August SAT 2024 Math Exam Predictions | What will be on the Test?

[@OverclockingCowboy]

For the percent problem 8:00, you can just divide 21 directly by 20 to get the value. t = (21/20) r = (1.05) r You get 5% greater.

[@epicexamprep]

You're absolutely right! Dividing 21 directly by 20 to get 1.05 is a quick and efficient way to see that t is 5% greater than r. This is a great shortcut when you're familiar with the relationship between the numbers! Thank you for sharing =)

[@IvetaShlepchyan]

23:52 To avoid calculation errors we can solve it using desmos 1) create table x1 y1 0, 0 10, 1600 2) y1~a(x1-10)^2+1600 will give us a = -16 3) f(x) =-16(x-10)^2+1600 f(15) will give us 1200 :)

[@IvetaShlepchyan]

44:33 to graph using desmos without finding equations first (let desmos find equation for us :) 1) table x1 y1 1 3 2 4 2) y1~mx1+b 3) y1~ab^x1

[@OliviaRossi-o1r]

Honestly this video is much better in comparison to the other SAT math videos I've watched! There are actually challenging questions. Thank you teacher!!

[@epicexamprep]

Wow, thank you! I am glad that you found the questions helpful! ☺️

[@IvetaShlepchyan]

33:22 we can also solve it without using inscribed angle theory 1) since KLM = 25° => LMN = 25° 2) draw OL => since MO and OL are both equal 6 (radius) we get isosceles triangle => so angle LMO = MLO = 25° 3) draw line OK we get isoscales traingle OKL in wich KLO = 25 +25 =50 so LKO is also 50 180-100=80° angle KOL 4) Arc length = radius * angle in Pi radians = 6 * 80/180 pi = 8/3 pi Or central angle/360 = arc length/ circumference 80/360 = x / 2*6*pi x= 8/3 pi

[@OverclockingCowboy]

On the linear and exponential equations 47:50, you can apply the test point r = 3/2 to the equations. For the exponential equation y = (9/4)(4/3)^3/2 = 3 root (4/3) = 6 / root(3) = 3.46 For the linear equation y = 3.50 Hence, s>t when 1

[@epicexamprep]

Your method of applying a test point like r= 3/2 is an excellent way to verify that s>t in the interval 1

[@OverclockingCowboy]

⁠​⁠@@epicexamprep For a comprehensive proof, it is not too bad to apply test points outside the interval. We can apply test points with rounded numbers. For r = 0 Linear y = 2 Exponential y = (9/4)(4/3)^0 = 9/4 For r = 3 Linear y = 5 y = (9/4)(4/3)^3 = 3 (4/3)^2 = 16/3 In both cases, t > s.

[@sanjanacm5082]

Thank you so much for this amazing video. I had been waiting for this one for so long. This August SAT prediction video covers all the important topics and questions, and it's really helpful and informative. Please keep posting awesome videos like this.😊

[@oyeitsmilangaming7386]

Can you explain me 41:42

[@epicexamprep]

You're welcome! Glad you found it helpful and informative! =) =)

[@chanelgreen4096]

I'm going to retake the SAT to try for a perfect score just because, I would also like to motivate and have cemented to kids and students of all ages that SAT is just a fun examination game, something that can be done regularly without moving aside your entire life to study for just the SAT. It really should be like having breakfast in the morning and brushing your teeth at night, and always score a descent average score.

[@epicexamprep]

Your determination to retake the SAT and your perspective on making it a less stressful part of life are admirable! I think this is a great attitude to have and can certainly help reduce the pressure and anxiety many students feel. Good luck on getting the perfect score!! I hope you get it! =) =)

[@Baddie8901]

Hey so you`ve written the SAT before. I need to ask you a question and please genuinely answer. Was the real SAT harder than the practice tests? I need like a break down please

[@garo8863]

you make math look so easy! Thank you!)

