# 2015 Junior Certificate Higher Level Official Sample Paper 1

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 2015 Junior Certificate Higher Level Official Sample Paper 1 Question 1 (Suggested maximum time: 5 minutes) The sets U, P, Q, and R are shown in the Venn diagram below. (a) Use the Venn diagram to list the elements of: (i) P Q P Q is the set containing everything in P and everything in Q, so {1,2,3,4,5,6}. (ii) Q R Q R is the set containing everything that is in both Q and R, so {5,6}. (iii) P (Q R)

2 P (Q R) is the set containing everything in P, and also everything in both Q and R, so {1,2,4,5,6}. (b) Miriam says: For all sets, union is distributive over intersection. Name a set that you would use along with P (Q R) to show that Miriam s claim is true for the sets P, Q, and R in the Venn diagram above. Miriam s statement means that P (Q R) = (P Q) (P R). To show this, we would need to use the sets P Q and P R. So either of these sets P Q or P R is a valid answer to this question. Question 2 (Suggested maximum time: 5 minutes) The sets U, A, and B are defined as follows, where U is the universal set: U = {2,3,4,5,...,30} A = {multiples of 2} B = {multiples of 3} C = {multiples of 5} (a) Find #[(A B C) ] the number of elements in the complement of the set A B C. The elements in the set (A B C) will be the whole numbers between 2 and 30 inclusive which are not multiples of 2, 3 or 5. These numbers are {7,11,13,17,19,23,29}. This means that the number of elements #[(A B C) ] = 7. (b) How many divisors does each of the numbers in (A B C) have?

3 Each of these numbers has exactly two divisors: itself and 1. (c) What name is given to numbers that have exactly this many divisors? These numbers are called prime numbers. Question 3 (Suggested maximum time: 10 minutes) A group of 100 students were surveyed to find out whether they drank tea (T ), coffee (C), or a soft drink (D) at any time in the previous week. These are the results: 24 had not drunk any of the three 8 drank tea and a soft drink, but not coffee 51 drank tea or coffee, but not a soft drink 9 drank a soft drink and coffee 41 drank tea 20 drank at least two of the three 4 drank all three. (a) Represent the above information on the Venn diagram.

4 We suggest you draw out the Venn diagram on paper, and complete it as you go through the following explanation, to make it easier to follow the solution. We can fill in the following data from the above statements: 24 students are in (T C D), 8 students are in (T D) \C and 4 students are in T C D. We are told that 9 students are in D C. Since there are 4 in T D C, this means that there are 5 students in (D C) \ T. We are told that 20 students are in (T C) (C D) (D T ), so there must be = 3 students in (T C) \ D. We are told that 41 students are in T. This means that in T \ (C D), there must be = 26 students. We are told that 51 students are in (T C)\D, so there must be = 22 students in C \ (T D). Finally, the total number of students is 100, so the remaining section D \ (T C) must have = 8 students. Our completed Venn diagram:

5 ALTERNATE SOLUTION We will label the number of students in the separate sections of our Venn diagram as follows: We know the following data directly from the above statements: 24 students are not in any of the three sets, y = 8 and q = 4. We are told that 9 students drank a soft drink and coffee, so q + r = 9. This means that r = 5. We are told that 20 students drank at least two of the three, so 20 = y + p + q + r. This means that p = = 3. We are told that 41 drank tea, so x + y + p + q = 41. This means that x = = 26. We are told that 51 students drank tea or coffee, but no a soft drink, so x + y + z = 51. This means that z = = 22. Finally, the total number of students is 100, so x + y + z + p + q + r + s + 24 = 100. This means that s = = 8. Our completed Venn diagram:

6 (b) Find the probability that a student chosen at random from the group had drunk tea or coffee. The total number of students who had drunk tea or coffee, i.e. the number of students in T C, is = 68. Thus, there is a probability of = 0.68 of choosing such a student (c) Find the probability that a student chosen at random from the group had drunk tea and coffee but not a soft drink. The total number of students who had drunk tea and coffee but not a soft drink, i.e. the number of students in (T C) \ D, is 3. Thus, there is a probability of = 0.03 of choosing such a student Question 4 (Suggested maximum time: 10 minutes) Dermot has e5,000 and would like to invest it for two years. A special savings account is offering a rate of 3% for the first year and a higher rate for the second year, if the money is retained in the account. Tax of 41% will be deducted each year from the interest earned. (a) How much will the investment be worth at the end of one year, after tax is deducted? At the end of the first year, the gross interest earned will be 3% of the sum invested, so = e150. We must deduct 41% tax from the gross interest, so 150 ( ) 41 = = e88.50 is the net interest earned. Thus at the end of the first year, the investment will contain the original amount plus the net interest earned: = e5,088.50

7 (b) Dermot calculates that, after tax has been deducted, his investment will be worth e5, at the end of the second year. Calculate the rate of interest for the second year. Give your answer as a percentage, correct to one decimal place. Firstly, note that instead of calculating 41% of the interest and then subtracting the tax, to find the net interest we can just find 59% of the gross interest. The value of the investment at the start of the second year is the same as the value at the end of the first year. Let our new rate of interest be denoted by i%. The gross interest earned at the end of the second year will be i. The net interest earned at the end of the second year will be i Thus the value of the investment at the end of the second year will be the net interest plus the value at the start of the second year: ( i ) 100 ( i ) 100 = = i = i = i = ( ) Thus, the rate of interest is i = 4.5% correct to one decimal place. Question 5 (Suggested maximum time: 10 minutes) A meal in a restaurant cost Jerry e The price included VAT at 9%. Jerry wanted to know the price of the meal before the VAT was included. He calculated 9% of e30.52 and subtracted it from the cost of the meal. (a) Explain why Jerry will not get the correct answer using this method.

