SAT Math: The Final Mile

SAT Math THE FINAL MILE Copyright © 20 21 Harsh Manghnani and Nathan Quadras All rights reserved. No parts of this book may be reproduced without explicit written permission from the authors. SAT is a trademark registered and owned by the College Board, which is not affiliated with, and does not endorse, this product. Contact thefinalmilesat@gmail.com for requests or details. throughout the entire process of formulating this book. Their help was truly invaluable to us. Special thanks to our mentors, Mr. Satya Prakash Singh and Mr Dilip Biswas, for their unparalleled guidance Contents 1. Quadratic Equations ................................................................. ......................................1 2. Systems of Equations ............................................................................... ......................4 3. Inequalities .............................. ................................................................. ............. ........7 4. Exponents and Radicals ........................................... .. ............................ ... .... ............... 1 1 5. Linear and Exponential Function s .............................................................. ... ............ .. 1 3 6. Percent age ................................................................................................ .................. 1 7 7. Ratio and Proportion ........................... ...................................................... ...... ........... 18 8. Rates .......................... .............................................................................. ................. ... 19 9. Straight Lines ....................................... ....................................................... ................. 2 0 10. Functions .................................................................................................... .. ............... 2 3 11. Absolute Value Functions ............................................................................ ................. 28 12. Complex Numbers ............................................................ .. ........................ .................3 0 13. Expressions ................................................................................ ................. .......... .......3 3 14. Solving Equations ..................................................... ............................. .. ................. .... 35 15. Angles .................................. .................................... ............................ .................. ..... 36 16. Trigonometry .................. .................................................. .............. .................. ........ 38 17. Triangles .................................................................... ................... ............................... 4 2 18. Circles ................................ ...................................... ................... ................................. 47 19. Volume .......................................................... ................... ........................................... 49 20. Statistics .......................................................... ................... .......................................... 5 1 21. Probability .......................................................... ............. ............................................ 5 4 22. Linear Models ......................................................... ...... ............... ...... ..... .............. ...... 5 5 1 Quadratic Equations Different Forms of Quadratic Functions: Standard form: 𝑓 ( 𝑥 ) = 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 Factored form: 𝑓 ( 𝑥 ) = 𝑎 ( 𝑥 − 𝑟 1 ) ( 𝑥 − 𝑟 2 ) Vertex form: 𝑓 ( 𝑥 ) = 𝑎 ( 𝑥 − ℎ ) 2 + 𝑘 a: It controls the degree of curvature of the graph and whether the graph opens upwards or downwards. If a < 0 , then the graph opens downwards, but if a > 0 , then the graph opens upwards. Larger the value of a, narrower the parabola. Conversely, smaller the value of a, wider the parabola. b: It controls the location of the axis of symmetry of the parabola. Along with the value of b, it also depends on the coefficient a. Let’s have a look at the following situation: 1.) a is positive and b is negati ve, it shifts to the right. 