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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : EXIT Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Trigonometry Please choose a question to attempt from the following: 1 EXIT 2 3 Back to Unit 2 Menu 4 5 6 TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. B 40cm A 25° C 50cm Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. B 40cm A 25° C 50cm What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT 1. For area of a 2. Always triangle questions remember to with an angle round use: your present answer if the 1 Aquestions 2 ab sinasks C you to. TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. B 40cm A 25° C 50cm What would you like to do now? Try another like this Go to full solution Go to Comments Go to Trigonometry Menu EXIT Area of = 423cm2 Question 1 1. For area of triangles questions where an angle is present use: Find the area of the following triangle Area of = ½ bcsinA° = 50 x 40 x sin25° 2 to the nearest cm2. B = 422.61… 40cm A 2. Remember to round if asked to. 25° C 50cm = 423 to nearest unit Area of = 423cm2 Continue Solution Try another like this Comments Trigonometry Menu Back to Home Markers Comments Check formulae list for: 1. For area of triangles questions where an angle is present use: 1 Area of triangle = 2 absinC Area of = ½ bcsinA° = 50 x 40 x sin25° 2 (Note: 2 sides and the included angle) = 422.61… Relate formula to labels being used. B 2. Remember to round if asked to. 40cm c = 423 to nearest unit Area of = 423cm2 A 25° b 50cm C Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 1B Find the area of the following triangle. K 6.5m Get hint L Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT 150° 8m M TRIGONOMETRY : Question 1B Find the area of the following triangle. K 6.5m L What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT 150° 8m 1. For area of a 2. Always triangle questions remember to with an angle round use: your present answer if the Aquestions 12 ab sinasks C you to. M TRIGONOMETRY : Question 1B Find the area of the following triangle. K 6.5m L 150° 8m What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT Area of = 13m2 M Question 1B 1. For area of triangles questions where an angle is present use: Find the area of the Area of = ½ kmsinL° = 8 x 6.5 x sin150° 2 following triangle. K = 13 6.5m L 150 ° Area of = 13m2 8m Begin Solution Continue Solution Comments Trigonometry Menu Back to Home M Markers Comments Check formulae list for: 1. For area of triangles questions where an angle is present use: Area of = ½ kmsinL° = 8 x 6.5 x sin150° 2 absinC (Note: 2 sides and the included angle) Relate formula to labels being used. = 13 Area of = Area of triangle = 13m2 K 6.5m m L 150 ° k 8m M Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer N to the nearest 10km. Get hint Reveal answer only 340° Go to full solution Go to Comments Go to Trigonometry Menu EXIT TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer N to the nearest 10km. What would you like to do now? Reveal answer only Go to full solution 1. Identify what you 2. Calculate as 3. Make a sketch to 340° 5.need Substitute known 4. Identify which trig to findofand many the clarify matters. values, remembering to rule to use: the missing information you angles as use brackets as Two sides + two angles have topossible. help you. appropriate. = sine rule Go to Comments Go to Trigonometry Menu EXIT Three sides + one angle = cosine rule TRIGONOMETRY : Question 2 Two helicopters leave an air base. The first flies on a bearing of 340° at 160km/hr. The second flies due east at 200km/hr. How far apart will they be after 21/2 hours? Answer N to the nearest 10km. What would you like to do now? Try another like this 340° Go to full solution Go to Comments Go to Trigonometry Menu Distance is 740km EXIT Question 2 1. Identify what needs to be found. 160km/hr. 2. Need distances travelled and angle between flight paths. N 340° How far apart will they be after 21/2 hours? 20° 90° d1 = speed1 x time = 160 x 2.5 = 400km d2 = speed2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise Full angle = 20° + 90° = 110° 3. Sketch triangle. 200km/hr b Continue Solution 110° Try another like this Comments a 400km A 500km c Trigonometry Menu Back to Home Question 2 4. Apply Cosine rule. 160km/hr. a2 = b2 + c2 – (2bccosA°) N 340° How far apart will they be after 21/2 hours? 20° 90° 200km/hr Continue Solution Try another like this Comments Trigonometry Menu Back to Home 5. Substitute known values and remember to use brackets. = 4002 + 5002 – (2 x 400 x 500 x cos110°) = 546808.05.. a = 546808.05.. = 739.46.... 6. Remember to round answer if asked to. = 740 Distance is 740km Markers Comments 1. Identify what needs to be found. 2. Need distances travelled and angle between flight paths. Note: Bearings are measured clockwise from N. d1 = speed1 x time = 160 x 2.5 = 400km d2 = speed2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise Full angle = 20° + 90° = 110° 3. Sketch triangle. 400km 110° 500km Next Comment Trigonometry Menu Back to Home Markers Comments 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA 5. Substitute known values and remember to use brackets. (2 sides and the included angle) = 4002 + 5002 – (2 x 400 x 500 x cos110°) Relate to variables used = 546808.05.. a2 = b2 + c2 – 2bc cosA a = 546808.05.. = 739.46.... 6. Remember to round answer if asked to. = 740 Distance is 740km Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. N How far apart will they be? Get hint Reveal answer only Go to full solution Go to Comments 195° Go to Trigonometry Menu EXIT SE TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. N How far apart will they be? What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT 2. Calculate 1. Identify what you as 3. Make a sketch toof the many need to find and 4. Identify which trig as clarify matters. missing angles therule information you to use: possible. have to help you. Two sides + two angles 5. Substitute known = sine remembering rule values, to use brackets as 195° + one angle Three sides SE appropriate. = cosine rule TRIGONOMETRY : Question 2B Two ships sail from a port. The first travels for two hours on a bearing of 195° at a speed of 18mph. The second travels south-east for three hours at a speed of 15mph. N How far apart will they be? Distance is 41.2 miles What would you like to do now? Go to full solution Go to Comments 195° Go to Trigonometry Menu EXIT SE Question 2B How far apart will they be? 1. Identify what needs to be found. N 2. Need distances travelled and angle between paths. 1350 3hr@15mph 2hr@18mph 195° d1 = speed1 x time = 18 x 2 = 36miles 600 SE d2 = speed2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 3. Sketch triangle. A Begin Solution Continue Solution Comments Trigonometry Menu Back to Home 60° 45miles c 36miles b a Question 2B How far apart will they be? 4. Apply Cosine rule. N a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. 1350 3hr@15mph 2hr@18mph = 362 + 452 – (2 x 36 x 45 x cos60°) = 1701 195° 600 SE a = 1701 = 41.243.... = 41.2 Begin Solution Continue Solution Comments Trigonometry Menu Back to Home Distance is 41.2 miles Comments 1. Identify what needs to be found. 2. Need distances travelled and angle between flight paths. Note: Bearings are measured clockwise from N. d1 = speed1 x time = 18 x 2 = 36miles d2 = speed2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 3. Sketch triangle. A 60° 45miles c 36miles b a Next Comment Trigonometry Menu Back to Home Comments 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. = 362 + 452 – (2 x 36 x 45 x cos60°) = 1701 Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA (2 sides and the included angle) Relate to variables used a2 = b2 + c2 – 2bc cosA a = 1701 = 41.243.... = 41.2 Next Comment Distance is 41.2 miles Trigonometry Menu Back to Home TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. (b) Hence find the area of the kite to the nearest square unit. P 10cm Get hint Reveal answer only Q S 22cm Go to full solution 15cm Go to Comments Go to Trigonometry Menu EXIT R TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. (b) Hence find the area of the kite to the nearest square unit. P 1. Identify which trig 3. For10cm area of a rule to use: What would you like to do now? triangle with an 4. Always Two sides + two angles angle present to use: remember 2. Substitute known Q = sine rule Reveal answer only round your 22cmto values, remembering 1 answer if the use brackets A sin C as Three sides + one 2 abangle Go to full solution questions appropriate. = cosine rule asks 15cm you to. Go to Comments Go to Trigonometry Menu EXIT R S TRIGONOMETRY : Question 3 In the kite shown below PQ = 10cm, QR = 15cm & diagonal PR = 22cm. (a) Find the size of angle QPR. angleQPR = 35.3° (b) Hence find the area of the kite to the nearest square unit. P What would you like to do now? Try another like this = 127cm2 10cm Q S 22cm Go to full solution 15cm Go to Comments Go to Trigonometry Menu EXIT R Question 3 P 1. To find angle when you have 3 sides use 2nd version of cosine rule: 10cm Q S 22cm cosP r2 + q2 - p2 = 2rq = 102 + 222 - 152 2 x 10 x 22 15cm = (102 + 222 - 152) (2 x 10 x 22) R (a) Find the size of angle QPR Continue Solution Try another like this Comments Trigonometry Menu Back to Home = 0.8159… 2. Remember to use inverse function to find angle. angleQPR = cos-1(0.8159..) = 35.3° Question 3 P 1. Kite = 2 identical triangles. For area of triangles where an angle is present use: 10cm Q S 22cm (b) Area QPR = ½ qrsinP° = 10 x 22 x sin35.3° 2 15cm = 63.56..cm2 2. Remember to double this and round answer. R (b) Hence find the area of the kite to the nearest square unit Continue Solution Try another like this Comments Trigonometry Menu Back to Home Area of kite = 2 x 63.56.. = 127.12..cm2 = 127cm2 Markers Comments 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP r2 q2 + = p22rq Check