INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables TrigonometryStatistics Simultaneous Equations EXIT.

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables TrigonometryStatistics Simultaneous Equations EXIT

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Trigonometry You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu 56

TRIGONOMETRY : Question 1 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Find the area of the following triangle to the nearest cm 2. A B C 40cm 50cm 25° Get hint

TRIGONOMETRY : Question 1 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Find the area of the following triangle to the nearest cm 2. A B C 40cm 50cm 25° What would you like to do now? 1. For area of a triangle questions with an angle present use: 2. Always remember to round your answer if the questions asks you to.

TRIGONOMETRY : Question 1 Go to full solution Go to Comments Go to Trigonometry Menu EXIT Find the area of the following triangle to the nearest cm 2. A B C 40cm 50cm 25° Try another like this What would you like to do now? Area of  = 423cm 2

Question 1 Trigonometry Menu Back to Home Find the area of the following triangle to the nearest cm 2. A B C 40cm 50cm 25° 1. For area of triangles questions where an angle is present use: Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = 422.61… = 423 to nearest unit Area of  = 423cm 2 2. Remember to round if asked to. Comments Continue Solution Try another like this

Markers Comments Trigonometry Menu Back to Home Next Comment 1. For area of triangles questions where an angle is present use: Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = 422.61… = 423 to nearest unit Area of  = 423cm 2 2. Remember to round if asked to. Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1212 A B C 40cm 50cm 25° b c

TRIGONOMETRY : Question 1B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Find the area of the following triangle. Get hint K L M 6.5m 8m 150°

TRIGONOMETRY : Question 1B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Find the area of the following triangle. K L M 6.5m 8m 150° 1. For area of a triangle questions with an angle present use: 2. Always remember to round your answer if the questions asks you to. What would you like to do now?

TRIGONOMETRY : Question 1B Go to full solution Go to Comments Go to Trigonometry Menu EXIT Find the area of the following triangle. What would you like to do now? K L M 6.5m 8m 150° Area of  = 13m 2

Comments Begin Solution Continue Solution Question 1B Trigonometry Menu Back to Home Find the area of the following triangle. 1. For area of triangles questions where an angle is present use: K L M 6.5m 8m 150 ° Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 Area of  = 13m 2

Markers Comments Trigonometry Menu Back to Home Next Comment 1. For area of triangles questions where an angle is present use: Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 Area of  = 13m 2 K L M 6.5m 8m 150 ° k m Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used.

TRIGONOMETRY : Question 2 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Get hint 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 2 1 / 2 hours? Answer to the nearest 10km. 340° N

TRIGONOMETRY : Question 2 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT 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 2 1 / 2 hours? Answer to the nearest 10km. 340° N What would you like to do now? 1. Identify what you need to find and the information you have to help you. 3. Make a sketch to clarify matters. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 5. Substitute known values, remembering to use brackets as appropriate. 2. Calculate as many of the missing angles as possible.

TRIGONOMETRY : Question 2 Go to full solution Go to Comments Go to Trigonometry Menu EXIT 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 2 1 / 2 hours? Answer to the nearest 10km. 340° N Try another like this What would you like to do now? Distance is 740km

Question 2 Trigonometry Menu Back to Home 2. Need distances travelled and angle between flight paths. 3. Sketch triangle. N 340° 200km/hr 160km/hr. How far apart will they be after 21/2 hours? d 1 = speed 1 x time = 160 x 2.5 = 400km d 2 = speed 2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise Full angle = 20° + 90° = 110° 400km 500km 110° A b c a 1. Identify what needs to be found. 20° 90° Comments Continue Solution Try another like this

Question 2 Trigonometry Menu Back to Home 5. Substitute known values and remember to use brackets. N 340° 200km/hr 160km/hr. How far apart will they be after 21/2 hours? 4. Apply Cosine rule. 20° 90° a 2 = b 2 + c 2 – (2bccosA°) = 400 2 + 500 2 – (2 x 400 x 500 x cos110°) = 546808.05.. a =  546808.05.. = 739.46.... = 740 Distance is 740km 6. Remember to round answer if asked to. Comments Continue Solution Try another like this

Markers Comments Trigonometry Menu Back to Home Next Comment 2. Need distances travelled and angle between flight paths. 3. Sketch triangle. d 1 = speed 1 x time = 160 x 2.5 = 400km d 2 = speed 2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise Full angle = 20° + 90° = 110° 400km 500km 110° 1. Identify what needs to be found. Note: Bearings are measured clockwise from N.

Markers Comments Trigonometry Menu Back to Home Next Comment 5. Substitute known values and remember to use brackets. 4. Apply Cosine rule. a 2 = b 2 + c 2 – (2bc cosA°) = 400 2 + 500 2 – (2 x 400 x 500 x cos110°) = 546808.05.. a =  546808.05.. = 739.46.... = 740 Distance is 740km 6. Remember to round answer if asked to. Check formulae list for the cosine rule: a 2 = b 2 + c 2 – 2bc cosA (2 sides and the included angle) Relate to variables used a 2 = b 2 + c 2 – 2bc cosA

TRIGONOMETRY : Question 2B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Get hint 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. How far apart will they be? 195° N SE

TRIGONOMETRY : Question 2B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT 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. How far apart will they be? 195° N SE What would you like to do now? 1. Identify what you need to find and the information you have to help you. 3. Make a sketch to clarify matters. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 5. Substitute known values, remembering to use brackets as appropriate. 2. Calculate as many of the missing angles as possible.

TRIGONOMETRY : Question 2B Go to full solution Go to Comments Go to Trigonometry Menu EXIT 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. How far apart will they be? 195° N SE What would you like to do now? Distance is 41.2 miles

Comments Begin Solution Continue Solution Question 2B Trigonometry Menu Back to Home 2. Need distances travelled and angle between paths. 3. Sketch triangle. How far apart will they be? 1. Identify what needs to be found. 195° N SE 2hr@18mph 3hr@15mph d 1 = speed 1 x time = 18 x 2 = 36miles d 2 = speed 2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 60 0 135 0 60° 36miles 45miles A b c a

Comments Begin Solution Continue Solution Question 2B Trigonometry Menu Back to Home How far apart will they be? 195° N SE 2hr@18mph 3hr@15mph 60 0 135 0 5. Substitute known values and remember to use brackets. 4. Apply Cosine rule. a 2 = b 2 + c 2 – (2bc cosA°) = 36 2 + 45 2 – (2 x 36 x 45 x cos60°) = 1701 a =  1701 = 41.243.... = 41.2 Distance is 41.2 miles

Comments Trigonometry Menu Back to Home Next Comment 2. Need distances travelled and angle between flight paths. 3. Sketch triangle. 1. Identify what needs to be found. d 1 = speed 1 x time = 18 x 2 = 36miles d 2 = speed 2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 60° 36miles 45miles A b c a Note: Bearings are measured clockwise from N.