[@epicexamprep]

Glad you think so! You're welcome! =) =)

[@YoussefMohammed4711E]

thank you so much you deserve much more credit

[@epicexamprep]

Thank you 😌

[@inorin143]

14:02 , if the meter in the question was like 2 or maybe 3 and above, would it change the vertex form at all? “Snowball is at a height of approximately 1 meter and reaches maximum at 2.8 meters”

[@epicexamprep]

If the initial height of the snowball changes, the form of the quadratic equation remains the same, but the coefficient a will change. Given Information: The snowball reaches a maximum height of 2.8 meters at 0.6 seconds after being thrown. We use the vertex form of a quadratic equation: h = a(t - 0.6)^2 + 2.8, where (0.6, 2.8) is the vertex. (So the vertex will no change) 1. If the initial height is 2 meters: Given: The initial height is 2 meters when thrown. Maximum height is 2.8 meters at 0.6 seconds. To find the coefficient a: Use the initial condition: at t = 0, h = 2 meters. Plug these values into the vertex form: 2 = a(0 - 0.6)^2 + 2.8. Simplify the equation: 2 = a(0.36) + 2.8. Solve for a: Subtract 2.8 from 2: 2 - 2.8 = -0.8. Divide by 0.36: a = -0.8 / 0.36 ≈ -2.22. So, the quadratic equation is: h = -2.22(t - 0.6)^2 + 2.8. 2. If the initial height is 3 meters: Given: The initial height is 3 meters when thrown. Maximum height is 2.8 meters at 0.6 seconds. To find the coefficient a: Use the initial condition: at t = 0, h = 3 meters. Plug these values into the vertex form: 3 = a(0 - 0.6)^2 + 2.8. Simplify the equation: 3 = a(0.36) + 2.8. Solve for a: Subtract 2.8 from 3: 3 - 2.8 = 0.2. Divide by 0.36: a = 0.2 / 0.36 ≈ 0.56. So, the quadratic equation is: h = 0.56(t - 0.6)^2 + 2.8. So in conclusion The general form of the equation is h = a(t - 0.6)^2 + 2.8. The value of a changes based on the initial height, but vertex will remain unchanged. Sorry for long explanation! hah..hope it helps to make better sense of the problem though!! Thank you! =) =)

[@inorin143]

@@epicexamprep really helpful, thank u!

[@江郁敏-o7u]

Thank you, it’s super helpful!!

[@epicexamprep]

You're welcome!! Glad that you found it helpful! =)

[@SayanRajaDahal]

IN THIS question (46:42) how can value of b be different for both equation. If we are using same variable in different equation then the value of b has to be equal in both equation. Right?

[@epicexamprep]

So, the same variable name can be used in different equations or contexts, but its value can differ depending on how it's defined within each specific equation. For the Linear Equation: The linear equation is given by y = mx + b. Here, b represents the y-intercept of the linear function, which is a constant value that shifts the line vertically. From the points given, we determined that b = 2 for the linear equation. Exponential Equation: The exponential equation is given by y = ab^x. In this equation, b is part of the base of the exponential function. It determines how the function grows or decays as x changes. The value of b for the exponential equation was found to be 4/3, which is completely independent of the linear equation's b. The value of b in the linear equation is a specific constant related to the intercept on the y-axis, while b in the exponential equation is a base related to the rate of growth or decay. These are fundamentally different roles, so they can have different values even though the same letter is used. The notation is simply a convention, and the meaning and value of a variable depend on the specific context of the equation it is used in. In this problem, the two "b" values are unrelated and should not be confused with each other. I hope this helps and answers your question! =)

[@Loveitmore1]

@@epicexamprep I found this to be very confusing as well. When I calculated b =2 in the linear equation, I substituted that in for b in the exponential. oof!

[@virendarhanumante597]

please, keep it up.

