Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations

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

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study: UNIT 2 : Trigonometry Please choose a question to attempt from the following: 1 2 3 4 5 6 Back to Unit 2 Menu EXIT

TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2. A B C 40cm 50cm 25° Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

1. For area of a triangle questions with an angle present use:
TRIGONOMETRY : Question 1 Find the area of the following triangle to the nearest cm2. A B C 40cm 50cm 25° 1. For area of a triangle questions with an angle present use: What would you like to do now? 2. Always remember to round your answer if the questions asks you to. Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Find the area of the following triangle to the nearest cm2. A B C 40cm 50cm 25° What would you like to do now? Area of  = 423cm2 Try another like this Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Question 1 Find the area of the following triangle to the nearest cm2.
1. For area of triangles questions where an angle is present use: Find the area of the following triangle to the nearest cm2. Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 A B C 40cm 50cm 25° = … 2. Remember to round if asked to. = to nearest unit Area of  = 423cm2 Continue Solution Try another like this Comments Trigonometry Menu Back to Home

Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = 422.61… A B C
Markers Comments Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. For area of triangles questions where an angle is present use: 1 2 Area of  = ½ bcsinA° = 50 x 40 x sin25°  2 = … A B C 40cm 50cm 25° b c 2. Remember to round if asked to. = to nearest unit Area of  = 423cm2 Next Comment Trigonometry Menu Back to Home

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

1. For area of a triangle questions with an angle present use:
TRIGONOMETRY : Question 1B Find the area of the following triangle. K L M 6.5m 8m 150° 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. Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Find the area of the following triangle. K L M 6.5m 8m 150° What would you like to do now? Area of  = 13m2 Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Question 1B Find the area of the following triangle.
1. For area of triangles questions where an angle is present use: Find the area of the following triangle. Area of  = ½ kmsinL° K L M 6.5m 8m 150° = 8 x 6.5 x sin150°  2 = 13 Area of  = 13m2 Begin Solution Continue Solution Comments Trigonometry Menu Back to Home

Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 K Area of  = 13m2
Markers Comments Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. For area of triangles questions where an angle is present use: Area of  = ½ kmsinL° = 8 x 6.5 x sin150°  2 = 13 K L M 6.5m 8m 150° k m Area of  = 13m2 Next Comment Trigonometry Menu Back to Home

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 to the nearest 10km. N Get hint Reveal answer only 340° 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 21/2 hours? Answer to the nearest 10km. N What would you like to do now? 1. Identify what you need to find and the information you have to help you. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 2. Calculate as many of the missing angles as possible. 3. Make a sketch to clarify matters. 5. Substitute known values, remembering to use brackets as appropriate. Reveal answer only 340° 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 21/2 hours? Answer to the nearest 10km. N 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 N 1. Identify what needs to be found.
160km/hr. N 2. Need distances travelled and angle between flight paths. 340° d1 = speed1 x time = 160 x 2.5 = 400km How far apart will they be after 21/2 hours? d2 = speed2 x time = 200 x 2.5 = 500km 340° clockwise = 20° anti-clockwise 20° Full angle = 20° + 90° = 110° 90° 3. Sketch triangle. 200km/hr A b c a 400km 500km 110° Continue Solution Try another like this Comments Trigonometry Menu Back to Home

Question 2 N 4. Apply Cosine rule. a2 = b2 + c2 – (2bccosA°) 340°
160km/hr. N a2 = b2 + c2 – (2bccosA°) 340° 5. Substitute known values and remember to use brackets. How far apart will they be after 21/2 hours? = – (2 x 400 x 500 x cos110°) 20° = 90° a =  = 200km/hr 6. Remember to round answer if asked to. Continue Solution = 740 Try another like this Distance is 740km Comments Trigonometry Menu Back to Home

1. Identify what needs to be found. Note: Bearings are measured
Markers Comments 1. Identify what needs to be found. Note: Bearings are measured clockwise from N. 2. Need distances travelled and angle between flight paths. 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 500km 110° Next Comment Trigonometry Menu Back to Home

Check formulae list for the cosine rule:
Markers Comments Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA (2 sides and the included angle) Relate to variables used 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. = – (2 x 400 x 500 x cos110°) = a =  = 6. Remember to round answer if asked to. = 740 Next Comment Trigonometry Menu Distance is 740km Back to Home

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 Get hint Reveal answer only 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 2. Calculate as many of the missing angles as possible. What would you like to do now? 1. Identify what you need to find and the information you have to help you. 4. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule 3. Make a sketch to clarify matters. Reveal answer only 5. Substitute known values, remembering to use brackets as appropriate. 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 Distance is 41.2 miles What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Question 2B N 1. Identify what needs to be found.
195° N SE How far apart will they be? 2. Need distances travelled and angle between paths. 1350 d1 = speed1 x time = 18 x 2 = 36miles d2 = speed2 x time = 15 x 3 = 45miles NB: SE = 135° Angle = 195° - 135° = 60° 600 3. Sketch triangle. A b c a 60° 36miles 45miles Begin Solution Continue Solution Comments Trigonometry Menu Back to Home