Comments Trigonometry Menu Back to Home Next Comment 5. Substitute known values and remember to use brackets. 4. Apply Cosine rule. a 2 = b 2 + c 2 – (2bc cosA°) = 36 2 + 45 2 – (2 x 36 x 45 x cos60°) = 1701 a =  1701 = 41.243.... = 41.2 Check formulae list for the cosine rule: a 2 = b 2 + c 2 – 2bc cosA (2 sides and the included angle) Relate to variables used a 2 = b 2 + c 2 – 2bc cosA Distance is 41.2 miles

TRIGONOMETRY : Question 3 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Get hint 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 Q R S 10cm 15cm 22cm

P Q R S 10cm 15cm 22cm TRIGONOMETRY : Question 3 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT What would you like to do now? 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. 3. For area of a triangle with an angle present use: 4. Always remember to round your answer if the questions asks you to. 1. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 2. Substitute known values, remembering to use brackets as appropriate.

TRIGONOMETRY : Question 3 Go to full solution Go to Comments Go to Trigonometry Menu EXIT Try another like this What would you like to do now? 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 Q R S 10cm 15cm 22cm angleQPR = 35.3° = 127cm 2

Question 3 Trigonometry Menu Back to Home 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. Comments Continue Solution Try another like this P Q R S 10cm 15cm 22cm (a) Find the size of angle QPR cosP = 2rq r 2 + q 2 - p 2 = 10 2 + 22 2 - 15 2 2 x 10 x 22 = (10 2 + 22 2 - 15 2 )  (2 x 10 x 22) = 0.8159… angleQPR = cos -1 (0.8159..) = 35.3°

Question 3 Trigonometry Menu Back to Home 2. Remember to double this and round answer. Comments Continue Solution Try another like this P Q R S 10cm 15cm 22cm (b) Hence find the area of the kite to the nearest square unit 1. Kite = 2 identical triangles. For area of triangles where an angle is present use: (b) Area  QPR = ½ qrsinP° = 10 x 22 x sin35.3°  2 = 63.56..cm 2 Area of kite = 2 x 63.56.. = 127.12..cm 2 = 127cm 2

Markers Comments Trigonometry Menu Back to Home Next Comment 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. cosP = 2rq r 2 + q 2 - p 2 = 10 2 + 22 2 - 15 2 2 x 10 x 22 = (10 2 + 22 2 - 15 2 )  (2 x 10 x 22) = 0.8159… angleQPR = cos -1 (0.8159..) = 35.3° Check the formulae list for the second form of the cosine rule: cosA = b 2 + c 2 – a 2 2bc ( 3 sides)

Markers Comments Trigonometry Menu Back to Home Next Comment 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. cosP = 2rq r 2 + q 2 - p 2 = 10 2 + 22 2 - 15 2 2 x 10 x 22 = (10 2 + 22 2 - 15 2 )  (2 x 10 x 22) = 0.8159… angleQPR = cos -1 (0.8159..) = 35.3° Relate to variables used: cosP = = q 2 + r 2 – p 2 2qr 22 2 + 10 2 – 15 2 2x22x10 P Q R 10cm 15cm 22cm p q r

Markers Comments Trigonometry Menu Back to Home Next Comment 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. cosP = 2rq r 2 + q 2 - p 2 = 10 2 + 22 2 - 15 2 2 x 10 x 22 = (10 2 + 22 2 - 15 2 )  (2 x 10 x 22) = 0.8159… angleQPR = cos -1 (0.8159..) = 35.3° Note: When keying in to calculator work out the top line and the bottom line before dividing or use brackets.

TRIGONOMETRY : Question 3B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Get hint 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. E F G H 15cm 25cm

TRIGONOMETRY : Question 3B Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT What would you like to do now? 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. E F G H 15cm 25cm 1. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 2. Substitute known values, remembering to use brackets as appropriate. 4. Always remember to round your answer if the questions asks you to. 3. For area of a triangle with an angle present use:

TRIGONOMETRY : Question 3B Go to full solution Go to Comments Go to Trigonometry Menu EXIT What would you like to do now? 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. E F G H 15cm 25cm = 415cm 2 angleEFH = 33.6°

Question 3B Trigonometry Menu Back to Home 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. Comments Continue Solution (a) Find the size of angle EFH E F G H 15 25 (a) cosF = 2eh e 2 + h 2 - f 2 = 15 2 + 25 2 - 15 2 2 x 25 x 15 = (15 2 + 25 2 - 15 2 )  (2 x 25 x 15) = 0.8333… angleEFH = cos -1 (0.8333..) = 33.6°

Question 3B Trigonometry Menu Back to Home 2. Remember to double this and round answer. Comments Continue Solution 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: E F G H 15 25 (b) Hence find the area to the nearest square unit (b) Area  EFH = ½ ehsinF° = 25 x 15 x sin33.6°  2 = 207.52....cm 2 Area of kite = 2 x 207.52...... = 415.04..cm 2 = 415cm 2 33.6 0

Markers Comments Trigonometry Menu Back to Home Next Comment 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. (a) cosF = 2eh e 2 + h 2 - f 2 = 15 2 + 25 2 - 15 2 2 x 25 x 15 = (15 2 + 25 2 - 15 2 )  (2 x 25 x 15) = 0.8333… angleEFH = cos -1 (0.8333..) = 33.6° Check the formulae list for the second form of the cosine rule: cosA = b 2 + c 2 – a 2 2bc ( 3 sides)

Markers Comments Trigonometry Menu Back to Home Next Comment 1. To find angle when you have 3 sides use 2 nd version of cosine rule: 2. Remember to use inverse function to find angle. (a) cosF = 2eh e 2 + h 2 - f 2 = 15 2 + 25 2 - 15 2 2 x 25 x 15 = (15 2 + 25 2 - 15 2 )  (2 x 25 x 15) = 0.8333… angleEFH = cos -1 (0.8333..) = 33.6° Relate to variables used: cosF = = e 2 + h 2 – f 2 2eh 22 2 + 10 2 – 15 2 2x22x10 E F H 15cm 25cm e f h

Markers Comments Trigonometry Menu Back to Home Next Comment 2. Remember to double this and round answer. 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....cm 2 Area of kite = 2 x 207.52...... = 415.04..cm 2 = 415cm 2 Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. E F H 15cm 25cm e f h 1212

TRIGONOMETRY : Question 4 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT Get hint In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. T U V 105° 35° 5.9cm 10cm

T U V 105° 35° 5.9cm 10cm TRIGONOMETRY : Question 4 Go to full solution Go to Comments Go to Trigonometry Menu Reveal answer only EXIT What would you like to do now? 1. For perimeter need all three sides. So must find TU. 5. Always remember to round your answer if the questions asks you to. 3. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 4. Substitute known values, remembering to use brackets as appropriate. In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. 2. Calculate unknown angles.

