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Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2009 and 2010

Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2009 and 2010. 1. Jonathan found this equation… 4(a 2 ) n x 3a 4 = 12a 16 What is the value of n? 2. Factorise a 2 + 7a – 60 3. Solve these equations… a) a / 3 – 4 = 5 b) 3a + 5 = 3 – 5a

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Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2009 and 2010

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  1. Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2009 and 2010 • 1. Jonathan found this equation… • 4(a2)n x 3a4 = 12a16 • What is the value of n? • 2. Factorise a2 + 7a – 60 • 3. Solve these equations… • a) a/3 – 4 = 5 • b) 3a + 5 = 3 – 5a • c) (1 – 2a)(a + 3) = 0 • 4. Solve this inequation… • 5a – 8 > 12 • 5. Make r the subject of this formula A = 4πr2 • 6. Solve for a… • a) (4a – 5)(a + 2) = 0 • b) 4(a + 3) = 11 • c) 4a – 7 = 8 + 2a • 7. Find the whole numbers that can replace a, b and c so that this equation is true… • x2 + ax – 8 = (x + 8)(x + c) • 8. Two square fields are side by side. They are the same length but one is 3m wider. Together they have an area of 860m2. Calculate the value of a • 9. Solve a2 – 10a – 39 = 0 • 10. Expand and simplify… • (2a + 3)2 = 11. if 8a9 = 2a4 what is n =? 4an 12. Expand and simplify… (a + 5)(a – 7) = 13. Pam sends Christmas cards to her friends. Stamps cost 50 cents for each. Cards cost $2.75 for each. Pam spends a total of $68.25. She writes this equation 0.50f + 2.75f = 68.25 How many friends has Pam got? 14. Anne was told that one factor of a2 + 48a – 100 is a – 2. What is the other factor? 15. Simplify 2a + 4a = 3 5 16. Solve for both a and b… 3a + 8b = 79 and a = b + 8 a a a 3

  2. Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2008, 2009 and 2010 • 1. Jonathan found this equation… • 4(a2)n x 3a4 = 12a16 • What is the value of n? • 2. Factorise a2 + 7a – 60 • 3. Solve these equations… • a) a/3 – 4 = 5 • b) 3a + 5 = 3 – 5a • c) (1 – 2a)(a + 3) = 0 • 4. Solve this inequation… • 5a – 8 > 12 • 5. Make r the subject of this formula A = 4πr2 • 6. Solve for a… • a) (4a – 5)(a + 2) = 0 • b) 4(a + 3) = 11 • c) 4a – 7 = 8 + 2a • 7. Find the whole numbers that can replace a, b and c so that this equation is true… • x2 + ax – 8 = (x + b)(x + c) • 8. Two square fields are side by side. They are the same length but one is 3m wider. Together they have an area of 860m2. Calculate the value of a • 9. Solve a2 – 10a – 39 = 0 • 10. Expand and simplify… • (2a + 3)2 = 11. if 8a9 = 2a4 what is n =? 4an 12. Expand and simplify… (a + 5)(a – 7) = 13. Pam sends Christmas cards to her friends. Stamps cost 50 cents for each. Cards cost $2.75 for each. Pam spends a total of $68.25. She writes this equation 0.50f + 2.75f = 68.25 How many friends has Pam got? 14. Anne was told that one factor of a2 + 48a – 100 is a – 2. What is the other factor? 15. Simplify 2a + 4a = 3 5 16. Solve for both a and b… 3a + 8b = 79 and a = b + 8 n = 5 5/4 -2 a is ___ or ___ = 4a2n x 3a4 so 2n + 4 = 16 a = -0.75 2n = 12 a + 3 = 2.75 a2 - 7a = 12a2n + 4 + 5a - 35 n = 6 = a2 – 2a – 35 2a = 15 4a – 2a = 8 + 7 a = 7.5 ( )( ) a a + 12 - 5 a/3 = 9 a = 27 What multiplies to – 8? -1x8 1x-8 -2x4 2x-4 because a is positive we can eliminate… 3a + 5a = 3 - 5 a = -1/4 therefore a = 7or2, b/c = -1or-2, 8or4 8a = -2 3.25f = 68.25 f = 21 1/2 -3 a is ___ or ___ what x – 2 is -100? 50 area = base x height 5a > 20 area = (a+a+3) x a so the other factor is (a + 50) a > 4 area = (2a+3) x a 10a + 15 12a 15 a area = 2a2+3a =22a 15 2a2 + 3a = 860 a a 3 4πr2 = A 2a2 + 3a – 860 = 0 a 2a ( )( ) = 0 - 20 + 43 r2 = A/4π -21.5 20 a is ___ or ___ r = ± √(A/4π) sub b + 8 in b = 5 ( )( ) a a + 3 - 13 3(b+8) + 8b = 79 a = 5 + 8 3b + 24 + 8b = 79 =4a2+12a+9 (2a+3)(2a+3) a = 13 11b = 55

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