Mathematics Higher Tier, June /1H (Paper 1, non-calculator)

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Warren Sharp
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Link to past paper on AQA website: www.aqa.org.uk The associated question paper is available to download freely from the AQA website.

1 Link to past paper on AQA website: The associated question paper is available to download freely from the AQA website. To navigate around the website, choose QUALIFICATIONS, GCSE, MATHS, MATHEMATICS, MATHEMATICS A (LINEAR), 4301 MATERIALS. Question 1 a) 58 x is very similar to the given statement so our answer will involve the digits The first figure 58 is unchanged but the second is smaller than 129 (the decimal place has been moved 3 places so our answer will be 7482 but with the decimal place moved 3 places to 7.482). Also we can apply common sense, 58 x x 0.1 = 6 so seems reasonable b) if we rearrange the given equation we have 7482/58 = 129 so our answer will involve the digits 129. The numerator is unchanged but the denominator is increased. Increasing the denominator will decrease the overall answer. We have increased the denominator by moving the decimal place 2 places so we need to move the decimal place on the answer by 2 places to Also common sense 7482/ / c) we know from the given equation that 58 x 129 = 7482 so our new equation simplifies to being 7482/7482 = 1 1 Question 2 An expression is where there are no equal signs, just a group of algebraic terms such as 3x 7 An equation has an equal sign ( ) and the letters stand for particular numbers. There will only be a finite number of solutions A formula will also have an equal sign but the letters stand for defined quantities or variables The equation is 20 3x = 11 The expression is x 2 + 7x 1 The formula is P = 2l + 2b Question 3 a) 803m could be anywhere between and (or ) The smallest possible height is 802.5m b) 1km = 1000m so 1km 2 will measure 1000m by 1000m so will be equal to 1,000,000m km 2 = 4,900,000m Page 1

2 Question 4 a) 128⁰ due to corresponding angles b) by corresponding angles the angle next to y (called Z on my diagram) will be 85⁰. Angles on a straight line make 180⁰ so y = 180⁰ 85⁰ = 95⁰ y Z 85⁰ Question 5 First put as a fraction 84/120 Then to turn into a percentage multiply the numerator (top) by 100 = = = = =70% Question Page 2

3 Question 7 a) angles in a quadrilateral always add up to 360⁰ x + x = 360 grouping terms x = 360 b) subtract 196 from both sides 4x = 164 divide both sides by 4 x = 41⁰ Question 8 a)to divide fractions simply change the to x and flip the second fraction over to give x now just multiply the top together and the bottom together to give = (if we had started with mixed fractions then we would have had to convert them both to top heavy fractions before we were able to divide them) b) we must first convert both fractions so that they have the same denominator. No need to convert to top heavy fractions. The lowest common denominator we could use is = 3, 1 = 1 We have 3-1 = 2 c) reciprocal of 0.5 is. it is hard to divide by decimals so multiply the numerator and the denominator by 10 = = = Page 3

4 Question 9 I would first put the data in the correct lines (never mind the correct order) 6 5, 7 7 2, 6, 0, 0 8 7, 5, 0, 4, Check still have 12 pieces of data. Now order each row. 6 5, 7 7 0, 0, 2, 6 Key , 2, 4, 5, Again check that there are still 12 pieces of data. represents 82 minutes Question 10 a) x y when x = -1 we have y = (-1) 2 (-1) 5 = = = -3 when x = 4 we have y = = = 7 also the symmetry in the y values convinces us that we have the right values b) c) x 2 x 5 = 0 is the same as y = 0 (because y = x 2 x 5) the line y =0 (which is the same as the x axis) crosses our curve where x = 2.8 (there would also be another solution as it crosses the x axis twice. The other solution would be x = -1.8) Page 4

5 Question 11 a) = 14 multiply both sides by 5 x = 70 b) 2(3y 1) = 13 expand the brackets 6y 2 = 13 add 2 to both sides 6y = 15 divide both sides by 6 y = 2.5 c) = 7 multiply both sides by 4 16 z = 28 add z to both sides 16 = z + 28 rewrite with z + 28 on the lhs of the equation z + 28 = 16 subtract 28 from both sides z = -12 d) 2(x + 1) 2 10(x + 1) we can cancel the 2 with the 10 to give (x + 1) 2 5(x + 1) we can cancel (x + 1) as well 1 5 Question 12 a) percentage increase is x 100 = x 100 = 90% x 100 b) In April 2000 the base was 100, in April 2006 the base was 190 so we divide by 100 and multiply by 190, = 800 x 190 = 152,000 (8 x 19 is 152 and then we need 3 zeros) Page 5

6 Question 13 a) we divide the numbers and subtract the powers = = = 10 4 So we have (2.8 x 10 9 ) (4 x 10 5 ) = 0.7 x 10 4 We need to adjust this so that is in standard form (with the initial number being between 1 and 9.999). We multiply the 0.7 by 10 so to compensate divide the 10 4 by 10 to get x 10 3 b) (5 x 10-3 ) x (5 x 10-3 ) = 25 x = 25 x 10-6 (but this is not standard form as the initial number needs to be between 1 and so divide 25 by 10 to get 2.5 and then to compensate multiply 10-6 by 10 to get 10-5 ) 2.5 x 10-5 Question 14 a) Volume of cylinder is given in formulae sheet. Volume is πr 2 h V = π x 8 2 x 5 = π x 64 x 5 = 320π cm 3 b)volume of cylinder B is πr 2 h = 320π, where h = 20 and r is unknown πr 2 x 20 = 320π divide both sides by π 20 x r 2 = 320 divide both sides by 20 r 2 = 16 square root both sides r = 4 (r = 4 but in the context of this question r cannot be negative so r = 4) Question 15 a) to get the four point moving average add up the first four values and divide by 4 so working backwards the first 4 values must add up to 114 (4 x 28.50) = 77.9 so the missing value must be = 36.1 b) to calculate the second four-point moving average we move along the line and add up the four values from to and then divide by 4. In effect we are dropping the first value of 34.7 and replacing it with a higher value of 37.6 so the four-point moving average will increase. It will now be greater than Page 6

