If we are comparing values of x and J, then why when x is negative and J is positive, why are saying that x>J?
For the last one, are you able to factor our 3 for K, since its the GCD of 219, 105, and 54, and then substitute the A=x^2?
Yes, you can do this too
For the first one, how do you know that B is wrong?
j corresponds to a y-value on our graph so the region x
All i know is you're goated
I used demos for the last one and it gave me -7/2 and -139/250 as roots. so obviously 139 times 250 is going to be bigger but should I worry about desmos not rounding up and giving me bigger numbers instead of the ones I should actually use?
Might be an input error. Desmos should be accurate for a quadratic like this. Try to solve by hand and use desmos as a tool to check your work.
how did you use desmos to find the roots?
Desmos is giving you the correct answer for your second root. That is equal to -5/9. Desmos rounded up on the 3rd digit giving you 0.556. 5/9 equals 0.555 with repeating 5. You need to find a way to deal with that when using Desmos.
On the last one why are you allowed to just make a be x^2? What would change if you made b It instead
We are making a substitution for a new variable of our choice. As long as we define our variable then we can make our equation in terms of the new variable (A, b, c, theta, basically anything) because they are equivalent
is there any ways to solve these using desmos?
yes, most of these can be made a lot easier with desmos
@13:12 how did you know A is a right angle
BC passes through the centre of the circle and has a central angle of 180 so angle BAC is half of that
@@DrewWerbowski that makes sense thanks
Is not the fifth question‘s answer 5/9?😶🌫You know 54(x^2+5/9)(x^2+7/2)
Not quite.
(x² + 5/9)(x² + 7/2) = 18x⁴ + 73x² + 35 so your 54 in the above expression should actually be a 3 (= k).
The question states that a, b, c, d and k are integers, so you'd need to clear the denominators: x² + 5/9 = 9x² + 5 and x² + 7/2 = 2x² + 7
how did you cancel out the negatives & the roots at 11 mins? (15 yr old btw)
what does root 2 * root 2 equal? it equals to 2. When you multiply the same root by itself, it just equals the inside of the root. A negative times a negative = a positive which is why the negatives cancel out.
@@chrisvfx3737 ohhrighhttttt.. thank youuu:)
There is another more difficult way to get the answer using sine rule. Question 3
132/sin 90 = sqrt(363)/sinx (x here is angle BCA)
This gives that angle BCA = 8.3
Hence angle CBA = 180-(90+8.3) = 81.7
To get side AC = Sin 90/133 = sin 81.7/AC
AC = 130.62
To get side DC
cos 8.3 = DC/130.62
Hence DC = 129.25
Therefore, BD = BC - DC = 130-129.25 = 2.75
Now back to the question, BC/BD = 132/2.75 = 48
Now, this method might be lengthy but it uses simple & common knowledge of trigonometry and geometry.
Your approach works, but you have to do quite a bit of calculations. If this is an exam question, I think you are better off working with corresponding sides of similar triangles ABC and ABD.
You calculated angle ACB as your first step. Angle BAD equals angle ACB. You can directly calculate BD on your second step. Not too bad, still using Sine Rule.
Using Sine Rule only:
AB / sin C = BC
BD / sin C = AB
Thus,
AB / BC = BD / AB
Same equation as the one in the video. Isn’t that strange?
@@OverclockingCowboy I know, that's why I introduced the method as 'More Difficult'
On the circle question, how do we know the triangle had a right angle?
property of a circle, "angle in a semicircle".
thales theorem also proves it
damm really gotta be creaetive with these
Bro you predicted the question no 5 but question was to find the a+c
jee aspirants dream questions in a paper
😅😅
shut up kid
84,105 can both just be plugged into desmos with no brain lol
You should try plugging your numbers and see how fast you can get the answer.
Way too much work for 10:00 , just use the product of solutions rule, product of solutions =c/a, barely need to do any work
Doesn’t that only work for quadratics? :0 (genuine question I’m not sure myself either)
plz turn your camera off in future videos. I'm trying to really focus so i can get a high sat score but your pretty face keeps distracting me