Question 2B N 4. Apply Cosine rule. SE a2 = b2 + c2 – (2bc cosA°)
195° N SE How far apart will they be? a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. 1350 = – (2 x 36 x 45 x cos60°) = 1701 600 a = 1701 = = 41.2 Begin Solution Continue Solution Distance is 41.2 miles Comments Trigonometry Menu Back to Home

1. Identify what needs to be found. Note: Bearings are measured
Comments 1. Identify what needs to be found. Note: Bearings are measured clockwise from N. 2. Need distances travelled and angle between flight paths. 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 b c a 60° 36miles 45miles Next Comment Trigonometry Menu Back to Home

Check formulae list for the cosine rule:
Comments Check formulae list for the cosine rule: a2 = b2 + c2 – 2bc cosA (2 sides and the included angle) Relate to variables used 4. Apply Cosine rule. a2 = b2 + c2 – (2bc cosA°) 5. Substitute known values and remember to use brackets. = – (2 x 36 x 45 x cos60°) = 1701 a = 1701 = = 41.2 Next Comment Distance is 41.2 miles Trigonometry Menu Back to Home

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 Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

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

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. angleQPR = 35.3° = 127cm2 P Q R S 10cm 15cm 22cm What would you like to do now? Try another like this Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Question 3 P Q R S 10cm 15cm 22cm 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP = 2rq r2 + q2 - p2 = 2 x 10 x 22 = ( )  (2 x 10 x 22) = … 2. Remember to use inverse function to find angle. (a) Find the size of angle QPR angleQPR = cos-1( ) = 35.3° Continue Solution Try another like this Comments Trigonometry Menu Back to Home

Question 3 P Q R S 10cm 15cm 22cm 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 = cm2 2. Remember to double this and round answer. Area of kite = 2 x (b) Hence find the area of the kite to the nearest square unit = cm2 Continue Solution = 127cm2 Try another like this Comments Trigonometry Menu Back to Home

= Markers Comments Check the formulae list for the
second form of the cosine rule: 1. To find angle when you have 3 sides use 2nd version of cosine rule: b2 + c2 – a2 2bc ( 3 sides) cosA = cosP = 2rq r2 + q2 - p2 = 2 x 10 x 22 = ( )  (2 x 10 x 22) = … 2. Remember to use inverse function to find angle. angleQPR = cos-1( ) = 35.3° Next Comment Trigonometry Menu Back to Home

= Markers Comments Relate to variables used: cosP = = q2 + r2 – p2 2qr
– 152 2x22x10 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP = 2rq r2 + q2 - p2 = 2 x 10 x 22 P Q R 10cm 15cm 22cm p q r = ( )  (2 x 10 x 22) = … 2. Remember to use inverse function to find angle. angleQPR = cos-1( ) = 35.3° Next Comment Trigonometry Menu Back to Home

= Markers Comments Note:
When keying in to calculator work out the top line and the bottom line before dividing or use brackets. 1. To find angle when you have 3 sides use 2nd version of cosine rule: cosP = 2rq r2 + q2 - p2 = 2 x 10 x 22 = ( )  (2 x 10 x 22) = … 2. Remember to use inverse function to find angle. angleQPR = cos-1( ) = 35.3° Next Comment Trigonometry Menu Back to Home

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 F G H 15cm 25cm Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

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

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. angleEFH = 33.6° = 415cm2 E F G H 15cm 25cm What would you like to do now? Go to full solution Go to Comments Go to Trigonometry Menu EXIT

Question 3B E F G H 15 25 1. To find angle when you have 3 sides use 2nd version of cosine rule: (a) cosF = 2eh e2 + h2 - f2 = 2 x 25 x 15 = ( )  (2 x 25 x 15) = … 2. Remember to use inverse function to find angle. (a) Find the size of angle EFH angleEFH = cos-1( ) = 33.6° Continue Solution Comments Trigonometry Menu Back to Home

Question 3B E F G H 15 25 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: 33.60 (b) Area EFH = ½ ehsinF° = 25 x 15 x sin33.6°  2 = cm2 2. Remember to double this and round answer. Area of kite = 2 x (b) Hence find the area to the nearest square unit = cm2 = 415cm2 Continue Solution Comments Trigonometry Menu Back to Home