the formulae list for the second form of the cosine rule: 2 + c 2 – a2 b cosA = 2bc ( 3 sides) = 102 + 222 - 152 2 x 10 x 22 = (102 + 222 - 152) (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. angleQPR = cos-1(0.8159..) = 35.3° Next Comment Trigonometry Menu Back to Home Markers Comments Relate to variables used: 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP cosP = r 2 + q2 = p22rq = = 102 + 222 - 152 2 x 10 x 22 = (102 + 222 - 152) (2 x 10 x 22) q2 + r2 – p2 2qr 222 + 102 – 152 2x22x10 P 10cm r Q q 22cm = 0.8159… 2. Remember to use inverse function to find angle. angleQPR = cos-1(0.8159..) = 35.3° 15cm p R Next Comment Trigonometry Menu Back to Home Markers Comments Note: 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP r 2 + q2 = p22rq = 102 + 222 - 152 2 x 10 x 22 When keying in to calculator work out the top line and the bottom line before dividing or use brackets. = (102 + 222 - 152) (2 x 10 x 22) = 0.8159… 2. Remember to use inverse function to find angle. angleQPR = cos-1(0.8159..) = 35.3° Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. Get hint E 15cm Reveal answer only Go to full solution 15cm 25cm H F Go to Comments 15cm 15cm Go to Trigonometry Menu G EXIT TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. What would you like to do now? E Reveal answer only 15cm 1. Identify which trig 15cm Go to full solution H Go to Comments Go to Trigonometry Menu EXIT 3. For area of a rule to use: triangle with an 4. Always Two sides + 25cm two angles angle present to use: 2. remember Substitute known = sine rule round your values, remembering to 1 answer if the use brackets A abangle sin C as Three sides + one 2 15cm 15cm questions appropriate. = cosine rule asks you to. G F TRIGONOMETRY : Question 3B The sides of a rhombus are each 15cm while the main diagonal is 25cm angleEFH = 33.6° Find the size of angle EFH and hence find the area of the rhombus to the nearest square unit. = 415cm2 E What would you like to do now? Go to full solution 15cm 15cm 25cm H F Go to Comments 15cm 15cm Go to Trigonometry Menu G EXIT E Question 3B 15 15 25 H 1. To find angle when you have 3 sides use 2nd version of cosine rule: F e 2 + h2 - f 2 (a) cosF = 2eh = 152 + 252 - 152 2 x 25 x 15 15 15 G (a) Find the size of angle EFH = (152 + 252 - 152) (2 x 25 x 15) = 0.8333… 2. Remember to use inverse function to find angle. angleEFH = cos-1(0.8333..) = 33.6° Continue Solution Comments Trigonometry Menu Back to Home E Question 3B 15 15 25 H 33.60 15 15 G (b) Hence find the area to the nearest square unit 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: (b) Area EFH = ½ ehsinF° F = 25 x 15 x sin33.6° 2 = 207.52....cm2 2. Remember to double this and round answer. Area of kite = 2 x 207.52...... = 415.04..cm2 = 415cm2 Continue Solution Comments Trigonometry Menu Back to Home Markers Comments 1. To find angle when you have 3 sides use 2nd version of cosine rule: (a) cosF = e2 h2 + 2eh f2 Check the formulae list for the second form of the cosine rule: 2 + c 2 – a2 b cosA = 2bc ( 3 sides) = 152 + 252 - 152 2 x 25 x 15 = (152 + 252 - 152) (2 x 25 x 15) = 0.8333… 2. Remember to use inverse function to find angle. angleEFH = cos-1(0.8333..) = 33.6° Next Comment Trigonometry Menu Back to Home Markers Comments Relate to variables used: 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosF = e2 + h 2 - f 2 (a) cosF = 2eh = 0.8333… 2. Remember to use inverse function H to find angle. 222 + 102 – 152 2x22x10 = = 152 + 252 - 152 2 x 25 x 15 = (152 + 252 - 152) (2 x 25 x 15) e 2 + h2 – f 2 2eh E 15cm h f 15cm e 25cm F angleEFH = cos-1(0.8333..) = 33.6° Next Comment Trigonometry Menu Back to Home Markers Comments Check formulae list for: 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: (b) Area EFH = ½ ehsinF° = 25 x 15 x sin33.6° 2 = 207.52....cm2 2. Remember to double this and round answer. 1 Area of triangle = 2 absinC (Note: 2 sides and the included angle) Relate formula to labels being used. E 15cm h f Area of kite = 2 x 207.52...... = 415.04..cm2 = 415cm2 15cm e H 25cm F Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. T Get hint 105° 5.9cm Reveal answer only Go to full solution 35° Go to Comments U Go to Trigonometry Menu EXIT V 10cm TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. What would you like to do now? Reveal answer only Go to full solution Go to Comments 35° U Go to Trigonometry Menu EXIT T 3. Identify which trig rule105° to use: 5. Always Two sides + two angles 5.9cm 1. For perimeter remember to =2.sine rule Calculate need all three round your unknown sides.ifSo answer themust find Three sides + one angle angles. TU. questions asks 4. Substitute known = cosine rule you to. values, remembering to use brackets as 10cm appropriate. V TRIGONOMETRY : Question 4 In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. = 22.6cm Find the perimeter to one decimal place. T 105° 5.9cm What would you like to do now? Go to full solution Go to Comments 35° U Go to Trigonometry Menu EXIT V 10cm T Question 4 1. Perimeter requires all three sides. 105° So we need to find TU . 5.9c m U 35 ° 2. Whether you use Sine or Cosine rule need angle V. 400 Angle V = 180° - 35° - 105° = 40° V 10cm Find the perimeter to one decimal place. 3. If we use Sine rule: v = sinV° t sinT° 4. Substitute known values: Continue Solution Comments Trigonometry Menu Back to Home v sin40° = 10 sin105° T Question 4 4. Substitute known values: 105° v = 10 sin40° sin105° 5.9c m 5. Cross multiply: U 35 ° 400 V v x sin105° = 10 x sin40° 10cm v = 10 x sin40° sin105° Find the perimeter to = 6.654… one decimal place. = 6.7cm 6. Answer the question: Continue Solution Comments Trigonometry Menu Back to Home Perim of = (6.7 + 10 + 5.9)cm = 22.6cm Markers Comments 1. Perimeter requires all three sides. So we need to find TU . 2. Whether you use Sine or Cosine rule need angle V. Since we can pair off two angles with the opposite sides Sine Rule Refer to the Formulae List : a b c = = Sine A Sine B Sine C T Angle V = 180° - 35° - 105° = 40° 3. If we use Sine rule: v = sinV° 105° t sinT° 4. Substitute known values: v = 10 sin40° sin105° 5.9cm v U 35° t 10cm Next Comment V Trigonometry Menu Back to Home Markers Comments 4. Substitute known values: v = 10 sin40° sin105° Go straight to values : 10 v = Sine 105˚ Sine 40˚ 5. Cross multiply: v x sin105° v = 10 x sin40° sin105° = 6.654… = 6.7cm 6. Answer the question: Perim of = (6.7 + 10 + 5.9)cm = 22.6cm Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 5 Lighthouse C is on a bearing of 050° from lighthouse A and northwest of lighthouse B. Lighthouse B is 12km due east of lighthouse A. A ship(S) is sailing directly from lighthouse B to lighthouse A. How close does it come to lighthouse C? Hints C N Answer only Full solution Comments 50° Trig Menu A EXIT 12km B S TRIGONOMETRY : Question 5 Lighthouse C is on a bearing of 050° from lighthouse A and northwest of lighthouse B. Lighthouse B is 12km due east of lighthouse A. A ship(S) is sailing directly from lighthouse B to lighthouse A. How close does it come to lighthouse C? C What would you like to do now? 2. Identify which trig 1.4.Calculate missing N rule to use: Closest Answer only Full solution Comments 50° Trig Menu A EXIT angles. Two sides distance = + two angles (NB. North= West = 3150) sine rule perpendicular distance. 3. Substitute known Three + one angle Create asides rightvalues, remembering to = cosine ruleas angleduse triangle brackets and use SOH CAH appropriate. TOA. 12km S B TRIGONOMETRY : Question 5 Lighthouse C is on a bearing of 050° from lighthouse A and northwest of lighthouse B. Lighthouse B is 12km due east of lighthouse A. A ship(S) is sailing directly from lighthouse B to lighthouse A. How close does it come to lighthouse C? C Closest distance is 5.48km What would you like to N do now? Full solution Comments 50° Trig Menu A EXIT 12km B S Question 5 C 1. Calculate missing angles. NW = 3150 AngleA = 90° - 50° = 40° N & angleB = 315° - 270° = 45° 950 So angle C = 180° - 40° - 45° = 95° A 5 0 ° 2. Draw a sketch 400 450 C b 12 km B A 95° 40° 45° 12km S 3. If we use Sine rule: Continue Solution Comments Trigonometry Menu Back to Home b = sinB° c sinC° B Question 5 4. Substitute known values: C N b = 12 sin45° sin95° 950 5. Cross multiply: A 5 0 ° 400 b x sin95° 450 = 12 x sin45° b = 12 x sin45° sin95° 12 km B S Continue Solution Comments Trigonometry Menu Back to Home = 8.5176.… = 8.52km Question 5 6. Closest point is perpendicular so sketch right angled triangle: C N C 950 8.52km = 8.52km A A 5 0 ° 400 a 40° S 450 7. Now using SOHCAH TOA: 12 km B sin40° = a 8.52 S a = 8.52 x sin40° Continue Solution Comments Trigonometry Menu Back to Home = 5.476…. = 5.48 Closest distance is 5.48km Markers Comments 1. Calculate missing angles. AngleA = 90° - 50° = 40° NW = 3150 Since we can pair off two angles with the opposite sides Sine Rule & angleB = 315° - 270° = 45° So angle C = 180° - 40° - 45° = 95° 2. Draw a sketch Refer to the Formulae List : a b c Sine A = Sine B = Sine C C 95° 40 45 ° 12km ° A B 3. If we use Sine rule: b sinB° = c sinC° Next Comment Trigonometry Menu Back to Home Markers Comments 6. Closest point is perpendicular so sketch right angled triangle: Note: the shortest distance from a point to a line is the perpendicular distance. C 8.52km A C a 40° S A B 7. Now using SOHCAH TOA: sin40° = a 8.52 a = 8.52 x sin40° = 5.476…. = 5.48 Closest distance is 5.48km Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 6 Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 2.9° while from the deck of ship B it is 3.8°. The ships are 400m apart. D How far from the platform is each vessel ? Hints Answer only 2.9° Full solution 3.8° 400m O Comments A Trig Menu EXIT B TRIGONOMETRY : Question 6 Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 2.9° while from the deck of ship B it is 3.8°. The ships are 400m apart. D How far from the platform is each vessel ? What would you like to do now? Answer only 