8 e30.52 represents the cost plus 9% VAT, so it is 109% of the price before VAT. 9% of e30.52 will therefore be = 9.81% of the price before VAT. When this value is subtracted from e30.52, Jerry will be left with = 99.19% of the cost of the meal before VAT. (b) Suppose that the rate of VAT was 13 5% instead of 9%. How much would Jerry have paid for the meal in that case? We divide e30.52 by 109%, giving = e28 which is the cost of the meal before VAT. We wish to add 13.5% to this figure, so we want 113.5% of e28 which is = e31.78 Question 6 (Suggested maximum time: 5 minutes) Niamh is in a clothes shop and has a voucher which she must use. The voucher gives a e10 reduction when buying goods to the value of at least e35. She also has e50 cash. (a) Write down an inequality in x to show the range of cash that she could spend in the shop. The voucher requires a minimum of e35 to be used. Since Niamh must use her voucher, she must spend a minimum of e35. If she spends this amount, she gets a e10 discount, meaning the minimum she can spend is e25. The maximum amount of cash she can spend is e50, so the range will be 25 x 50. (b) Niamh buys one item of clothing in the shop, using the voucher as she does so. Write an inequality in y to show the range of possible prices that this item could have, before the e10 reduction is applied.

9 As before, the minimum value Niamh can spend is e35. The maximum amount of cash she can spend is e50, and including the discount, the item of clothing could cost up to e60. Thus the range will be 35 y 60. Question 7 (Suggested maximum time: 15 minutes) A square with sides of length 10 units is shown in the diagram. A point A is chosen on a diagonal of the square, and two shaded squares are constructed as shown. By choosing different positions for A, it is possible to change the value of the total area of the two shaded squares. (a) Find the minimum possible value of the total area of the two shaded squares. Justify your answer fully.

10 The total area of the shaded region will be the sum of the areas of the two squares. The square to the left has sides of length A, and the square to the right has sides of length 10 A. Thus the total area will be: Area = A 2 + (10 A) 2 = A 2 + A 2 20A = 2A 2 20A We begin by substituting some values for A: This function looks like: A 2A 2 20A Area 0 2(0) 2 20(0) (1) 2 20(1) (2) 2 20(2) (3) 2 20(3) (4) 2 20(4) (5) 2 20(5) (6) 2 20(6) (7) 2 20(7) (8) 2 20(8) (9) 2 20(9) (10) 2 20(10) By inspecting the graph, we can see that the function is at its minimum when A = 5. Substituting this into our function for the area, we get the minimum possible value of the area is 50 units 2

11 (b) The diagram below shows the same square. The diagonal of one of the rectangles is also marked. The length of this diagonal is d. Show that the value of the total area of the two shaded squares is equal to d 2. Given that the overall shape is a square, the triangle formed in the bottom right corner must be a right-angled triangle. This triangle has sides of length A and 10 A, with hypotenuse d. Pythagoras Theorem tells us that d 2 = A 2 + (10 A) 2 and so the area of the shaded squares is the same as d 2. Question 8 (Suggested maximum time: 20 minutes) The first three stages of a pattern are shown below. Each stage of the pattern is made up of small squares. Each small square has an area of one square unit.

12 (a) Draw the next two stages of the pattern. Stage 4 Stage 5 (b) The perimeter of Stage 1 of the pattern is 4 units. The perimeter of Stage 2 of the pattern is 12 units. Find a general formula for the perimeter of Stage n of the pattern, where n N

13 We ll consider the perimeters of the first five Stages: Stage Perimeter 4 units 12 units 20 units 28 units 36 units From one stage to another, there is a difference of 8 in the perimeter. Thus, the perimeter is an arithmetic sequence with first element a = 4 and common difference d = 8. The general formula for the perimeter of Stage n is therefore P n = a + (n 1)d = 4 + 8(n 1) = 8n 4 (c) Find a general formula for the area of Stage n of the pattern, where n N

14 We will test to see if the sequence is linear by considering the first differences: d 1 = A 2 A 1 = 5 1 = 4 d 2 = A 3 A 2 = 13 5 = 8 d 3 = A 4 A 3 = = 12 d 4 = A 5 A 4 = = 16 The first differences are not constant, so the sequence is not arithmetic. We will look at the second differences: s 1 = d 2 d 1 = 8 4 = 4 s 2 = d 3 d 2 = 12 8 = 4 s 3 = d 4 d 3 = = 4 The second differences are constant, so the sequence is quadratic, and so has the general formula A n = an 2 + bn + c where a, b and c need to be determined. For a quadratic sequence of this form, the constant second difference is equal to 2a, so we have 2a = 4, or a = 2. This means our sequence has the form A n = 2n 2 + bn + c Now, we know that A 1 = 1 and A 2 = 5, so 1 = A 1 = 2(1) 2 + b(1) + c = 2 + b + c b + c = 1 5 = A 2 = 2(2) 2 + b(2) + c = 8 + 2b + c 2b + c = 3 We will solve these equations simultaneously: b + c = 1 2b + c = 3 b = 2 Now, from the first of our two simultaneous equations to see that ( 2) + c = 1 or c = 1+2 = 1. This means that our sequence has the form A n = 2n 2 2n+1

