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## Year 9 – End of Year Revision

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**ζ**Dr Frost Year 9 – End of Year Revision**Percentages**Be careful: Are you trying to find the new value or the old value? In the first case, you multiply, in the second case, you divide. Percentage change is based on the old value. A jumper is bought in for £30 and marked up by 40%. What is it sold for? Answer: 30 x 1.4 = £42 I put £15,000 into a savings account. It accrues 2.6% interest. What is in my account in one year’s time? Answer: £15390 ? ? After one year the value of a care fell by 20% to £9600. What was its original value? Answer: £12000 Lucy made 20% profit on the picture frame she sold at £35. What did she buy it in for? Answer: £29.17 ? ?**Percentages**The interest rate for a savings account is 2.5% p.a. with compound interest. The principal is £1500. How much do I have in 10 years time? Answer: £1500 x 1.02510 = £1920.13 My Bentley depreciates in value 10% each year. It is bought new for £150,000. How much is it worth in 5 years time? Answer: £150,000 x 0.95 = £88573.50 ? ?**Compound Measures**A cat travels at 15km/s. It races around a 50km track. How much time did it take him? Answer = 3.33s ? The density of a hamster is 1.3kg/m3. Its volume is 0.03m3. What is the hamster’s mass? Answer = 0.039kg ?**Graphs**Match the graphs with the equations, and identify what type of equation it is. 1 5 9 y = -2x3 + x2 + 6x y = 4x y = 2x - 3 y = x2 + x – 2 y = 5 – 2x2 y = 2x3 y = 5 – x y = x3 – 7x + 6 y = -3x3 6 Cubic 11 Exponential 9 Straight Line 1 Quadratic 2 Quadratic 8 Reciprocal 5 Cubic 10 Straight Line 3 Cubic 4 Cubic 7 Reciprocal ? ? ? 6 10 ? 2 ? ? ? 3 7 11 ? ? ? ? 8 4**Graphs**y = x3 – 2x2 - 5x +6 ? ? ? ? ? ? ? ? When sketching, ensure you sketch a curvy line (i.e. don’t join up your points with lines), or you’ll lose a mark.**Changing the Subject**Change the subject of the formula to the indicated letter. ? (b) ? ? ? ? ? ? ? ? ?**Changing the Subject**? ? ? ? ? ? ? ? ? ?**Changing the Subject**The following require you to factorise at some point. Make a the subject of the formula: n2-Pn P-1 ? _3a_ a+1 ? a = n = ? ?**Simultaneous Equations**You can either use elimination or substitution. 3x + 2y = 10 5x – 2y = 14 3x + 2y = 4 4x + 3y = 7 ? ? x = 3, y = 0.5 x = -2, y = 5**Probability**Question: Give there’s 5 red balls and 2 blue balls. What’s the probability that after removing two balls from the bag, we have a red ball and a blue ball? ? 4 6 R ? 5 7 R B ? 2 6 R 5 6 ? 2 7 ? B B 1 6 ? 10 21 ? Answer =**Probability**? Matching outcomes What’s the probability that when I roll 10 dice, I see the same number on every die? Total outcomes ? What’s the probability that when I roll 10 dice, the total of the dice is 10? ? Difficult: What’s the probability that when I roll 3 dice, I see exactly two sixes.**Probability**If I have two dice, one numbered 1, 2, 3 and the other numbered 2, 3, 4, what’s the probability the sum is at least 5? Second Die ? First Die ? 62 9 3 p(sum ≥ 5) = =**Sequences**Determine the formula for the following sequences. 5, 8, 11, 14, 17, ... 10, 8, 6, 4, 2, 0, -2, ... Un = 3n + 2 ? Un = 12 – 2n ? 3, 9, 17, 27, 39, ... 1, 3, 6, 10, 15, ... Un = n2 + 3n - 1 ? Un = 0.5n(n+1) ?**Expanding brackets**Expand the following. (x+1)(x-2) = x2 – x – 2 (x-4)(x-8) = x2 – 12x + 32 (x+1)(y+1) = xy + x + y + 1 (x2+1)(y2-1) = x2y2 – x2 + y2 – 1 (2x+1)(2x-1) = 4x2 – 1 (x + 1)(x + y + 1) = x2 + xy + 2x + y + 1 x(y-x)-y(x-y) = y2 – x2 ? ? ? This is known as the: difference of two squares ? ? ? ? ?**Factorisation**Factorise the following x2 + 7x + 12 = (x + 4)(x + 3) x2 + 2x – 3 = (x – 1)(x + 3) x2 – 10x + 24 = (x – 4)(x – 6) 2x2 – 5x – 12 = (2x + 3)(x – 4) 12x2 + 5x – 3 = (4x + 3)(3x – 1) x2 – 9 = (x + 3)(x – 3) 4 – y2 = (2 + y)(2 – y) x3 – x = x(x + 1)(x – 1) 16x2y2 – 9z4 = (4xy + 3z2)(4xy – 3z2) x4 + 2x2 – 143 = (x2 + 13)(x2 – 11) ? ? ? ? ? ? ? ? ? ?