= Markers Comments Check the formulae list for the
second form of the cosine rule: 1. To find angle when you have 3 sides use 2nd version of cosine rule: b2 + c2 – a2 2bc ( 3 sides) (a) cosF = 2eh e2 + h2 - f2 cosA = = 2 x 25 x 15 = ( )  (2 x 25 x 15) = … 2. Remember to use inverse function to find angle. angleEFH = cos-1( ) = 33.6° Next Comment Trigonometry Menu Back to Home

= Markers Comments Relate to variables used: cosF =
e2 + h2 – f2 2eh – 152 2x22x10 1. To find angle when you have 3 sides use 2nd version of cosine rule: (a) cosF = 2eh e2 + h2 - f2 = 2 x 25 x 15 E F H 15cm 25cm e f h = ( )  (2 x 25 x 15) = … 2. Remember to use inverse function to find angle. angleEFH = cos-1( ) = 33.6° Next Comment Trigonometry Menu Back to Home

Check formulae list for: Area of triangle = absinC
Markers Comments Check formulae list for: Area of triangle = absinC (Note: 2 sides and the included angle) Relate formula to labels being used. 1. Rhombus = 2 identical triangles. For area of triangles where an angle is present use: 1 2 (b) Area EFH = ½ ehsinF° = 25 x 15 x sin33.6°  2 = cm2 E F H 15cm 25cm e f h 2. Remember to double this and round answer. Area of kite = 2 x = cm2 = 415cm2 Next Comment Trigonometry Menu Back to Home

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 Get hint Reveal answer only Go to full solution Go to Comments Go to Trigonometry Menu EXIT

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

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

Question 4 Angle V = 180° - 35° - 105° = 40° v t = sinV° sinT° v 10 =
5.9cm 10cm Perimeter requires all three sides. So we need to find TU . 2. Whether you use Sine or Cosine rule need angle V. 400 Angle V = 180° - 35° - 105° = 40° 3. If we use Sine rule: v t = sinV° sinT° Find the perimeter to one decimal place. 4. Substitute known values: v = sin40° sin105° Continue Solution Comments Trigonometry Menu Back to Home

Question 4 v 10 = sin40° sin105° Perim of  = (6.7 + 10 + 5.9)cm
35° 5.9cm 10cm 4. Substitute known values: v = sin40° sin105° 5. Cross multiply: 400 v x sin105° = 10 x sin40° v = 10 x sin40°  sin105° Find the perimeter to one decimal place. = … = 6.7cm 6. Answer the question: Continue Solution Perim of  = ( )cm Comments = 22.6cm Trigonometry Menu Back to Home

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

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

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 Hints Answer only 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 2. Identify which trig rule to use: Two sides + two angles = sine rule Three sides + one angle = cosine rule What would you like to do now? 4. Closest distance = perpendicular distance. Create a right-angled triangle and use SOH CAH TOA. Calculate missing angles. (NB. North West = 3150) Answer only 3. Substitute known values, remembering to use brackets as appropriate. 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 Closest distance is 5.48km What would you like to do now? Full solution Comments Trig Menu EXIT

N A B C S Question 5 b c = sinB° sinC° 12km Calculate missing angles.
50° 12km S Question 5 Calculate missing angles. AngleA = 90° - 50° = 40° NW = 3150 & angleB = 315° - 270° = 45° 950 So angle C = 180° - 40° - 45° = 95° 2. Draw a sketch 400 450 40° 45° 12km 95° A B C b 3. If we use Sine rule: b c = sinB° sinC° Continue Solution Comments Trigonometry Menu Back to Home

Question 5 N A B C S b 12 = sin45° sin95° = 8.5176.… = 8.52km 12km
50° 12km S 4. Substitute known values: b = sin45° sin95° 950 5. Cross multiply: b x sin95° = 12 x sin45° 400 450 b = 12 x sin45°  sin95° = … = 8.52km Continue Solution Comments Trigonometry Menu Back to Home

Question 5 C N = 8.52km A B a sin40° = 8.52 S
50° 12km S 6. Closest point is perpendicular so sketch right angled triangle: A C S 40° 8.52km a 950 = 8.52km 400 450 7. Now using SOHCAH TOA: sin40° = a 8.52 a = 8.52 x sin40° = 5.476…. Continue Solution = 5.48 Comments Closest distance is 5.48km Trigonometry Menu Back to Home

b c = sinB° sinC° Markers Comments Since we can pair off two angles
with the opposite sides Sine Rule Calculate missing angles. AngleA = 90° - 50° = 40° NW = 3150 & angleB = 315° - 270° = 45° Refer to the Formulae List : = = So angle C = 180° - 40° - 45° = 95° a Sine A b Sine B c Sine C 2. Draw a sketch 40° 45° 12km 95° A B C 3. If we use Sine rule: b c = sinB° sinC° Next Comment Trigonometry Menu Back to Home