2.9° Full solution 400m Comments A Trig Menu EXIT 4. Then use SOH 2. Identify trig 1. Calculate missing CAH TOA to which find rulefrom to angles. use to find length rig to 3. Substitute known common side to both first ship. Second values, remembering to triangles: ship is then 400 m 3.8° use brackets as Two further. sides + two angles appropriate. = sine rule O Three sides + one angle = cosine rule B TRIGONOMETRY : Question 6 Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 2.9° while from the deck of ship B it is 3.8°. The ships are 400m apart. D How far from the platform is each vessel ? What would you like to do now? Try another like this 2.9° Full solution 3.8° 400m O Comments A Trig Menu EXIT B ShipA is 1685m away & ShipB is 1285m away D Question 6 1. Calculate missing angles. Angle B = 180° - 3.8° = 176.2° Angle D = 180° - 2.9° - 176.2° = 0.9° 0.9° 2. Draw a sketch 2.9 ° 400 m D 3.8 ° 0.9° O A 2.9° 176.2° 400m A B Continue Solution Comments Trigonometry Menu Back to Home B 3. BD is common link to both triangles so find it using sine rule. a = sinA° d sinD° D Question 6 4. Substitute known values: a = 400 sin2.9° sin0.9° 0.9° 5. Cross multiply: 2.9 ° 400 m 3.8 ° a x sin0.9° = 400 x sin2.9° a = 400 x sin2.9° sin0.9° O = 1288m A B Continue Solution Comments Trigonometry Menu Back to Home to nearest metre D Question 6 6. Sketch right angled triangle: D 1288m 0.9° 3.8° B d 2.9 ° 400 m 3.8 ° 7. Now using SOHCAH TOA: cos3.8° 1285 m A B Continue Solution O O = d 1288 d = 1288 x cos3.8° = 1285m to nearest metre Try another like this OA is 1285m+400m = 1685m Comments ShipA is 1685m away & ShipB is 1285m away Trigonometry Menu Back to Home Markers Comments Since we can pair off two angles with the opposite sides 1. Calculate missing angles. Angle B = 180° - 3.8° = 176.2° Angle D = 180° - 2.9° - 176.2° = 0.9° Sine Rule Refer to the Formulae List : 2. Draw a sketch D a b c = = Sine A Sine B Sine C 0.9° A 2.9° 176.2° 400m B 3. BD is common link to both triangles so find it using sine rule. a = sinA° d sinD° Next Comment Trigonometry Menu Back to Home Markers Comments 6. Sketch right angled triangle: D Since angle BOD is right - angled use SOHCAHTOA in triangle BOD 1288m 3.8° B d O 7. Now using SOHCAH TOA: cos3.8° = d 1288 d = 1288 x cos3.8° = 1285m to nearest metre OA is 1285m+400m = 1685m ShipA is 1685m away & ShipB is 1285m away Next Comment Trigonometry Menu Back to Home TRIGONOMETRY : Question 6B Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 27° while from the deck of vessel B it is 35°. The ships are 80m apart and the height of their decks is 5m. D How high is the drill tower? Hints Answer only 27° Full solution 35° 80m O Comments A Trig Menu EXIT B TRIGONOMETRY : Question 6B Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 27° while from the deck of vessel B it is 35°. The ships are 80m apart and the height of their decks is 5m. D How high is the drill tower? What would you like to do now? Answer only 27° Full solution 80m Comments A Trig Menu EXIT 4. Then use SOH 2. Identify trig 1. Calculate missing CAH TOA to which find ruleoftorig use to find angles. Height from 3. Substitute known common side to both deck level. values, remembering to triangles: Remember decks use brackets as 35° Two sides + two angles are 5m above sea appropriate. = sine rule level. O Three sides + one angle = cosine rule B TRIGONOMETRY : Question 6B Two supply vessels are approaching an oil platform. From the deck of ship A the angle of elevation of the drill tower is 27° while from the deck of vessel B it is 35°. The ships are 80m apart and the height of their decks is 5m. D How high is the drill tower? What would you like to do now? 27° Full solution 35° 80m O Comments A Trig Menu EXIT B Height of tower is 123m D Question 6B 1. Calculate missing angles. Angle B = 180° - 35° = 145° Angle D = 180° - 145° - 27° = 8° 8° 2. Draw a sketch D 27 ° 80m 8° 35 ° O A B Continue Solution Comments Trigonometry Menu Back to Home A 27° 80m 145° B 3. BD is common link to both triangles so find it using sine rule. a = sinA° d sinD° D Question 6B 4. Substitute known values: a = 80 sin27° sin8° 8° 5. Cross multiply: a x sin8° 27 ° 80m 35 ° a = 80 x sin27° sin8° O A B Continue Solution Comments Trigonometry Menu Back to Home = 80 x sin27° = 261m to nearest metre D Question 6B 6. Sketch right angled triangle: D 261m 8° b 27° B O 7. Now using SOHCAH TOA: 27 ° 80m 35 ° sin27° O = b 261 b = 261 x sin27° A B Continue Solution Comments Trigonometry Menu Back to Home = 118m to nearest metre Height = OD + deck height = 118 + 5 = 123m Height of tower is 123m Comments 1. Calculate missing angles. Since we can pair off two angles with the opposite sides Angle B = 180° - 35° = 145° Angle D = 180° - 145° - 27° = 8° Sine Rule Refer to the Formulae List : 2. Draw a sketch D a b c = = Sine A Sine B Sine C 8° A 27° 80m 145° B 3. BD is common link to both triangles so find it using sine rule. a = sinA° d sinD° Next Comment Trigonometry Menu Back to Home Markers Comments 6. Sketch right angled triangle: D 261m Since angle BOD is right - angled use SOHCAHTOA in triangle BOD b 27° B O 7. Now using SOHCAH TOA: sin27° = b 261 b = 261 x sin27° = 118m to nearest metre Height = OD + deck height Next Comment = 118 + 5 = 123m Trigonometry Menu Height of tower is 123m Back to Home INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Statistics Please choose a question to attempt from the following: 1 EXIT 2 3 4 Back to Unit 2 Menu 5 STATISTICS : Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 35 10 7 7 15 9 (a) Find the mean and standard deviation for this data. (b) A similar experiment was conducted with a second company. The results for this were…. mean = 12 and standard deviation = 1.88 How does the second company compare to the first? Get hint Reveal answer only Go to full solution EXIT Go to Comments Go to Statistics Menu STATISTICS : Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 35 (a) SD (b) EXIT 10 15 9 Draw a7table 7 n data deviation for this data. 2 comparing Find the mean and standard mean x x to mean. Then square A similar experiment was conducted with a second n Thevalues. company. results for this were…. n 1 When comparing data sets= always mean = 12 and standard deviation 1.88 make comment on How does the second companythe compare to(which the first? average is bigger etc.) and the spread of the data. Reveal answer only Go to Comments Go to full solution Go to Statistics Menu What would you like to do now? STATISTICS : Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 35 10 7 7 15 9 (a) Find the mean and standard deviation for this data. (b) A similar experiment was conducted with a second company. The results for this were…. mean = 12 and standard deviation = 1.88 How does the second company compare to the first? (a) Mean = 8 SD 3.87 (b)The second company has a longer average response time. The smaller standard deviation means their arrival time is more predictable. Go to Comments What would you like to do now? EXIT Go to full solution Go to Statistics Menu Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 3 5 10 7 7 15 9 (a) Find the mean and standard deviation for this data. Begin Solution Continue Solution Comments Statistics Menu Back to Home 1. Calculate mean. (a) Mean = (3+5+10+7+7+15+9)7 = 8 So x = 8 and no.pieces of data = n = 7 Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 3 5 10 7 7 15 9 (a) Find the mean and standard deviation for this data. x 8 Begin Solution Continue Solution Comments Statistics Menu Back to Home 2. Draw table comparing data to mean. x 3 5 10 7 7 15 9 -5 -3 2 -1 -1 7 1 25 9 4 1 1 49 1 (x - x )2 = 90 xx xx 2 Question 1 A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 3 5 10 7 7 15 9 (a) Find the mean and standard deviation for this data. Begin Solution Continue Solution Comments Statistics Menu Back to Home 3. Use formula to calculate standard deviation. xx SD n 1 90 SD 6 SD 3.87 2 Just found!! Question 1 (b) A similar experiment was conducted with a second company. The results for this were…. mean = 12 and standard deviation = 1.88 How does the second company compare to the first? x 8 1. Always compare mean and standard deviation. (b)The second company has a longer average response time. The smaller standard deviation means their arrival time is more predictable. SD 3.87 What would you like to do now? Begin Solution Continue Solution Comments Statistics Menu Back to Home Comments 3. Use formula to calculate standard deviation. xx SD n 1 90 SD 6 SD 3.87 2 Check the list of Formulae for the Standard Deviation Formula: ( x x) n 1 2 or 2 ( x ) x 2 n n 1 The second formula can be used in the calculator paper. The calculator must be in Stats. Mode to allow the data to be entered. Next Comment Statistics Menu Back to Home STATISTICS : Question 2 Weeks of training A TV personality takes part in a 20 week training schedule to copy a 100m sprinter. Her times are recorded every 2 weeks and plotted in the graph then a line of best fit is drawn. EXIT Continue question STATISTICS : Question 2 Get hint Reveal answer only Go to full solution Go to Comments Go to Stats Menu Weeks of training (i) Find the equation of the line in terms of T and W. (ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs EXIT STATISTICS : Question 2 Find gradient Use Use your your and note intercept equation equation and and of T axis. substitute substitute WT= =12. 