15 ALTERNATE SOLUTION It will be easier to split the patterns into an upper and a lower section. The first three Stages are: Stage 1 Stage 2 Stage 3 We will count the number of small squares in each section for the first five Stages Stage Upper 1 square 4 squares 9 squares 16 squares 25 squares Lower 0 squares 1 squares 4 squares 9 squares 16 squares Total Area 1 units 2 5 units 2 13 units 2 25 units 2 41 units 2 From this table, we can see that for stage n, there will be n 2 squares in the upper section and (n 1) 2 squares in the lower section. The general formula for the perimeter of Stage n is therefore A n = n 2 + (n 1) 2 = n 2 + n 2 2n + 1 = 2n 2 2n + 1 units 2 (d) What kind of sequence (linear, quadratic, exponential, or none of these) do the areas follow? Justify your answer. This sequence is quadratic because the highest power in the general formula is n 2. It cannot be exponential because there is no term of the form a n.

16 Question 9 (Suggested maximum time: 20 minutes) A plot consists of a rectangular garden measuring 8 m by 10 m, surrounded by a path of constant width, as shown in the diagram. The total area of the plot (garden and path) is 143 m 2. Three students, Kevin, Elaine, and Tony, have been given the problem of trying to find the width of the path. Each of them is using a different method, but all of them are using x to represent the width of the path. Kevin divides the path into eight pieces. He writes down the area of each piece in terms of x. He then forms an equation by setting the area of the path plus the area of the garden equal to the total area of the plot. (a) Write, in terms of x, the area of each section into Kevin s diagram below. We have four different types of sections in Kevin s diagram: the four corner squares, the rectangles at the top and bottom, the rectangles at the sides and the rectangle in the middle. the areas are given as follows: Corners: The area of these squares is x x = x 2 m 2 Top and Bottom: The area of these rectangles is x 8 = 8x m 2 Sides: The area of these rectangles is x 10 = 10x m 2 Centre: The area of this rectangle is 8 10 = 80 m 2 (b) Write down and simplify the equation that Kevin should get. Give your answer in the form ax 2 + bx + c = 0.

17 If we add the areas from all of Kevin s sections, we will get the total area, which is 143 m 2. 4(x 2 ) + 2(8x) + 2(10x) + 1(80) = 143 4x x + 20x + 80 = 143 4x x 63 = 0 Elaine writes down the length and width of the plot in terms of x. She multiplies these and sets the answer equal to the total area of the plot. (c) Write, in terms of x, the length and the width of the plot in the spaces on Elaine s diagram. The overall width of the plot consists of the width of the garden, plus the width of the path on both sides, so 8 + 2x. Similarly, the overall length of the plot consists of the length of the garden, plus the width of the path on both sides, so x. (d) Write down and simplify the equation that Elaine should get. Give your answer in the form ax 2 + bx + c = 0.

18 The overall width times the overall length will give the total area of the plot. So: (8 + 2x)(10 + 2x) = x + 20x + 4x 2 = 143 4x x 63 = 0 As we might have expected, this agrees with Kevin s calculations. (e) Solve an equation to find the width of the path. We can factorise our equation as follows: 4x x 63 = 0 (2x 3)(2x + 21) = 0 2x 3 = 0 or 2x + 21 = 0 Thus, either x = or x = 2. Since our path has to have a positive width, we take x = 1.5 m as the width of the path. (f) Tony does not answer the problem by solving an equation. Instead, he does it by trying out different values for x. Show some calculations that Tony might have used to solve the problem.

19 We will test some values for x. The simplest approach is to take x = 1,2,3,... until we find two points close the correct area. x = 1 Area = (8 + 2)(10 + 2) = 120 m 2 x = 2 Area = (8 + 4)(10 + 4) = 168 m 2 x = 3 Area = (8 + 6)(10 + 6) = 224 m 2 Since the total area is actually 143, the correct value of x should be between 1 and 2 because 120 < 143 < 168. We can try x = 1.5 as a guess (and in fact we already know this is the correct answer). x = 1.5 Area = (8 + 3)(10 + 3) = 143 m 2 (g) Which of the three methods do you think is best? Give a reason for your answer. Answer: Elaine s method is best. Reason: Although all three methods give the correct answer, Elaine s method is the quickest and simplest. Tony s method involves estimating the correct answer, which could give rise to some errors. Kevin s method requires several calculations before we arrive at the equation to solve. Question 10 (Suggested maximum time: 20 minutes) Part of the graph of the function y = x 2 + ax + b, where a,b Z, is shown below

20 The points R(2,3) and S( 5, 4) are on the curve. (a) Use the given points to form two equations in a and b. We know that if a point (x 1,y 1 ) is on a line, then x 1 and y 1 must satisfy the equation of that line. Thus: 3 = (2) 2 + a(2) + b 3 = 4 + 2a + b 2a + b = 1 Similarly, 4 = ( 5) 2 + a( 5) + b 4 = 25 5a + b 5a b = 29 (b) Solve your equations to find the value of a and the value of b.