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

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 Hints Answer only 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 4. Then use SOH CAH TOA to find length from rig to first ship. Second ship is then 400 m further. 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 Calculate missing angles. What would you like to do now? 3. Substitute known values, remembering to use brackets as appropriate. Answer only 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 What would you like to do now? Try another like this Full solution Comments Trig Menu ShipA is 1685m away & ShipB is 1285m away EXIT

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

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

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

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

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

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 Hints Answer only 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 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 4. Then use SOH CAH TOA to find Height of rig from deck level. Remember decks are 5m above sea level. 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 Calculate missing angles. What would you like to do now? 3. Substitute known values, remembering to use brackets as appropriate. Answer only 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 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 What would you like to do now? Full solution Comments Trig Menu Height of tower is 123m EXIT

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

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

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

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

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

You have chosen to study: UNIT 2 : Statistics Please choose a question to attempt from the following: 1 2 3 4 5 Back to Unit 2 Menu EXIT

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 …. (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 = and standard deviation = 1.88 How does the second company compare to the first? Get hint Go to Comments Reveal answer only Go to Statistics Menu Go to full solution EXIT

Draw a table comparing data to mean.
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 …. (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 = 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. Go to Comments Reveal answer only Go to Statistics Menu Go to full solution EXIT What would you like to do now?

STATISTICS : Question 1 (a) Mean = 8
A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. (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 = and standard deviation = 1.88 How does the second company compare to the first? (a) Mean = 8 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? Go to full solution Go to Statistics Menu EXIT

and no.pieces of data = n = 7
Question 1 Calculate mean. A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. (a) Find the mean and standard deviation for this data. (a) Mean = ( )7 = 8 So = 8 and no.pieces of data = n = 7 Begin Solution Continue Solution Comments Statistics Menu Back to Home

Question 1 x  (x - )2 = 2. Draw table comparing data to mean. 3 -5 25
A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. (a) Find the mean and standard deviation for this data. x  (x - )2 = 90 Begin Solution Continue Solution Comments Statistics Menu Back to Home

Question 1 3. Use formula to calculate standard deviation.
A taxi company was phoned each night of the week and the response time in minutes of their cars were noted. They were …. (a) Find the mean and standard deviation for this data. Just found!! Begin Solution Continue Solution Comments Statistics Menu Back to Home

Question 1 1. Always compare mean and standard deviation.
(b) A similar experiment was conducted with a second company. The results for this were…. mean = and standard deviation = 1.88 How does the second company compare to the first? 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? Begin Solution Continue Solution Comments Statistics Menu Back to Home

Check the list of Formulae for the Standard Deviation Formula:
Comments Check the list of Formulae for the Standard Deviation Formula: 3. Use formula to calculate standard deviation. 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 Time(s) 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. Continue question EXIT

STATISTICS : Question 2 Time(s) Weeks of training
Get hint Reveal answer only Time(s) 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

Find gradient and note intercept of T axis.
Use your equation and substitute W = 12. Use your equation and substitute T = 11.5 STATISTICS : Question 2 Reveal answer only Time(s) 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? Time(s)
Go to full solution Go to Comments Go to Stats Menu Weeks of training T = -¼ W + 15 (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 12secs. EXIT 14 weeks

Question 2 = -¼ Find gradient and note intercept of T axis.
(i) Find the equation of the line in terms of T and W. (a) m = (y2 – y1) (x2 – x1) 20 - 0 = -¼ = Intercept at 15 Equation is T = -¼ W + 15 Go to graph Begin Solution Continue Solution Comments Statistics Menu Back to Home

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

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

= -¼ Comments To find the equation of a line from the graph:
Must Learn: y = mx + c Find gradient and note intercept of T axis. m = (y2 – y1) (x2 – x1) (a) 20 - 0 = = -¼ Intercept at 15 Equation is T = -¼ W + 15 gradient intercept So you need to find these!! Next Comment Statistics Menu Back to Home

= -¼ Comments vertical horizontal m =
Find gradient and note intercept of T axis. m = (y2 – y1) (x2 – x1) (a) 20 - 0 = = -¼ 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 Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 21 15 ? What percentage liked Shrek but not the Simpsons? 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 Full solution Statistics Menu EXIT Reveal Comments

Remember all entries in table must add to total sample size
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 Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 21 15 ? Remember all entries in table must add to total sample size To find probabilities use: P = no of favourable / no of data What percentage liked Shrek but not the Simpsons? If someone is picked at random what is the probability that (i) they liked the Simpsons but not Shrek? (ii) they liked neither? What would you like to do now? Full solution Statistics Menu EXIT Comments Reveal

126 21 15 ? STATISTICS : Question 3 10% 1/12 7/60
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 21 15 ? 10% What percentage liked Shrek but not the Simpsons? If someone is picked at random what is the probability that (i) they liked the Simpsons but not Shrek? (ii) they liked neither? 1/12 7/60 Full solution Statistics Menu EXIT Comments What now?