11.5 Reveal answer only Go to full solution Go to Comments Go to Stats Menu Weeks of training (i) Find the equation of the line in terms of T and W. (ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs EXIT What would you like to do now? STATISTICS : Question 2 What would you like to do now? Go to full solution Go to Comments Go to Stats Menu Weeks of training (i) Find the equation of the line in terms of T and W. (ii) Use answer (i) to predict EXIT T = -¼ W + 15 (a) her time after 12 weeks of training 12secs. (b) the week when her time was 11.5secs 14 weeks Question 2 (i) Find the equation of the line in terms of T and W. 1. Find gradient and note intercept of T axis. (a) m = (y2 – y1) 10 - 15 = -¼ = (x2 – x1) 20 - 0 Intercept at 15 Equation is Go to graph Begin Solution Continue Solution Comments Statistics Menu Back to Home T = -¼ W + 15 Question 2 (ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs 2. Use your equation and substitute W = 12. (b)(i) If w = 12 then T = -¼ W + 15 becomes T = (-¼ x 12) + 15 = -3 + 15 = 12 Time at 12 weeks is 12secs. Begin Solution Continue Solution Comments Statistics Menu Back to Home Question 2 (ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs 3. Use your equation and substitute T = 11.5 (b)(i) If t = 11.5 then becomes T = -¼ W + 15 -¼ W + 15 = 11.5 (-15) (-15) -¼ W = -3.5 x (–4) W = 14 Reach a time of 11.5sec after 14 weeks. What would you like to do now? Begin Solution Continue Solution Comments Statistics Menu Back to Home Comments 1. Find gradient and note intercept of T axis. (a) m = (y2 – y1) 10 - 15 = -¼ = (x2 – x1) 20 - 0 To find the equation of a line from the graph: Must Learn: y = mx + c Intercept at 15 Equation is T = -¼ W + 15 gradient intercept So you need to find these!! Next Comment Statistics Menu Back to Home Comments 1. Find gradient and note intercept of T axis. (a) m = (y2 – y1) 10 - 15 = -¼ = (x2 – x1) 20 - 0 vertical m = horizontal Note: Always draw the horizontal before the vertical: Intercept at 15 horizontal (+ve) Equation is T = -¼ W + 15 vertical (-ve) Next Comment Statistics Menu Back to Home STATISTICS : Question 3 A sample of 180 teenagers were asked their opinions on the TV series the “Simpsons” & the movie “Shrek” and their responses were displayed in the following table… Like Don’t like Simpsons Like Shrek Don’t like Shrek Simpsons 126 ? 15 21 (a) What percentage liked Shrek but not the Simpsons? (b) If someone is picked at random what is the probability that (i) they liked the Simpsons but not Shrek? (ii) they liked neither? Get hint EXIT Reveal Full solution Comments Statistics Menu STATISTICS : Question 3 A sample of 180 teenagers were asked their opinions on the TV series the “Simpsons” & the movie “Shrek” and their responses were displayed in the following table… Like Don’t like Simpsons Like Shrek Don’t like Shrek 126 15 Simpsons Remember To find probabilities all entries in use: ? P = no oftable must add to total favourable / no of sample size data 21 (a) What percentage liked Shrek but not the Simpsons? (b) If someone is picked at random what is the probability that (i) they liked the Simpsons but not Shrek? (ii) they liked neither? EXIT What would you like to do now? Full solution Statistics Menu Comments Reveal STATISTICS : Question 3 A sample of 180 teenagers were asked their opinions on the TV series the “Simpsons” & the movie “Shrek” and their responses were displayed in the following table… Like Don’t like Simpsons Like Shrek Don’t like Shrek Simpsons 126 ? 15 21 (a) What percentage liked Shrek but not the Simpsons? 10% (b) If someone is picked at random what is the probability that (i) they liked the Simpsons but not Shrek? (ii) they liked neither? Full solution EXIT Comments 7/ 60 Statistics Menu What now? 1/ 12 Question 3 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Shrek Don’t like Simpsons 126 ? 15 21 (a) What percentage liked Shrek but not the Simpsons? Begin Solution Continue Solution Comments Statistics Menu Back to Home NB: There are 180 in survey!! (a) Like Shrek but not Simpsons = 180 – 126 – 15 – 21 = 18 % = 18/180 = 1/ 10 = 10% Question 3 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Shrek Don’t like Simpsons 126 ? 15 21 what is the probability that (i) they liked the Simpsons but not Shrek? Begin Solution Continue Solution Comments Statistics Menu Back to Home (b)(i) Prob = 15/ 180 = 1/ 12 Question 3 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Shrek Don’t like Simpsons 126 ? 15 21 (b)(ii) Prob = 21/ 180 = 7/ 60 What would you like to do now? (ii) they liked neither? Begin Solution Continue Solution Comments Statistics Menu Back to Home Comments 1. Use P = no of favourable / no of data NB: There are 180 in survey!! (a) Like Shrek but not Simpsons Note: To change a fraction to a % multiply by 100% 18 180 = 18 180 x 100% = 180 – 126 – 15 – 21 = 18 % = 18/180 = 1/ 10 = 10% Next Comment Statistics Menu Back to Home Comments To calculate simple probabilities: 1. Use P = no of favourable / no of data Probability = (b)(i) Prob = 15/ 180 = 1/ 12 Number of favourable outcomes Number of possible outcomes Next Comment Statistics Menu Back to Home STATISTICS : Question 4 Get hint On a college course you have to pick a language plus a Reveal ans leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING Full solution Comments Stats Menu (a) Make a list of all possible combinations of courses. (b) If a combination is selected at random what is the probability that it is …… (i) Includes Spanish? (ii) Includes swimming? EXIT (iii) Doesn’t include French but includes music? What would you like to do now? STATISTICS : Question 4 On a college course you have to pick a language plus a Reveal ans leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING Full solution Comments Stats Menu Use a tree (a) Make a list of all possible combinations of courses. diagram & “branch out” (b) If a combination is selected at random what is the with Noweach list language. probability that it is …… the pairs Use your From list eachof of language “branch (i) Includes Spanish? possible subjects. out” with each combinations to find (ii) Includes swimming? leisure activity. probabilities. EXIT (iii) Doesn’t include French but includes music? STATISTICS : Question 4 On a college course you have to pick a language plus a leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH Full solution LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING (a) Make a list of all possible combinations of courses. Comments Stats Menu CLICK (b) If a combination is selected at random what is the probability that it is …… (i) Includes Spanish? (ii) Includes swimming? EXIT 1/ 3 1/ 4 1/ (iii) Doesn’t include French but includes music? 6 (a) French/Music French/Video French/Bsktbll French/Swim German/Music GERMAN German/Video German/Bsktbll German/Swim Spanish/Music Spanish/Video Spanish/Bsktbll Spanish/Swim Hints Use a tree diagram & “branch out” with each language. From each language “branch out” with each leisure activity. Now list the pairs of subjects. Hints French/Music Have list of combinations handy. French/Video French/Bsktbll (b)(i) Prob = 4/12 = 1/ 3 Remember to simplify. French/Swim What now? German/Music German/Video Comments (b)(ii) Prob = 3/12 = 1/ 4 Stats Menu German/Bsktbll What is probability: German/Swim (i) Includes Spanish? Spanish/Music Spanish/Video Spanish/Bsktbll Spanish/Swim (ii) Includes swimming? (b)(iii) Prob = 2/12 = 1/6 (iii)Doesn’t include French but includes music? Comments (a) French/Music French/Video No. of possible outcomes French/Bsktbll French/Swim GERMAN = 3 x 4 = 12 German/Music To calculate simple probabilities: German/Video Probability = German/Bsktbll German/Swim Number of favourable Number of possible Spanish/Music Spanish/Video Spanish/Bsktbll Next Comment Spanish/Swim Statistics Menu Back to Home STATISTICS : Question 5 The delivery times for a fast food company are shown in the following cumulative frequency table. Time No. Deliveries up to 4 mins up to 8 mins up to 12 mins 5 13 37 up to 16 mins up to 20 mins 53 60 Get hint Reveal ans Full solution Comments Stats Menu (a) How many deliveries took longer than 16 mins? (b) Use the data to construct a cumulative frequency graph. (c) Use the graph to find the median and semi-interquartile range for this data. STATISTICS : Question 5 The delivery times for a fast food company are shown in the following cumulative frequency table. What would you like to do now? For how many greater Time than 16 Median is findup difference to 4 mins middle between end valueso 8we upofto mins value values SIQR = ½(Q3 – Q1) want up to 16. up to 12 mins halfwayQ1 – 25% point & point on up–to 16point mins Q3 75% frequency up to 20 mins axis. No. Deliveries 5 13 37 53 60 Reveal ans Full solution Comments Stats Menu (a) How many deliveries took longer than 16 mins? (b) Use the data to construct a cumulative frequency graph. (c) Use the graph to find the median and semi-interquartile range for this data. STATISTICS : Question 5 The delivery times for a fast food company are shown in the following cumulative frequency table. Time No. Deliveries up to 4 mins up to 8 mins up to 12 mins 5 13 37 up to 16 mins up to 20 mins 53 60 Full solution Comments Stats Menu =7 (a) How many deliveries took longer than 16 mins? (b) Use the data to construct a cumulative frequency graph. (c) Use the graph to find the median and semi-interquartile range for this data. Median = 11 mins SIQR = 2.75 mins (a) Deliveries taking longer than 16 mins = 60 – 53 = 7 Time Deliveries Cum freq 4 5 ¾ of 60 = 45 Q3 Q2 8 13 12 37 Comments ½ of 60 = 30 16 53 20 60 Stats Menu ¼ of 60 = 15 What would you like to do now? Q1 8.5 (C) Median = 11mins. 11 14 Delivery time(mins) (c) SIQR = ½(Q3 – Q1) = (14 – 8.5) 2 = 2.75mins Comments Note: In a Cumulative Frequency Diagram: Cumulative Frequency On y-axis (cumulative frequency) 60 Q1 at 25%, 50% Q2 at 50%, Q3 at 75% 11 20 Delivery Time And read values from the delivery time scale (x-axis). End of Statistics Next Comment Statistics Menu Back to Home INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Graphs, Charts & Tables Please choose a question to attempt from the following: 1 Stem & Leaf EXIT 2 Dot Plot 3 Cum Freq Table 4 Dot to boxplot Back to Unit 2 Menu 5 Stem to boxplot 6 Piechart GRAPHS, CHARTS, TABLES : Question 1 The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 19 1 2 8 20 1 5 5 21 8 6 7 n = 25 17 4 = £174 (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? Go to full solution Get hint EXIT Reveal answer Go to Comments GRAPHS, CHARTS, TABLES : Question 1 The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 16 2 3 6 17 1 1 1 18 2 3 3 19 1 2 8 20 1 5 5 21 8 (a) Use this information to find the 9 Use median Q1 is midpoint position 8 8= (n+1)9/ from start to 2 to find median median 5 6 7 7 6 Q3 is midpoint nfrom = 25median to end 17 4 = £174 (i) median (ii) lower & upper quartiles What would you like to do now? (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? Graphs etc Menu Go to full solution EXIT Reveal answer Go to Comments GRAPHS, CHARTS, TABLES : Question 1 The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 19 1 2 8 20 1 5 5 21 8 6 7 n = 25 17 4 = £174 (a) Use this information to find the What would you like to do now? median = £183 (ii) lower & upper quartilesQ1 = £171 (i) median Q3 = £195 (iii) the semi-interquartile range = £12 (b)What is the probability that someone chosen at random earns less than £180? = 2/5 Go to full solution EXIT Graphs etc Menu Go to Comments Question 1 16 17 18 19 20 21 2 1 2 1 1 8 3 1 3 2 5 6 1 3 8 5 1. Use median = (n+1) / 2 to find median 9 8 5 6 (a)(i) Since n = 25 then the median is 8 6 9 7 7 n = 25 17 4 = £174 (i) Median (ii) lower & upper quartiles (iii) the semi-interquartile range Begin Solution Continue Solution Comments Menu Back to Home 13th value ie median = £183 (NOT 3!!!) 