21 We can solve the two above equations simultaneously. Adding both equations together we get: 2a + b = 1 5a b = 29 7a = 28 This means that a = 28 7 = 4. We can now go to the first equation: 2(4)+b = 1, or b = 1 8 = 9. This means that the function is of the form y = x 2 + 4x 9. (c) Write down the co-ordinates of the point where the curve crosses the y-axis. The curve will cross the y-axis when x = 0, so when y = (0) 2 + 4(x) 9 = 9 (d) By solving an equation, find the points where the curve crosses the x-axis. Give each answer correct to one decimal place.

22 The curve will cross the x-axis when y = 0, so when x 2 +4x 9 = 0. We will use the quadratic formula to solve for an equation of the form ax 2 + bx + c = 0 x = b ± b 2 4ac 2a x = (4) ± (4) 2 4(1)( 9) 2(1) x = 4 ± x = 4 ± 52 2 x = 4 ± x = 4 ± x = 2 ± 13 Thus, the curve crosses the x-axis at the points x = 1.6 and x = 5.6 correct to one decimal place. Question 11 (Suggested maximum time: 15 minutes) The graphs of two functions, f and g, are shown on the co-ordinate grid below. The functions are: f : x (x + 2) 2 4 g : x (x 3) 2 4

23 (a) Match the graphs to the functions by writing f or g beside the corresponding graph on the grid.

24 We will work out the values of f (0) and g(0) and use them to determine which graph is which. f (0) = (0 + 2) 2 4 = 4 4 = 0 and g(0) = (0 3) 2 4 = 9 4 = 5 Thus, examining the graph we can see that f is the graph on the left and g is the dashed graph to the right. (b) Write down the roots of f and the roots of g. By inspection of the graph, the roots of f are 4 and 0. Similarly, the roots of g are 1 and 5.

25 ALTERNATE SOLUTION Roots of f : we will manipulate the equation of f to determine its roots: (x + 2) 2 4 = 0 (x + 2) 2 = 4 x + 2 = ± 4 = ±2 Thus, the roots of f are x = 0 and x = 4. Roots of g: we will manipulate the equation of g to determine its roots: (x 3) 2 4 = 0 (x 3) 2 = 4 x 3 = ± 4 = ±2 Thus, the roots of g are x = 1 and x = 5. (c) Sketch the graph of h : x (x 1) 2 4 on the co-ordinate grid above, where x R.

26 Consider the functions f and g. Both of these functions are of the form (x c) 2 4. For f we have c = 2 and for g we have c = 3. The shapes of the two graphs are the same, but f is centred about the line x = 2, whereas g is centred about the line x = 3. If we consider the graph of f, to get the graph of g, we shift the axis of symmetry of f from x = 2 to x = 3, a shift of 3 ( 2) = 5 units to the right. Thus, changing the value of c shifts the axis of symmetry along the x-axis. Now, since h(x) = (x 1) 2 4, so c = 1. Again, if we consider the graph of f, to get the graph of h, we shift the axis of symmetry of f by 1 ( 2) = 3 units to the right. This gives:

27 ALTERNATE SOLUTION We begin by substituting some values for x: Plotting these points, we get x h(x) y 2 ( 2 1) ( 1 1) (0 1) (1 1) (2 1) (3 1) (4 1) (d) p is a natural number, such that (x p) 2 2 = x 2 10x Find the value of p.

28 Firstly, we ll rearrange our equation to read (x p) 2 = x 2 10x + 25 Now, we can factorise the right hand side as x 2 10x+25 = (x 5) 2, so we have (x p) 2 = (x 5) 2 and thus p = 5. (e) Write down the equation of the axis of symmetry of the graph of the function: k(x) = x 2 10x + 23 We know from part (d) that we can write k(x) = (x 5) 2 2. Now, the functions of f,g and h, are all of the form (x c) 2 4, and we can see that the axis of symmetry for f is x = 2, for g it is x = 3 and for h it is x = 1, i.e. in all cases x = c is the axis of symmetry. Now, we know that k(x) = (x 5) 2 2, so in the same way, the axis of symmetry for k(x) will be x = 5. Question 12 (Suggested maximum time: 5 minutes) Give a reason why the graph below does not represent a function of x. For any function, each x value has no more than one value for y. However, we can see from this graph that almost all x points have two y values. So this graph does not represent a function.

### Coimisiún na Scrúduithe Stáit State Examinations Commission. Junior Certificate Examination 2015 Sample Paper. Mathematics

2015. S34 S Coimisiún na Scrúduithe Stáit State Examinations Commission Junior Certificate Examination 2015 Mathematics Paper 1 Higher Level Time: 2 hours, 30 minutes 300 marks Examination number Running

### Algebra Revision Sheet Questions 2 and 3 of Paper 1

Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the

### Park Forest Math Team. Meet #5. Algebra. Self-study Packet

Park Forest Math Team Meet #5 Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number

### Teaching & Learning Plans. Quadratic Equations. Junior Certificate Syllabus

Teaching & Learning Plans Quadratic Equations Junior Certificate Syllabus The Teaching & Learning Plans are structured as follows: Aims outline what the lesson, or series of lessons, hopes to achieve.

