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Parents Pricing Home NSC Mathematical Literacy Perimeter, area and volume Justin has an injury that requires him to use a wheelchair for a while
Justin has an injury that requires him to use a wheelchair for a while - NSC Mathematical Literacy - Question 2 - 2016 - Paper 1 Question 2
View full question Justin has an injury that requires him to use a wheelchair for a while. He uses the diagrams below to determine certain dimensions of the wheelchair.
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View marking scheme Worked Solution & Example Answer:Justin has an injury that requires him to use a wheelchair for a while - NSC Mathematical Literacy - Question 2 - 2016 - Paper 1
Determine the length of the outer diameter (rounded off to the nearest millimeter) of the big wheel. Only available for registered users.
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To find the outer diameter of the big wheel, we first calculate 54% of the length of the wheelchair:
ext{Outer Diameter} = rac{54}{100} imes 121.92 ext{ cm} = 65.368 ext{ cm}
ightarrow 65.368 imes 10 = 653.68 ext{ mm}
ightarrow 654 ext{ mm (rounded)}
Calculate how far apart the wheel spokes are spaced from each other on the rim. Only available for registered users.
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First, we calculate the circumference of the rim using the inner diameter:
e x t C i r c u m f e r e n c e = e x t π i m e s 584 e x t m m = 3.142 i m e s 584 e x t m m i g h t a r r o w 1834.94 e x t m m ext{Circumference} = ext{π} imes 584 ext{ mm} = 3.142 imes 584 ext{ mm}
ightarrow 1834.94 ext{ mm} e x t C i rc u m f ere n ce = e x t π im es 584 e x t mm = 3.142 im es 584 e x t mm i g h t a rro w 1834.94 e x t mm The part of the circumference filled by spokes:
e x t P a r t f i l l e d = 24 i m e s 2 e x t m m = 48 e x t m m ext{Part filled} = 24 imes 2 ext{ mm} = 48 ext{ mm} e x t P a r t f i ll e d = 24 im es 2 e x t mm = 48 e x t mm Distance between spokes:
ext{Distance Apart} = rac{1834.94 - 48}{24}
ightarrow rac{1786.94}{24}
ightarrow 74.87 ext{ mm}
Determine, showing ALL calculations, how wide (in mm) the available gap on both sides of the wheelchair is. Only available for registered users.
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The width of the wheelchair is 60.96 cm, which we convert to mm:
e x t W i d t h = 60.96 i m e s 10 i g h t a r r o w 609.6 e x t m m ext{Width} = 60.96 imes 10
ightarrow 609.6 ext{ mm} e x t Wi d t h = 60.96 im es 10 i g h t a rro w 609.6 e x t mm The available gap on both sides when passing through the doorway opening:
ext{Gap} = rac{750 - 609.6}{2} = rac{140.4}{2} = 70.2 ext{ mm}
Determine the total width, in metres, of the two doors. Only available for registered users.
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To find the total width, we add up the contributions from all parts:
e x t T o t a l W i d t h = ( 80 i m e s 4 ) + ( 640 i m e s 2 ) = 320 + 1280 = 1600 e x t m m i g h t a r r o w 1.6 e x t m ext{Total Width} = (80 imes 4) + (640 imes 2) = 320 + 1280 = 1600 ext{ mm}
ightarrow 1.6 ext{ m} e x t T o t a l Wi d t h = ( 80 im es 4 ) + ( 640 im es 2 ) = 320 + 1280 = 1600 e x t mm i g h t a rro w 1.6 e x t m
Calculate the value of e, the length of the rectangular glass-panel inserts. Only available for registered users.
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We need to first find the total panel height:
e x t T o t a l = 2485 − ( 80 + 640 + 95 + 220 ) + 2 e i g h t a r r o w 2485 − 1035 + 2 e = 0 ext{Total} = 2485 - (80 + 640 + 95 + 220) + 2e
ightarrow 2485 - 1035 + 2e = 0 e x t T o t a l = 2485 − ( 80 + 640 + 95 + 220 ) + 2 e i g h t a rro w 2485 − 1035 + 2 e = 0 Solving for e:
2 e = 677.5 i g h t a r r o w e = 338.75 e x t m m 2e = 677.5
ightarrow e = 338.75 ext{ mm} 2 e = 677.5 i g h t a rro w e = 338.75 e x t mm
Calculate the total area (in mm²) of ALL the glass-panel inserts. Only available for registered users.
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Area of glass-panel inserts:
ext{Area} = (640 imes 481) + 2 imes rac{3.142}{8} imes 640^2
ightarrow 307200 + 640 ext{ mm}^2 = 2377,881.6 ext{ mm}^2
Calculate (in kg) the total mass of the glass panels. Only available for registered users.
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The total mass is given by the formula:
ext{Mass} = ext{Volume} imes ext{Density}
ightarrow rac{15985408 ext{ cm}^3}{1000} imes 2.5 = 39.6 ext{ kg} Join the NSC students using SimpleStudy...97% of StudentsReport Improved Results
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