National 5 Mathematics. Practice Paper E. Worked Solutions

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1 National 5 Mathematics Practice Paper E Worked Solutions Paper One: Non-Calculator Copyright All rights reserved.

2 SQA Past Papers & Specimen Papers Working through SQA Past Papers & Specimen Papers are probably the best practice you can get for the actual exam so you should plan to do as many as possible. Make sure you practice doing a whole paper in the allocated time so you can get use to the pace. The best way to use this guide is for checking your answers after you have tried the questions yourself. Please don t just read the solutions whenever you get stuck! If you find the resources helpful to your deeper understanding of National 5 Mathematics, as well as helping you better prepare for the final exam, then please let others know about us at thank you! The contents of these worked solutions have not been checked or approved by the Scottish Qualifications Authority. They reflect the authors opinions of good answers to exam questions and where possible have been checked against publicly available marking instructions. Copyright You may use this resource for personal use only. No part of these resources may be copied, reproduced in any format (including electronic), put on the World Wide Web or in any way shared without the express permission of the authors in accordance with the Copyright, Design and Patents Act of Any person or organisation who makes unauthorised copies of any part of these resources may be liable to prosecution and possible civil claims for damages [2]

3 Change 1 st & 3 rd fraction to improper (top heavy): Cancel diagonally the 5 ss to give 1 ss: BIDMAS (calculate multiply before add): Multiply the first fraction by 2 2 : Denominators now the same: Add numerators, denominators stay same: 21 6 Change to a mixed number: Alternative Method With reference to line 4 above (sometimes known as the Smile & Kiss method): Multiply the 3 6 and place 18 on the denominator Multiply 6 7 and place 42 on the numerator (top left) Multiply 3 7 and place 21 on the numerator (top right) as below: = = = 7 3 = [3]

4 2. (4xx + 2)(xx 5) + 3xx Multiply out the brackets: 4xx(xx 5) + 2(xx 5) + 3xx Multiply out the brackets again: 4xx 2 20xx + 2xx xx Simplify: 4xx 2 15xx Taking two sets of points on the line: (1, 4) & (2, 2) (xx1, yy1) & ( xx2, yy2) Substitute xx1 = 1, yy1 = 4, xx2 = 2, yy2 = 2 into the gradient formula: mm = yy2 yy1 = 2 4 = 2 = 2 xx2 xx Take two points on the line (1, 4), let aa = 1, bb = 4. Substitute aa, bb & mm into: yy bb = mm(xx aa) yy 4 = 2(xx 1) Multiply out the brackets: yy 4 = 2xx + 2 Take 4 to the RHS: yy = 2xx Simplify: yy = 2xx + 6 Alternative Method Plot the points from the table onto a graph, find the yy intercept (cc) and substitute mm & cc into yy = mmmm + cc to obtain the same answer [4]

5 4. Multiply by xx: Take 16xx to the LHS & 2 to the RHS: Simplify: 2 xx + 9 = xx = 16xx 16xx + 9xx = 2 7xx = 2 Multiply by 1: 7xx = 2 Divide by 7: xx = Compare: 2xx 2 2xx 1 = 0 with: aaxx 2 + bbbb + cc = 0 Therefore aa = 2, bb = 2, cc = 1. Substitute aa, bb & cc into the quadratic formula given in the Formulae List: xx = bb ± bb2 4aaaa 2aa Substituting aa, bb & cc: xx = ( 2 )± ( 2)2 4 2 ( 1) 2 2 Simplify: = 4 ± Simplify again: = 4 ± 12 4 Substitute 12 = 4 3 = 2 3 into the above: = 4 ± Factorise 4 ± 2 3: = 2(1 ± 3 ) 4 Cancel down the 2 & 4: = 1 ± [5]

6 6. (a) yy = 36 (xx 2) 2 Swap positions of 36 & (xx 2) 2 : = (xx 2) Turning point = (2, 36) (b) xx = 2 is the equation of the axis of symmetry (c) Since the turning point is (2, 36) and SS is (6, 20) then the difference along the xx-axis between SS and the turning point is 6 2 = [6]