2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2 (ii) Both 6th & 7th values are £171 so Q1 = £171 3. There are 12 values after median so Q3 position = 13 + (12 + 1) / 2 19th is £192 & 20th is £198 so Q3 = £195 What would you like to do now? Question 1 16 17 18 19 20 21 2 1 2 1 1 8 3 1 3 2 5 6 1 3 8 5 4. Use SIQR = ½ (Q3 – Q1 ) / 2 9 8 5 6 8 6 9 7 (iii) SIQR = ½(Q3 – Q1) = (£195 - £171) 2 7 n = 25 17 4 = £174 (i) Median (ii) lower & upper quartiles (iii) the semi-interquartile range Begin Solution Continue Solution Comments Menu Back to Home = £12 Question 1 16 17 18 19 20 21 2 1 2 1 1 8 3 1 3 2 5 6 1 3 8 5 5. Use P = no of favourable / no of data 9 8 5 6 No of favourable ( under £180) = 10 8 6 9 7 7 n = 25 17 4 = £174 (b)What is the probability that someone chosen at random earns less than £180? Begin Solution Continue Solution Comments Menu Back to Home No of data = n = 25 (b) Prob(under £180) = 10/ 25 = 2/ 5 . Comments 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13th value ie median = £183 2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2 (ii) Both 6th & 7th values are £171 so Q1 = £171 Median: the middle number in the ordered list. 25 numbers in the list. 1 – 12 13 14 - 25 12 numbers on either side of the median median is the 13th number in order. 3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2 19th is £192 & 20th is £198 so Q3 = £195 Next Comment Menu Back to Home Comments 1. Use median = (n+1) / 2 to find median To find the upper and lower quartiles deal with the numbers on either side of the median separately. (a)(i) Since n = 25 then the median is 13th value ie median = £183 2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2 Q1 12 numbers before median. 6 numbers either side of Q1 is midway between the 6th and 7th number. (ii) Both 6th & 7th values are £171 so Q1 = £171 3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2 19th is £192 & 20th is £198 so Q3 = £195 Next Comment Menu Back to Home Comments 1. Use median = (n+1) / 2 to find median To find the upper and lower quartiles deal with the numbers on either side of the median separately. (a)(i) Since n = 25 then the median is 13th value ie median = £183 2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2 Q3 12 numbers after median. 6 numbers either side of Q3 is midway between the 19th and 20th number. (ii) Both 6th & 7th values are £171 so Q1 = £171 3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2 19th is £192 & 20th is £198 so Q3 = £195 Next Comment Menu Back to Home Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Get hint EXIT Reveal answer Go to full solution Go to Comments Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows Establish lowest & 28 29 29 30 31 30 28 30highest 29 28 values and Plot a dot for draw line with use: scale. For probability each piece of 29 30 30 28 28 29 29 29 29 28 data and label P = no of favourable diagram. / no of data a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? What would you like to do now? EXIT Graphs etc Menu Go to full solution Reveal answer Go to Comments Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 a) Illustrate this using a dot plot. CLICK b) What type of distribution does this show? Tightly clustered c) If a bag is chosen at random what is the probability it 3/10 will be heavier than the modal weight? Graphs etc Menu EXIT Go to full solution Go to Comments Question 2 28 30 29 29 29 28 30 29 29 30 30 29 30 29 28 29 31 28 28 28 1. Establish lowest & highest values and draw line with scale. (a) Lowest = 28 & highest = 31. Weights in g Illustrate this using a dot plot. 26 Begin Solution Continue Solution Comments Menu Back to Home 28 30 32 2. Plot a dot for each piece of data and label diagram. Question 2 28 30 29 29 29 28 30 29 29 30 30 29 30 29 28 29 31 28 28 28 3. Make sure you know the possible descriptions of data. Weights in g What type of distribution does this show? 26 28 30 Begin Solution Continue Solution Comments Menu Back to Home (b) Tightly clustered distribution. 32 Question 2 28 30 29 29 29 28 30 29 29 30 30 29 4. Use P = no of favourable / no of data 30 29 28 29 31 28 28 28 Mode! Weights in g If a bag is chosen at random what is the probability it will be heavier than the modal weight? 26 Begin Solution 28 30 No of favourable ( bigger than 29) = 6 Continue Solution Comments Menu Back to Home 32 No of data = n = 20 (c) Prob(W > mode) = 6/ 3/ = 20 10 . What would you like to do now? Comments Other types of distribution: 3. Make sure you know the possible descriptions of data. Weights in g 26 28 30 (b) Tightly clustered distribution. 32 Next Comment Menu Back to Home Comments Other types of distribution: 3. Make sure you know the possible descriptions of data. Weights in g 26 28 30 (b) Tightly clustered distribution. 32 Next Comment Menu Back to Home Comments Other types of distribution: 3. Make sure you know the possible descriptions of data. Weights in g 26 28 30 (b) Tightly clustered distribution. 32 Next Comment Menu Back to Home Comments 4. Use P = no of favourable / no of data Mode! To calculate simple probabilities: Probability = Weights in g Number of favourable outcomes Number of possible outcomes 2 2 3 3 6 8 0 2 No of favourable ( bigger than 29) = 6 Next Comment No of data = n = 20 Menu (c) Prob(W > mode) = 6/ 20 = 3/10 . Back to Home Charts, Graphs & Tables : Question 3 The results for a class test were 18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? Get hint Graphs etc Menu EXIT Reveal answer Go to full solution Go to Comments Charts, Graphs & Tables : Question 3 The results for a class test were lowest & 18 14 16 17 14 16 13 11 13 13 16 14Establish 13 18 Use15median = highest values and (n+1) / 2 to 10 14 17 13 15 15 18 14 17 13 16 10 14draw 13 17 Complete each table. For probability use: establish row 1 step in at a which row time, P = no of favourable lies./ (a) Construct a cumulative frequency table formedian this data. calculating no of data running total as (b) What is the median for this data? you go. (c) What is the probability that a pupil selected at random scored under 14? What would you like to do now? EXIT Graphs etc Menu Go to full solution Reveal answer Go to Comments Charts, Graphs & Tables : Question 3 The results for a class test were 18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. CLICK (b) What is the median for this data? Median = 14 (c) What is the probability that a pupil selected at random scored under 14? 1/3 Graphs etc Menu EXIT Go to full solution Go to Comments Question 3 1. Establish lowest & highest values and draw a table. 18 14 16 17 14 16 13 11 13 (a) Lowest = 10 & highest = 18 13 16 14 13 18 15 10 14 17 Mark 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. Begin Solution Continue Solution 10 11 12 13 14 15 16 17 18 Frequency Cum Frequency 2 1 0 7 6 3 4 4 3 2 3 3 10 16 19 23 27 30 Comments Menu Back to Home 2. Complete each row 1 step at a time, calculating running total as you go. Question 3 3. Use median = (n+1) / 2 to establish in which row median lies. 18 14 16 17 14 16 13 11 13 Mark Frequency Cum Frequency 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (b) What is the median for this data? What would you like to do now? Begin Solution Continue Solution Comments Menu Back to Home 10 11 12 13 14 15 16 17 18 2 1 0 7 6 3 4 4 3 2 3 3 10 16 19 23 27 30 For 30 values median is between 15th & 16th both of which are in row 14. Median Mark = 14 Question 3 4. Use P = no of favourable / no of data 18 14 16 17 14 16 13 11 13 Mark Frequency Cum Frequency 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (c) What is the probability that a pupil selected at random scored under 14? What would you like to do now? Begin Solution Continue Solution Comments Menu Back to Home 10 11 12 13 14 15 16 17 18 2 1 0 7 6 3 4 4 3 2 3 3 10 16 19 23 27 30 No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10/ 30 = 1/ 3 . Comments Median: Mark 10 11 12 13 14 15 16 17 18 Freq 2 1 0 7 6 3 4 4 3 Cum Freq 2 3 3 10 16 19 23 27 30 For 30 values median is between 15th & 16th both of which are in row 14. Median = 14 1 – 15 Q2 16 - 30 Median = 14 Find the mark at which the cumulative frequency first reaches between 15th and 16th number. Next Comment Menu Back to Home Comments To calculate simple probabilities: Mark Freq Cum Freq Probability = 10 11 12 13 14 15 16 17 18 2 1 0 7 6 3 4 4 3 2 3 3 10 16 19 23 27 30 Number of favourable outcomes Number of possible outcomes No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10/ 30 = 1/ 3 Next Comment . Menu Back to