2.) a is negative and b is positive it shifts to the right. 3.) a is positive and b is positive it shifts to the left. 4.) a is negative and b is positive it shifts to the left. c: This represents the y - intercept of the parabola. h: x coordinate of the vertex. This can also be viewed as the coordinate for the vertical line of symmetry k: y coordinate of the vertex. This can also be seen as the maximum or minimum value depending on the value of a. if a < 0 then k represents the maximum value of the parabola while if a > 0 , k represents the minimum val u e of the parabola 𝑟 1 and 𝑟 2: These represent the roots of the parabola. 𝑟 1 + 𝑟 2 2 : gives the x coordinate of the vertex. In other words, the midpoint of the roots represents the x coordinate of the vertex. 𝑓 ( 𝑟 1 + 𝑟 2 2 ) : This represents the y coordinate of the parabola. 𝑓 ( 0 ) : This represent the y - intercept. 2 Quadratic Formula: The quadratic formula, which is − 𝑏 ± √ 𝑏 2 − 4 𝑎𝑐 2 𝑎 , allows us to find the roots of quadratic equations O ne root is represented by − 𝑏 + √ 𝑏 2 − 4 𝑎𝑐 2 𝑎 and the second root is represented by − 𝑏 − √ 𝑏 2 − 4 𝑎𝑐 2 𝑎 Relation between roots and coefficients : Sum of roots: For the equation 𝑓 ( 𝑥 ) = 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 , the sum of the roots is − 𝑏 𝑎 = − 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡 𝑜𝑓 𝑥 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡 𝑜𝑓 𝑥 2 Product of roots: For the equation 𝑓 ( 𝑥 ) = 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 , the product of the roots is 𝑐 𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑟𝑚 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡 𝑜𝑓 𝑥 2 Therefore, a quadratic equation can be expressed in the following form 𝑥 2 − ( 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 ) 𝑥 + 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 =0 𝑥 2 − ( − 𝑏 𝑎 ) + 𝑐 𝑎 =0 𝑥 2 − 𝑆𝑥 + 𝑃 =0 Discriminant: The number of roots of a parabola depends on the value of the discriminant. The discriminant of a quadratic in the form of 𝑓 ( 𝑥 ) = 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 is 𝛥 = 𝑏 2 − 4 𝑎𝑐 Let’s look at the following situations: Discriminant Value Roots Graph 𝛥 > 0 It has 2 real and distinct roots/solutions 𝛥 < 0 I t has no real roots/Solutions 3 𝛥 = 0 I t has one real solution 4 System s of Equations Equation with No Solutions A system of equation is said to have no solution if there is no common solution for both equations. This is the case because these lines are parallel lines and will thus never intersect. How to Identify a system of equation that has No Solutions ? Systems with No Solutions have these characteri stics. 1. The Lines have the same slope 2. They have different y intercepts In such a situation convert both the equations into standard form. The ratio of the coefficients should be as follows : 𝒂 𝟏 𝒂 𝟐 = 𝒃 𝟏 𝒃 𝟐 ≠ 𝒄 𝟏 𝒄 𝟐 . Let’s look at an example. How many Solutions does the equation below have? 3x + 4y = 6 6x + 8y = 9 A. One B. Two C. No solution D. Infinitely many solutions 3 𝑥 − 4 𝑦 − 6 = 0 6 𝑥 + 8 𝑦 − 9 = 0 3 6 = 4 8 1 2 = 1 2 In the case of the y intercept: 6 9 = 1 3 ≠ 1 2 System with i nfinitely many S olutions A system of equations is said to have infinitely many solutions if the 2 equation s are equivalent to each other and are represented by the same line when plotted on a graph. 5 How to identify a s ystem of e quations with i nfinitely many s olution s : 1) They have the same slope 2) They have the same y - intercept In other words, 𝑎 1 𝑎 2 = 𝑏 1 𝑏 2 = 𝑐 1 𝑐 2 Look at this example : How many solutions does this system of equations have? 2x + 3y = 5 4x + 6y =10 2 𝑦 = 3 6 = 5 10 1 2 = 1 2 = 1 2 These 2 equations represent the same line and thus have infinitely many solutions. Equations with one S olution These are the equation of graphs which intersect at one point and thus have only one solution. How to Identify a s ystem of e quations with o ne s olution 1. They have different slopes 2. They have different y intercepts. 𝑎 1 𝑎 2 ≠ 𝑏 1 𝑏 2 ≠ 𝑐 1 𝑐 3 Let’s look at an example: How many solutions do es this system of equations have? 