Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2009 and Jonathan found this equation… 4(a 2 ) n x 3a 4 = 12a 16 What is the value of n? 2. Factorise a 2 + 7a – 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 > Make r the subject of this formula A = 4πr 2 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… x 2 + 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 860m 2. Calculate the value of a 9. Solve a 2 – 10a – 39 = Expand and simplify… (2a + 3) 2 = 3 a aa 11. if 8a 9 = 2a 4 what is n =? 4a n 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 $ She writes this equation 0.50f f = How many friends has Pam got? 14. Anne was told that one factor of a a – 100 is a – 2. What is the other factor? 15. Simplify 2a + 4a = Solve for both a and b… 3a + 8b = 79 and a = b + 8

Yr 11 MCAT Algebra Practice 3 – based on NCEA externals 2008, 2009 and Jonathan found this equation… 4(a 2 ) n x 3a 4 = 12a 16 What is the value of n? 2. Factorise a 2 + 7a – 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 > Make r the subject of this formula A = 4πr 2 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… x 2 + 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 860m 2. Calculate the value of a 9. Solve a 2 – 10a – 39 = Expand and simplify… (2a + 3) 2 = 3 a aa 11. if 8a 9 = 2a 4 what is n =? 4a n 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 $ She writes this equation 0.50f f = How many friends has Pam got? 14. Anne was told that one factor of a a – 100 is a – 2. What is the other factor? 15. Simplify 2a + 4a = Solve for both a and b… 3a + 8b = 79 and a = b + 8 = 4a 2n x 3a 4 = 12a 2n + 4 so 2n + 4 = 16 2n = 12 n = 6 ( )( ) a a a / 3 = 9 a = 27 3a + 5a = a = -2 a = -1 / 4 a is ___ or ___ 1/21/2 -3 5a > 20 a > 4 4πr 2 = A r 2 = A / 4π r = ± √( A / 4π ) a is ___ or ___ 5/45/4 -2 a + 3 = 2.75 a = a – 2a = a = 15 a = 7.5 What multiplies to – 8?-1x81x-8 -2x42x-4 because a is positive we can eliminate… therefore a = 7or2,b/c = -1or-2, 8or4 area = base x height area = (a+a+3) x a area = (2a+3) x a area = 2a 2 +3a 2a 2 + 3a = 860 2a 2 + 3a – 860 = 0 ( )( ) = 0 2a a a is ___ or ___ ( )( ) a a (2a+3)(2a+3) =4a 2 +12a+9 n = 5 a2a2 - 7a + 5a - 35 = a 2 – 2a – f = f = 21 what x – 2 is -100? 50 so the other factor is (a + 50) 10a a 15 =22a 15 sub b + 8 in 3(b+8) + 8b = 79 3b b = 79 11b = 55 b = 5 a = a = 13