### Chapter 8. Quadratic Equations and Functions

Chapter 8. Quadratic Equations and Functions 8.1. Solve Quadratic Equations KYOTE Standards: CR 0; CA 11 In this section, we discuss solving quadratic equations by factoring, by using the square root property

### MATHEMATICS SOLUTIONS Junior Certificate Higher Level Contents

MATHEMATICS SOLUTIONS Junior Certificate Higher Level Contents Paper 1 015 SEC Sample Paper 1 (Phase )... Sample Paper 1 Educate.ie Paper 1 (Phase )... 7 Sample Paper Educate.ie Paper 1 (Phase )... 14

### KS4 Curriculum Plan Maths FOUNDATION TIER Year 9 Autumn Term 1 Unit 1: Number

KS4 Curriculum Plan Maths FOUNDATION TIER Year 9 Autumn Term 1 Unit 1: Number 1.1 Calculations 1.2 Decimal Numbers 1.3 Place Value Use priority of operations with positive and negative numbers. Simplify

### Foundation. Scheme of Work. Year 10 September 2016-July 2017

Foundation Scheme of Work Year 10 September 016-July 017 Foundation Tier Students will be assessed by completing two tests (topic) each Half Term. PERCENTAGES Use percentages in real-life situations VAT

### 2014 Leaving Cert Ordinary Level Official Sample Paper 1

2014 Leaving Cert Ordinary Level Official Sample Paper 1 Section A Concepts and Skills 150 marks Question 1 (25 marks) (a) Write 6 2 and 81 1 2 without using indices. Given the relation x 1 = 1 x, we have

### Individual Round Arithmetic

Individual Round Arithmetic (1) A stack of 100 nickels is 6.25 inches high. To the nearest \$.01, how much would a stack of nickels 8 feet high be worth? 8 feet = 8 12 inches. Dividing 96 inches by 6.25

### The Not-Formula Book for C1

Not The Not-Formula Book for C1 Everything you need to know for Core 1 that won t be in the formula book Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes

### Math 181 Spring 2007 HW 1 Corrected

Math 181 Spring 2007 HW 1 Corrected February 1, 2007 Sec. 1.1 # 2 The graphs of f and g are given (see the graph in the book). (a) State the values of f( 4) and g(3). Find 4 on the x-axis (horizontal axis)

### CAMI Education linked to CAPS: Mathematics

- 1 - TOPIC 1.1 Whole numbers _CAPS Curriculum TERM 1 CONTENT Properties of numbers Describe the real number system by recognizing, defining and distinguishing properties of: Natural numbers Whole numbers

### 3.2. Solving quadratic equations. Introduction. Prerequisites. Learning Outcomes. Learning Style

Solving quadratic equations 3.2 Introduction A quadratic equation is one which can be written in the form ax 2 + bx + c = 0 where a, b and c are numbers and x is the unknown whose value(s) we wish to find.

### MATHS LEVEL DESCRIPTORS

MATHS LEVEL DESCRIPTORS Number Level 3 Understand the place value of numbers up to thousands. Order numbers up to 9999. Round numbers to the nearest 10 or 100. Understand the number line below zero, and

### 1.6 Powers of 10 and standard form Write a number in standard form. Calculate with numbers in standard form.

Unit/section title 1 Number Unit objectives (Edexcel Scheme of Work Unit 1: Powers, decimals, HCF and LCM, positive and negative, roots, rounding, reciprocals, standard form, indices and surds) 1.1 Number

### Exam 2 Review. 3. How to tell if an equation is linear? An equation is linear if it can be written, through simplification, in the form.

Exam 2 Review Chapter 1 Section1 Do You Know: 1. What does it mean to solve an equation? To solve an equation is to find the solution set, that is, to find the set of all elements in the domain of the

### Factoring Polynomials

UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

Definition of the Quadratic Formula The Quadratic Formula uses the a, b and c from numbers; they are the "numerical coefficients"., where a, b and c are just The Quadratic Formula is: For ax 2 + bx + c

Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function

### Lesson 3.2 Exercises, pages

Lesson 3.2 Exercises, pages 190 195 Students should verify all the solutions. A 4. Which equations are quadratic equations? Explain how you know. a) 3x 2 = 30 b) x 2-9x + 8 = 0 This equation is quadratic

### FACTORING QUADRATICS 8.1.1 and 8.1.2

FACTORING QUADRATICS 8.1.1 and 8.1.2 Chapter 8 introduces students to quadratic equations. These equations can be written in the form of y = ax 2 + bx + c and, when graphed, produce a curve called a parabola.

9.3 Solving Quadratic Equations by Using the Quadratic Formula 9.3 OBJECTIVES 1. Solve a quadratic equation by using the quadratic formula 2. Determine the nature of the solutions of a quadratic equation

### ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite

ALGEBRA Pupils should be taught to: Generate and describe sequences As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: sequence, term, nth term, consecutive, rule,

### Frogs, Fleas and Painted Cubes: Homework Examples from ACE

Frogs, Fleas and Painted Cubes: Homework Examples from ACE Investigation 1: Introduction to Quadratic Functions, ACE #7, 14 Investigation 2: Quadratic Expressions, ACE #17, 24, 26, 36, 51 Investigation

### 1.1 Practice Worksheet

Math 1 MPS Instructor: Cheryl Jaeger Balm 1 1.1 Practice Worksheet 1. Write each English phrase as a mathematical expression. (a) Three less than twice a number (b) Four more than half of a number (c)

### Decide how many topics you wish to revise at a time (let s say 10).

1 Minute Maths for the Higher Exam (grades B, C and D topics*) Too fast for a first-time use but... brilliant for topics you have already understood and want to quickly revise. for the Foundation Exam

### Mathematics Chapter 8 and 10 Test Summary 10M2

Quadratic expressions and equations Expressions such as x 2 + 3x, a 2 7 and 4t 2 9t + 5 are called quadratic expressions because the highest power of the variable is 2. The word comes from the Latin quadratus