𝑦 = 2 𝑥 + 2 𝑦 = − 3 𝑥 + 7 On arranging the equation into standard form, we get: 2 𝑥 − 𝑦 + 2 3 𝑥 − 𝑦 − 7 Now on arranging the coefficients and constants into their respective ratios, we get: 2 3 ≠ 1 ≠ 2 7 Therefore, the above system of equations has only one solution 6 Number of Equations Expression Graph No Solutions 𝑎 1 𝑎 2 = 𝑏 1 𝑏 2 ≠ 𝑐 1 𝑐 2 Infinite number of solutions 𝑎 1 𝑎 2 = 𝑏 1 𝑏 2 = 𝑐 1 𝑐 2 One solution 𝑎 1 𝑎 2 ≠ 𝑏 1 𝑏 2 ≠ 𝑐 1 𝑐 3 7 Inequalities Inequalities resemble normal equations, except for the fact that when you multiply or divide by a negative sign, you need to reverse the inequality. − 2 𝑥 < 2 − 𝑥 < 1 𝑥 > − 1 As you can see, we first divide both sides by 2, and then to get the range of the values of x, we mu ltiply both sides by - 1, reversing the sign in the process. Try doing this question and see if you get the correct answer: Question: 𝟕𝒙 + 𝟖 ≤ 𝟗𝒙 + 𝟐𝟒 Ans: 𝑥 ≥ − 8 Note: Only reverse the sign when multiplying or dividing both sides by a negative number, NO T if a negative number simply exists in the inequality. If 2 𝑥 < − 2 , 𝑥 < − 1 , not 𝑥 > − 1 If a question asks you to pick which of the given options is a solution to the inequality, try to plot it on a number line and find out. For example, if the inequality y ou get is 𝑥 > − 5 , and the given options are A. - 7 B. - 6 C. - 5 D. - 4 You can try plotting it on a number line, and you’ll see that the only number lying in the shaded region is - 4. - 5 is not a solution to the given inequality as it is a ‘strictly greater than inequality’ and has an open circle at - 5 indicating that it is not inclusive. Thus, the answer is D. - 4 Another type of problem quite common in the SAT is when a set of information is given, and a student must pick the right inequality. For example, it might be given that a man burns 5 calories for every mile he bikes and burns 10 calories for every mile he runs, and that he must burn at least 100 calories a day. In this case, if the miles he bikes in a day is given by p and the miles he runs in a day is given by q, then the inequality is 5 𝑝 + 10 𝑞 ≥ 100 < Less than > Greater than ≤ Less than or equal to ≥ G reater than or equal to = Equal to 8 Note: If a<b: (i) a+c<b+c (ii) a - c<b - c (iii) ka<kb, for k>0 (iv) ka>kb, for k<0 y ≤ 𝟓 y < 5 Here, the gray region represents the region Here, the gray region represents the region where the y value is less than or equal to 5. where the y value is less than 5. T he dotted line indicates that coordinates where y=5 are not solutions to the inequality. x ≤ 5 Here, the gray region represents the region where the x value is less than or equal to 5. 9 2x - 3y > 6 Here, the gray region represents the region where the value of 2x - 3y is greater than 6. The dotted line indicates that values of x and y for which 2x - 3y=6 are not solutions to the inequality. 2x - 3y ≥ 6 Here, the gray region represents the region where the value of 2x - 3y is greater than or equal to 6. 10 x+y ≤ 𝟔 x - y ≥ 7 H ere, there are two inequalities on the graph: x+y≤6 and x - y≥7 . The light gray regions represent the regions where either x+y≤6 or x - y≥7 , and the dark gray region represents the intersection of both the inequalities, i.e., where x+y≤6 and x - y≥ 7 11 Exponents and Radicals 1. Any number to the power 1 is the number itself, and any number to the power 0 is 1. a. 𝑥 1 = 𝑥 b. 𝑥 0 = 1 2. 𝑥 𝑚 ⋅ 𝑥 𝑛 = 𝑥 𝑚 + 𝑛 3. 𝑥 𝑚 𝑥 𝑛 = 𝑥 𝑚 − 𝑛 4. ( 𝑥 𝑚 ) 𝑛 = ( 𝑥 𝑛 ) 𝑚 = 𝑥 𝑚𝑛 5. ( 𝑥𝑦 ) 𝑚 = 𝑥 𝑚 ⋅ 𝑦 𝑚 6. ( 𝑥 𝑦 ) 𝑚 = 𝑥 𝑚 𝑦 𝑚 7. 𝑥 − 𝑚 = 1 𝑥 𝑚 8. √ 𝑥 × √ 𝑦 = √ 𝑥𝑦 9. If you have a negative sign INSIDE the parentheses, you must raise it to the power of the parentheses as well, i.e., ( − 𝑥 ) 𝑚 is written as ( − 1 ⋅ 𝑥 ) 𝑚 = ( − 1 ) 𝑚 ⋅ 𝑥 𝑚 10. If the bases are equal, the powers can be equated, i.e., If 𝑥 𝑚 = 𝑥 𝑛 , 𝑚 = 𝑛 , 𝑥 ≠ 1 Note: A square root, i.e., √ 𝑥 , can be written as 𝑥 1 2 . Similarly, √ 𝑥 3 = 𝑥 1 3 , √ 𝑥 𝑛 = 𝑥 1 𝑛 and √ 𝑥 𝑚 𝑛 = 𝑥 𝑚 𝑛 To simplify a radical, find its prime factors, pair them up, and take them out of the root. For instance, √ 192 = √ 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2 × 2 × 2 × √ 3 = 8 √ 3 Example: If 𝑥 𝑚𝑥 ⋅ 𝑥 𝑛𝑥 = 𝑥 45 and 𝑚 + 𝑛 = 15 , we can find the value of x by using the laws of exponents. 