TRIGONOMETRY : Question 4 Go to full solution Go to Comments Go to Trigonometry Menu EXIT What would you like to do now? In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm and UV = 10cm. Find the perimeter to one decimal place. T U V 105° 35° 5.9cm 10cm = 22.6cm

Question 4 Trigonometry Menu Back to Home 1.Perimeter requires all three sides. So we need to find TU. 2. Whether you use Sine or Cosine rule need angle V. Comments Continue Solution Angle V = 180° - 35° - 105° = 40° T U V 105° 35 ° 5.9c m 10cm Find the perimeter to one decimal place. 3. If we use Sine rule: v t = sinV°sinT° 4. Substitute known values: v 10 = sin40°sin105° 40 0

Question 4 Trigonometry Menu Back to Home Comments Continue Solution T U V 105° 35 ° 5.9c m 10cm Find the perimeter to one decimal place. 4. Substitute known values: v 10 = sin40° sin105° v x sin105° v = 10 x sin40°  sin105° = 6.654… = 6.7cm 5. Cross multiply: Perim of  = (6.7 + 10 + 5.9)cm = 22.6cm 6. Answer the question: = 10 x sin40° 40 0

Markers Comments Trigonometry Menu Back to Home Next Comment 1.Perimeter requires all three sides. So we need to find TU. 2. Whether you use Sine or Cosine rule need angle V. Angle V = 180° - 35° - 105° = 40° 3. If we use Sine rule: v t = sinV°sinT° 4. Substitute known values: v 10 = sin40°sin105° Since we can pair off two angles with the opposite sides Sine Rule Refer to the Formulae List : = = a Sine A c Sine C b Sine B T U V 105° 35° 5.9cm 10cm t v

Markers Comments Trigonometry Menu Back to Home Next Comment 4. Substitute known values: v 10 = sin40° sin105° v x sin105° v = 10 x sin40°  sin105° = 6.654… = 6.7cm 5. Cross multiply: Perim of  = (6.7 + 10 + 5.9)cm = 22.6cm 6. Answer the question: Go straight to values : 10 Sine 105˚ v Sine 40˚ =

TRIGONOMETRY : Question 5 Full solution Comments Trig Menu Answer only EXIT Hints 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? N A B C 50° 12km S

TRIGONOMETRY : Question 5 Full solution Comments Trig Menu Answer only EXIT 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? N A B C 50° 12km S What would you like to do now? 1.Calculate missing angles. (NB. North West = 315 0 ) 2. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 3. Substitute known values, remembering to use brackets as appropriate. 4. Closest distance = perpendicular distance. Create a right- angled triangle and use SOH CAH TOA.

TRIGONOMETRY : Question 5 Full solution Comments Trig Menu EXIT 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? N A B C 50° 12km S What would you like to do now? Closest distance is 5.48km

Question 5 Trigonometry Menu Back to Home 1.Calculate missing angles. 2. Draw a sketch Comments Continue Solution 3. If we use Sine rule: N A B C 50°50° 12 km S AngleA = 90° - 50° = 40° 40 0 45 0 NW = 315 0 So angle C = 180° - 40° - 45° = 95° 95 0 40°45° 12km 95° A B C b b c = sinB°sinC° & angleB = 315° - 270° = 45°

Question 5 Trigonometry Menu Back to Home Comments Continue Solution 4. Substitute known values: N A B C 50°50° 12 km S b 12 = sin45°sin95° 40 0 45 0 95 0 5. Cross multiply: b x sin95°= 12 x sin45° b = 12 x sin45°  sin95° = 8.5176.… = 8.52km

Question 5 Trigonometry Menu Back to Home Comments Continue Solution 6. Closest point is perpendicular so sketch right angled triangle: N A B C 50°50° 12 km S 40 0 45 0 95 0 7. Now using SOHCAH TOA: A C S 40° 8.52km a sin40° = a 8.52 a = 8.52 x sin40° = 5.476…. = 5.48 Closest distance is 5.48km = 8.52km

Markers Comments Trigonometry Menu Back to Home Next Comment 1.Calculate missing angles. 2. Draw a sketch 3. If we use Sine rule: AngleA = 90° - 50° = 40°NW = 315 0 So angle C = 180° - 40° - 45° = 95° 40 ° 45 ° 12km 95° A B C b c = sinB°sinC° & angleB = 315° - 270° = 45° Since we can pair off two angles with the opposite sides Sine Rule Refer to the Formulae List : = = a Sine A c Sine C b Sine B

Markers Comments Trigonometry Menu Back to Home Next Comment Note: the shortest distance from a point to a line is the perpendicular distance. C AB 6. Closest point is perpendicular so sketch right angled triangle: 7. Now using SOHCAH TOA: A C S 40° 8.52km a sin40° = a 8.52 a = 8.52 x sin40° = 5.476…. = 5.48 Closest distance is 5.48km

TRIGONOMETRY : Question 6 Full solution Comments Trig Menu Answer only EXIT Hints 2.9°3.8° 400m A B D O 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. How far from the platform is each vessel ?

TRIGONOMETRY : Question 6 Full solution Comments Trig Menu Answer only EXIT 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. How far from the platform is each vessel ? 2.9°3.8° 400m A B D O 1.Calculate missing angles. 2. Identify which trig rule to use to find common side to both triangles: Two sides + two angles = sine rule Three sides + one angle = cosine rule 3. Substitute known values, remembering to use brackets as appropriate. 4. Then use SOH CAH TOA to find length from rig to first ship. Second ship is then 400 m further. What would you like to do now?

TRIGONOMETRY : Question 6 Full solution Comments Trig Menu EXIT 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. How far from the platform is each vessel ? 2.9°3.8° 400m A B D O ShipA is 1685m away & ShipB is 1285m away What would you like to do now? Try another like this

Question 6 Trigonometry Menu Back to Home Comments Continue Solution 1.Calculate missing angles. Angle B = 180° - 3.8° = 176.2° 2.9 ° 3.8 ° 400 m A B D O Angle D = 180° - 2.9° - 176.2° = 0.9° 176.2° 0.9° A B D 400m 2.9° 0.9° 176.2° 2. Draw a sketch 3. BD is common link to both triangles so find it using sine rule. a d = sinA°sinD°

Question 6 Trigonometry Menu Back to Home Comments Continue Solution 2.9 ° 3.8 ° 400 m A B D O 176.2° 0.9° 5. Cross multiply: 4. Substitute known values: a 400 = sin2.9°sin0.9° a x sin0.9°= 400 x sin2.9° a = 400 x sin2.9°  sin0.9° = 1288m to nearest metre

Question 6 Trigonometry Menu Back to Home Comments Continue Solution Try another like this 2.9 ° 3.8 ° 400 m A B D O 176.2° 0.9° 6. Sketch right angled triangle: 7. Now using SOHCAH TOA: B D O 3.8° 1288m d 7. Now using SOHCAH TOA: d = 1288 x cos3.8° = 1285m to nearest metre cos3.8° = d 1288 OA is 1285m+400m = 1685m ShipA is 1685m away & ShipB is 1285m away 1285 m

Markers Comments Trigonometry Menu Back to Home Next Comment 1.Calculate missing angles. Angle B = 180° - 3.8° = 176.2° Angle D = 180° - 2.9° - 176.2° = 0.9° A B D 400m 2.9° 0.9° 176.2° 2. Draw a sketch 3. BD is common link to both triangles so find it using sine rule. a d = sinA°sinD° Since we can pair off two angles with the opposite sides Sine Rule Refer to the Formulae List : = = a Sine A c Sine C b Sine B