### In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data.

MATHEMATICS: THE LEVEL DESCRIPTIONS In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data. Attainment target

### Time Topic What students should know Mathswatch links for revision

Time Topic What students should know Mathswatch links for revision 1.1 Pythagoras' theorem 1 Understand Pythagoras theorem. Calculate the length of the hypotenuse in a right-angled triangle. Solve problems

### THE DISTANCE FORMULA

THE DISTANCE FORMULA In this activity, you will develop a formula for calculating the distance between any two points in a coordinate plane. Part 1: Distance Along a Horizontal or Vertical Line To find

### a) x 2 8x = 25 x 2 8x + 16 = (x 4) 2 = 41 x = 4 ± 41 x + 1 = ± 6 e) x 2 = 5 c) 2x 2 + 2x 7 = 0 2x 2 + 2x = 7 x 2 + x = 7 2

Solving Quadratic Equations By Square Root Method Solving Quadratic Equations By Completing The Square Consider the equation x = a, which we now solve: x = a x a = 0 (x a)(x + a) = 0 x a = 0 x + a = 0

### Introduction. The Aims & Objectives of the Mathematical Portion of the IBA Entry Test

Introduction The career world is competitive. The competition and the opportunities in the career world become a serious problem for students if they do not do well in Mathematics, because then they are

### Solving Quadratic Equations by Completing the Square

9. Solving Quadratic Equations by Completing the Square 9. OBJECTIVES 1. Solve a quadratic equation by the square root method. Solve a quadratic equation by completing the square. Solve a geometric application

### Squares and Square Roots

SQUARES AND SQUARE ROOTS 89 Squares and Square Roots CHAPTER 6 6.1 Introduction You know that the area of a square = side side (where side means the length of a side ). Study the following table. Side

### Pre-Algebra - MA1100. Topic Lesson Objectives. Variables and Expressions

Variables and Expressions Problem Solving: Using a Problem-Solving Plan Use a four-step plan to solve problems. Choose an appropriate method of computation. Numbers and Expressions Use the order of operations

### Introduction to Modular Arithmetic, the rings Z 6 and Z 7

Introduction to Modular Arithmetic, the rings Z 6 and Z 7 The main objective of this discussion is to learn modular arithmetic. We do this by building two systems using modular arithmetic and then by solving

### Transformations Packet Geometry

Transformations Packet Geometry 1 Name: Geometry Chapter 14: Transformations Date Due Section Topics Assignment 14.1 14.2 14.3 Preimage Image Isometry Mapping Reflections Note: Notation is different in

### 3.1/3.2: Solving Inequalities and Their Graphs

3.1/3.2: Solving Inequalities and Their Graphs - Same as solving equations - adding or subtracting a number will NOT change the inequality symbol. - Always go back and check your solution - Solutions are

### CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA

We Can Early Learning Curriculum PreK Grades 8 12 INSIDE ALGEBRA, GRADES 8 12 CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA April 2016 www.voyagersopris.com Mathematical

### Key Topics What will ALL students learn? What will the most able students learn?

2013 2014 Scheme of Work Subject MATHS Year 9 Course/ Year Term 1 Key Topics What will ALL students learn? What will the most able students learn? Number Written methods of calculations Decimals Rounding

### REVISED GCSE Scheme of Work Mathematics Higher Unit T3. For First Teaching September 2010 For First Examination Summer 2011

REVISED GCSE Scheme of Work Mathematics Higher Unit T3 For First Teaching September 2010 For First Examination Summer 2011 Version 1: 28 April 10 Version 1: 28 April 10 Unit T3 Unit T3 This is a working

### Geometry and Arithmetic

Geometry and Arithmetic Alex Tao 10 June 2008 1 Rational Points on Conics We begin by stating a few definitions: A rational number is a quotient of two integers and the whole set of rational numbers is

### CAMI Education linked to CAPS: Mathematics

- 1 - TOPIC 1.1 Whole numbers _CAPS curriculum TERM 1 CONTENT Mental calculations Revise: Multiplication of whole numbers to at least 12 12 Ordering and comparing whole numbers Revise prime numbers to

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### 2. INEQUALITIES AND ABSOLUTE VALUES

2. INEQUALITIES AND ABSOLUTE VALUES 2.1. The Ordering of the Real Numbers In addition to the arithmetic structure of the real numbers there is the order structure. The real numbers can be represented by

### 7-2 Solving Exponential Equations and Inequalities. Solve each equation y 3 = 4 y + 1 SOLUTION:

Solve each equation. 2. 16 2y 3 = 4 y + 1 4. 49 x + 5 = 7 8x 6 6. A certificate of deposit (CD) pays 2.25% annual interest compounded biweekly. If you deposit \$500 into this CD, what will the balance be

### Numeracy and mathematics Experiences and outcomes

Numeracy and mathematics Experiences and outcomes My learning in mathematics enables me to: develop a secure understanding of the concepts, principles and processes of mathematics and apply these in different

### Mathematics Improving your Key Stage 3 test result

Mathematics Improving your Key Stage 3 test result A booklet of helpful hints, tips and a revision checklist Contents Page Why are KS3 tests important? 2 The Mathematics test explained 2 How can you support

### Year 9 mathematics test

Ma KEY STAGE 3 Year 9 mathematics test Tier 6 8 Paper 1 Calculator not allowed First name Last name Class Date Please read this page, but do not open your booklet until your teacher tells you to start.

