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# CEER 2012 Math Lecture

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# CEER 2012 Math Lecture

Presentation during the Math lecture of the UP Aguman CEER 2012 at Angeles City, Pampanga last 21 July 2012.

Presentation during the Math lecture of the UP Aguman CEER 2012 at Angeles City, Pampanga last 21 July 2012.

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### CEER 2012 Math Lecture

1. 1. The lecture shall begin shortly…
2. 2. Mathematics Review
3. 3. Umpisa na!
4. 4. 5M sa math lecture 1. MAKINIG 2. MAG-BEHAVE 3. MAGTANONG 4. MAGSAGOT 5. MAG-ENJOY 
5. 5. Bago ang lahat.... (a) 0 (b) 1 (c) 5 (d) 7
6. 6. Bago ang lahat.... (a) 0 (b) 1 (c) 5 (d) 7
7. 7. QUESTION 1 Determine the domain and range of (a) (b) (c) (d)
8. 8. QUESTION 1 Solution: The domain of y excludes values of x that will make the denominator zero. Thus, the domain is To solve for the range, we first solve for x in terms of y: x 5 y y x 1 x 5 Therefore, the x 7 range is xy 7y x 5 xy x 7y 5 x y 1 7y 5 7y 5 x y 1
9. 9. QUESTION 2 If and find (a) (c) (b) (d)
10. 10. QUESTION 2 Solution: Recall: for functions F and G: and
11. 11. QUESTION 2 Solution: QUESTION 2 Alternative Solution: SUBSTITUTE a value of x and test which choice will give the same value. Para madali, let x = 0.
12. 12. QUESTION 2 Solution: QUESTION! Which will give a value of 3 at x = 0? (a) (c) (b) (d) Astig, ‘di ba? 
13. 13. QUESTION 3 Which of the following is a linear function? (a) (b) (c) (d)
14. 14. QUESTION 3 Solution: Recall that a linear function is a polynomial function wherein the highest power of the independent variable is 1. Is LINEAR, so the answer is (a) (a) Is QUADRATIC because of (b) the terms 3x2 WAIT! This is (c) also linear! cannot be a linear function since x and are (d) in the denominator The answers are BOTH (a) & (c)! Weh, ‘di nga?!
15. 15. QUESTION 4 What is the equation of the linear function y whose graph passes through the point (2, 4) and has the given slope m = 5/7? (a) (c) (b) (d)
16. 16. QUESTION 4 Solution: We use the slope-intercept form STRATEGY: Substitute x = 2, y = 4 and m = 5/7 then solve for b. Hence, the equation of the line is or
17. 17. QUESTION 4 Solution: QUESTION 4 Alternative Solution: Check the choices! Which among the choices… CLUE: 5 ang nasa unahan 1. Has slope 5/7? ng x at 7 ang nasa denominator 2. Has a value y = 4 when x = 2?
18. 18. QUESTION 5 Determine the distance from the point ( 2, 9) to the line 3x + 4y = 2.
19. 19. QUESTION 5 Solution: "No choice" Solution: We have NO CHOICE but use the following formula for the distance D of a point (x0, y0) from a line with equation Ax + By + C = 0: Before doing anything, rewrite 3x + 4y = 2 as 3x + 4y 2 = 0 Then, substitute the values A = 3, B = 4, C = 2, x0 = 2, and y0 = 9.
20. 20. QUESTION 5 Solution:
21. 21. QUESTION 6 If ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0, which of the following statements is true about the discriminant D? (a) If D < 0, the two roots are real and equal. (b) If D < 0, the two roots are imaginary and unequal. (c) If D > 0, the two roots are real and unequal. (d) If D < 0, the two roots are imaginary and equal.
22. 22. QUESTION 6 Solution: Recall: the solutions or ROOTS of the quadratic equation ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0, can be solved using the QUADRATIC FORMULA: The DISCRIMINANT D of ax2 + bx + c = 0 is the value INSIDE THE SQUARE ROOT; i.e.,
23. 23. QUESTION 6 Solution: The DISCRIMINANT D determines the type or NATURE of solutions or roots a quadratic equation with real coefficients has.