7 Due to symmetry, there will also be a difference of 4 between the turning point and RR The xx coordinate of RR will therefore be 2 4 = 2 The yy coordinate of RR will be the same as the yy coordinate of SS since the line is horizontal, yy coordinate of SS = 20 Coordinates RR ( 2, 20) Notes In the equation yy = (xx 2) The negative at the front indicates the parabola is a maximum shape (as opposed to mimimum) The first number ( 2) in the equation is the opposite sign to the xx on the turning point: xx = ( 2) = 2 The second number in the equation (36) is the same sign to the yy on the turning point: yy = 36 The equation of axis of symmetry is the dotted line above: xx = [7]

8 7. With reference to the triangle at the top right in the above shape: 4 cm xx 5 cm (radius) Pythagoras Theorem: xx 2 = Simplify: xx 2 = Simply again: xx 2 = 9 Take the square root: xx = 9 With reference to the triangle at the bottom left in the above shape: yy 2 = cm 4 cm (from 7-3) yy 2 = yy 2 = 9 yy = 3 Width of base = 2 yy = 2 3 = 6 cccc [8]

9 8. yy = ssssss 2xx 0 Notes yy = ssssss 2xx 0 represents a sine wave of amplitude 1 (number of front of the ssssss 2xx 0 ) 2 (number in front of the xx) is the number of complete cycles in ff(xx) = 4 xx + 2 (a) Substitute xx = 72 into the equation: ff(72) = Swap 72 for 36 2 Simplify 36 = 6: Simplify again: Simplify again: = = = = [9]

10 (b) Substitute xx for tt: ff(tt) = 4 tt + 2 Substitute ff(tt) for 3 2: 3 2 = 4 tt + 2 Swap sides: 4 tt + 2 = 3 2 Take 2 to the RHS: 4 tt = Simplify: 4 tt = 2 2 Divide by 4: tt = Square both sides: tt = Multiply numerators & denominators: tt = Simplify: tt = [10]

11 10. (2xx 5) cccc Area = 7 cccc 2 2xx cccc Area of a triangle: AA = 1 2 bb h Substitute AA, bb & h: 7 = 1 2 2xx (2xx 5) Swap sides: 1 2 2xx (2xx 5) = 7 Cancel out the 2 ss: xx (2xx 5) = 7 Multiply out the brackets: 2xx 2 5xx = 7 Take all terms to the LHS: 2xx 2 5xx 7 = 0 Factorise: (2xx 7)(xx + 1) = 0 Split the brackets: 2xx 7 = 0 or xx + 1 = 0 Solve: 2xx = 7 or xx = 1 (discard since vvvv) Divide by 2: xx = 7 2 cccc [11]

12 National 5 Mathematics Practice Paper E Worked Solutions Paper Two: Calculator [12]

13 1. EE = mmcc 2 Substitute mm & cc: = ( ) 2 Simplify: = Simplify again: = Notes Use 10 xx (or equivalent) button on your calculator. When writing the answer in scientific natation, place the decimal point between the first two numbers cm 19 cm From the Formulae List: Area of a triangle = 1 2 aaaaaaaaaaaa = ssssss110 = cccc 2 tttt 1 dddd [13]

14 3. Final Amount = Initial value 100 ± % n, nn = number of hours 100 Final Temperature = = 20 C (nearest degree) Alternative Method Hour One: Temperature = = C Hour Two: Temperature = = C Hour Three: Temperature = = = 20 C (nearest degree) Notes nn = 3 because the difference between 8 pm & 11 pm is 3 hours Only use the alternative method if you are not confident with the formula method as the alternative method can be time consuming [14]

15 4. (a) CC = (4, 3, 4), DD = (6, 2, 2) (b) BB = (6, 4, 2) Notes When determining the coordinates, use A (2, 4, 6) as your reference point and think how many along xx, along yy and along zz. zz yy xx [15]

16 5. V = 200 ml V = 1600 ml Volume Scale Factor = = 8 Volume Scale Factor = (Linear Scale Factor)³ Substitute 8 into the equation: 8 = (Linear Scale Factor)³ Swap sides: (Linear Scale Factor) 3 = 8 Take the cubed root : 3 Linear Scale Factor = 8 = 2 Height of salon bottle = Linear Scale Factor Height of travel bottle = 2 12 = 24 cccc Notes Calculating an area use: Area Scale Factor = (Linear Scale Factor)² Calculating a volume use: Volume Scale Factor = (Linear Scale Factor)³ [16]