Markers Comments Trigonometry Menu Back to Home Next Comment 6. Sketch right angled triangle: B D O 3.8° 1288m d 7. Now using SOHCAH TOA: d = 1288 x cos3.8° = 1285m to nearest metre cos3.8° = d 1288 OA is 1285m+400m = 1685m ShipA is 1685m away & ShipB is 1285m away Since angle BOD is right - angled use SOHCAHTOA in triangle BOD

TRIGONOMETRY : Question 6B Full solution Comments Trig Menu Answer only EXIT Hints 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. How high is the drill tower? 27° 35° 80m A B D O

27° 35° 80m A B D O TRIGONOMETRY : Question 6B Full solution Comments Trig Menu Answer only EXIT 1.Calculate missing angles. 2. Identify which trig rule to use to find common side to both triangles: Two sides + two angles = sine rule Three sides + one angle = cosine rule 3. Substitute known values, remembering to use brackets as appropriate. 4. Then use SOH CAH TOA to find Height of rig from deck level. Remember decks are 5m above sea level. What would you like to do now? 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. How high is the drill tower?

TRIGONOMETRY : Question 6B Full solution Comments Trig Menu EXIT What would you like to do now? 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. How high is the drill tower? 27° 35° 80m A B D O Height of tower is 123m

Question 6B Trigonometry Menu Back to Home Comments Continue Solution 1.Calculate missing angles. Angle B = 180° - 35° = 145° Angle D = 180° - 145° - 27° = 8° 145° 8° 2. Draw a sketch 3. BD is common link to both triangles so find it using sine rule. a d = sinA°sinD° 27 ° 35 ° 80m A B D O A B D 27° 8°8° 145°

Question 6B Trigonometry Menu Back to Home Comments Continue Solution 5. Cross multiply: 4. Substitute known values: a 80 = sin27°sin8° a x sin8°= 80 x sin27° a = 80 x sin27°  sin8° = 261m to nearest metre 145° 8° 27 ° 35 ° 80m A B D O

Question 6B Trigonometry Menu Back to Home Comments Continue Solution 6. Sketch right angled triangle: 7. Now using SOHCAH TOA: b = 261 x sin27° = 118m to nearest metre sin27° = b 261 Height = OD + deck height = 118 + 5 = 123m 145° 8° 27 ° 35 ° 80m A B D O B D O 27° 261m b Height of tower is 123m

Comments Trigonometry Menu Back to Home Next Comment 1.Calculate missing angles. Angle B = 180° - 35° = 145° Angle D = 180° - 145° - 27° = 8° 2. Draw a sketch 3. BD is common link to both triangles so find it using sine rule. a d = sinA°sinD° A B D 80m 27° 8°8° 145° Since we can pair off two angles with the opposite sides Sine Rule Refer to the Formulae List : = = a Sine A c Sine C b Sine B

Markers Comments Trigonometry Menu Back to Home Next Comment 6. Sketch right angled triangle: 7. Now using SOHCAH TOA: b = 261 x sin27° = 118m to nearest metre sin27° = b 261 Height = OD + deck height = 118 + 5 = 123m B D O 27° 261m b Height of tower is 123m Since angle BOD is right - angled use SOHCAHTOA in triangle BOD

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Statistics You have chosen to study: Please choose a question to attempt from the following: 1 EXIT Back to Unit 2 Menu 2345

STATISTICS : Question 1 Go to full solution Go to Comments Go to Statistics MenuReveal answer only EXIT A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 351077159 (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

STATISTICS : Question 1 Go to full solution Go to Comments Go to Statistics Menu Reveal answer only EXIT A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 351077159 (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? Draw a table comparing data to mean. Then square values. When comparing data sets always make comment on the average (which is bigger etc.) and the spread of the data. What would you like to do now?

(b)The second company has a longer average response time. The smaller standard deviation means their arrival time is more predictable. STATISTICS : Question 1 Go to full solution Go to Comments Go to Statistics Menu EXIT A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. 351077159 (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 What would you like to do now?

Comments Begin Solution Continue Solution Question 1 Statistics Menu Back to Home 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. (a) Mean = (3+5+10+7+7+15+9)  7 = 8 1.Calculate mean. So = 8 and no.pieces of data = n = 7

Comments Begin Solution Continue Solution Question 1 Statistics Menu Back to Home 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. 2. Draw table comparing data to mean. x 3 -5 25 5 -3 9 10 2 4 7 - 1 1 15 7 49 9 1 1  (x - ) 2 = 90

Comments Begin Solution Continue Solution Question 1 Statistics Menu Back to Home 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. 3. Use formula to calculate standard deviation. Just found!!

Comments Begin Solution Continue Solution Question 1 Statistics Menu Back to Home 1. Always compare mean and standard deviation. (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? (b)The second company has a longer average response time. The smaller standard deviation means their arrival time is more predictable. What would you like to do now?

Comments Statistics Menu Back to Home Next Comment 3. Use formula to calculate standard deviation. Check the list of Formulae for the Standard Deviation Formula: The second formula can be used in the calculator paper. The calculator must be in Stats. Mode to allow the data to be entered.

STATISTICS : Question 2 EXIT Weeks of training Time(s) 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. Continue question

STATISTICS : Question 2 Go to full solution Go to Comments Go to Stats Menu Reveal answer only EXIT Get hint Weeks of training Time(s) (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

(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 Use your equation and substitute W = 12. STATISTICS : Question 2 Go to full solution Go to Comments Go to Stats Menu Reveal answer only EXIT Weeks of training Time(s) Find gradient and note intercept of T axis. Use your equation and substitute T = 11.5 What would you like to do now?

STATISTICS : Question 2 Go to full solution Go to Comments Go to Stats Menu EXIT Weeks of training Time(s) (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 T = -¼ W + 15 12secs. 14 weeks What would you like to do now?

Comments Begin Solution Continue Solution Question 2 Statistics Menu Back to Home 1.Find gradient and note intercept of T axis. (i) Find the equation of the line in terms of T and W. Go to graph m = (y 2 – y 1 ) (x 2 – x 1 ) (a)10 - 15 20 - 0 = Intercept at 15 Equation isT = -¼ W + 15 = -¼

(ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs Comments Begin Solution Continue Solution Question 2 Statistics Menu Back to Home 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.

(ii) Use answer (i) to predict (a) her time after 12 weeks of training (b) the week when her time was 11.5secs Comments Begin Solution Continue Solution Question 2 Statistics Menu Back to Home 3. Use your equation and substitute T = 11.5 (b)(i) If t = 11.5 then T = -¼ W + 15 becomes -¼ W + 15 = 11.5 (-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?