[@epicexamprep]

Plan to! =)

[@oyeitsmilangaming7386]

@@epicexamprepcan you explain 41:42

[@khasannishanov5830]

56:07, the answer is 4/3

[@epicexamprep]

For the question, "Two numbers, a and b, are each greater than zero, and 4 times the square root of a is equal to 9 times the cube root of b. If a = 2/3, for what value of x is a^x equal to b?" ... The correct answer is 15/2 or 7.5. If you want to explain to me how you got 4/3 and I can try to see where you made a misstep! Thank you. =)

[@irmakaygunoglu]

Thank you so much, it was great! More video for verbal section till August exam please 😊

[@epicexamprep]

Yes! Working on it now! It will drop sometime this week =)

[@aidenwang5283]

thank you so much. it's so helpful

[@epicexamprep]

You're welcome! Glad it was helpful! =)

[@omers8497]

Thank you very much 🥺💚

[@epicexamprep]

You're welcome! Thank you for watching =)

[@apsidalsolid4607]

Hello! Thank you for this awesome video; its really nice having harder problems that actually challenge you! I do know how to solve a majority of these questions but sometimes I end up making careless mistakes, do you have any advice for this? On the real test I believe I will have plenty of time, but I would rather not make those mistakes in the first place.

[@epicexamprep]

Thank you for watching and for your compliment! =) ... Yes to avoid careless mistakes, I recommend the following: Always Revisit the Question: After solving a problem, make it a habit to go back and read the final question carefully. This step ensures you're answering exactly what's being asked. For instance, if the question asks for x−3 but you calculated x, you'll catch that discrepancy. Similarly, if you're working with inequalities and the problem asks for the "least" or "greatest" integer, make sure your answer reflects that, avoiding common traps like if you get x

[@TheSatSlayers]

please make more videos in preparation for the august exam

[@epicexamprep]

I will be! Stay tuned ;)

[@no-vm1jv]

hi! im kind of struggling on topics relating to linear/linear inequalities/system of equations word problems, some statistics problems, and also some quadratics word problems. so is it possible for you to make a video on it with difficult level questions?? love your videos!!

[@epicexamprep]

Yes, definitely! I also recommend to watch the May predictions: kzbin.info/www/bejne/o362eX9vbqx-l80 and June Predictions: kzbin.info/www/bejne/gWnSfqasYsl1grs I cover advanced concepts in those! =) =)

[@oyeitsmilangaming7386]

Thanks beb ❤.

[@epicexamprep]

You're welcome 😊

[@haticemeryemoeztuerk6537]

goddess of dsat!❤

[@epicexamprep]

haha! Thank you!! =) =)

[@aarya-cy4ws]

at 49:31, how did you get those intervals for a and b? trying to understand, but cannot see it

[@OverclockingCowboy]

a is between f(1) and f(4), b is between f(4) and f(7) Parabola is above x axis at f(1), below at f(4), above at f(7) You are looking at a parabola that lies between 1 and 7 with vertex somewhere around 4.

[@epicexamprep]

The curve crosses the x-axis at two points, a and b. These are the roots of the function, where the output of the function is zero. At x = a and x = b, f(x) = 0. What do the conditions mean? The problem gives us three clues: f(1) > 0: This means when we put x = 1 into the function, the output is positive. f(4) < 0: This means when we put x = 4 into the function, the output is negative. f(7) > 0: This means when we put x = 7 into the function, the output is positive again. 4. How do these clues help us? First Clue f(1) > 0: When the output of the function is positive, it means that x = 1 is outside the two roots a and b. The parabola is U-shaped, so the positive areas are on either side of the parabola, outside the two roots. This tells us that 1 is either to the left of both roots or to the right of both roots. Second Clue f(4) < 0: When the output is negative, it means that x = 4 is between the two roots. The negative area is in the middle of the U-shape, between the two roots. Third Clue f(7) > 0: When the output is positive again, it means that x = 7 is outside the two roots, on the opposite side of where x = 1 is. So, 7 must be to the right of both roots. From these clues, we can figure out where the roots a and b must be: Since 4 is between the roots, a must be less than 4 and b must be greater than 4. 1 is less than both roots, so 1 < a. 7 is greater than both roots, so b < 7. This gives us the intervals: a must be between 1 and 4 (not including 1 and 4). b must be between 4 and 7 (not including 4 and 7). So then to find the values of a and b: a can be 2 or 3, since these are the integers between 1 and 4. b can be 5 or 6, since these are the integers between 4 and 7. For example, if a = 2 and b = 5, the function behaves in exactly the way the clues described.... Does it make more sense now? I hope so! Thank you for your question! =)

[@胡泽睿]

thank you❤❤

[@epicexamprep]

You're welcome 😊

[@oljzzaluadin774]

32:33 can you explain please, how did you understand that angle KOM is 50 degrees?

