12 CSEC Maths Answer Key

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1 1 CSEC Maths Answer Key 1 Computation No. Answers Further explanations 1 D In order to write a number in standard form it must be written in the form A 10 ±n, where 1 A < 10. B 3 B 4 D Therefore, to write in standard form we move the decimal point three places to the right. The number of decimal places moved gives us the value of n. If we move to the right, n will have a negative sign in front of it. The answer is, therefore, That is to say B A 7 D 8 C 9 A 3 8 Move decimal point 3 places to the right Move decimal point 3 places to the right

2 10 A 11 A 1 C D 14 D 15 C 16 D 17 B Answer Answer ( 5 4 ) 3 ( 1 8 ) A A 3.65 written to three significant figures is C The ratio is given as 3 : 5 : 7. There are, therefore, parts to share. 1 B Daniel and Victoria received 36 tokens. This represents a combined share of parts. 1 parts represents 36 tokens Therefore, 15 parts represents tokens. 1

3 D 3 A % 8 4 B Let the unknown number be x. Therefore, 40% of x is A 6 B 7 D 0.4x 10 x x % C 9 C 30 B Sandra s score 90% B A 33 B Then Therefore, A US$1.00 TT$6.50. Therefore, US$ TT$ D.5 metres centimetres 50 centimetres. 36 A 3500 millimetres 3500/1000 metres 3.5 metres. 3

4 37 C. tonnes kilograms 00 kilograms. 38 A ( ) 3 + ( 1) ( 1) 9 39 B ( 3 ) ( ) 40 A ( 3) + ( 1) ( 1) 8 Number Theory No. Answers Further explanations 1 C The first step is to find all the prime factors of 36, 60 and Therefore, The highest common factor (HCF) is found by multiplying the factors that repeat in all three numbers: HCF 3 1 4

5 A The first thing to do is to find the lowest common multiple (LCM) of, 6, 4 and 1 (the denominator of each fraction given). 3 B, 6, 4, 1 1, 3,, 6 3 1, 3, 1, 3 1, 1, 1, 1 The LCM of, 6, 4 and The next step is to write each fraction such that the denominator of each is the same: 1, 5 6, 3 4, 1 1 6, 10, 9, 1 1 Therefore, Arranging the fractions in ascending order: 1 1, 6 1, 9 1, Therefore, the final answer is 1 1, 1, 3 4, D 6, 9, 1 3, 9, 6 3 3, 9, 3 3 1, 3, 1 1, 1, 1 The lowest common multiple of 6, 9 and

6 5 A Multiples of :, 4, 6, 8, 10, 1, 14, 16, 18, 0,, 4, 6, 8, 30, 3, 34, 36, Multiples of 3: 3, 6, 9, 1, 15, 18, 1, 4, 7, 30, 33, 36, 39, Multiples of 4: 4, 8, 1, 16, 0, 4, 8, 3, 36, 40, The first three common multiples of, 3 and 4 are 1, 4 and A Using the distributive law: (18 + ) D 8 C In order to identify a prime number you can eliminate the ones that are not. 45 is divisible by 5 because the last digit is is divisible by because the last digit is 6 (even). 49 is divisible by 7. 9 C Therefore, 47 is the prime number. 10 D 11 B 1 B 13 C 14 A The given sequence is 4, 7, 11, 16. The difference between the consecutive terms is as follows: The difference between 4 and 7 is 3. The difference between 7 and 11 is 4. The difference between 11 and 16 is 5. Notice that the difference increases by 1. Therefore, the difference between 16 and the next term should be 6. So the next term in the sequence 4, 7, 11, 16 is ( ). 6

7 15 D Digit 3 is 3 hundredths A (00 + 5) (37 00) + (37 5) 17 C 18 D 365 can be written as Therefore, Consumer Arithmetic No. Answers Further explanations 1 B A Loan amount $ Total amount paid back on loan $500 1 $1 000 Interest $1 000 $ $000 3 B R 100 I % P T