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

126 21 15 ? Question 3 (b)(i) Prob = 15/180 = 1/12
1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 21 15 ? (b)(i) Prob = 15/180 = 1/12 what is the probability that they liked the Simpsons but not Shrek? Begin Solution Continue Solution Comments Statistics Menu Back to Home

126 21 15 ? Question 3 = 7/60 (b)(ii) Prob = 21/180
1. Use P = no of favourable / no of data Like Simpsons Like Shrek Don’t like Simpsons Don’t like Shrek 126 21 15 ? = 7/60 (b)(ii) Prob = 21/180 What would you like to do now? (ii) they liked neither? Begin Solution Continue Solution Comments Statistics Menu Back to Home

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

Number of favourable outcomes Number of possible outcomes
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 leisure activity from the following lists Reveal ans Full solution LEISURE MUSIC VIDEO PRODUCTION BASKETBALL SWIMMING LANGUAGE FRENCH GERMAN SPANISH Comments Stats Menu (a) Make a list of all possible combinations of courses. 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?

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

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

(a) Hints Use a tree diagram & “branch out” with each language.
BASKETBALL VIDEO MUSIC SWIMMING French/Music Use a tree diagram & “branch out” with each language. FRENCH French/Video French/Bsktbll From each language “branch out” with each leisure activity. French/Swim BASKETBALL VIDEO MUSIC SWIMMING German/Music Now list the pairs of subjects. German/Video GERMAN German/Bsktbll SPANISH German/Swim BASKETBALL VIDEO MUSIC SWIMMING Spanish/Music Spanish/Video Spanish/Bsktbll Spanish/Swim

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

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

(a) How many deliveries took longer than 16 mins?
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 up to 16 mins up to 20 mins 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.

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

(a) How many deliveries took longer than 16 mins?
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 up to 16 mins up to 20 mins Full solution Comments Stats Menu (a) How many deliveries took longer than 16 mins? = 7 (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 4 5 8 13 12 37 16 53 20 60 Delivery time(mins) Cum freq ¾ of 60 = 45 Q3 Comments ½ of 60 = 30 Q2 Stats Menu ¼ of 60 = 15 What would you like to do now? Q1 8.5 11 14 (C) Median = 11mins. (c) SIQR = ½(Q3 – Q1) = (14 – 8.5)  2 = 2.75mins

In a Cumulative Frequency Diagram: Cumulative Frequency
Comments Note: In a Cumulative Frequency Diagram: Cumulative Frequency On y-axis (cumulative frequency) Q1 at 25%, Q2 at 50%, Q3 at 75% 60 20 50% 11 And read values from the delivery time scale (x-axis). Delivery Time End of Statistics Next Comment Statistics Menu Back to Home

You have chosen to study: UNIT 2 : Graphs, Charts & Tables Please choose a question to attempt from the following: 1 2 3 4 5 6 Stem & Leaf Dot Plot Cum Freq Table Dot to boxplot Stem to boxplot Piechart Back to Unit 2 Menu EXIT

GRAPHS, CHARTS, TABLES : Question 1
The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. n = 25 = £174 (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range 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

Q1 is midpoint from start to median Q3 is midpoint from median to end
GRAPHS, CHARTS, TABLES : Question 1 The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. n = 25 = £174 Q1 is midpoint from start to median Q3 is midpoint from median to end Use median position = (n+1) / 2 to find median (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range What would you like to do now? 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 ….. n = 25 = £174 (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range median = £183 Q1 = £171 What would you like to do now? Q3 = £195 = £12 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 1. Use median = (n+1) / 2 to find median
n = 25 = £174 (a)(i) Since n = 25 then the median is 13th value ie median = £183 (NOT 3!!!) 2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2 Median lower & upper quartiles (iii) the semi-interquartile range (ii) Both 6th & 7th values are £ so Q1 = £171 3. There are 12 values after median so Q3 position = 13 + (12 + 1) / 2 Begin Solution 19th is £192 & 20th is £ so Q3 = £195 Continue Solution Comments Menu What would you like to do now? Back to Home

Question 1 4. Use SIQR = ½ (Q3 – Q1 ) / 2 (iii) SIQR = ½(Q3 – Q1)
n = 25 = £174 = (£195 - £171)  2 = £12 Median lower & upper quartiles (iii) the semi-interquartile range Begin Solution Continue Solution Comments Menu Back to Home