Comments Statistics Menu Back to Home Next Comment 1.Find gradient and note intercept of T axis. m = (y 2 – y 1 ) (x 2 – x 1 ) (a)10 - 15 20 - 0 = Intercept at 15 Equation isT = -¼ W + 15 = -¼ To find the equation of a line from the graph: Must Learn: y = mx + c gradientintercept So you need to find these!!

Comments Statistics Menu Back to Home Next Comment 1.Find gradient and note intercept of T axis. m = (y 2 – y 1 ) (x 2 – x 1 ) (a)10 - 15 20 - 0 = Intercept at 15 Equation isT = -¼ W + 15 = -¼ m = vertical horizontal Note:Always draw the horizontal before the vertical: horizontal (+ve) vertical (-ve)

(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? STATISTICS : Question 3 Full solution Comments Statistics Menu Reveal EXIT Get hint 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 Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ?

(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? STATISTICS : Question 3 Full solution Comments Statistics Menu Reveal EXIT 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 Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ? Remember all entries in table must add to total sample size To find probabilities use: P = no of favourable / no of data What would you like to do now?

(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? STATISTICS : Question 3 Full solution Comments Statistics Menu EXIT 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 Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ? 10% 1 / 12 7 / 60 What now?

Comments Begin Solution Continue Solution Question 3 Statistics Menu Back to Home 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ? (a)What percentage liked Shrek but not the Simpsons? (a) Like Shrek but not Simpsons = 180 – 126 – 15 – 21 = 18 % = 18 / 180 = 1 / 10 = 10% NB: There are 180 in survey!!

Comments Begin Solution Continue Solution Question 3 Statistics Menu Back to Home 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ? what is the probability that (i)they liked the Simpsons but not Shrek? (b)(i) Prob = 15 / 180 = 1 / 12

Comments Begin Solution Continue Solution Question 3 Statistics Menu Back to Home 1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 2115 ? (ii) they liked neither? (b)(ii) Prob = 21 / 180 = 7 / 60 What would you like to do now?

Comments Statistics Menu Back to Home Next Comment 1. Use P = no of favourable / no of data (a) Like Shrek but not Simpsons = 180 – 126 – 15 – 21 = 18 % = 18 / 180 = 1 / 10 = 10% NB: There are 180 in survey!! Note: To change a fraction to a % multiply by 100% 18 180 = x 100% 18 180

Comments Statistics Menu Back to Home Next Comment 1. Use P = no of favourable / no of data (b)(i) Prob = 15 / 180 = 1 / 12 Probability = Number of favourable outcomes Number of possible outcomes To calculate simple probabilities:

(b)If a combination is selected at random what is the probability that it is …… STATISTICS : Question 4 Full solution Comments Stats Menu Reveal ans EXIT Get hint On a college course you have to pick a language plus a leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING (a) Make a list of all possible combinations of courses. (i) Includes Spanish? (ii) Includes swimming? (iii) Doesn’t include French but includes music?

(b)If a combination is selected at random what is the probability that it is …… STATISTICS : Question 4 Full solution Comments Stats Menu Reveal ans EXIT On a college course you have to pick a language plus a leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING (a) Make a list of all possible combinations of courses. (i) Includes Spanish? (ii) Includes swimming? (iii) Doesn’t include French but includes music? Use a tree diagram & “branch out” with each language. From each language “branch out” with each leisure activity. Now list the pairs of subjects. Use your list of possible combinations to find probabilities. What would you like to do now?

(b)If a combination is selected at random what is the probability that it is …… STATISTICS : Question 4 Full solution Comments Stats Menu EXIT On a college course you have to pick a language plus a leisure activity from the following lists LANGUAGE FRENCH GERMAN SPANISH LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING (a) Make a list of all possible combinations of courses. (i) Includes Spanish? (ii) Includes swimming? (iii) Doesn’t include French but includes music? 1/31/3 1/41/4 1/61/6 CLICK

GERMAN FRENCH SPANISH BASKETBALL VIDEO MUSIC SWIMMING BASKETBALL VIDEO MUSIC SWIMMING BASKETBALL VIDEO MUSIC SWIMMING (a) French/Music French/Video French/Bsktbll French/Swim German/Music 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 Have list of combinations handy. Remember to simplify. French/Music French/Video French/Bsktbll French/Swim German/Music German/Video German/Bsktbll German/Swim Spanish/Music Spanish/Video Spanish/Bsktbll Spanish/Swim = 1 / 3 = 1 / 4 (b)(i) Prob (b)(ii) Prob = 4 / 12 = 3 / 12 (b)(iii) Prob = 2 / 12 = 1 / 6 Comments Stats Menu What now? (i) Includes Spanish? (ii) Includes swimming? (iii)Doesn’t include French but includes music? What is probability:

Comments Statistics Menu Back to Home Next Comment GERMAN FRENCH SPANISH BASKETBALL VIDEO MUSIC SWIMMING BASKETBALL VIDEO MUSIC SWIMMING BASKETBALL VIDEO MUSIC SWIMMING (a) French/Music French/Video French/Bsktbll French/Swim German/Music German/Video German/Bsktbll German/Swim Spanish/Music Spanish/Video Spanish/Bsktbll Spanish/Swim No. of possible outcomes = 3 x 4 = 12 Probability = Number of favourable Number of possible To calculate simple probabilities:

The delivery times for a fast food company are shown in the following cumulative frequency table. Time No. Deliveries up to 4 mins 5 up to 8 mins 13 up to 12 mins 37 up to 16 mins 53 up to 20 mins 60 (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. (a)How many deliveries took longer than 16 mins? STATISTICS : Question 5 Full solution Comments Stats Menu Reveal ans Get hint

The delivery times for a fast food company are shown in the following cumulative frequency table. Time No. Deliveries up to 4 mins 5 up to 8 mins 13 up to 12 mins 37 up to 16 mins 53 up to 20 mins 60 (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. (a)How many deliveries took longer than 16 mins? STATISTICS : Question 5 Full solution Comments Stats Menu Reveal ans For how many greater than 16 find difference between end value of values up to 16. Median is middle value- so we want halfway- point on frequency axis. SIQR = ½(Q3 – Q1) Q1 – 25% point & Q3 – 75% point What would you like to do now?

The delivery times for a fast food company are shown in the following cumulative frequency table. Time No. Deliveries up to 4 mins 5 up to 8 mins 13 up to 12 mins 37 up to 16 mins 53 up to 20 mins 60 (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. (a)How many deliveries took longer than 16 mins? STATISTICS : Question 5 Full solution Comments Stats Menu = 7 Median = 11 minsSIQR = 2.75 mins

Delivery time(mins) Cum freq Q2 11 Q1 8.5 Q3 14 ½ of 60 = 30 ¼ of 60 = 15 ¾ of 60 = 45 (C) Median = 11mins. (c) SIQR = ½(Q3 – Q1) = (14 – 8.5)  2 = 2.75mins (a) Deliveries taking longer than 16 mins = 60 – 53= 7 Comments Stats Menu TimeDeliveries 45 813 1237 1653 2060 What would you like to do now?