[@epicexamprep]

Yes! Because angle KOM is the center angle for Arc KM....the inscribed angle of Arc KM is angle KLM, which is 25º...Through the inscribed angle theorem (mentioned in video), I know that the central angle (cut from same arc KM) will be twice that of its inscribed angle (also cut from same arc...in this case arc KM)...Does that make sense! Let me know! Thank you =) =)

[@snehaganatra8194]

56:56 how did you factor that?

[@snehaganatra8194]

nvm got it!

[@epicexamprep]

For me, I fiind factoring the easiest way...Another way that it could have been done is to recognize that 5^(x-3) can be rewritten as 5^x / 5^3 (Law of Exponents). This changes the equation to: 5^x - (5^x / 5^3) Factor out 5^x from the expression on the left side: 5^x * (1 - 1/5^3) Simplify the expression inside the parentheses: 1 - 1/5^3 can be written as (5^3 - 1) / 5^3 So, the equation becomes: 5^x * (5^3 - 1) / 5^3 = 124 * 5^y Since 5^3 = 125, the expression simplifies further: (5^3 - 1) = 124, so we have: 5^x * 124 / 125 = 124 * 5^y Cancel out 124 from both sides: 5^x / 125 = 5^y Recognize that 125 = 5^3, so the equation simplifies to: 5^(x-3) = 5^y Since the bases are the same, equate the exponents: x - 3 = y Final answer: y = x - 3 Just another way to see and solve the problem! =) Hope it helps.

[@snehaganatra8194]

@@epicexamprep Thank you!!

[@tanner1376]

56:42

[@epicexamprep]

Do you have a doubt on that exponent question?

[@tanner1376]

@@epicexamprep No, I just had difficulty on this one so I wanted to save the timestamp to go back later! Thanks for reaching out though!

[@fedal5956]

For the problem at 12:21 wouldn't the answer be 22

[@epicexamprep]

No...It's 20...How did you get 22? If you tell me your steps I can try to see where you misstepped. Thank you! =)

[@fedal5956]

@@epicexamprep so when I put in .88=1.1*1-p/100 into demos and replace p with x it says it is 22. I used Desmos to double check my working usually. I honestly don't know where I got it wrong.

[@NihadFarajov]

Please keep it up

[@epicexamprep]

Thank you! I will =)

[@itsyoboiasu1441]

41:39 i dont get this one

[@epicexamprep]

I explain it more here (I do 2 problems that are highly similar to each other): kzbin.info/www/bejne/gGG2h2R6fcydqs0

[@SamanthaPerez-j2y]

In 5:50, why did you add 1 to the percent?

[@epicexamprep]

When you hear that something has increased by a certain percentage, you're adding that percentage to the original amount. To do this in a simple calculation, you start with the whole (which is 100% or 1) and then add the extra percentage as a decimal. Example: 128% greater means that the new amount is 128% more than the original amount. Think of it like this: you have the original amount (100% or "1"), and then you add the extra 128% (which is written as 1.28 in decimal form). So, you end up with 1 + 1.28 = 2.28 times the original amount. 29% greater works the same way: You start with the whole (100% or "1") and add the 29% (which is 0.29 in decimal). So, 1 + 0.29 = 1.29 times the original amount. In short, the "1" just represents the original amount, and the percentage part is how much more you’re adding to it. Does this help??? I hope so. Thank you for your question =)

[@inorin143]

should i expect this kind of difficulty on questions for module 1 and 2? Or do these appear in like module 2 only. btw keep it up!

