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Parents Pricing Home NSC Mathematical Literacy Measuring length 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) of the big wheel. Only available for registered users.
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To find the outer diameter of the bigger wheel, use the given information:
The length of the wheelchair is 121,92 cm.
Calculate the outer diameter:
ext{Outer diameter} = 54 rac{100}{ ext{Length of wheelchair}} imes 121,92 ext{ cm}
= 658 , 368 e x t m m e x t ( a p p r o x i m a t e l y 658 m m r o u n d e d o f f ) = 658,368 ext{ mm} ext{ (approximately 658 mm rounded off)} = 658 , 368 e x t mm e x t ( a pp ro x ima t e l y 658 mm ro u n d e d o ff )
Calculate how far apart the wheel spokes are spaced from each other on the rim. Only available for registered users.
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The circumference of the rim can be calculated using:
e x t C i r c u m f e r e n c e = e x t π i m e s e x t d i a m e t e r = 3 , 142 i m e s 584 e x t m m = 1834 , 634 e x t m m ext{Circumference} = ext{π} imes ext{diameter} = 3,142 imes 584 ext{ mm} \\ = 1834,634 ext{ mm} e x t C i rc u m f ere n ce = e x t π im ese x t d iam e t er = 3 , 142 im es 584 e x t mm = 1834 , 634 e x t mm With 24 spokes, each having a width of 2 mm, the part filled by spokes is:
e x t P a r t f i l l e d b y s p o k e s = 24 i m e s 2 e x t m m = 48 e x t m m ext{Part filled by spokes} = 24 imes 2 ext{ mm} = 48 ext{ mm} e x t P a r t f i ll e d b ys p o k es = 24 im es 2 e x t mm = 48 e x t mm Therefore, the remaining circumference is:
e x t R e m a i n i n g c i r c u m f e r e n c e = 1834 , 634 e x t m m − 48 e x t m m = 1786 , 634 e x t m m ext{Remaining circumference} = 1834,634 ext{ mm} - 48 ext{ mm} = 1786,634 ext{ mm} e x t R e mainin g c i rc u m f ere n ce = 1834 , 634 e x t mm − 48 e x t mm = 1786 , 634 e x t mm The distance between the spokes:
ext{Distance between spokes} = rac{1786,634}{24} ext{ mm} \\ ext{(divide by the number of spokes)} \\
= 74,44 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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To determine the gap between the wheelchair and the doorway:
Width of wheelchair and hands:
e x t W i d t h = 60 , 96 e x t c m i m e s 10 = 609 , 6 e x t m m ext{Width} = 60,96 ext{ cm} imes 10 = 609,6 ext{ mm} e x t Wi d t h = 60 , 96 e x t c m im es 10 = 609 , 6 e x t mm
The width of the doorway is 750 mm.
Calculate the gap:
ext{Gap} = rac{750 - 609,6}{2} ext{ mm}
= 70,2 ext{ mm}
Determine the total width, in metres, of the two doors. Only available for registered users.
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Calculate the total width:
= ( 80 i m e s 4 ) + ( 640 i m e s 2 ) e x t m m = 320 e x t m m + 1280 e x t m m = 1600 e x t m m = 1 , 6 e x t m = (80 imes 4) + (640 imes 2) ext{ mm} \\
= 320 ext{ mm} + 1280 ext{ mm} \\
= 1600 ext{ mm} \\
= 1,6 ext{ m} = ( 80 im es 4 ) + ( 640 im es 2 ) e x t mm = 320 e x t mm + 1280 e x t mm = 1600 e x t mm = 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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Using the formulae provided:
e = rac{(2 485 + (640 + 80 + 80 + 80 + 80 - 95 - 95 - 220)}{2} ext{ mm}
= rac{2 485 - 220}{2} ext{ mm} \
= 677,5 ext{ mm}
Calculate the total area (in mm²) of ALL the glass-panel inserts. Only available for registered users.
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For the total area:
Calculate the area for both shapes:
e x t T o t a l a r e a = ( 640 i m e s 220 ) + ( 3 , 142 i m e s 64 0 2 / 2 ) e x t m m 2 = 1600000 e x t m m 2 + 3143481 , 6 e x t m m 2 = 4743881 , 6 e x t m m 2 ext{Total area} = (640 imes 220) + (3,142 imes 640^2 / 2) ext{ mm²} \\
= 1 600 000 ext{ mm²} + 3 143 481,6 ext{ mm²} \\
= 4 743 881,6 ext{ mm²} e x t T o t a l a re a = ( 640 im es 220 ) + ( 3 , 142 im es 64 0 2 /2 ) e x t m m 2 = 1600000 e x t m m 2 + 3143481 , 6 e x t m m 2 = 4743881 , 6 e x t m m 2
Calculate (in kg) the total mass of the glass panels. Only available for registered users.
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To find the mass:
Use the formula:
e x t T o t a l m a s s = e x t T o t a l v o l u m e i m e s e x t D e n s i t y = 15985408 e x t c m 3 i m e s 2 , 5 e x t g / c m 3 = 39 , 6 e x t k g ext{Total mass} = ext{Total volume} imes ext{Density} \\
= 15 985 408 ext{ cm³} imes 2,5 ext{ g/cm³} \\
= 39,6 ext{ kg} e x t T o t a l ma ss = e x t T o t a l v o l u m e im ese x t De n s i t y = 15985408 e x t c m 3 im es 2 , 5 e x t g / c m 3 = 39 , 6 e x t k g
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