7 Question 16 3(a b) = 2b + 7 expand the brackets 3a 3b = 2b + 7 add 3b to both sides 3a = 5b + 7 divide by 3 a = Question 17 We have the two corresponding sides of 6cm and 21cm. We want to find a side on the larger triangle so our scale factor will be. PQ = 5 x = 5 x = 17.5cm Question 18 a) ABCD forms a cyclic quadrilateral. Opposite angles in a cyclic quadrilateral add up to 180⁰ angle BAD = 180⁰ 105⁰ = 75⁰ b) angles on a straight line make 180⁰ so angle PAB is 42⁰ (180⁰ (75⁰ + 63⁰) = 42⁰) then by alternate segment theorem, angle ADB = angle PAB = 42⁰ (alternate segment theorem states that 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) Question 19 a) 28 4 x 7 = x 7 = = = 5 7 p = 5 b) multiply the top and the bottom by 5 = = Page 7

8 Question 20 a) h 1/ 2 which can also be written as h = k/r 2 we have h = 4.5 when r = 4 substituting these values we get 4.5 = k/ = k/16 multiply both sides by x 16 = k 72 = k going back to our original equation and replacing k with 72 we have h = 72/r 2 b) when h = 8 we have 8 = 72/r 2 multiply both sides by r 2 8r 2 = 72 divide both sides by 8 r 2 = 9 square root both sides r = 3cm (r = 3 but in the context of this question r cannot be negative so r = 3) Question 21 Most probability questions can be answered using tree diagrams Bill Ben 0.4 correct 0.7 correct 0.6 incorrect 0.4 correct 0.3 incorrect 0.6 incorrect The probability that exactly one of them gets it correct is the probability that only Bill gets it correct plus the probability that only Ben gets it correct. Probability that only Bill gets it correct is p(bill correct) x p (Ben incorrect) = 0.7 x 0.6 = 042 Probability that only Ben gets it correct is p(bill incorrect) x p(ben correct) = 0.3 x0.4 = 0.12 So overall probability is = Page 8

9 Question 22 a) 9 3/2 = (9 1/2 ) 3 = ( 9) 3 = 3 3 = 27 b) 9 x = 27 4 now we know from a) that 27 = 9 3/2 so replacing 27 with 9 3/2 we have 9 x = (9 3/2 ) 4 = 9 6 so x = 6 Question 23 we can work out as + = 5a + 2b + a + 7b = 6a + 9b we are told that is parallel to so we must be able to write = c x where c is just an ordinary number (a constant). we have 6a + 9b = c x (4a + kb) 6a + 9b = 4ca + ckb by comparing coefficients for the a 6 = 4c divide both sides by 4 c = 1.5 comparing coefficients for the b 9 = ck we know c = = 1.5 x k divide both sides by 1.5 k = Page 9

10 Question 24 a) if we start with y = x 2 or y = f(x) our new equation will be y = f(x-7). This is adding or subtracting a number so is a translation. The change is inside the brackets so affects the x values (and when it affects the x it does it in the opposite way to what you would expect). So we have a translation of 7 units in the positive x direction. 7 b) if we start with y = x 2 or y = f(x) our new equation (y = 3x 2 ) will be y = 3f(x). This is multiplying by a number so is a stretch. The change is outside the brackets so affects the y values. So we have a stretch of scale factor 3 parallel to the y axis Page 10

11 Question 25 This is effectively two simultaneous equations x 2 + y 2 = 26 (eq n 1) and y = x + 4 (eq n 2) we can only solve by substitution. elimination won t work here as we have x 2 and y 2 in one equation and x and y in the other) substitute the value for y (from eq n 2) into eq n 1 x 2 + (x + 4) 2 = 26 x 2 + (x + 4)(x + 4) = 26 expand the brackets x 2 + (x + 4)(x + 4) = 26 x 2 + x x + 4x = 26 grouping terms 2x 2 + 8x + 16 = 26 subtract 26 from both sides 2x 2 + 8x - 10 = 0 divide both sides by 2 x 2 + 4x - 5 = 0 b)we have a quadratic to solve, we can factorise, complete the square or use the quadratic formula. Factorise It will factorise if we can find two numbers that multiply to make -5 and add to make +4. The two numbers are +5 and -1 so we have (x + 5)(x 1) = 0 If two things multiply to give 0 then one of them must be equal to 0 so x + 5 = 0 or x 1 = 0. either x = -5 or x = 1 Completing the square (x + 2) = 0 (x + 2) = 0 (x + 2) 2-9 = 0 add 9 to both sides (x + 2) 2 = 9 square root both sides x + 2 = 3 subtract 2 from both sides x = 1 or x = -5 Quadratic Formula Page 11

12 a = 1, b = 4 and c = x1x 5 2x or = 1 or x = -5 when x = 1, from y = x + 4, y = = 5 when x = -5, y = = -1 our two co-ordinates are (1, 5) and (-5, -1) From the diagram A must be (1,5) and B must be (-5, -1) It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by AQA. In addition these solutions may not necessarily constitute the only possible solutions. If you found these solutions helpful then visit where you will find many more Page 12

Mathematics Higher Tier, June /1H (Paper 1, non calculator)

Mathematics Higher Tier, June /1H (Paper 1, non calculator) Link to past paper on AQA website: www.aqa.org.uk The question paper associated with these worked answers is available to download for free from the AQA website. You can navigate around the website in