Question 1 5. Use P = no of favourable / no of data
No of favourable ( under £180) = 10 n = 25 = £174 No of data = n = 25 (b) Prob(under £180) = 10/25 = 2/5 . What is the probability that someone chosen at random earns less than £180? Begin Solution Continue Solution Comments Menu Back to Home

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

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

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

P = no of favourable / no of data
Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 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 What would you like to do now? Graphs etc Menu Go to full solution EXIT Reveal answer Go to Comments

The weights in grams of 20 bags of crisps were as follows 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? CLICK Tightly clustered 3/10 Graphs etc Menu Go to full solution EXIT Go to Comments

Question 2 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 28 30 32 Begin Solution 2. Plot a dot for each piece of data and label diagram. Continue Solution Comments Menu Back to Home

(b) Tightly clustered distribution.
Question 2 3. Make sure you know the possible descriptions of data. Weights in g What type of distribution does this show? 26 28 30 32 Begin Solution Continue Solution (b) Tightly clustered distribution. Comments Menu Back to Home

Question 2 4. Use P = no of favourable / no of data Mode! Weights in g If a bag is chosen at random what is the probability it will be heavier than the modal weight? 26 28 30 32 Begin Solution No of favourable ( bigger than 29) = 6 Continue Solution No of data = n = 20 Comments (c) Prob(W > mode) = 6/20 = 3/10 . Menu What would you like to do now? Back to Home

(b) Tightly clustered distribution.
Comments Other types of distribution: 3. Make sure you know the possible descriptions of data. Weights in g 26 28 30 32 Next Comment (b) Tightly clustered distribution. Menu Back to Home

Number of favourable outcomes Number of possible outcomes
Comments To calculate simple probabilities: 4. Use P = no of favourable / no of data Mode! Probability = Weights in g Number of favourable outcomes Number of possible outcomes 26 28 30 32 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

The results for a class test were (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 Go to full solution EXIT Reveal answer Go to Comments

P = no of favourable / no of data
Charts, Graphs & Tables : Question 3 The results for a class test were (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. Use median = (n+1) / 2 to establish in which row median lies. Complete each row 1 step at a time, calculating running total as you go. For probability use: P = no of favourable / no of data What would you like to do now? Graphs etc Menu Go to full solution EXIT Reveal answer Go to Comments

The results for a class test were (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 Graphs etc Menu Go to full solution EXIT Go to Comments

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

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

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

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

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

The dot plot below shows the number of matches per box in a sample of 23 boxes. Find the (i) median (ii) lower quartile (iii) upper quartile Construct a boxplot using this data. In a second sample the semi-interquartile range was How does this distribution compare to the above sample? Get hint Graphs etc Menu Go to full solution EXIT Reveal answer Go to Comments

Q1 is midpoint from start to median Q3 is midpoint from median to end
Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. Q1 is midpoint from start to median Q3 is midpoint from median to end Use median position = (n+1) / 2 to find median remember bigger SIQR means more variation (spread) in data. Find the (i) median (ii) lower quartile (iii) upper quartile Construct a boxplot using this data. In a second sample the semi-interquartile range was How does this distribution compare to the above sample? What would you like to do now? Graphs etc Menu Go to full solution EXIT Reveal answer Go to Comments

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 Find the (i) median (ii) lower quartile (iii) upper quartile Construct a boxplot using this data. In a second sample the semi-interquartile range was How does this distribution compare to the above sample? CLICK the data is distributed more widely than (or not as clustered as) the above data EXIT Full solution Comments Menu

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

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

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

1. Use median = (n+1) / 2 to find median
Comments The median: 1. Use median = (n+1) / 2 to find median 23 numbers in the list: (i) Sample size = 23 so median position is 12. ie (23+1)2 Q2 Median = 50 2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2 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 Next Comment (iii) Middle of 2nd 11 is position 18. Menu So Q3 = 52 Back to Home

1. Use median = (n+1) / 2 to find median
Comments For quartiles: 1. Use median = (n+1) / 2 to find median (i) Sample size = 23 so median position is 12. ie (23+1)2 Q1 Q2 Median = 50 12 2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2 Q3 Q2 (ii) Middle of 1st 11 is position 6. Now count through the list until you reach the 6th, 12th,and 18th number in the list. So Q1 = 49 3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2 Next Comment (iii) Middle of 2nd 11 is position 18. Menu So Q3 = 52 Back to Home

The semi-interquartile range is a measure of the range of the
Comments The semi-interquartile range is a measure of the range of the “middle” 50%. 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. S.I.R. = (Q3 - Q1) 1 2 For above sample SIQR = ( )  2 = 1.5 It is a measure of how spread-out and so how “consistent” or “reliable” the data is. In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data Remember: when asked to compare data always consider average and spread. Next Comment Menu Back to Home