8 4 C 5 A Interest received by Adam I P R T P R T T Interest received by Ann I Since Adam and Ann received the same interest: T T T 6000 T 6000/000 T 3 years $60 6 C Cash price $000 Total amount paid on hire purchase $900 + (1 $10) $900 + $1440 $340 Amount of money that can be saved if the television is bought at the cash price $340 $000 $340 7 A Amount paid in tax 500 $0.80 $000 8 B Taxable income $ Tax payable 5% $ $ C 10 C 10 Tax paid 10% $ $ Amount Alex has to pay the delivery man $150 + $15 $165 Change received $00 $165 $35 11 A Original price of tool $800 Price is increased by 1.5% Final price (1.5% 800) ( ) $900 1 A 8

9 13 D Discount 5% $ $70 Amount paid at the cashier $1400 $70 $ C Insurance for home $100 Insurance for contents of home $600 Insurance that will have to be paid $100 + $600 $ B Initial value of car $ Depreciation rate 10% per annum After the first year the value of the car ( ) $ After the second year the value of the car ( ) $ C Let the rate of depreciation be r%. Original price $x r r Value of car on December 31st x x ( x) x( ) The value of the car on December 31st is given as 0.95x. r Therefore, ( ) r 100 r r r 5% 17 C Let the number of units of electricity be x. Adam s bill $ (00 0.1) (x 00) x x x 45 x 45/0.09 x 500 units 18 C Profit 100% 0% 000 9

10 19 D Loan $ Amount paid back on loan $ B 1 C A Profit 100% 8% D Let the marked price be x. Discount 0% ($00) Therefore, 0% of x $00 4 B 5 B x 00 0x x 0 x $ D Fixed charge $50 Hourly rate $0 Total charge $130 Let the number of hours be x. Therefore, 0x x x 80 x 80/0 x 4 hours 7 C 10

11 4 Sets No. Answers Further explanations 1 B C 3 D 4 B 5 C There are 3 members of set A. Therefore, the number of subsets n D 7 B 8 C 9 B A B C represents all the members that are common to all sets: A B A B C C A B C {3} 11

12 10 D n(u) 35 n(c) 5 n(p) 5 x x x 11 B Let the number of students who study both subjects be x. Therefore: The number of students who study chemistry only 5 x The number of students who study physics only x 5 x + x + x x 35 x x 1 1 C 13 D 14 B 15 A 16 D 17 A n(s) 10 n(t) A From the diagram: n(s T) C 0 D 1

13 1 B P {, 4, 6, 8, 10, 1} Q {1,, 3, 4, 6, 8, 1, 4} A 3 B 4 D 5 B 6 D 7 C The diagram shows P Q Therefore, P Q {, 4, 6, 8, 1} 8 C If we look at the Venn diagram we can infer that: all students who study physics study maths, because physics is a subset of maths 9 B 30 C there are some students who study both accounts and maths, since these two sets intersect there are no students who study accounts and physics, because these two sets do not intersect. Therefore, the answer is I and III. 13

14 5 Measurement No. Answers Further explanations 1 B The perimeter of the triangle is 4 cm and the lengths of its sides are x, x and 3x. Therefore, x + x + 3x 4 6x 4 x 4 6 x 4 cm D The perimeter of the triangle is 6 cm and the lengths of its sides are (x 1), (x + 3) and (x + 6). Therefore, (x 1) + (x + 3) + (x + 6) 6 3x x x 6 8 3x 18 x A 4 B x 6 cm Length of the longest side (x + 6) cm. π 3 ( ) Perimeter of shape π 14

15 5 C Area of rectangle l b Given that l b 48 New length 1 l New breadth 3b New area 1 l 3b 3 lb But lb 48 Therefore, new area cm 6 D 7 C Area of outer circle with radius 8 cm ( ) Area of inner circle with radius 6 cm ( ) π 8 64π π 6 36π 8 A Area of shaded region 64π 36π 8π 9 A Area of parallelogram length DC perpendicular distance between lines AB and DC 10 C Perpendicular distance between the lines AB and DC 4 sin 60 Therefore, area of parallelogram 6 4 sin 60 cm 11 B A cube has six faces. The area of one face cm Therefore, the length of one side 9 3 cm 1 B Volume of a cube (side) cm 3 13 D 15

16 14 B Given that the circumference 44 cm The circumference of a circle is given by C πr Therefore, π r r π r π 15 D 16 B 17 A 18 D The length of an arc of a sector θ 60 π r B 0 C θ The area of a sector πr π 1 1π cm 360 AX 1 AB cm : 3 cm 1 C Using Pythagoras theorem: AX + OX OA 8 + OX 10 OX 10 8 OX OX 36 OX 36 OX 6 cm A 16