[@aarya-cy4ws]

usually module two is harder if you get over half the questions from module one right, but these questions seem harder than the module questions on bluebook 😭

[@inorin143]

@@aarya-cy4wsikr, are questions on the real sat usually like this? Im scared for august lol

[@gamerboyblade6881]

does anyone have resources for rearranging/manipulating equations like in the first question? im good at everything but i struggle so much with those 💔💔

[@epicexamprep]

Hi! I would recommend to go here: satsuitequestionbank.collegeboard.org/digital/search In the Search choose assessment SAT, Test: Math, and then choose Domain: Advanced Math and Skill: Equivalent Expressions =) =) =)

[@gamerboyblade6881]

@@epicexamprep aahh youre the best, thank you!!!

[@prprpr3108]

For the first problem, I don’t understand how i can just put 5xy under xyz and leave the 1/5xy

[@epicexamprep]

Let's go through the steps to understand the simplification process in this equation: The original equation you have is: 1/(5xy) + xyz/(5xy) = 1/(4yz). Step 1: Simplifying the Second Term Look closely at the second term on the left side: xyz/(5xy). In this term, we can simplify it by canceling out the "xy" from both the top (numerator) and the bottom (denominator). When you do that, you're left with: xyz/(5xy) = z/5. So now, the equation looks like this: 1/(5xy) + z/5 = 1/(4yz). Step 2: Combining the Terms on the Left Side To combine the terms on the left side, notice that the first term has a denominator of 5xy, but the second term (z/5) doesn’t. To make the second term have the same denominator (5xy), you multiply both the top and bottom of z/5 by xy, which gives you: z/5 = (z * xy)/(5 * xy) = xyz/(5xy). Now, you can rewrite the equation as: (1 + xyz)/(5xy) = 1/(4yz). When you simplify xyz/(5xy) to z/5, you’re just canceling out the common "xy" terms. This is a basic algebraic step that helps in simplifying the equation. By rewriting the equation, it becomes easier to work with and solve for y. Does this help to clear your doubt? I hope so! =)

[@EtiekikereUdoma]

Sat is not this tough though Questions with too many steps It helps though 💕

[@epicexamprep]

These concepts are geared more towards the hard questions that can appear on the exam and those usually require a few more steps than the easy and medium level questions. =)

[@aarya-cy4ws]

@@epicexamprep oh ok, i was wondering why the questions were so difficult 😭

[@onurkaandinler2802]

are these hard questions or mixxed? i dont think any of them is easy tho.

[@epicexamprep]

Most of them are hard, so that you can ace that difficult module 2! =)

[@xkieh]

where exactly did you got this pdf?Is it official? thank you for the video!

[@epicexamprep]

I made them! Inspired by past official questions. =)

[@kenyenpinyosuk2094]

Excuse me, is this like the old past paper questions?

[@epicexamprep]

They are similar to current Digital SAT questions! However, a lot of the paper-based concepts from math are still relevant on DSAT =)

[@josmithajoseph1352]

Hey, at 45.11 is the answer b because that's where both of the lines intersect?

[@epicexamprep]

The answer B (1 < r < 2) is correct because this is the range where the conditions for the points (r, s) on the linear equation and (s, t) on the exponential equation are satisfied. Specifically, within this range, the relationship s = r + 2 and t = (9/4)(4/3)^s with s > t holds true. So yes, the intersection helps to visualize and understand where the linear function's values s are greater than the exponential function's values t. In this case, the range 1

[@josmithajoseph1352]

@@epicexamprep Thank you, Also do you think there are going to have surface area questions?

[@Hehehehehoho]

@@josmithajoseph1352 bro idk if it;s going to be or not but it was in june sat. so yh better learn. also don;t forget know how to solve questions that involve pyramids.

[@hazelnut2.8]

can I get the pdf please

[@ajiiiibano]

please make the vertex form video!