The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. Get hint = 114kg Reveal answer Full solution Comments Find the median, lower & upper quartiles for this data. Use the data to construct a boxplot. The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. EXIT

What now? The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. Q1 is midpoint from start to median Q3 is midpoint from median to end Use median position = (n+1) / 2 to find median position Menu When comparing two data sets comment on spread and average = 114kg Reveal answer Full solution Comments Find the median, lower & upper quartiles for this data. Use the data to construct a boxplot. The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. EXIT

The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. = 114kg median = 87 Menu Q1 = 77 Q3 = 99 Full solution Comments Find the median, lower & upper quartiles for this data. Use the data to construct a boxplot. The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. CLICK CLICK EXIT

(a)(i) Since n = 26 then the median is between 13th & 14th value ie median = 87 = 114kg 2. There are 13 values before median so Q1 position is 6th value 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 so Q3 = 99 Continue Solution Comments Menu Back to Home

Question 5 4. Draw number line with scale.
Make sure you note highest & lowest as well as Q1, Q2, Q3. Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. = 114kg (b) Use the data to construct a boxplot. Begin Solution Continue Solution Comments 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. 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. Begin Solution Continue Solution What would you like to do now? Comments Menu Back to Home

To draw a boxplot you need a “five-figure summary”:
Comments Remember: To draw a boxplot you need a “five-figure summary”: 4. Draw number line with scale. Make sure you note highest & lowest as well as Q1, Q2, Q3. Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. Box Plot : Lowest Q1 Q2 Q3 Highest five-figure summary Next Comment Menu Back to Home

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

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

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

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

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

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

(b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = 84
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 84 = x 630 x 360 angle 360° = 84 630 630 x = 84 x 360 630 x angle = 360° x 84 x = 360 630 x angle = 360° x 84  630 End of graphs, charts etc. = 48° Next Comment Menu Back to Home

Simultaneous Equations
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Simultaneous Equations Please choose a question to attempt from the following: 1 2 3 4 Back to Unit 2 Menu EXIT

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

Plot and join points. Solution is where lines cross
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 To draw graph: Construct a table of values with at least 2 x-coordinates. Plot and join points. Solution is where lines cross 3x + 2y = 17. What would you like to do now? Reveal answer only Go to Comments EXIT Go to full solution Go to Sim Eq Menu

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 + 2y = 17. What would you like to do now? Try another like this Go to Comments EXIT Go to full solution Go to Sim Eq Menu

Question 1 1. Construct a table of values with at least 2 x coordinates. The diagram shows the graph of 3x + 2y = 17. y = 2x - 2 x Copy the diagram and on your diagram draw the graph of y = 2x – 2 , hence solve 3x + 2y = 17 y = 2x – 2 y Begin Solution Continue Solution Comments Sim Eq Menu Back to Home

Question 1 2. Plot and join points. Solution is where lines cross.
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 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

There are two ways of drawing the line y = 2x - 1 Method 1
Markers Comments There are two ways of drawing the line y = 2x - 1 Method 1 Finding two points on the line: x = y = 2 x 0 – 1 = -1 x = y = 2 x = 3 First Point (0,-1) Second Point (-1,3) Plot and join (0,-1), and (-1,3). 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2 x y Next Comment Sim Eqs Menu Back to Home

gradient y - intercept y = 2x - 1 gradient m = 2 y - intercept c = -1
Markers Comments Method 2 Using y = mx + c form: y = mx + c gradient y - intercept 2. Plot and join points. Solution is where lines cross. y = 2x - 1 gradient m = 2 y - intercept c = -1 Plot C(0, -1) and draw line with m = 2 Next Comment Sim Eqs Menu Solution is x = 3 & y = 4 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 Reveal answer only Go to Comments Get hint EXIT Go to full solution Go to Sim Eq Menu

Plot and join points. Solution is where lines cross
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 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? Reveal answer only Go to Comments EXIT Go to full solution Go to Sim Eq Menu

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 What would you like to do now? Solution is x = 2 & y = 4 Go to Comments EXIT Go to full solution Go to Sim Eq Menu

Question 1B 1. Construct a table of values with at least 2 x coordinates. The diagram shows the graph of 2x + y = 8. y = 1/2x + 3 x y 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

Question 1B 2. Plot and join points. Solution is where lines cross.
The diagram shows the graph of 2x + y = 8. y = 1/2x + 3 x y Copy the diagram and on your 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? Begin Solution Continue Solution Comments Sim Eq Menu Solution is x = 2 & y = 4 Back to Home