24. 24. As an ASIDE… Some "UPCAT-level" problems that can be solved using the discriminant:
25. 25. QUESTION 7 Determine the radius of the circle whose equation is (a) 2 y (b) 3 (c) 4 r (d) 5 x
26. 26. QUESTION 7 Solution The CENTER-RADIUS FORM of the equation of a circle with radius r and center at (h, k) is To write x2 + y2 􀀀 8x + 6y = 0 in center-radius form, complete the square: The radius is
27. 27. QUESTION 8 Find the quotient of
28. 28. QUESTION 8 Solution
29. 29. QUESTION 9
30. 30. QUESTION 10 What is x in the equation ? (a) 5 (b) 3 (c) 3 (d) 2
31. 31. QUESTION 11 Evaluate (a) 3/2 (b) 2/3 (c) 3 (d) 6
32. 32. QUESTION 11 Solution By definition, the LOGARITHM of a positive number x to the base b, denoted by logb x, is the POWER y of b equal to x; i.e., Example: log3 9 = 2 since 32 = 9. Simple, ‘di ba? CHALLENGE: What is the value of ?
33. 33. QUESTION 12 Solve for all possible values of x in the equation (a) 3 and 2 (b) 2 and 3 (c) 6 and 9 (d) 9 and 6
34. 34. QUESTION 12 Solution A property of logarithm is that Shortest solution: SUBSTITUTE the choices to the original equation!
35. 35. QUESTION 13 Solve for q in the equation (a) (c) (b) (d)
36. 36. QUESTION 13 Solution NOSE BLEEEED!
37. 37. Naku, m atagal pa ‘to….
38. 38. QUESTION 14 Faye is 5 greater than twice the age of Luigi. 5 years from now, Faye will be twice as old as Luigi. How old is Faye 3 years ago? (a) 41 (c) 39 (b) 38 (d) 37
39. 39. QUESTION 14 Solution AGE PROBLEM: Let x = Luigi’s age 2x+5 = Faye’s age Age 5 years Age now from now Luigi x x+5 (2x + 5) + 5 = Faye 2x + 5 2x + 10
40. 40. QUESTION 15 Paolo can finish compiling the books in library in 25 minutes. Kevin can finish it in 25 minutes while Carmela took her 50 minutes. How many minutes will it take them if they were to compile the books altogether? (a) 10 (c) 20 (b) 25 (d) 33
41. 41. QUESTION 15 Solution WORK PROBLEM: Let x = no. of min they can finish the job together No. of Rate per EQUATION: minutes minute Paolo 25 1/25 Kevin 25 1/25 Carmela 50 1/50 Together x 1/x
42. 42. QUESTION 16 There are 570 students in a school. If the ratio of female to male is 7:12, how many male students are there? (a) 300 (c) 380 (b) 370 (d) 390
43. 43. QUESTION 16 Solution 570 students in the ratio 7:12 MALES FEMALES One block =
44. 44. As an ASIDE…
45. 45. QUESTION 17 When each side of a square lot was decreased by 3m, the area of the lot was decreased by 105 sq. m. What was the length of each side of the original lot? (a) 18 (c) 20 (b) 19 (d) 21
46. 46. QUESTION 17 Solution Let x = length of the side of the square EQUATION: Length of a Area side Original x x2 New x 3 (x 3)2
47. 47. QUESTION 18 The difference of 2/3 of an even integer and one- half of the next consecutive even integers is equal to 5. What is the odd integer between these two even integers? (a) 26 (c) 36 (b) 27 (d) 37
48. 48. QUESTION 18 Solution Let x = 1st even integer x + 2 = 1st even integer EQUATION: The ODD integer in between is the one AFTER 36, which is 37 
49. 49. QUESTION 19 Find the average of all numbers from 1 to 100 that end in 8. (a) 53 (c) 51 (b) 52 (d) 45
50. 50. QUESTION 19 Solution The average looks like this: The numerator is actually a sum of an ARITHMETIC PROGRESSION with first term a1 = 8 and tenth term a10 = 98, given by The average is then 530/10 = 53
51. 51. As an ASIDE… FACT: The average of the first n terms of an arithmetic progression is just actually the AVERAGE of the FIRST AND LAST TERM!
52. 52. QUESTION 20 A tank is 7/8 filled with oil. After 75 liters of oil are drawn out, the tank is still half-full. How many liters can the tank hold? (a) 200 (c) 240 (b) 220 (d) 260
53. 53. CAPACITY QUESTION 20 Solution = 25(8) = 200 L 25 L 25 L 75 L drawn out 25 L 25 L 25 L 25 L 25 L 25 L 7/8 full 1/2 full
54. 54. QUESTION 21 Two new aquariums are being set up. Each one starts with 150 quarts of water. The first fills at the rate of 15 quarts per minute. The second one fills at the rate of 20 quarts per minute. When would the first tank contain 6/7 as much as the second tank? (a) After 7 min (c) After 9 min (b) After 8 min (d) After 10 min
55. 55. QUESTION 21 Solution Let x = no. of minutes EQUATION:
56. 56. QUESTION 22 In a classroom, chairs are arranged so that each row has the same number. If Ana sits 4th from the front and 6th from the back, 7th from the left and 3rd from the right. How many chairs are there? (a) 49 (b) 64 (c) 81 (d) 100
57. 57. QUESTION 22 Solution FRONT LEFT anna RIGHT NO. OF CHAIRS: 9 X 9 = 81 BACK
58. 58. QUESTION 23 A circle with radius of 5 m and a square of 10 m are arranged so that a vertex of the square is at the center of the circle. What is the area common to the figures?