### MAT 080-Algebra II Applications of Quadratic Equations

MAT 080-Algebra II Applications of Quadratic Equations Objectives a Applications involving rectangles b Applications involving right triangles a Applications involving rectangles One of the common applications

### Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if

1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c

### TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers

TYPES OF NUMBERS When two or more integers are multiplied together, each number is a factor of the product. Nonnegative integers that have exactly two factors, namely, one and itself, are called prime

### 3. Polynomials. 3.1 Parabola primer

3. Polynomials 3.1 Parabola primer A parabola is the graph of a quadratic (degree 2) polynomial, that is, a parabola is polynomial of the form: standard form: y = ax 2 + bx + c. When we are given a polynomial

### REVIEW SHEETS INTERMEDIATE ALGEBRA MATH 95

REVIEW SHEETS INTERMEDIATE ALGEBRA MATH 95 A Summary of Concepts Needed to be Successful in Mathematics The following sheets list the key concepts which are taught in the specified math course. The sheets

### C A R I B B E A N E X A M I N A T I O N S C O U N C I L REPORT ON CANDIDATES WORK IN THE SECONDARY EDUCATION CERTIFICATE EXAMINATION MAY/JUNE 2011

C A R I B B E A N E X A M I N A T I O N S C O U N C I L REPORT ON CANDIDATES WORK IN THE SECONDARY EDUCATION CERTIFICATE EXAMINATION MAY/JUNE 2011 MATHEMATICS GENERAL PROFICIENCY EXAMINATION Copyright

### Actually, if you have a graphing calculator this technique can be used to find solutions to any equation, not just quadratics. All you need to do is

QUADRATIC EQUATIONS Definition ax 2 + bx + c = 0 a, b, c are constants (generally integers) Roots Synonyms: Solutions or Zeros Can have 0, 1, or 2 real roots Consider the graph of quadratic equations.

### Algebra C/D Borderline Revision Document

Algebra C/D Borderline Revision Document Algebra content that could be on Exam Paper 1 or Exam Paper 2 but: If you need to use a calculator for *** shown (Q8 and Q9) then do so. Questions are numbered

### Infinite Algebra 1 supports the teaching of the Common Core State Standards listed below.

Infinite Algebra 1 Kuta Software LLC Common Core Alignment Software version 2.05 Last revised July 2015 Infinite Algebra 1 supports the teaching of the Common Core State Standards listed below. High School

Name: Period: 10.1 Notes-Graphing Quadratics Section 1: Identifying the vertex (minimum/maximum), the axis of symmetry, and the roots (zeros): State the maximum or minimum point (vertex), the axis of symmetry,

### Edexcel AS and A-level Modular Mathematics

Core Mathematics Edexcel AS and A-level Modular Mathematics Greg Attwood Alistair Macpherson Bronwen Moran Joe Petran Keith Pledger Geoff Staley Dave Wilkins Contents The highlighted sections will help

### The x-intercepts of the graph are the x-values for the points where the graph intersects the x-axis. A parabola may have one, two, or no x-intercepts.

Chapter 10-1 Identify Quadratics and their graphs A parabola is the graph of a quadratic function. A quadratic function is a function that can be written in the form, f(x) = ax 2 + bx + c, a 0 or y = ax

### Functions and Equations

Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c

### Lines That Pass Through Regions

: Student Outcomes Given two points in the coordinate plane and a rectangular or triangular region, students determine whether the line through those points meets the region, and if it does, they describe

### Vocabulary Words and Definitions for Algebra

Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms

### AQA Level 2 Certificate FURTHER MATHEMATICS

AQA Qualifications AQA Level 2 Certificate FURTHER MATHEMATICS Level 2 (8360) Our specification is published on our website (www.aqa.org.uk). We will let centres know in writing about any changes to the

### Section 2.1 Rectangular Coordinate Systems

P a g e 1 Section 2.1 Rectangular Coordinate Systems 1. Pythagorean Theorem In a right triangle, the lengths of the sides are related by the equation where a and b are the lengths of the legs and c is

### Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.

Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve word problems that call for addition of three whole numbers

### Grade 6 Math Circles March 24/25, 2015 Pythagorean Theorem Solutions

Faculty of Mathematics Waterloo, Ontario NL 3G1 Centre for Education in Mathematics and Computing Grade 6 Math Circles March 4/5, 015 Pythagorean Theorem Solutions Triangles: They re Alright When They

### Second Year Algebra. Gobel, Gordon, et. al. Walla Walla High School

Second Year Algebra Gobel, Gordon, et al Walla Walla High School c 010 Contents 1 Quadratic Equations and Functions 1 11 Simple quadratic equations 1 Solving quadratic equations by factoring 3 13 Completing

### Chapter 8 Graphs and Functions:

Chapter 8 Graphs and Functions: Cartesian axes, coordinates and points 8.1 Pictorially we plot points and graphs in a plane (flat space) using a set of Cartesian axes traditionally called the x and y axes

### Algebra: Real World Applications and Problems

Algebra: Real World Applications and Problems Algebra is boring. Right? Hopefully not. Algebra has no applications in the real world. Wrong. Absolutely wrong. I hope to show this in the following document.