Home Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 (a) Find the 50 52 54 56 58 (i) median (ii) lower quartile (iii) upper quartile (b) Construct a boxplot using this data. (c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Get hint EXIT Graphs etc Menu Reveal answer Go to full solution Go to Comments Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. median Q1Use is midpoint position = (n+1) from start to / 2 remember tomedian find median bigger SIQR means more Q3 is midpoint variation from median to (spread) in end 48 50 52 54 56 58 data. (a) Find the (i) median (ii) lower quartile (iii) upper quartile (b) Construct a boxplot using this data. (c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? What would you like to do now? Graphs etc Menu EXIT Reveal answer Go to full solution Go to Comments Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. Median = 50 So Q1 = 49 So Q3 = 52 48 (a) Find the 50 52 54 56 58 (i) median (ii) lower quartile (iii) upper quartile (b) Construct a boxplot using this data. CLICK (c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? the data is distributed more widely than (or not as clustered as) the above data EXIT Menu Full solution Comments 1. Use median = (n+1) / 2 to find median Question 4 (a) (i) Sample size = 23 so median position is 12. ie (23+1)2 48 50 52 (a) Find the 54 56 (i) median (ii) lower quartile (iii) upper quartile 58 Median = 50 2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2 (ii) Middle of 1st 11 is position 6. So Q1 = 49 Begin Solution Continue Solution Comments Menu Back to Home 3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2 (iii) Middle of 2nd 11 is position 18. So Q3 = 52 Question 4 4. Draw number line with scale. Make sure you note highest & lowest as well as Q1, Q2, Q3. (b)Lowest = 48, Q1 = 49, Q2 = 50, Q3 = 52 & Highest = 58. 48 50 52 54 56 58 (b) Construct a boxplot using this data. Begin Solution Continue Solution Comments Menu Back to Home 48 50 52 54 56 58 Question 4 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c) For above sample SIQR = (52 - 49) 2 = 1.5 48 50 52 54 56 58 (c) In a second sample the semi-interquartile range was In a sample where the SIQR is 2.5 the data is distributed more widely than (or not as clustered as) the above data 2.5. How does this compare? What would you like to do now? Begin Solution Continue Solution Comments Menu Back to Home Comments 1. Use median = (n+1) / 2 to find median The median: 23 numbers in the list: (a) (i) Sample size = 23 so median position is 12. 1 - 11 12 13 - 23 ie (23+1)2 Median = 50 2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2 Q2 11 numbers on either side of the median (ii) Middle of 1st 11 is position 6. So Q1 = 49 3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2 (iii) Middle of 2nd 11 is position 18. So Q3 = 52 Next Comment Menu Back to Home Comments 1. Use median = (n+1) / 2 to find median (a) (i) Sample size = 23 For quartiles: 1 - 5 so median position is 12. 2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2 (ii) Middle of 1st 11 is position 6. So Q1 = 49 3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2 (iii) Middle of 2nd 11 is position 18. So Q3 = 52 7 - 11 12 Q2 12 Q2 Q1 ie (23+1)2 Median = 50 6 13 - 17 18 19 - 23 Q3 Now count through the list until you reach the 6th, 12th,and 18th number in the list. Next Comment Menu Back to Home Comments 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c) For above sample SIQR = (52 - 49) 2 = 1.5 In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data The semi-interquartile range is a measure of the range of the “middle” 50%. S.I.R. = 1 (Q3 - Q1) 2 It is a measure of how spread-out and so how “consistent” or “reliable” the data is. Remember: when asked to compare data always consider average and spread. Next Comment Menu Back to Home Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. Get hint 6 7 8 9 10 11 12 0 1 2 4 5 2 3 2 4 7 5 7 7 2 5 6 6 8 9 6 9 9 7 Reveal answer Full solution 11 4 = 114kg 1 1 3 Comments (a) Find the median, lower & upper quartiles for this data. (b) Use the data to construct a boxplot. (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. EXIT 60 70 80 90 100 110 120 Charts, Graphs & Tables : Question 5 What now? The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. Menu median Q1Use is midpoint 6 0 2 position = (n+1) from start to / 7 1 3 5 7 7 When Reveal answer 2 tomedian find median 8 2 2 2 5 6 6 8 9comparing position 9 4 4 6 9 9 two data sets Q3 is midpoint Full solution 10 5 7 7 comment on from11median to = 114kg 11 spread4 and end 12 1 1 3 Comments average (a) Find the median, lower & upper quartiles for this data. (b) Use the data to construct a boxplot. (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. EXIT 60 70 80 90 100 110 120 Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. 6 7 8 9 10 11 12 0 1 2 4 5 2 3 2 4 7 median = 87 5 7 7 2 5 6 6 8 9 6 9 9 7 Menu Q1 = 77 Q3 = 99 Full solution 11 4 = 114kg 1 1 3 Comments (a) Find the median, lower & upper quartiles for this data. (b) Use the data to construct a boxplot. CLICK (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. EXIT 60 70 80 90 100 110 120 CLICK Question 5 6 7 8 9 10 11 12 0 1 2 4 5 2 3 2 4 7 5 7 7 2 5 6 6 8 9 6 9 9 7 11 4 = 114kg 1 1 3 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 26 then the median is between 13th & 14th value ie median = 87 2. There are 13 values before median so Q1 position is 6th value (a) Find the median, lower & upper quartiles for this data. (ii) so Q1 = 77 3. There are 13 values after median so Q3 position is 20th position Begin Solution Continue Solution Comments Menu Back to Home so Q3 = 99 Question 5 6 7 8 9 10 11 12 0 1 2 4 5 2 3 2 4 7 5 7 7 2 5 6 6 8 9 6 9 9 7 11 4 = 114kg 1 1 3 4. Draw number line with scale. Make sure you note highest & lowest as well as Q1, Q2, Q3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. (b) Use the data to construct a boxplot. Begin Solution Continue Solution Comments Menu Back to Home 60 70 80 90 100 110 120 5. Compare spread and relevant average. Question 5 (c) The boxplot below shows the weight distribution for (c) Lightest has put on weight – these people after several lowest now 65, months. heaviest 3 have lost weight – Compare the two & highest now 115, comment on the results. median same but overall spread of weights has decreased as Q3-Q1 was 22 60 70 80 90 100 110 120 but is now only 15. Begin Solution Continue Solution Comments Menu Back to Home What would you like to do now? Comments Remember: 4. Draw number line with scale. Make sure you note highest & lowest as well as Q1, Q2, Q3. (b)Lowest = 60, Q1 = 77, Q2 = 87, To draw a boxplot you need a “five-figure summary”: Box Plot : Q3 = 99 & Highest = 123. Lowest 60 70 80 90 100 110 Q1 Q2 Q3 Highest 120 five-figure summary Next Comment Menu Back to Home Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. (a) How many people Watching TV went clubbing? 144° (b) If 84 people went to the cinema x° theatre clubbing theatre then how big is x°? Get hint EXIT Graphs etc Menu Go to full solution Reveal answer Go to Comments Charts, Graphs & Tables : Question 6 The pie chart below showsangle the breakdown of how a sample of amount = 630 people spent their Saturday 360° nights. 630 (a) How many people Watching TV went clubbing? 144° (b) If 84 people went to the cinema x° theatre clubbing theatre then how big is x°? What would you like to do now? EXIT Graphs etc Menu Go to full solution Reveal answer Go to Comments Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. (a) How many people Watching TV cinema = 252 went clubbing? 144° (b) If 84 people went to the x° theatre clubbing theatre then how big is x°? = 48° What would you like to do now? Graphs etc Menu EXIT Go to full solution Go to Comments Question 6 Watching TV 144° 1. Set up ratio of angles and sectors and cross multiply. cinema x° theatre clubbin g How many people went clubbing? Begin Solution Continue Solution Comments Menu Back to Home (a) The angle is 144° so ….. angle = amount 360° 630 144° = amount 360° 630 360 x amount = 144 x 630 amount = 144 x 630 360 = 252 Question 6 Watching TV 144° 2. Set up ratio of angles and sectors and cross multiply. cinema x° theatre clubbin g (b) If 84 people went to the theatre then how big is x°? Begin Solution Continue Solution Comments Menu Back to Home (b) The amount is 84 so ….. angle = amount 360° 630 angle = 84 360° 630 630 x angle = 360° x 84 angle = 360° x 84 630 = 48° Comments 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle = amount 360° 630 144° = amount 360° 630 360 x amount = 144 x 630 Can also be tackled by using proportion: amount angle at centre total sample 360 Amount = 144 360 x 630 amount = 144 x 630 360 = 252 Next Comment Menu Back to Home Comments 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle = amount 360° 630 angle = 84 360° 630 Can also be tackled by using proportion: amount angle at centre total sample 360 84 = x 360 x 630 630 x = 84 x 360 630 x angle = 360° x 84 angle = 360° x 84 630 = 48° x = 84 x 360 630 End of graphs, charts etc. Next Comment Menu Back to Home INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Simultaneous Equations Please choose a question to attempt from the following: 1 EXIT 2 3 Back to Unit 2 Menu 4 Simultaneous Equations : Question 1 The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 3x + 2y = 17. EXIT Reveal answer only Go to Comments Go to full solution Go to Sim Eq Menu Get hint Simultaneous Equations : Question 1 The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of To draw graph: Construct a join Plot and table ofSolution points. values with at is where lines least 2 xcross y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 coordinates. 