59. 59. QUESTION 23 Solution The area common to the figures is 10 m equal to ¼ the area of the circle: 10 m 5m 5m
60. 60. QUESTION 24 How many liters of 20% chemical solution must be mixed with 30 liters of 60% solution to get a 50% mixture? (a) 5 L (b) 10 L (c) 15 L (d) 20 L
61. 61. QUESTION 24 Solution Let x = no. of L of 20% chemical sol’n % Amount of Vol (L) concen- chemical tration Sol 1 x 20% 0.2x Sol 2 30 60% 30(0.6) = 18 mixture (x + 30) 50% 0.5(x + 30)
62. 62. QUESTION 24 Solution EQUATION:
63. 63. ANG TSALAP- TSALAP!
64. 64. QUESTION 25 A URent-A-Car rents an intermediate-size car at a daily rate of 349.50 Php plus 1.00 Php per km. a business person is not to exceed a daily rental budget of 800.00 Php. What mileage will allow the business person to stay within the budget? (a) 300 (c) 400 (b) 350 (d) 450
65. 65. QUESTION 25 Solution Let x = mileage EQUATION:
66. 66. Rules of Counting The Fundamental Principle of Counting: If an operation can be performed in n1 ways, and for each of these a second operation can be performed in n2 ways,then the two operations can be performed in n1n2 ways. Extension: The Multiplication Rule If an operation can be performed in n1 ways, and for each of these a second operation can be performed in n2 ways, a third operation in n3 ways,…, and a kth operation in nk ways, then the k operations can be performed in n1n2n3…nk ways.
67. 67. Permutations Rules of Counting PERMUTATION – based on arrangement of objects, with order being considered Permutation of n objects: n(n – 1)(n – 2)… (3)(2)(1) = n! (n factorial) Permutation of n objects taken r at a time: n! n Pr n r ! Permutation of n objects with repetition: n! n1 ! n2 !...nk !
68. 68. Combinations Rules of Counting Combination – based on arrangement of objects, without considering order Combination of n objects taken r at a time: n n! n Cr r r! n r !
69. 69. 13,983,816 possible combinations
70. 70. QUESTION 26 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 , if each digit can be used only once? (a) 100 (c) 120 (b) 110 (d) 130
71. 71. QUESTION 26 Solution 6 5 4 1st digit: 2nd digit: 3rd digit: 6 choices 5 choices 4 choices By the Multiplication Rule:
72. 72. QUESTION 27 The basketball girls are having competition for inter-colleges. There are 15 players but the coaches can choose only five. How many ways can five players be chosen from the 15 that are present? (a) 3,103 (c) 3,000 (b) 2,503 (d) 3,003
73. 73. QUESTION 27 Solution Since order is NOT important in choosing the five players out of 15, we use the Combination rule with n = 15 and r = 5:
74. 74. QUESTION 28 A coach must choose first five players from a team of 12 players. How many different ways can the coach choose the first five? (a) 790 (c) 800 (b) 792 (d) 752
75. 75. QUESTION 28 Solution Same as no. 27 
76. 76. QUESTION 30 What is the perimeter of the triangle defined by the points (2 , 1), (4 , 5) and (2 , 5)?
77. 77. QUESTION 30 Solution We can use the DISTANCE FORMULA to compute the perimeter of a triangle in the Cartesian plane (i.e., the sum of the lengths of the sides of the triangle). Kaya lang, ‘di ko na realize na easy lang ang case sa problem kasi RIGHT TRIANGLE na! (See the board for the solution) :p
78. 78. QUESTION 32 If arcs AB and CD measure 4s - 9o and s + 3o respectively and angle X is 24o, find the value of s. See figure below. (a) 6 (c) 18 (b) 12 (d) 20
79. 79. QUESTION 32 Solution GEOMETRY FACT:
80. 80. QUESTION 33 How many possible chords can you form given 20 points lying on a circle? (a) 380 (c) 382 (b) 190 (d) 191
81. 81. QUESTION 33 Solution The number of chords can be obtained using the Combination Rule with n = 20, r = 2
82. 82. QUESTION 34 Which of the following sets of numbers cannot be the measurements of the sides of a triangle? (a) 1, 2, 2 (b) (c) 3, 4, 5 (d) 1, 2, 3
83. 83. QUESTION 34 Solution Use the TRIANGLE INEQUALITY: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. (d) 1, 2, 3 1 + 2 = 3 – should be GREATER!