Comments Statistics Menu Back to Home Next Comment Cumulative Frequency Delivery Time 60 20 50% 11 Note: In a Cumulative Frequency Diagram: On y-axis (cumulative frequency) Q 1 at 25%, Q 2 at 50%, Q 3 at 75% And read values from the delivery time scale (x-axis). End of Statistics

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables You have chosen to study: Please choose a question to attempt from the following: 12 34 EXIT Back to Unit 2 Menu 56 Stem & Leaf Dot Plot Cum Freq Table Dot to boxplot Stem to boxplot Piechart

(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? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Reveal answer EXIT Get hint

(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? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end What would you like to do now?

(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? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments EXIT Graphs etc Menu median = £183 Q1 = £171 Q3 = £195 = £12 = 2/5 What would you like to do now?

Comments Begin Solution Continue Solution Question 1 Menu Back to Home 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 position = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 What would you like to do now? (NOT 3!!!)

Comments Begin Solution Continue Solution Question 1 Menu Back to Home 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 4. Use SIQR = ½ (Q 3 – Q 1 ) / 2 (iii) SIQR = ½(Q3 – Q1) = (£195 - £171)  2 = £12

Question 1 5. Use P = no of favourable / no of data 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (b)What is the probability that someone chosen at random earns less than £180? No of favourable ( under £180) = 10 No of data = n = 25 (b) Prob(under £180) = 10 / 25 = 2 / 5. Comments Begin Solution Continue Solution Menu Back to Home

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 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 medianmedian is the 13th number in order.

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q1Q1 12 numbers before median. 6 numbers either side of Q 1 is midway between the 6th and 7th number.

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q3Q3 12 numbers after median. 6 numbers either side of Q 3 is midway between the 19th and 20th number.

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? Go to full solution Go to Comments Reveal answer Get hint EXIT

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? Establish lowest & highest values and draw line with scale. Plot a dot for each piece of data and label diagram. For probability use: P = no of favourable / no of data Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT What would you like to do now?

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? Tightly clustered 3/10 EXIT Go to full solution Go to Comments Graphs etc Menu CLICK

Question 2 1. Establish lowest & highest values and draw line with scale. 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 Illustrate this using a dot plot. 26283032 (a) Lowest = 28 & highest = 31. Weights in g 2. Plot a dot for each piece of data and label diagram. Comments Begin Solution Continue Solution Menu Back to Home

Question 2 3. Make sure you know the possible descriptions of data. 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 26283032 Weights in g What type of distribution does this show? (b) Tightly clustered distribution. Comments Begin Solution Continue Solution Menu Back to Home

Question 2 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 26283032 Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. If a bag is chosen at random what is the probability it will be heavier than the modal weight? Mode! Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 3. Make sure you know the possible descriptions of data. 26283032 Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

Comments Probability = Number of favourable outcomes Number of possible outcomes Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. Mode! 2626 2828 3030 3232 To calculate simple probabilities: Menu Back to Home Next Comment

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? Go to full solution Go to Comments Reveal answer Get hint EXIT Graphs etc Menu

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? Establish lowest & highest values and draw table. Complete each row 1 step at a time, calculating running total as you go. For probability use: P = no of favourable / no of data Use median = (n+1) / 2 to establish in which row median lies. EXIT Go to full solution Go to Comments Graphs etc Menu Reveal answer What would you like to do now?

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? CLICK Median = 14 1/3 EXIT Go to full solution Go to Comments Graphs etc Menu

Question 3 1. Establish lowest & highest values and draw a table. (a) Lowest = 10 & highest = 18 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 (a)Construct a cumulative frequency table for this data. Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 2. Complete each row 1 step at a time, calculating running total as you go. Comments Begin Solution Continue Solution Menu Back to Home

Question 3 3. Use median = (n+1) / 2 to establish in which row median lies. 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 (b) What is the median for this data? For 30 values median is between 15th & 16th both of which are in row 14. Median Mark = 14 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Question 3 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 (c) What is the probability that a pupil selected at random scored under 14? No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / 3. 4. Use P = no of favourable / no of data Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments For 30 values median is between 15th & 16th both of which are in row 14. Median = 14 Mark Freq Cum Freq 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 4 23 4 27 3 30 7 10 3 19 Median: 1 – 15 Q 2 16 - 30 Find the mark at which the cumulative frequency first reaches between 15 th and 16 th number. Median = 14 Menu Back to Home Next Comment

Comments Mark Freq Cum Freq 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 4 23 4 27 3 30 7 10 3 19 No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / 3. Probability = Number of favourable outcomes Number of possible outcomes To calculate simple probabilities: Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (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? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (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? Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end remember bigger SIQR means more variation (spread) in data. EXIT Go to full solution Go to CommentsReveal answer Graphs etc Menu What would you like to do now?

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (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? Median = 50 So Q1 = 49 So Q3 = 52 the data is distributed more widely than (or not as clustered as) the above data EXIT Menu Full solutionComments CLICK

Question 4 48 50 52 54 56 58 (a)Find the (i) median (ii) lower quartile (iii) upper quartile (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 Comments Begin Solution Continue Solution Menu Back to Home

Question 4 48 50 52 54 56 58 (b) Construct a boxplot using this data. 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. 48 50 52 54 56 58 (b)Lowest = 48, Q1 = 49, Q2 = 50, Q3 = 52 & Highest = 58. Comments Begin Solution Continue Solution Menu Back to Home

Question 4 48 50 52 54 56 58 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c) In a second sample the semi-interquartile range was 2.5. How does this compare? (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 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 23 numbers in the list: 1 - 1112 13 - 23 Q2Q2 11 numbers on either side of the median The median: Menu Back to Home Next Comment

Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 For quartiles: 1 - 5 6 7 - 11 12 Q2Q2 Q1Q1 Q3Q3 13 - 17 18 19 - 23 12 Q2Q2 Now count through the list until you reach the 6 th, 12 th,and 18 th number in the list. Menu Back to Home Next Comment

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. = (Q 3 - Q 1 ) 1212 Remember: when asked to compare data always consider average and spread. It is a measure of how spread-out and so how “consistent” or “reliable” the data is. Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (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. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 Full solution Comments Reveal answer Get hint EXIT

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (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. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 Use median position = (n+1) / 2 to find median position Q1 is midpoint from start to median Q3 is midpoint from median to end When comparing two data sets comment on spread and average Full solution Comments Reveal answer Menu EXIT What now?

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (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. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 median = 87 Q1 = 77 Q3 = 99 Full solution Comments Menu EXIT CLICK

Question 5 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 26 then the median is between 13 th & 14 th value ie median = 87 (ii) so Q1 = 77 2. There are 13 values before median so Q 1 position is 6 th value 3. There are 13 values after median so Q 3 position is 20 th position so Q3 = 99 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg (a)Find the median, lower & upper quartiles for this data. Comments Begin Solution Continue Solution Menu Back to Home

Question 5 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. 60 70 80 90 100 110 120 (b) Use the data to construct a boxplot. Comments Begin Solution Continue Solution Menu Back to Home

Question 5 5. Compare spread and relevant average. (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. 60 70 80 90 100 110 120 (c)Lightest has put on weight – lowest now 65, heaviest 3 have lost weight – highest now 115, median same but overall spread of weights has decreased as Q3-Q1 was 22 but is now only 15. Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. 60 70 80 90 100 110 120 Box Plot : LowestHighestQ1Q1 Q2Q2 Q3Q3 Remember: To draw a boxplot you need a “five-figure summary”: five-figure summary Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? angle 360° = amount 630 EXIT Go to full solution Go to Comments Reveal answer Graphs etc Menu What would you like to do now?