17 6. In the below, the P coefficients have been scaled to have the same value although it does not matter which is chosen. Let BB = BBBBBBBB, PP = PPPPPPPPPP: 2BB + 5PP = (1) 2 to give (3) below: 3BB + 2PP = (2) 5 to give (4) below: 4BB + 10PP = (3) 15BB + 10PP = (4) Equation (4) (3) 11BB = 17.6 Divide by 11: BB = 1.6 Substitute BB = 1.6 into equation (1) above: PP = 5.2 Simplify: PP = 5.2 Take 3.2 to the RHS: 5PP = Simplify: 5PP = 2 Divide by 5: PP = 0.4 The lengths are 1 bead = 1.6 cccc & 1 pearl = 0.4 cccc [17]

18 7. (a) Volume of a sphere (Formulae List): V = 4 3 ππr3 = 4 3 ππ 0.53 = = cccc 3 (3 SF) (b) Volume of a cylinder: V = ππr 2 h = ππ h Simplify: = 1.54 h Swap sides: 1.54 h = Divide by 1.54: h = 0.34 cccc (2 decimal places) Notes The volume of a cylinder (VV = ππrr 2 h) is not given on the Formulae List To help remember the formula, think circle area (AA = ππrr 2 ) multiplied by the height since a cylinder is a prism and VVVVVVVVVVVV = AA h [18]

19 8. V km A 24 B Point V 50 is the zz aaaaaaaaaa from A 66 is the zz aaaaaaaaaa from B Point B Bearing of V from B = 294 (from the question) = 66 Since B is due east of A: = 24 Sine Rule With two sets of opposites (angles & lengths) and only one unknown length then the sine rule can be used to find the missing length AB [19]

20 From the Formulae List: aa sin AA = bb sin BB = cc SSSSSS CC Substitute values in: AAAA = 5 ssssss 116 ssssss 24 Multiply by sin 116: AAAA = 5 ssssss116 ssssss24 AAAA = kkkk (2 dddd) The distance between the two hostels is kkkk correct to 2 decimal places Notes The Sine Rule is used to calculate angles & lengths in non-right angled triangles and requires two sets of opposites (angles & lengths) with only one unknown. To calculate: a length use: aa SSSSSS AA = bb SSSSSS BB = cc SSSSSS CC an angle use: SSSSSS AA aa = SSSSSS BB bb = SSSSSS CC cc [20]

21 9. (a) With reference to the triangle above, use SOH-CAH-TOA to find xx: cccccccc = Take the inverse cos: xx = cccccc = 23 The angle the chain makes with the vertical is approximately (b) Arc Length = Angle Fraction ππππ = 46 ππ 25 = = 10.0 mm (3 SSSS) The maximum length of arc through which the end of the chain swings is 10.0 m correct to 3 significant figures Notes Double 23 to obtain the 46 in the above as this is the sector angle [21]

22 10. kkxx 2 4xx + 2 = 0 Real roots means that the discriminant is greater than, or equal, to 0 bb 2 4aaaa 0 Compare kkxx 2 4xx + 2 with aaxx 2 + bbbb + cc therefore: aa = kk, bb = 4, cc = 2 Substitute aa, bb & cc into: bb 2 4aaaa 0 ( 4) 2 4 kk 2 0 Simplify: 16 8kk 0 Take the 16 to the RHS: 8kk 16 Multiply both sides by 1 (inequality changes direction): 8kk 16 Divide by 8: kk 2 Note With reference to the above, multiplying both sides of the inequality by -1 means that the direction of the inequality sign changes [22]

23 11.(a) 3ssssss xx 1 = 0 0 xx 360 Take 1 to the RHS; 3ssssss xx = 1 Divide by 3: ssssss xx = 1 3 Take the inverse sign: xx = ssssss = 35.3 For second angle: Consider the All-Sine-Tan-Cos diagram shown below Since Sin, then the two angles are in Quadrants 1 & 2 Quadrant 2 (below) shows: 2nd Angle = 180 xx = = xx = & correct to 1 decimal place [23]

24 (b) tttttttt cccccccc Substitute tttttttt = ssssssss cccccccc into the above: ssssssss cccccccc cccccccc Cancel out the cccccccc : ssssss xx Therefore: tttttttt cccccccc = ssssssss Notes The two trig identities to remember (not given in the Formulae List) are: tttttttt = ssssss xx cccccc xx ssssss 2 + cccccc 2 xx = 1 It is also important to note that ssssss 2 xx + cccccc 2 xx = 1 can be written as: ssssss 2 xx = 1 cccccc 2 xx cccccc 2 xx = 1 ssssss 2 xx [24]

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