17 3 B 4 B 8 litres cm cm 3 5 B Each cup has a capacity of 50 cm 3 Therefore, the number of cups A Start time 06:0 End time 07:05 7 C Duration 07:05 06:0 45 minutes Converting 45 minutes into hours: hour 60 Distance speed time km 8 B 8 cm on the map represents an actual distance of 0 km. Therefore, the scale is determined as follows: 8 cm : 0 km 8 cm : m (Convert kilometres to metres by multiplying by 1000) 8 cm : cm (Convert metres to centimetres by multiplying by 100) 8 : (Divide both sides by 8) 1 :

18 9 C Each exterior angle in a regular polygon 360/n, where n is the number of sides in the regular polygon. It is given that each exterior angle of a regular polygon is Therefore, 60 n 60n 360 n n 6 (i.e. a hexagon) 6 Statistics No. Answers Further explanations 1 A A 3 A 4 A Length, x (cm) Frequency, f xf a b 8 16 Total 96 The sum of the xf column is 96. From the table, b 7a Summing the xf column and equating it to a a a a 14 a 14 7 a So 7 cm occurred twice. 18

19 5 A The mean of 5 numbers is x + 14 Therefore, x x x 35 9 x 6 6 B 7 7 D The mean of 6 numbers is 8. Therefore, 8 sum 6 8 B 9 A 10 B 11 C 1 C 13 B 14 A 15 B sum 8 6 sum 48 The number 15 is added to the 6 numbers. New sum , and there are now 7 numbers. New mean

20 16 D Rent 30%, food 0%, transportation 10%, savings? Savings 100% (30% + 0% + 10%) 100% 60% 40% Angle of sector representing savings 40% A 18 D 19 B 0 B 1 C Cumulative frequency Score In order to determine the median, read from the graph the value of the score when the cumulative frequency 50. Median score 5 B In order to determine the lower quartile, read from the graph the value of the score when the cumulative frequency 5. Lower quartile Q D In order to determine the upper quartile, read from the graph the value of the score when the cumulative frequency 75. Lower quartile Q A The interquartile range 1 (Q 3 Q 1 ) 1 (3 3) 4.5 0

21 5 C 6 B 7 D 8 C 9 B Height (cm) Number of seedlings The number of seedlings having a height less than 14 cm Total number of seedlings P(height less than 14 cm) C Since only one face of a die has a 5 on it, P(5) 1 6 There are three odd numbers on the faces of a die. They are 1, 3 and 5. Therefore, P(odd number) 3 6 P(getting a 5 followed by an odd number) P(5) P(odd number) C

22 7 Algebra No. Answers Further explanations 1 D 3(x ) ( 3 x) ( 3 ) 3x + 6 B x(x + 3y) y(x 4y) x + 6xy xy + 4y (Use the distributive law to remove the brackets) x + 4xy + 4y 3 C 7x x 18 9x A a 5(p q) a 5p 5q 5p a + 5q (5p) (a + 5q) (Multiplying both sides by ) 10p (a + 5q) 5 A x( + 3y) 3x(1 y) 4x + 6xy 3x + 6xy 4x 3x + 6xy + 6xy x + 1xy 6 A 7 A 3 * ab b a ( ) ( 3) B ( ) 3(ab ) 3 3ab 3ab x 3 x 7x 9 B ( ) + + 3pq p q 3 p q 6p q 10 A ( )( ) ( ) ( ) ( ) B ( ( ))

23 1 B 6(x ) 3(x 1) 6 (Use the distributive law to expand the left-hand side) 6x 1 3x x 9 6 3x x 15 3 (Adding 9 to both sides of the equation) (Dividing both sides of the equation by 3) 13 D 3x x x 5 3x 700 x x D 3x 5 x x x x 4 x 4 x 1 15 C (x 1) 6(x 1) x 1 6x x 6x x x 1 x 1 4 x 3 3

