WIUT 2009 February solutions - Exam paper solutions - Question 1-6

Aug 25, 2026

WIUT February 2009 exam paper solutions covering six math problems across algebra, functions, arithmetic progressions, and geometry. The document provides multiple solution approaches for a quadratic difference equation, simplifies a matrix-related expression through stepwise algebra, and solves a word problem using unit prices to compute the total eggs and profit. It also derives parabola intercepts and turning point, determines the 50th term of an arithmetic progression, and calculates specified angles in a parallel-lines configuration using pentagon angle sums.

LanguageEnglish

CategoryExam

FormatPDF

Pages4

Reading Time10 min

AccessFree to download

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

February 2009 WIUT exam paper solutions Question 1 Solve (x2 + 4x-4)2- (x2 – 4x-4)2 = 0. Solution. This problem can be solved using several methods: Method A: We can simplify the expression using the identity a2 -b2 =(a+b)(a-b) . Then we will have (x2 + 4x-4)2- (x2 – 4x-4) 2= [(x2 + 4x – 4) + (x2 – 4x-4)] [(x2 + 4x – 4) - (x2 – 4x-4)] == [2x2 – 8] [8x]=0 As the product will be 0 when one of the factors is zero, we will examine two cases where 1) 8x=0 => x1=0 2) 2x2 – 8=0 => x2=–2, x3=2 Method B: (x2 + 4x-4)2 –(x2 – 4x-4) 2=0 => (x2 + 4x-4)2 = (x2 – 4x-4) 2 As we know, if a2 = b2 then a=±b We have (x2 + 4x – 4)2 = (x2 – 4x – 4) 2 and thus two cases should be examined: 1) (x2 + 4x – 4) = (x2 – 4x – 4) => 4x= –4x => 8x = 0 => x1=0 2) (x2 + 4x – 4) = -(x2 – 4x – 4) => 2x2 – 8=0 => x2=–2, x3=2 Method C: We can just open the brackets. This method is a bit confusing, thus requires more attention and accuracy: (x2 + 4x-4)2- (x2 – 4x-4) 2 = (x2 + 4x – 4)(x2 + 4x – 4) - (x2 – 4x – 4)(x2 – 4x – 4) == (x4 + 4x3–4x2+4x3+16x2–16x – 4x2–16x +16)–(x4 – 4x3–4x2–4x3+16x2+16x – 4x2+16x+16) = 16x3–64x = 16x(x2–4)=0 As the product will be 0 when one of the factors is zero, we will examine two cases where 1) 16x=0 => x1=0 2) x2 – 4=0 => x2=-2, x3=2 You can use any method and be awarded a full mark provided that you show the calculations and solve it correctly. Answer: x = –2 , x =0 or x =2 Question 2 Find the value of A− B + 3⎢⎡⎨⎧A− B + AB⎜⎛ 1 + 1 ⎞⎟⎫⎬2 − 4A2 + B⎤⎥ when A = 2 , B = − 1 . ⎢⎣⎩ ⎝ A B ⎠⎭ ⎦⎥ 3 3 Solution: Before calculating its value for A = ~~ ~~23,B = −~~ ~~13 , it is better to simplify the expression: Step 1: AB⎜⎛ 1 + 1 ⎞⎟ = AB( A+ B ) = A+ B ⎝ A B ⎠ AB Step 2: ⎨A− B + AB ⎜1A + ~~ ~~1B⎟⎬2 = {A− B + (A+ B) }2 = (2A) 2 = 4A2 Step 3: A− B + 3⎢⎨A− B + AB⎜1A + ~~ ~~1B⎟⎬2 − 4A2 + B⎥ = A− B + 3[4A2 − 4A2 + B] = A− B + 3B = A+ 2B So, the expression has become A+2B after simplification. Now, it is very easy to find its value for A = 2 , B = − 1 which is A+ 2B = 2 + 2( −1) = 2 − 2 = 0 3 3 3 3 3 3 Answer: The value is = 0 Question 3 An egg dealer bought a number of eggs at 60p for 6, and five times that number for 900p for 100. He sold them all at 72p per 6 eggs and made a profit of 1020p. How many eggs did he buy? p is a ‘penny’ [British money] Solution: When solving such kind of problems, it is very important to rewrite its mathematical expression. 1) Here, we are not given the number of eggs. Let the number of eggs bought first time be X (You can use any other letter if you want) . Then the number of eggs bought second time will be 5X. 2) In his first purchase, the dealer bought the eggs at a price of 60pennies/6 eggs => 10 pennies per an egg. Second time the price was 900 pennies/100 eggs => 9 pennies per an egg. 3) Now, we can calculate how much money he spent to buy the eggs: X eggs at a price of 10 pennies per egg + 5X eggs at a price of 9 pennies per an egg. That is, 10X +9*5X=55X pennies. 4) Now we calculate his income: he sold X+5X=6X eggs at a price of 72 pennies per 6 eggs, that is 12 pennies per an egg. His income 12*6X=72X pennies 5) The profit = Income-Expenditure is given to be 1020. => 72X pennies-55X pennies= 17X pennies=1020 pennies => 17X=1020 => X=60 eggs So, the dealer bought 60 eggs at the first purchase. In total he bought 60+5*60=360 eggs. Answer: He bought 360 eggs Question 4 Draw the curve represented by y = x2 + 4x-21 stating and showing clearly the point where the curve turns and the points where it meets the axes. Solution: The function is quadratic, it means the graph will be a parabola. The coefficient of x2 is positive (a=1), it means the branches of parabola will be “looking upwards”. We find where the graph intersects the X and Y axes: 1) The graph intersects Y axis when x=0 => y=02+4*0–21=–21. That is, at the point (0; –21) 2) The graph intersects X axis when y=0 => x2+4x–21=0 => x1=–7 and x2=3 Tha

FAQs

How are the solutions for Question 1 organized, and why are there multiple methods?

Question 1 presents three methods (A, B, and C) to solve the same equation, showing factorization and equivalent equality steps. Full marks depend on correct calculations and clear work, regardless of method.

What simplification is used in Question 2 before substituting A=2 and B=-1?

Question 2 first simplifies AB multiplied by the expression involving A and B, then reduces the larger bracketed term to 4A^2. The overall expression becomes A+2B, making substitution straightforward.

How is the profit egg problem in Question 3 solved?

Question 3 defines X as the first number of eggs, then calculates per-egg costs for the two purchases using penny-per-egg rates. It computes income from total eggs sold and sets profit (income minus expenditure) equal to 1020p to solve for X and the total eggs.

Aug 25, 2026

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