3x + 2y = 17.would you like to do now? What EXIT Reveal answer only Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 1 The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 Solution is x=3 & y=4 3x + What 2y = 17. would you like to do now? Try another like this EXIT Go to full solution Go to Comments Go to Sim Eq Menu Question 1 The diagram shows the graph of 3x + 2y = 17. y = 2x - 2 Copy the diagram and on your diagram draw the graph of y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 1. Construct a table of values with at least 2 x coordinates. x y 0 -2 5 8 Question 1 The diagram shows the graph 2. Plot and join points. Solution is where lines cross. of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 Begin Solution Try another like this Comments Solution is x=3 & y=4 Sim Eq Menu What would you like to do now? Back to Home Markers Comments 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2 x y 0 -2 5 8 There are two ways of drawing the line y = 2x - 1 Method 1 Finding two points on the line: x=0 y = 2 x 0 – 1 = -1 x=2 y =2x2-1 = 3 First Point (0,-1) Second Point (-1,3) Plot and join (0,-1), and (-1,3). Next Comment Sim Eqs Menu Back to Home Markers Comments 2. Plot and join points. Solution is where lines cross. Method 2 Using y = mx + c form: y = mx + c gradient y - intercept y = 2x - 1 gradient m=2 y - intercept c = -1 Plot C(0, -1) and draw line with m = 2 Next Comment Solution is x=3 & y=4 Sim Eqs Menu Back to Home Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1/2x + 3 , hence solve 2x + y = 8 y = 1/2x + 3 EXIT Reveal answer only Go to Comments Go to full solution Go to Sim Eq Menu Get hint Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your To draw graph: Construct a join Plot and table ofSolution points. values with at is where lines least 2 xcross coordinates. diagram draw the graph of y = 1/2x + 3 , hence solve 2x + y = 8 y = 1/2x + 3 What would you like to do now? EXIT Reveal answer only Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1/2x + 3 , hence solve 2x + y = 8 y = 1/2x + 3 Solution is x=2 & y=4 What would you like to do now? Go to Comments EXIT Go to full solution Go to Sim Eq Menu Question 1B 1. Construct a table of values with at The diagram shows the graph of least 2 x coordinates. 2x + y = 8. y = 1/2x + 3 Copy the diagram and on your diagram draw the graph of y = 1/2x + 3 , hence solve 2x + y = 8 y = 1/2x + 3 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home x y 0 3 6 6 Question 1B The diagram shows the graph of 2x + y = 8. 2. Plot and join points. Solution is where lines cross. y = 1/2x + 3 Copy the diagram and on your x y 0 3 6 6 diagram draw the graph of y = 1/2x + 3 , What would you like to do now? hence solve 2x + y = 8 y = 1/2x + 3 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home Solution is x=2 & y=4 Markers Comments 1. Construct a table of values with at least 2 x coordinates. y = 1/2x + 3 x y 0 3 6 6 There are two ways of drawing the line y = ½ x + 3 Method 1 Finding two points on the line: x=0 y =½ x0+3= 3 x=2 y = ½ x 2 +3 = 4 First Point (0, 3) Second Point (2, 4) Plot and join (0, 3), and (2, 4). Next Comment Sim Eqs Menu Back to Home Markers Comments 2. Plot and join points. Solution is where lines cross. y= 1/ x 2 +3 x y 0 3 6 6 Method 2 Using y = mx + c form: y = mx + c gradient y - intercept y=½x+3 gradient m=½ y - intercept c = +3 Plot C(0, 3) and draw line with m = ½ Next Comment Sim Eqs Menu Solution is x=2 & y=4 Back to Home Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Get hint Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Eliminate either variable by making coefficient same. Substitute found value into either of original equations. Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Solution is u=2 & v=1 What would you like to do now? Try another like this Go to full solution Go to Comments Go to Sim Eq Menu EXIT Question 2 Solve 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. 3u - 2v = 4 2u + 5v = 9 (x5) (x2) 1 2 Now get: 15u 10v 20 4u 10v 18 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 19u 38 u 2 3 = 1 (x5) 4 = 2 (x2) 3 + 4 Question 2 Solve 3u - 2v = 4 2u + 5v = 9 2. Substitute found value into either of original equations. 3u - 2v = 4 2u + 5v = 9 (x5) (x2) 1 2 Substitute 2 for u in equation 2 4 + 5v = 9 5v = 5 v=1 Begin Solution Try another like this Comments Sim Eq Menu Back to Home Solution is u = 2 & v = 1 What would you like to do now? Markers Comments Note: 1. Eliminate either u’s or v’s by making coefficient same. 3u - 2v = 4 2u + 5v = 9 (x5) (x2) When the “signs” are the same subtract to eliminate. 1 2 When the “signs” are different add to eliminate. Now get: 15u 10v 20 4u 10v 18 19u 38 u 2 3 = 1 (x5) 4 = 2 (x2) e.g. 2x + 3y = 4 6x + 3y = 4 Subtract the equations 3 + 4 Next Comment Sim Eqs Menu Back to Home Markers Comments Note: 1. Eliminate either u’s or v’s by making coefficient same. 3u - 2v = 4 2u + 5v = 9 (x5) (x2) When the “signs” are the same subtract to eliminate. 1 2 When the “signs” are different add to eliminate. Now get: 15u 10v 20 4u 10v 18 19u 38 u 2 3 = 1 (x5) 4 = 2 (x2) e.g. 2x + 3y = 4 6x - 3y = 4 Add the equations 3 + 4 Next Comment Sim Eqs Menu Back to Home Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Get hint Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Eliminate either variable by making coefficient same. Substitute found value into either of original equations. Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Solution is q = -1 & q = 0.6 What would you like to do now? Go to full solution Go to Comments Go to Sim Eq Menu EXIT Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) 5p + 3q = 0 4p + 5q = -2.6 (x5) 1 2 Now get: 20 p 12q 0 20 p 25q 13 13q 13 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home q 1 3 = 1 (x4) 4 = 2 (x5) 4 - 3 Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 2. Substitute found value into either of original equations. (x4) 5p + 3q = 0 4p + 5q = -2.6 (x5) 1 2 Substitute -1 for q in equation 1 5p + (- 3) = 0 5p = 3 p =3/5 = 0.6 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home Solution is q = -1 & p = 0.6 What would you like to do now? Markers Comments 1. Eliminate either p’s or q’s by making coefficient same. (x4) 5p + 3q = 0 4p + 5q = -2.6 (x5) 13q 13 q 1 When the “signs” are the same subtract to eliminate. 1 2 When the “signs” are different add to eliminate. Now get: 20 p 12q 0 20 p 25q 13 Note: 3 = 1 (x4) 4 = 2 (x5) e.g. 2x + 3y = 4 6x + 3y = 4 Subtract the equations 4 - 3 Next Comment Sim Eqs Menu Back to Home Markers Comments 1. Eliminate either p’s or q’s by making coefficient same. (x4) 5p + 3q = 0 4p + 5q = -2.6 (x5) 13q 13 q 1 When the “signs” are the same subtract to eliminate. 1 2 When the “signs” are different add to eliminate. Now get: 20 p 12q 0 20 p 25q 13 Note: 3 = 1 (x4) 4 = 2 (x5) 4 - 3 e.g. 2x + 3y = 4 6x - 3y = 4 Add the equations Next Comment Sim Eqs Menu Back to Home Simultaneous Equations : Question 3 If two coffees & three doughnuts cost £2.90 while three coffees & one doughnut cost £2.60 then find the cost of two coffees & five doughnuts. Get hint Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 3 If two coffees & three doughnuts cost £2.90 while three coffees & one doughnut cost £2.60 then find the cost of two coffees & five doughnuts. What would you like to do now? Reveal answer only Go to full solution Go to Comments Go to Sim Eq Menu EXIT Form two Eliminate equations, either c’s orcosts keeping Substitute d’s by Remember to to inmaking pence found value coefficient answer the of avoid decimals. into either same. question!!! original equations. Simultaneous Equations : Question 3 If two coffees & three doughnuts cost £2.90 while three coffees & one doughnut cost £2.60 then find the cost of two coffees & five doughnuts. Two coffees & five doughnuts = £3.90 Try another like this Go to full solution Go to Comments Go to Sim Eq Menu EXIT Question 3 If two coffees & 1. Form two equations, keeping costs in pence to avoid decimals. three doughnuts cost £2.90 Let coffees cost c pence while three coffees & doughnuts d pence then we have & one doughnut cost £2.60 then find the cost of two coffees & five doughnuts. Begin Solution Try another like this Comments Sim Eq Menu Back to Home 2c + 3d = 290 3c + 1d = 260 (x1) (x3) 1 2 2. Eliminate either c’s or d’s by making coefficient same. 2c 3d 290 9c 3d 780 7c 490 c 70 3 = 1 (x1) 4 = 2 (x3) 4 - 3 Question 3 If two coffees & 3. Substitute found value into either of original equations. three doughnuts cost £2.90 while three coffees & one doughnut cost £2.60 then find the cost of two coffees & five doughnuts. Let coffees cost c pence & doughnuts d pence then we have 2c + 3d = 290 3c + 1d = 260 (x1) (x3) Substitute 70 for c in equation What would you like to do now? 1 2 2 210 + d = 260 d = 50 Comments Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = £1.40 + £2.50 Sim Eq Menu = £3.90 Begin Solution Try another like this Back to Home Comments 1. Form two equations, keeping costs in pence to avoid decimals. Let coffees cost c pence & doughnuts d pence then we have 2c + 3d = 290 3c + 1d = 260 (x1) (x3) 1 2 i.e. 2. Eliminate either c’s or d’s by making coefficient same. 2c 3d 290 9c 3d 780 7c 490 c 70 Step 1 Form the two simultaneous equations first by introducing a letter to represent the cost of a coffee ( c) and a different letter to represent the cost of a doughnut (d). 3 = 1 (x1) 4 = 2 (x3) 4 - 3 2c + 3d 1d + 2c = = 290 260 Note change to pence eases working. Next Comment Sim Eqs Menu Back to Home Comments 1. Form two equations, keeping costs in pence to avoid decimals. Let coffees cost c pence & doughnuts d pence then we have 2c + 3d = 290 3c + 1d = 260 (x1) (x3) Step 2 Solve by elimination. Choose whichever variable it is easier to make have the same coefficient in both equations. 