24 16 D 1 < x 3 < < x < < x < 7 17 A 3x 13 Therefore, 16 lies within this range. 3x x 15 x 15 3 x 5 18 D 15 < 5 3x 19 D 0 B 15 5 < 3x 10 < 3x 10 > 3x > x x < 40 1 A Nyla s current age (x 4) + Since Nyla is 6 times the age of Anya, C 3 A Anya s current age ( 4 ) ( 1 ) x + x x x

25 4 C Let one number be x and another number be y. The square of each number is x and y. 5 C 6 A 7 C The difference of two square numbers is x y. Two times the difference of two square numbers is (x y ). Given that the difference of two square numbers is negative, this means that (x y ) < 0. 8 B 5x 3y + x 5 y (The lowest common multiple of 3y and 5y is 15y) 15y 3y 5 15y 5y 3 9 B 5(5 x) + 3( x) 15y a 5b 5x + 6x 31x 15y 15y 7c d (The lowest common multiple of 5b and d is 10bd) 10bd 5b d 10bd d 5b d( a) 5 b(7 c) 10bd 4ad 35bc 10bd 30 C 5

26 31 A 1 y x 1 y x When x and y 9: 9 k 9 4k k 36 The equation relating x and y is y When x 3, y x 3 B 33 D 34 B 35 D 36 B 37 D y k y k x m y k y k x m x m x m k x (Divide both sides by k) (Square both sides) my (Cross-multiply) m kx (Divide both sides by y ) y 38 B 6

27 39 B 40 B 8 Relations, Functions and Graphs No. Answers Further explanations 1 D If an equation of a line is in the form y mx + c, then m is the gradient of the line. 1 y 6x y 6x 1 A y 3x + 1 Therefore, the gradient of the line is 3. 3 B 4 A 5 C 6 B If a point lies on the line y 3x + the coordinates must satisfy the equation. The point ( 1, 1) lies on the line because x 1 and y 1 satisfy the equation. 1 3( 1) C If the gradient of the line P is, then the gradient of any line parallel to P must be. 3y 6x + y x + 3 This line is parallel to P since its gradient m (from y mx + c) is also. 7

28 8 A The equation for the line R is y 3 x. y x + 3 y 1 x + 3 The gradient of R is therefore 1. When two lines are perpendicular, the product of their gradients is 1. Therefore, any line perpendicular to R must have a gradient m of 1 m 1 1 m 1 m. 9 C The line y x + 1, has a gradient of, and is therefore perpendicular to R. 10 D 11 C 1 A The line y 0 is the x-axis. The curve intersects the x-axis at x 3 and x C The minimum point of the curve is ( 1, 4). 14 B Reading off the values of x when y 3: x and x 0 15 D 16 D 17 A 18 C 19 B 0 B 8

29 1 C B 3 B 4 A 3x + 6x a(x + h) + k 3x + 6x a(x + hx + h ) + k 5 C 6 C 7 C 3x + 6x ax + ahx + ah + k Equating coefficients: a 3 ah 6 (3)h 6 h 6 6 h 1 ah + k (3)(1) + k 3 + k k 3 k 5 8 C A vertical line test is used to identify a function. When a vertical line is drawn it must intersect the curve at only one point in order to be a function. This is true for all the curves except C. 9 A 30 A If f(x) 3x Then, f( ) 3( ) 3(4) B 9

30 3 C 33 A 34 B 35 B 36 A 37 D 38 A 39 B f( 3) ( 3) A Let y x + 3 Interchange x and y: 41 B x y + 3 y x 3 x 3 y Therefore, f 1 (x) f 1 ( 1) 1 3 x A In order to find fg(), first find g(): g() () 4 Now find f(): f() () Therefore, fg() 7 30

31 43 C First find fg(x): f(x) x + 3 g(x) x fg(x) (x ) + 3 x x 1 Let y x 1 Interchange x and y: x y 1 y x + 1 y x + 1 y x + 1 (fg) 1 (x) x D In order to find the inverse of the function f(x), perform the following steps. Given that f(x) 3x Let y 3x Interchange x and y: x 3y 3y x 45 B 3y x + y x + 3 Therefore, f 1 (x) x A 47 B 31