### Year 11 - GCSE Higher Tier. Revision Solutions

Year 11 - GCSE Higher Tier Revision 2011 - Solutions Transformations Reflect the triangle in the line x = - 2 Translate this triangle 2 right and 3 up Reflect the triangle in the line y = 5 Rotate this

### MATH 100 PRACTICE FINAL EXAM

MATH 100 PRACTICE FINAL EXAM Lecture Version Name: ID Number: Instructor: Section: Do not open this booklet until told to do so! On the separate answer sheet, fill in your name and identification number

### IOWA End-of-Course Assessment Programs. Released Items ALGEBRA I. Copyright 2010 by The University of Iowa.

IOWA End-of-Course Assessment Programs Released Items Copyright 2010 by The University of Iowa. ALGEBRA I 1 Sally works as a car salesperson and earns a monthly salary of \$2,000. She also earns \$500 for

### Understanding Basic Calculus

Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other

### Higher Education Math Placement

Higher Education Math Placement Placement Assessment Problem Types 1. Whole Numbers, Fractions, and Decimals 1.1 Operations with Whole Numbers Addition with carry Subtraction with borrowing Multiplication

### Step 1 MATHS. To achieve Step 1 in Maths students must master the following skills and competencies: Number. Shape. Algebra

MATHS Step 1 To achieve Step 1 in Maths students must master the following skills and competencies: Number Add and subtract positive decimal numbers Add and subtract negative numbers in context Order decimal

### GCSE Mathematics Calculator Foundation Tier Free Practice Set 6 1 hour 30 minutes ANSWERS. Marks shown in brackets for each question (2)

MathsMadeEasy 3 GCSE Mathematics Calculator Foundation Tier Free Practice Set 6 1 hour 30 minutes ANSWERS Marks shown in brackets for each question Typical Grade Boundaries C D E F G 76 60 47 33 20 Legend

476 CHAPTER 7 Graphs, Equations, and Inequalities 7. Quadratic Equations Now Work the Are You Prepared? problems on page 48. OBJECTIVES 1 Solve Quadratic Equations by Factoring (p. 476) Solve Quadratic

### CHAPTER 35 GRAPHICAL SOLUTION OF EQUATIONS

CHAPTER 35 GRAPHICAL SOLUTION OF EQUATIONS EXERCISE 143 Page 369 1. Solve the simultaneous equations graphically: y = 3x 2 y = x + 6 Since both equations represent straight-line graphs, only two coordinates

### Simultaneous Equations

Paper Reference(s) 666/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Simultaneous Equations Calculators may NOT be used for these questions. Information for Candidates A booklet Mathematical Formulae

### Bridging Documents for Mathematics

Bridging Documents for Mathematics 5 th /6 th Class, Primary Junior Cycle, Post-Primary Primary Post-Primary Card # Strand(s): Number, Measure Number (Strand 3) 2-5 Strand: Shape and Space Geometry and

### RIT scores between 191 and 200

Measures of Academic Progress for Mathematics RIT scores between 191 and 200 Number Sense and Operations Whole Numbers Solve simple addition word problems Find and extend patterns Demonstrate the associative,

### Solutions for 2015 Algebra II with Trigonometry Exam Written by Ashley Johnson and Miranda Bowie, University of North Alabama

Solutions for 0 Algebra II with Trigonometry Exam Written by Ashley Johnson and Miranda Bowie, University of North Alabama. For f(x = 6x 4(x +, find f(. Solution: The notation f( tells us to evaluate the

### 2. THE x-y PLANE 7 C7

2. THE x-y PLANE 2.1. The Real Line When we plot quantities on a graph we can plot not only integer values like 1, 2 and 3 but also fractions, like 3½ or 4¾. In fact we can, in principle, plot any real

### Expression. Variable Equation Polynomial Monomial Add. Area. Volume Surface Space Length Width. Probability. Chance Random Likely Possibility Odds

Isosceles Triangle Congruent Leg Side Expression Equation Polynomial Monomial Radical Square Root Check Times Itself Function Relation One Domain Range Area Volume Surface Space Length Width Quantitative

### Instructions for SA Completion

Instructions for SA Completion 1- Take notes on these Pythagorean Theorem Course Materials then do and check the associated practice questions for an explanation on how to do the Pythagorean Theorem Substantive

### Biggar High School Mathematics Department. National 4 Learning Intentions & Success Criteria: Assessing My Progress

Biggar High School Mathematics Department National 4 Learning Intentions & Success Criteria: Assessing My Progress Expressions and Formulae Topic Learning Intention Success Criteria I understand this Algebra

### Curriculum Overview YR 9 MATHS. SUPPORT CORE HIGHER Topics Topics Topics Powers of 10 Powers of 10 Significant figures

Curriculum Overview YR 9 MATHS AUTUMN Thursday 28th August- Friday 19th December SUPPORT CORE HIGHER Topics Topics Topics Powers of 10 Powers of 10 Significant figures Rounding Rounding Upper and lower

### Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress

Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation

### Remember that the information below is always provided on the formula sheet at the start of your exam paper

Maths GCSE Linear HIGHER Things to Remember Remember that the information below is always provided on the formula sheet at the start of your exam paper In addition to these formulae, you also need to learn

### Specimen paper MATHEMATICS HIGHER TIER. Time allowed: 2 hours. GCSE BITESIZE examinations. General Certificate of Secondary Education

GCSE BITESIZE examinations General Certificate of Secondary Education Specimen paper MATHEMATICS HIGHER TIER 2005 Paper 1 Non-calculator Time allowed: 2 hours You must not use a calculator. Answer all