84. 84. QUESTION 35 The figure shows a square inside a circle that is inside the bigger square. If the diagonal of the bigger square is units, what is the area of the shaded region?
85. 85. As an ASIDE… The Pythagorean Theorem and Special Right Triangles
86. 86. QUESTION 35 Solution Note that: •The side of the larger square is 2 (special right triangle) 2 •The side of the square is the diameter of the circle, so the radius of the circle is 1
87. 87. QUESTION 35 Solution Note that: •The diagonal of the smaller square is also 2. s •If s is the side of the smaller square, then 2 •The area of the shaded area is then
88. 88. QUESTION 36 Which of the following statements is NOT true about the figure? Parallel lines a and b are intersected by line x forming the angles 1, 2, 3, 4, 5 and 6. (a) Angles 1 and 6 are congruent. (b) Angles 1 and 5 are supplementary with each other. (c) Angles 3 and 4 are congruent. (d) Angles 2 and 4 are supplementary with each other.
89. 89. QUESTION 36 Solution (a) Angles 1 and 6 are congruent. (alt. ext.) (b) Angles 1 and 5 are supplementary with each other. (ext.) (c) Angles 3 and 4 are congruent. (alt. int.) (d) Angles 2 and 4 are NOT supplementary with each other – they are CONGRUENT (corresponding angles)
90. 90. QUESTION 38 How many sides does a polygon have if the sum of the measurements of the interior angles is 1980o? (a) 11 (b) 12 (c) 13 (d) 14
91. 91. QUESTION 38 Solution The sum of the interior angles of a triangle is given by http://www.mathopenref.com/polygoninteriorangles.html
92. 92. QUESTION 39 An ore sample containing 300 milligrams of radioactive material was discovered. It was known that the material has a half-life of one day. Find the amount of radioactive material in the sample at the beginning of the 5th day. (a) 9.375 mg (b) 18.75 mg (c) 37.5 mg (d) 75
93. 93. QUESTION 39 Solution This can be solved using a geometric progression with first term a1 = 300 common ratio r = ½, and n = 5 days
94. 94. QUESTION 40 A survey of 60 senior students was taken and the following results were seen: 12 students applied for UST and UP only, 6 students applied for ADMU only, 29 students applied for UST, 2 students applied for UST and ADMU only, 10 students applied for UST, ADMU and UP, 33 students applied for UP and only 1 applied for ADMU and UP only. How many of the surveyed students did not apply in any of the three universities (UP, UST, ADMU)? (a) 0 (b) 8 (c) 14 (d) 20
95. 95. QUESTION 40 Solution Using Venn Diagram 12 – UP & UST only UP UST 6 – ADMU only 12 2 – UST and ADMU only 10 10 – all three 1 2 1 – UP and ADMU only 6 ADMU
96. 96. QUESTION 40 Solution 33 (12 + 10 + 1) = 10 UP UST 29 (12 + 10 + 2) = 5 12 Add all numbers in the 10 5 circles: 46 10 1 2 What’s outside: 6 60 46 = 14 ADMU 14
97. 97. BRIEF TIPS AND TRICKS
98. 98. BRIEF TIPS AND TRICKS 1. READ EACH QUESTION CAREFULLY. 2. Take each solution one step at a time. Some seemingly difficult questions are really just a series of easy questions. 3. Remember thy formulas and important facts (especially in Geometry) 4. Answer the easy items first. If you can’t solve a problem right away, SKIP it and proceed to the next.
99. 99. BRIEF TIPS AND TRICKS 4. Try the PROCESS OF ELIMINATION. A little guessing might work. 5. Employ the EASIEST way as possible (e.g., substitution, shortcuts, tricks, etc.) 6. Use your scratch paper wisely… 7. If you still have time, CHECK your answers, ESPECIALLY your shaded ovals! 8. RELAX…. Don’t panic!
100. 100. PRACTICE PROBLEMS! 1. If x + y = 4 and xy = 2, find the value of x2 + y2 2. If 1/3 of the liquid contents of a can evaporates on the first day and 3/4 of the remaining contents evaporates on the second day, what is the fractional part of the original contents remaining at the end of the second day? 3. What is the smallest three-digit number that leaves a remainder of 1 when divided by 2, 3, or 5? 4. The average of 4 numbers is 12. What is the new average if 10 is added to the numbers? 5.