1 2 2. Eliminate either c’s or d’s by making coefficient same. 2c 3d 290 9c 3d 780 7c 490 c 70 3 = 1 (x1) 4 = 2 (x3) 4 - 3 Next Comment Sim Eqs Menu Back to Home Comments 3. Substitute found value into either of original equations. Let coffees cost c pence & doughnuts d pence then we have 2c + 3d = 290 3c + 1d = 260 Step 3 Once you have a value for one variable you can substitute this value into any of the equations to find the value of the other variable. It is usually best to choose an equation that you were given in question. Substitute 70 for c in equation 210 + d = 260 d = 50 Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = £1.40 + £2.50 Step 4 Remember to answer the question!!! Next Comment Sim Eqs Menu = £3.90 Back to Home Simultaneous Equations : Question 3B A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? Get hint EXIT Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 3B A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwiEliminate syrup Form two either c’s or and costs 71p per litre to produce. Substitute equations, d’s by making Remember to found value eliminating coefficient answer the of into either decimals A third blend is made using 75% banana syrupquestion!!! and 25% kiwi. same. original wherever How does its cost compare to the other two blends? equations. possible. What would you like to do now? EXIT Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 3B A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? So this blend is more expensive than the other two. EXIT Go to full solution Go to Comments Go to Sim Eq Menu Question 3B Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 1. Form two equations, keeping costs in pence to avoid decimals. Let one litre of banana syrup cost B pence & one litre of kiwi syrup cost K pence. 0.70B + 0.30K = 74 (x10) 0.55B + 0.45K = 71 (x100) 1 2 2. Get rid of decimals 7 B 3K 740 55B 45K 7100 3 = 1 (x10) 4 = 2 (x100) Question 3B Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 3. Eliminate either B’s or K’s by making coefficient same. 7 B 3K 740 55B 45K 7100 105B 45K 11100 55B 45K 7100 50B 4000 B 80 (x15) 3 4 5 = (x15) 3 4 5 - 4 Question 3B Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. 4. Substitute found value into an equation without decimals. 7 B 3K 740 55B 45K 7100 3 (x15) 4 Substitute 80 for B in equation 3 560 + 3K = 740 A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? What would you like to do now? Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 3K = 180 K = 60 B 80 5. Use these values to answer question. 75%B+25%K = (0.75 x 80p)+(0.25 x 60p) = 60p + 15p = 75p So this blend is more expensive than the other two. Markers Comments 1. Form two equations, keeping costs in pence to avoid decimals. Let one litre of banana syrup cost B pence & one litre of kiwi syrup cost K pence. 0.70B + 0.30K = 74 (x10) 0.55B + 0.45K = 71 (x100) 1 2 2. Get rid of decimals 7 B 3K 740 55B 45K 7100 3 = 1 4 = 2 Step 1 Form the two simultaneous equations first by introducing a letter to represent the cost per litre of banana syrup ( B) and a different letter to represent the cost per litre of kiwis fruit (K). i.e. 0.70 B + 0.30K = 0.55B + 0.45K = 74 71 (x10) Multiply all terms Note change to (x100) by 100 to remove pence eases decimals. working. Next Comment Sim Eqs Menu Back to Home Markers Comments Step 2 Solve by elimination. 3. Eliminate either B’s or K’s by making coefficient same. 7 B 3K 740 55B 45K 7100 105B 45K 11100 55B 45K 7100 50B 4000 Choose whichever variable it is easier to make have the same coefficient in both equations. (x15) 3 4 5 = (x15) 3 4 5 - 4 B 80 Next Comment Sim Eqs Menu Back to Home Markers Comments Step 3 Once you have a value for one variable you can substitute this value into any of the equations to find the value of the other variable. 4. Substitute found value into an equation without decimals. 7 B 3K 740 55B 45K 7100 3 (x15) 4 Substitute 80 for B in equation 3 560 + 3K = 740 3K = 180 K = 60 B 80 5. Use these values to answer question. It is usually best to choose an equation that you were given in question. Step 4 Remember to answer the question!!! 75%B+25%K = (0.75 x 80p)+(0.25 x 60p) = 60p + 15p = 75p So this blend is more expensive than the other two. Next Comment Sim Eqs Menu Back to Home Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two terms add first two to obtain 3rd add 2nd & 3rd to obtain 4th add 3rd & 4th to obtain 5th etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a) If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. Get hint EXIT Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two terms Write down Solveexpressions and two To find Establish values ofusing P &found Qprevious : rd substitute add firstMatch two to obtain 3 equations. term fromEliminate (a) with value into either two to form values given in question. P’s or Q’s by rd to obtain add 2nd & 3either 4th next. of original making coefficient equations. rd th same.5th add 3 & 4 to obtain etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a) If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. What would you like to do now? EXIT Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two terms add first two to obtain 3rd add 2nd & 3rd to obtain 4th add 3rd & 4th to obtain 5th etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a) If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. P + Q, P + 2Q 2P + 3Q 3P + 5Q P=7 Q 2 (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. Try another like this EXIT Go to full solution Go to Comments Go to Sim Eq Menu Question 4 (a) If the first two terms of such a sequence are P and Q then find 1. Write down expressions using previous two to form next. (a) First term = P & second term = Q expressions for the next 4 terms 3rd term = P + Q in their simplest form. 4th term = Q + (P + Q) = P + 2Q 5th term = (P + Q) + (P + 2Q) = 2P + 3Q 6th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q Begin Solution Try another like this Comments Sim Eq Menu Back to Home Question 4 (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. 1. Match term from (a) with values given in question. 2P 3Q 8 3P 5Q 11 Q 2 Try another like this Comments Sim Eq Menu Back to Home 2 (x3) (x2) 2. Eliminate either P’s or Q’s by making coefficient same. 6 P 9Q 24 6 P 10Q 22 Begin Solution 1 3 = 1 (x3) 4 = 2 (x2) 4 - 3 Question 4 (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. 3. Substitute found value into either of original equations. 2P 3Q 8 3P 5Q 11 1 2 (x3) (x2) Substitute -2 for Q in equation 2 2P + (-6) = 8 2P = 14 P=7 Begin Solution First two terms are 7 and –2 respectively. Try another like this Comments Sim Eq Menu Back to Home What would you like to do now? Comments 1. Write down expressions using previous two to form next. (a) First term = P & second term = Q 3rd term = P + Q For problems in context it is often useful to do a simple numerical example before attempting the algebraic problem. Fibonacci Sequence: 4th term = Q + (P + Q) = P + 2Q 3, 7, 10, 17, 27, 44, …… 5th term = (P + Q) + (P + 2Q) P = 2P + 3Q 6th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q Q P + Q P + 2Q 4, 6, 10, 16, 26, 42, …… Then introduce the variables: P, Q, P + Q, P + 2Q, 2P + 3Q, … Next Comment Sim Eqs Menu Back to Home Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 1 7 8 5 -6 -1 3 -5 -4 -1 The two number pyramids below have the middle two rows missing. Find the values of v and w. (B) -18 11 (A) 3v 2w w – 2v v + w v + w v – 3w 6w v-w Get hint EXIT Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the Form 2 underneath equations ….. sum of the values in the two blocks 8 5 andSubstitute eliminate found either Work your way to 1 v’svalue or w’sinto by making either of ansame. expression for 7 -6 coefficient original equations. the top row by -1 -5 filling in the 3 -4 -1 middle rows. The two number pyramids below have the middle two rows missing. Find the values of v and w. -18 (A) 3v EXIT 2w w – 2v v + w (B) 11 v + w v – 3w 6w v-w Reveal answer Go to Comments Go to full solution Go to Sim Eq Menu Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 1 7 8 5 -6 -1 3 -5 -4 -1 The two number pyramids below have the middle two rows missing. Find the values of v and w. -18 (A) 3v W 1 EXIT 2w w – 2v v + w V=4 Go to full solution (B) 11 v + w v – 3w 6w v-w Go to Comments Go to Sim Eq Menu Question 4B 1. Work your way to an expression for the top row by filling in the middle rows. -18 (A) 3v w – 2v v + w 2w Pyramid (A) 2nd row 3v + 2w, -2v + 3w, -v + 2w (B) 3rd row v + 5w, -3v + 5w 11 Top row v + w v – 3w 6w -2v + 10w = - 18 v-w Pyramid (B) 2nd row 2v - 2w, v + 3w, v + 5w Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 3rd row 3v + w, 2v + 8w Top row 5v + 9w = 11 2. Form 2 equations and eliminate either v’s or w’s by making coefficient same. Question 4B -18 (A) 3v w – 2v v + w 2w (B) 6w v-w 10V 50W 90 10V 18W 22 68W 68 Begin Solution Continue Solution Comments Sim Eq Menu Back to Home 1 (x5) (x2) 2 Now get: 11 v + w v – 3w 2V 10W 18 5V 9W 11 3 = 1 (x5) 4 = 2 (x2) 4 + 3 W 1 What would you like to do now? 3. Substitute found value into either of original equations. Question 4B -18 (A) 3v w – 2v v + w 2w 2V 10W 18 5V 9W 11 1 2 (x5) (x2) Substitute -1 for W in equation (B) 11 5V + (-9) = 11 5V = 20 v + w v – 3w 6w Begin Solution Continue Solution Comments Sim Eq Menu Back to Home v-w V=4 2 Comments 1. Work your way to an expression for the top row by filling in the middle rows. Use diagrams given to organise working: -18 (A) Pyramid (A) 2nd 5w + v row 3v + 2w, -2v + 3w, -v + 2w 3v + 2w 3v 5w - 3v 3w - 2v 2w 2w - v w – 2v v+w 3rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 (B) 11 3v + w Pyramid (B) 2nd row 2v - 2w, v + 3w, v + 5w 3rd row 3v + w, 2v + 8w Top row 5v + 9w = 11 8w + 2v 2v - 2w 3w + v 5w + v v+w v – 3w 6w v-w Next Comment Sim Eqs Menu Back to Home Markers Comments 1. Work your way to an expression for the top row by filling in the middle rows. Hence equations: 10w - 2v = -18 9w + 5v = 11 Pyramid (A) 2nd row 3v + 2w, -2v + 3w, -v + 2w 3rd row v + 5w, -3v + 5w Top row -2v + 10w Solve by the method of elimination. = - 18 End of simultaneous Equations Pyramid (B) 2nd row 2v - 2w, v + 3w, v + 5w 3rd row 3v + w, 2v + 8w Top row 5v + 9w = 11 Next Comment Sim Eqs Menu Back to Home