𝑥 𝑚𝑥 ⋅ 𝑥 𝑛𝑥 = 𝑥 45 𝑥 𝑚𝑥 + 𝑛𝑥 = 𝑥 45 ( 𝑚 + 𝑛 ) 𝑥 = 45 (Equating the exponents as bases are equal) 15 𝑥 = 45 𝑥 = 3 12 Note : On the SAT, we always take the positive value of a root, i.e., √ 4 is only equal to 2, not - 2. However, if x 2 =4, then x= +2 or - 2. 𝑥 2 = 64 has two solutions, +8 and - 8. However, √ 64 has just one solution, +8. 13 Linear and Exponential Functions What is Linear Growth ? When the independent variable undergoes a change in constant increments and the dependent variable grows each time by a constant amount, the growth is referred to as linear growth. In other words, for an equal change in the intervals of in the x value, the y value changes by a constant amount. Linear growth can be expressed through the means of a linear function. 𝑓 ( 𝑥 ) = 𝑚𝑥 + 𝑏 Linear Growth Linear Decay Let us determine what each component of a linear equation means : m: This represents the slope of the linear function. It shows the change in the value of y for a change in the value of x. In other words, it is the c onstant rate of change in the value of y. b: This represents the y coordinate of the y intercept. In other words, it is the value of the linear function when the value of x is 0. It shows the initial value of the function, therefore when the question asks you to give the starting value of the function, find the y intercept of the function. Note: If you see c instead of b, don’t worry! Both b and c represent the y - intercept of the function. 14 How do you find the equation for a linear function when given a table? 𝒙 𝒇 ( 𝒙 ) 1 7 3 11 4 13 First, we must determine the slope of the linear function which can be done by simply choosing any two pairs from the table. consider points (1,7) and (3,11). On calculating the slope, we get: 𝑚 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 11 − 7 3 − 1 = 4 2 = 2 So far, our equation is 𝑦 = 2 𝑥 + 𝑐 . In order to determine the y intercept of the linear function, we must insert any point into the equation. Consider point (1,7) which when inserted into the function, gives us: 7 = 2 + 𝑐 𝑐 Now that we have the y intercep t value, we can form our equation which is: 𝑦 = 2 𝑥 + 2 How to find the Linear equation from a graph. To solve this question, we must make use of the slope intercept formula. On looking at the graph we can see that the line l passes through the point (0,5) on the y axis. There fore, for line l, b=5. To calculate the slope, we must select to points on the graph through which line l passes through. Take points (6,8) and (8,9). Using these points, the slope of the graph comes to 𝑚 = 9 − 8 8 − 6 = 1 2 Therefore, the equation of the line l is 𝑦 = 1 2 𝑥 + 5 5 = ρ 15 What is Exponential Growth ? When the independent variable undergoes a change in constant increments and the dependent variable grows each time by the same rate. In other words, the y value increase at a rate relative to the previous value. 𝑓 ( 𝑥 ) = 𝑎 ( 𝑏 ) 𝑥 Exponential Growth Exponential Decay Let us analyze each part of the exponential function : a: This represents the initial value of the function. In other words, it is the y intercept of the exponential function. The point is usually expressed as 𝑓 ( 0 ) . In order to determine the value of a, you have to simpl y look at the point on the graph where the function intersects the y axis or where the x value is 0 in the case of tables b: This represents the growth/decay factor of the exponential function. If the value of b is 0 < 𝑏 < 1 then the function is experiencing exponential decay, but if the value of b is 𝑏 < 1 , then the function is experiencing exponential growth. In order to calculate the value of b, the growth factor, you must first find the percent growth or decay rate, which is expressed as a decimal. For example, if the function increases by 20% for every increase in the x value, then b is written as follows: 𝑏 = ( 1 ± 𝑟 ) 𝑏 = ( 1 + 0 20 ) 𝑏 = 1 20 However, if the function decreased by 20% for every increase in the x value, the function ex periences exponential decay and is expressed as follows: 𝑏 = ( 1 ± 𝑟 ) 𝑏 = ( 1 − 0 20 ) 𝑏 = 0 ⋅ 80 16 Finding an exponential function from a table x 𝒇 ( 𝒙 ) 1 c 2 d 3 e 𝑏 = 𝑑 𝑐 = ⅇ 𝑑 On finding the value of b and inserting it into the expression, you have to simply insert a given point into the expression to find the value of a