**y**4 Enlargement 3 2 Image 1 x-5 -4 -3 -2 -1 0 1 2 3 4 5 6 Object -1 -2 -3 Enlarge the shape by a scale factor of 2 about the point (0,-2) -4**y**4 Enlargement 3 2 1 x-5 -4 -3 -2 -1 0 1 2 3 4 5 6 Object -1 -2 -3 Enlarge the shape by a scale factor of -1 about the point (0,2) -4**y**4 Enlargement 3 2 1 x-5 -4 -3 -2 -1 0 1 2 3 4 5 6 Object -1 -2 -3 Enlarge the shape by a scale factor of -0.5 about the point (0,2) -4**Trigonometry**x 4 60 ° 65 ° x x = 13.96 ? x = 3.46 ? 30 ° 12 x 15 x = 6.99 ?**Trigonometry**θ = 45° ? b c 1 1 1 a θ θ 2 3 d θ 6 3 8 θ θ = 33.69° ? θ = 70.53° ? θ = 48.59° ?**Trigonometry**What is the cosine of the angle between the internal diagonal of a cube and the bottom face of the cube? ? √2 √ 3 Answer =**Solving Equations**Solve the following equations for x. x(2x + 1)(x – 2) = 0 x = 0 or -0.5 or 2 x2 = 4 x = 2 or -2 x2 = 3x x = 0 or 3 x2 + 5x – 6 = 0 x = -6 or 1 x3 = x x = -1, 0 or 1 x2 + 32 = 12x x = 4 or 8 25x2 – 4 = 0 x = 2/5 or -2/5 ? ? ? ? ? ? ?**Solving Equations**Determine x Answer: By Pythagoras, x2 + (2x+2)2 = (3x-2)2 Expanding and simplifying, we get 4x2 – 20x = 0 Solving, x = 5 (we reject the 0 solution). ? 3x - 2 x 2x + 2**Similarity**3 ? x = 16 5 x 10**Similarity**A square is inscribed in a right-angled triangle with length 4 and height 3. Find the width of the square. ? 12 7 Length of square = 3 4**Loci**A Spotted StuddertSheep is known to be within 3km of A and 4km of B. What region could the sheep be in? 3km 4km A B**Loci**Now the sheep is also known to be of equal distance from A and B. Where can it be? 3km 4km A B**Loci**Now the sheep is within 3km of A, but at least 4km away from B. Where could it be? 3km 4km A B**Loci**I’m equidistance from two lines AB and AC. Where could I be? B A C**Loci**I’m the indicated distance away from the walls of a building. Where could I be? Circular corners. Straight corners.**Loci**A My sheep is attached to a fixed point A on a square building, of 10m x 10m, by a piece of rope 20m in length. Both the sheep and rope are fire resistant. What region can he reach? 20m 10m**Dimensional Analysis**Click your choice. (all variables are lengths) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?**Ratio**Method 2: of the tank is black and is yellow. The plague leaves us with of the fish, so the tank is now full of black fish and of yellow fish. Now the of the tank wiped out is replenished with black fish, so that’s black fish (and still yellow fish). This ratio is 5:1. Method 1: Suppose a full tank has 12 fish. Then 9 fish are black and 3 yellow. The plague leaves 6 black fish and 2 white. Then if we fill up the rest of the tank with black fish, we have 10 black fish and 2 yellow. This ratio is 5:1. My fish tank has black and yellow fish in the ratio 3:1. A fish plague, Sanjotitus, wipes out a third of my fish. I then restock my fish tank with just black fish, so that I have the same number of fish as before. What’s the new ratio of black to yellow fish? Answer = 5 : 1 ?**Proportion**Given that y is proportional to x, find the missing values. ? ?**Inverse proportion**Given that L is inversely proportional to √x, fill in the missing values in this table. ? ?**Inequalities**Solve the following. 2x > x - 6 x > - 6 ? -x + 1 ≤ 6 x ≥ -5 ? 1 ≤ 2x + 3 < 9 -1 ≤ x < 3 ? ?**Inequalities on a number line.**x < -1 or x > 4 2 ≤ x < 4 0 1 2 3 4 5 -1 0 1 2 3 4 ? ?**Inequalities on a number line.**2 ≤ x < 5 Draw the range of x on the number line given that both of these inequalities hold. x < 3 or x > 4 0 1 2 3 4 5 ?**8**6 4 2 -2 -4 -6 -4 < y ≤ -2 -10 -8 -6 -4 -2 2 4 6 8 10**8**6 4 2 -2 -4 -6 y ≤ x + 1 and x ≤ 6 and y > 2 -10 -8 -6 -4 -2 2 4 6 8 10**Inequalities**When would ? When all of x, y and z are negative, or one of x, y and z are negative.**Arcs and Sectors**Sector area = 10.91 ? Area = 20 50° Arc length = 4.36 ? 5 135° (Hint: Plug values into your formula and rearrange) Sector area = 4.04cm2 ? 105° Radius = 4.12 ? ? Arc length = 3.85cm 2.1cm**Arcs and Sectors**The shape PQR is a minor sector. The area of a sector is 100cm2. The length of the arc QR is 20cm. Determine the length PQ.Answer: 10cm Determine the angle QPRAnswer: 114.6° Q P ? R ?**Volume of a prism**Volume = 400cm3 ? 4cm 8cm 10cm 6cm 3m 6m Volume = 17m2 x 6 = 102m3 1m ? 5m 5m**Surface Area**2m Surface Area = 112m3 4m ? 8m