32 48 D 49 D The shaded region lies between the lines y and y 1. Therefore, {(x, y): 1 y } 50 A 51 D 5 A 53 C 54 B 55 B 9 Geometry and Trigonometry No. Answers Further explanations 1 C B Use Pythagoras theorem to find the length AC: AC + BC AB AC AC AC AC 5 AC 5 AC 5 adjacent AC Therefore, cosθ hypotenuse AB

33 3 C 4 B sin ( 180 θ ) opposite hypotenuse PR PQ 5 C 6 D Use Pythagoras theorem to find the length AC: AC + BC AB 7 B AC + p ( p + 9 ) AC + p p + 9 AC p + 9 p AC 9 AC 9 AC 3 adjacent Therefore, cosθ hypotenuse AC AB 8 B Use Pythagoras theorem to find BC: AC + BC AB 9 A 8 + BC BC 100 BC BC 36 BC 36 BC 6 cm tan CDB opposite adjacent BC BD 6 5 p

34 10 B Use Pythagoras theorem to find BD: BD BC + CD 11 C 1 B 13 D 14 D 15 C 16 A 17 B 18 C 19 A 0 C 1 A BD BD BD 61 cos CBD adjacent hypotenuse BC BD APRQ 180 ( ) (Also, PSQ 35, as the angles subtended by an arc in the same segment of a circle are equal.) 34

35 3 D The angle subtended at the centre of a circle is twice the angle subtended at the circumference: θ A 5 C MOX LOM OML B 7 B 8 A DEB D Opposite angles of a cyclic quadrilateral: BCD C The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment: ABC A The angle subtended at the centre of a circle is twice the angle subtended at the circumference: AOC D PQ is a tangent to the circle at the point A. Therefore, PAO 90 : BAP D 34 B 35 C 36 C 37 B 38 A 35

36 39 A 40 A 41 D 4 B 43 A 44 C 45 C 46 A 47 D 48 B 49 B 50 B 51 C 5 D 53 A 54 B 55 B 56 C 57 D 58 A 59 A 36

37 60 C 10 Vectors and Matrices No. Answers Further explanations 1 D A + B ( ) A 3 A PQ (( 1 ) + ( 3 1) ) (( 1 1) + ( 3 4) ) (( 1 ) + ( 1) ) (( 1 1) + ( 4) ) ( ) ( ) A 5 B 6 B 37

38 7 D 8 A If a matrix is singular, the determinant 0. Therefore, (p ) (6 ) 0 4p ( 1) 0 4p p 1 9 C p 1 4 p 3 10 B Given that A a c b d The inverse A 1 1 A d c b a Therefore, the inverse of the matrix A 1 3 is

39 11 B Let A x y Therefore, A 1 ( 1 5 ) ( 3 ) Equation (1) ( 5 ) ( 6 ) Multiply both sides of Equation (1) by A 1 : x y x y (( 5 4) ( 1) ) (( 3 4) + ( 1 1) ) x y x y 1 Therefore, x and y 1. 1 B 13 A Given that P (, 3) P undergoes a translation of 1 3, and the image is P : P Therefore, the coordinates of P are ( 3, 6). 39

40 14 B 15 C 16 C 17 A P (, 1) and M P (( 0 ) + ( 1 1) ) (( 1 ) + ( 0 1) ) 18 D 19 A 0 B 1 The coordinates of the image P are ( 1, ) Let A be a rotation of 90 in a clockwise direction Let B be a reflection in the y-axis The magnitude of OA is OA units. 1 D AB AO + OB B 3 B From O to A, move 3 units to the right and 4 units upwards: 3 OA 4 40

41 4 C From O to B, move 4 units to the right and 1 unit upwards: 4 OB 1 5 C OA units 6 B AO is in the opposite direction of OA 3 AO 4 3 4, so you multiply by 1: 7 D AB AO + OB C 9 B 30 D 31 B 3 D Parallel vectors are scalar multiples of each other. Given the vector a b, the only parallel vectors are II a 4b, because this can be written as (a b). III 6b + 3a, because this can be written as 3(a b). 41

UNIT CSEC Multiple Choice Items Sample Paper 01

UNIT CSEC Multiple Choice Items Sample Paper 01 UNIT 0 Sample CSEC Multiple Choice Items and Revision Questions UNIT 0.. CSEC Multiple Choice Items Sample Paper 0 This paper consists of 60 Multiple Choice items from the Core Syllabus according to the