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? = 252 = 48° EXIT Go to full solution Go to Comments Graphs etc Menu What would you like to do now?

Question 6 clubbin g 144° x°x° theatre cinemaWatching TV How many people went clubbing? 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount 630 144° 360° = amount 630 360 x amount = 144 x 630 amount = 144 x 630  360 = 252 Comments Begin Solution Continue Solution Menu Back to Home

Question 6 clubbin g 144° x°x° theatre cinemaWatching TV 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = 84 630 630 x angle = 360° x 84 angle = 360° x 84  630 = 48° (b) If 84 people went to the theatre then how big is x°? Comments Begin Solution Continue Solution Menu Back to Home

Comments 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount 630 144° 360° = amount 630 360 x amount = 144 x 630 amount = 144 x 630  360 = 252 Can also be tackled by using proportion: Amount = x 630 144 360 Menu Back to Home Next Comment

Comments Can also be tackled by using proportion: 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = 84 630 630 x angle = 360° x 84 angle = 360° x 84  630 = 48° 84 = x 630 x 360 630 x = 84 x 360 x = 84 360 630 x Menu Back to Home Next Comment End of graphs, charts etc.

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu

Simultaneous Equations : Question 1 Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT 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. Get hint

Simultaneous Equations : Question 1 Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT 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. To draw graph: Construct a table of values with at least 2 x- coordinates. Plot and join points. Solution is where lines cross What would you like to do now?

Simultaneous Equations : Question 1 Go to full solution Go to Comments Go to Sim Eq Menu EXIT 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. Solution is x = 3 & y = 4 Try another like this What would you like to do now?

Comments Begin Solution Continue Solution Question 1 Sim Eq Menu Back to Home 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 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2x 0 5 y -2 8

Begin Solution Try another like this Question 1 Back to Home 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 2. Plot and join points. Solution is where lines cross. Solution is x = 3 & y = 4 Comments Sim Eq Menu What would you like to do now?

Markers Comments Sim Eqs Menu Back to Home Next Comment 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2x 0 5 y -2 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 = 2 x 2 - 1 = 3 First Point (0,-1) Second Point (-1,3) Plot and join (0,-1), and (-1,3).

Markers Comments Sim Eqs Menu Back to Home Next Comment 2. Plot and join points. Solution is where lines cross. Solution is x = 3 & y = 4 Method 2 Using y = mx + c form: y = mx + c gradient y - intercept Plot C(0, -1) and draw line with m = 2 y = 2x - 1 gradient m = 2 y - intercept c = -1

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT 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 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 Get hint

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT 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 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 To draw graph: Construct a table of values with at least 2 x- coordinates. Plot and join points. Solution is where lines cross What would you like to do now?

Go to full solution Go to Comments Go to Sim Eq Menu EXIT 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 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 Solution is x = 2 & y = 4 What would you like to do now?

Comments Begin Solution Continue Solution Question 1B Sim Eq Menu Back to Home The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 1. Construct a table of values with at least 2 x coordinates. x 0 6 y 3 6

Comments Begin Solution Continue Solution Question 1B Sim Eq Menu Back to Home The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 x 0 6 y 3 6 2. Plot and join points. Solution is where lines cross. Solution is x = 2 & y = 4 What would you like to do now?

Markers Comments Sim Eqs Menu Back to Home Next Comment 1. Construct a table of values with at least 2 x coordinates. y = 1 / 2 x + 3x 0 6 y 3 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 x 0 + 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).

Markers Comments Sim Eqs Menu Back to Home Next Comment y = 1 / 2 x + 3x 0 6 y 3 6 2. Plot and join points. Solution is where lines cross. Solution is x = 2 & y = 4 Method 2 Using y = mx + c form: y = mx + c gradient y - intercept Plot C(0, 3) and draw line with m = ½ y = ½ x + 3 gradient m = ½ y - intercept c = +3

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Get hint

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Eliminate either variable by making coefficient same. Substitute found value into either of original equations. What would you like to do now?

Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Solution is u = 2 & v = 1 Try another like this What would you like to do now?

Comments Begin Solution Continue Solution Question 2 Sim Eq Menu Back to Home Solve 3u - 2v = 4 2u + 5v = 9 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 +

Comments Begin Solution Question 2 Sim Eq Menu Back to Home Solve 3u - 2v = 4 2u + 5v = 9 3u - 2v = 4 2u + 5v = 9 2. Substitute found value into either of original equations. (x5) (x2) 2 1 Substitute 2 for u in equation 4 + 5v = 9 5v = 5 v = 1 Solution is u = 2 & v = 1 2 Try another like this What would you like to do now?

Markers Comments Sim Eqs Menu Back to Home Next Comment 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 + Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x + 3y = 4 Subtract the equations

Markers Comments Sim Eqs Menu Back to Home Next Comment 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 + Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x - 3y = 4 Add the equations

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Get hint

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Eliminate either variable by making coefficient same. Substitute found value into either of original equations. What would you like to do now?

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 Solution is q = -1 & q = 0.6 What would you like to do now?

Comments Begin Solution Continue Solution Question 2B Sim Eq Menu Back to Home 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Solve 5p + 3q = 0 4p + 5q = -2.6

Comments Begin Solution Continue Solution Question 2B Sim Eq Menu Back to Home 2. Substitute found value into either of original equations. Substitute -1 for q in equation 5p + (- 3) = 0 5p = 3 p =3/5 = 0.6 Solution is q = -1 & p = 0.6 1 Solve 5p + 3q = 0 4p + 5q = -2.6 5p + 3q = 0 4p + 5q = -2.6 (x4) (x5) 2 1 What would you like to do now?

Markers Comments Sim Eqs Menu Back to Home Next Comment 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x + 3y = 4 Subtract the equations

Markers Comments Sim Eqs Menu Back to Home Next Comment 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x - 3y = 4 Add the equations

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only 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. Get hint

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only 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. Form two equations, keeping costs in pence to avoid decimals. Eliminate either c’s or d’s by making coefficient same. Substitute found value into either of original equations. Remember to answer the question!!! What would you like to do now?

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. Try another like this Two coffees & five doughnuts = £3.90

Comments Begin Solution Question 3 Sim Eq Menu Back to Home 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in pence to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - 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. Let coffees cost c pence & doughnuts d pence then we have 2. Eliminate either c’s or d’s by making coefficient same. Try another like this

Comments Begin Solution Question 3 Sim Eq Menu Back to Home 3. Substitute found value into either of original equations. Substitute 70 for c in equation 210 + d = 260 d = 50 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. 2c + 3d = 290 3c + 1d = 260 (x1) (x3) 2 1 Let coffees cost c pence & doughnuts d pence then we have 2 Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = £1.40 + £2.50 = £3.90 Try another like this What would you like to do now?