There are two ways of drawing the line y = ½ x + 3 Method 1
Markers Comments There are two ways of drawing the line y = ½ x + 3 Method 1 Finding two points on the line: x = y = ½ x = 3 x = y = ½ x = 4 First Point (0, 3) Second Point (2, 4) Plot and join (0, 3), and (2, 4). 1. Construct a table of values with at least 2 x coordinates. y = 1/2x + 3 x y Next Comment Sim Eqs Menu Back to Home

gradient y - intercept y = ½ x + 3 gradient m = ½ y - intercept c = +3
Markers Comments Method 2 Using y = mx + c form: y = mx + c gradient y - intercept 2. Plot and join points. Solution is where lines cross. y = 1/2x + 3 x y y = ½ x + 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

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

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

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

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

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

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

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

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

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

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

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

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

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

Remember to answer the question!!!
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. What would you like to do now? Substitute found value into either of original equations. Reveal answer only Remember to answer the question!!! Go to full solution Go to Comments Go to Sim Eq Menu EXIT

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

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

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

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

Two coffees & five doughnuts = (2 x 70p) + (5 x 50p)
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. It is usually best to choose an equation that you were given in question. 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 Substitute 70 for c in equation Step 4 Remember to answer the question!!! 210 + d = 260 d = 50 Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) Next Comment = £ £2.50 Sim Eqs Menu = £3.90 Back to Home

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 Reveal answer Go to Comments EXIT Go to full solution Go to Sim Eq Menu

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

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. Go to Comments EXIT Go to full solution Go to Sim Eq Menu

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

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

Question 3B 4. Substitute found value into an equation without 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? 3 (x15) 4 Substitute 80 for B in equation 3 K = 740 3K = 180 K = 60 5. Use these values to answer question. What would you like to do now? 75%B+25%K = (0.75 x 80p)+(0.25 x 60p) Begin Solution = 60p p Continue Solution = 75p Comments So this blend is more expensive than the other two. Sim Eq Menu Back to Home

Multiply all terms by 100 to remove decimals.
Markers Comments 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). 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 K = 74 0.55B K = 71 (x10) (x100) 1 2 i.e B K = 74 0.55B K = 71 2. Get rid of decimals (x10) 3 = 1 Multiply all terms by 100 to remove decimals. Note change to pence eases working. (x100) 4 = 2 Next Comment Sim Eqs Menu Back to Home

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

(x15) 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. It is usually best to choose an equation that you were given in question. 4. Substitute found value into an equation without decimals. 3 (x15) 4 Substitute 80 for B in equation 3 K = 740 3K = 180 K = 60 Step 4 Remember to answer the question!!! 5. Use these values to answer question. 75%B+25%K = (0.75 x 80p)+(0.25 x 60p) = 60p p Next Comment = 75p Sim Eqs Menu So this blend is more expensive than the other two. Back to Home

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. 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. 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 Reveal answer Go to Comments EXIT Go to full solution Go to Sim Eq Menu

Write down expressions using previous two to form next. Solve and substitute found value into either of original equations. 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. 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. 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. Establish two equations. Eliminate either P’s or Q’s by making coefficient same. To find values of P & Q: Match term from (a) with values given in question. What would you like to do now? Go to Comments Reveal answer EXIT Go to full solution Go to Sim Eq Menu

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. 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. 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. P + Q, P + 2Q 2P + 3Q 3P + 5Q P = 7 Go to Comments Try another like this EXIT Go to full solution Go to Sim Eq Menu

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

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

Then introduce the variables:
Comments 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, … 1. Write down expressions using previous two to form next. (a) First term = P & second term = Q 3rd term = P + Q 4th term = Q + (P + Q) = P + 2Q 5th term = (P + Q) + (P + 2Q) = 2P + 3Q 6th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q Next Comment Sim Eqs Menu Back to Home

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

In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. Form 2 equations and eliminate either v’s or w’s by making coefficient same. Substitute found value into either of original equations. Work your way to an expression for the top row by filling in the middle rows. 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. (B) (A) -18 11 3v 2w w – 2v v + w v + w v – 3w w v - w Reveal answer Go to Comments EXIT Go to full solution Go to Sim Eq Menu

In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. (B) (A) -18 11 3v 2w w – 2v v + w v + w v – 3w w v - w V = 4 Go to Comments EXIT Go to full solution Go to Sim Eq Menu

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

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

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

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

Hence equations: 10w - 2v = -18 9w + 5v = 11 Solve by the method
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 Solve by the method of elimination. 3rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 End of simultaneous Equations Pyramid (B) 2nd row 2v - 2w, v + 3w, v + 5w 3rd row 3v + w, 2v + 8w Next Comment Top row 5v + 9w = 11 Sim Eqs Menu Back to Home

Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations

Similar presentations

Presentation on theme: "Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations

Similar presentations

Presentation on theme: "Trigonometry Statistics Graphs, Charts & Tables Simultaneous Equations"— Presentation transcript:

Similar presentations

About project

Feedback