[@oyeitsmilangaming7386]

Can u explain 41:42

[@epicexamprep]

I've added it to my "to-do soon" list! =)

[@epicexamprep]

Yes!...I explain this problem more thoroughly (with another example) in this video: kzbin.info/www/bejne/gGG2h2R6fcydqs0 Let me know if that helps better =)

[@oyeitsmilangaming7386]

@@epicexamprep thanks Beb ❤️

[@chuckc3665]

#1 you should first subtract 1/(5xy) on both sides then the problem becomes much easier

[@josmithajoseph1352]

Hey, at 39.39 what formula is it

[@epicexamprep]

The x^2 + y^2 =r^2? That is the formula for the Circle Equation when it is centered at origin (0,0). =)

[@daltang]

49:27 i cant seem to understand how you got the equation 1

[@epicexamprep]

Let me see if I can shine some light: First, this problem is a Quadratic Function: A quadratic function is any function that can be written in the form: f(x) = ax^2 + bx + c. In this problem, the quadratic function is given in factored form: f(x) = (x - a)(x - b). This form is another way of writing a quadratic function, where a and b are the roots (or solutions) of the equation f(x) = 0. The roots are the values of x where the function equals zero. For example, when x = a, f(a) = 0, and when x = b, f(b) = 0. The Problem: The function is given as f(x) = (x - a)(x - b), where a and b are integer constants. The conditions provided are: f(1) > 0 f(4) < 0 f(7) > 0 We need to determine the possible values of a + b. What Do These Conditions Mean? Condition 1: f(1) > 0 means that when x = 1, the function is positive. Graphically, this means the point x = 1 on the graph is above the x-axis. Condition 2: f(4) < 0 means that when x = 4, the function is negative. Graphically, this means the point x = 4 on the graph is below the x-axis. Condition 3: f(7) > 0 means that when x = 7, the function is positive. Graphically, this means the point x = 7 on the graph is above the x-axis. Finding the Range of a and b: Analyze f(1) > 0: Plugging x = 1 into the function gives us: f(1) = (1 - a)(1 - b). For f(1) to be positive, the product (1 - a)(1 - b) must be positive. This can happen in two cases: Case 1: Both factors are positive, meaning a < 1 and b < 1. Case 2: Both factors are negative, meaning a > 1 and b > 1. Since the next conditions suggest that a and b are around 4 and 7, the relevant case is when a > 1 and b > 1. Analyze f(4) < 0: Plugging x = 4 into the function gives us: f(4) = (4 - a)(4 - b). For f(4) to be negative, the product (4 - a)(4 - b) must be negative. This happens when one factor is positive and the other is negative: If 4 - a > 0 (which means a < 4), then 4 - b < 0 (which means b > 4). Conversely, if 4 - a < 0 (which means a > 4), then 4 - b > 0 (which means b < 4). Since we know from the next condition that f(7) > 0, the most consistent scenario is when a < 4 and b > 4. Analyze f(7) > 0: Plugging x = 7 into the function gives us: f(7) = (7 - a)(7 - b). For f(7) to be positive, the product (7 - a)(7 - b) must be positive. This happens in two cases: Case 1: Both 7 - a > 0 and 7 - b > 0, meaning a < 7 and b < 7. Case 2: Both 7 - a < 0 and 7 - b < 0, meaning a > 7 and b > 7. Given our previous analysis that one root is less than 4 and the other is greater than 4, the scenario that works best here is when a < 7 and b < 7. From the analysis: f(1) > 0 tells us a > 1 and b > 1. f(4) < 0 tells us a < 4 and b > 4, or vice versa. f(7) > 0 tells us a < 7 and b < 7. Putting it all together: 1 < a < 4 4 < b < 7 The possible integer values for a and b based on the inequalities are: a = 2, b = 5 a = 2, b = 6 a = 3, b = 5 a = 3, b = 6 The corresponding values for a + b are: If a = 2 and b = 5, then a + b = 7. If a = 2 and b = 6, then a + b = 8. If a = 3 and b = 5, then a + b = 8. If a = 3 and b = 6, then a + b = 9. Therefore, the possible values for a + b are 7, 8, or 9. Does this help to understand it better?? I hope so!! =) =)