Comments Sim Eqs Menu Back to Home Next Comment 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). 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in pence to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c pence & doughnuts d pence then we have 2. Eliminate either c’s or d’s by making coefficient same. i.e.2c +3d=290 1d +2c=260 Note change to pence eases working.

Comments Sim Eqs Menu Back to Home Next Comment Step 2 Solve by elimination. Choose whichever variable it is easier to make have the same coefficient in both equations. 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in pence to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c pence & doughnuts d pence then we have 2. Eliminate either c’s or d’s by making coefficient same.

Comments Sim Eqs Menu Back to Home Next Comment 3. Substitute found value into either of original equations. Substitute 70 for c in equation 210 + d = 260 d = 50 2c + 3d = 290 3c + 1d = 260 Let coffees cost c pence & doughnuts d pence then we have Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = £1.40 + £2.50 = £3.90 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. Step 4 Remember to answer the question!!!

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? Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 3B Get hint

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? Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 3B Form two equations, eliminating decimals wherever possible. Eliminate either c’s or d’s by making coefficient same. Substitute found value into either of original equations. Remember to answer the question!!! What would you like to do now?

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? Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 3B So this blend is more expensive than the other two.

Comments Begin Solution Continue Solution Question 3B Sim Eq Menu Back to Home 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in pence to avoid decimals. (x10) (x100) 3 4 2 1 = (x10) (x100) 1 = 2 2. Get rid of decimals 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? Let one litre of banana syrup cost B pence & one litre of kiwi syrup cost K pence.

Comments Begin Solution Continue Solution Question 3B Sim Eq Menu Back to Home 3 4 = (x15) 3 54 - 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? 3. Eliminate either B’s or K’s by making coefficient same. 5 4

Comments Begin Solution Continue Solution Question 3B Sim Eq Menu Back to Home 3 4 (x15) 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? 4. Substitute found value into an equation without decimals. Substitute 80 for B in equation 560 + 3K = 740 3K = 180 3 K = 60 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. What would you like to do now?

Markers Comments Sim Eqs Menu Back to Home Next Comment 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in pence to avoid decimals. (x10) (x100) 3 4 2 1 = (x10) (x100) 1 = 2 2. Get rid of decimals Let one litre of banana syrup cost B pence & one litre of kiwi syrup cost K pence. 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 =74 0.55B+ 0.45K =71 Multiply all terms by 100 to remove decimals. Note change to pence eases working.

Markers Comments Sim Eqs Menu Back to Home Next Comment 3 4 = (x15) 3 54 3. Eliminate either B’s or K’s by making coefficient same. 5 4 - Step 2 Solve by elimination. Choose whichever variable it is easier to make have the same coefficient in both equations.

Markers Comments Sim Eqs Menu Back to Home Next Comment 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. Step 4 Remember to answer the question!!! 3 4 (x15) 4. Substitute found value into an equation without decimals. Substitute 80 for B in equation 560 + 3K = 740 3K = 180 3 K = 60 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.

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th 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 5 th & 6 th 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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th 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 5 th & 6 th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. Write down expressions using previous two to form next. To find values of P & Q : Match term from (a) with values given in question. Establish two equations. Eliminate either P’s or Q’s by making coefficient same. Solve and substitute found value into either of original equations. What would you like to do now?

Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th 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 5 th & 6 th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. P + Q,P + 2Q 2P + 3Q 3P + 5Q P = 7 Try another like this

Comments Begin Solution Question 4 Sim Eq Menu Back to Home 1. Write down expressions using previous two to form next. (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. (a) First term = P & second term = Q 3 rd term = P + Q 4 th term = Q + (P + Q) = P + 2Q 5 th term = (P + Q) + (P + 2Q) = 2P + 3Q 6 th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q Try another like this

Comments Begin Solution Question 4 Sim Eq Menu Back to Home 1 2 = (x3) 1 43 - 1. Match term from (a) with values given in question. 3 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. (x2) 2. Eliminate either P’s or Q’s by making coefficient same. = 2 Try another like this

Comments Begin Solution Question 4 Sim Eq Menu Back to Home 3. Substitute found value into either of original equations. Substitute -2 for Q in equation 2P + (-6) = 8 2P = 14 2 P = 7 (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 2 (x3) (x2) First two terms are 7 and –2 respectively. Try another like this What would you like to do now?

Comments Sim Eqs Menu Back to Home Next Comment 1. Write down expressions using previous two to form next. (a) First term = P & second term = Q 3 rd term = P + Q 4 th term = Q + (P + Q) = P + 2Q 5 th term = (P + Q) + (P + 2Q) = 2P + 3Q 6 th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q For problems in context it is often useful to do a simple numerical example before attempting the algebraic problem. Fibonacci Sequence: 3, 7,10, 17, 27,44, …… P Q P + Q P + 2Q 4, 6,10, 16, 26,42, …… Then introduce the variables: P, Q,P + Q, P + 2Q, 2P + 3Q, …

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -18 11 Get hint

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -1811 Work your way to an expression for the top row by filling in the middle rows. Form 2 equations and eliminate either v’s or w’s by making coefficient same. Substitute found value into either of original equations.

Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -1811 V = 4

Comments Begin Solution Continue Solution Question 4B Sim Eq Menu Back to Home 1. Work your way to an expression for the top row by filling in the middle rows. 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11

Comments Begin Solution Continue Solution Question 4B Sim Eq Menu Back to Home 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 2. Form 2 equations and eliminate either v’s or w’s by making coefficient same. 1 2 (x5) (x2) = (x5) 1 43 + 3 4 = 2 (x2) Now get: What would you like to do now?

Comments Begin Solution Continue Solution Question 4B Sim Eq Menu Back to Home 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 1 2 (x5) (x2) 3. Substitute found value into either of original equations. Substitute -1 for W in equation 5V + (-9) = 11 5V = 20 V = 4 2

Comments Sim Eqs Menu Back to Home Next Comment 1. Work your way to an expression for the top row by filling in the middle rows. Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11 Use diagrams given to organise working: 3v + 2w 3w - 2v 2w - v -18 (A) 3v 2w w – 2v v + w 5w + v 5w - 3v (B) v + w v – 3w 6w v - w 11 2v - 2w 3w + v 5w + v 3v + w 8w + 2v

Markers Comments Sim Eqs Menu Back to Home Next Comment 1. Work your way to an expression for the top row by filling in the middle rows. Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11 Hence equations: 10w - 2v = -18 9w + 5v = 11 Solve by the method of elimination. End of simultaneous Equations

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables TrigonometryStatistics Simultaneous Equations EXIT.

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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables TrigonometryStatistics Simultaneous Equations EXIT.

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Presentation on theme: "INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables TrigonometryStatistics Simultaneous Equations EXIT."— Presentation transcript:

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