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Parents Pricing Home NSC Mathematical Literacy Reading information from a graph The Bambanani Crèche in Bethlehem bought the cubic blocks below from an auction
The Bambanani Crèche in Bethlehem bought the cubic blocks below from an auction - NSC Mathematical Literacy - Question 3 - 2019 - Paper 1 Question 3
View full question The Bambanani Crèche in Bethlehem bought the cubic blocks below from an auction. They have a side length of 45 cm. On two opposite sides of the block is a circular h... show full transcript
View marking scheme Worked Solution & Example Answer:The Bambanani Crèche in Bethlehem bought the cubic blocks below from an auction - NSC Mathematical Literacy - Question 3 - 2019 - Paper 1
3.1.1 (a) Calculate the area (in cm²) of ONE of the faces of the block that does not have a circular hole. Only available for registered users.
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To calculate the area of one face of the cube, we can apply the formula:
A r e a = s i d e i m e s s i d e = 45 e x t c m i m e s 45 e x t c m = 2025 e x t c m 2 Area = side imes side = 45 ext{ cm} imes 45 ext{ cm} = 2025 ext{ cm}² A re a = s i d e im ess i d e = 45 e x t c m im es 45 e x t c m = 2025 e x t c m 2 Thus, the area of one face without a circular hole is 2025 cm² .
3.1.1 (b) Show that the total surface area of the faces with circular holes = 11 582,869 cm². Only available for registered users.
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First, calculate the area of the circular holes using the formula:
A r e a c i r c l e = e x t π i m e s r 2 = 3 , 142 i m e s ( 9 , 5 e x t c m ) 2 = 283 , 655 e x t c m 2 Area_{circle} = ext{π} imes r^2 = 3,142 imes (9,5 ext{ cm})^2 = 283,655 ext{ cm}² A re a c i rc l e = e x t π im es r 2 = 3 , 142 im es ( 9 , 5 e x t c m ) 2 = 283 , 655 e x t c m 2 The area for one face with a hole includes the area of the face without the hole minus the area of the hole:
A r e a f a c e e x t w i t h h o l e = A r e a f a c e − A r e a h o l e = 2025 e x t c m 2 − 283 , 655 e x t c m 2 = 1741 , 345 e x t c m 2 Area_{face ext{ with hole}} = Area_{face} - Area_{hole} = 2025 ext{ cm}² - 283,655 ext{ cm}² = 1741,345 ext{ cm}² A re a f a cee x t w i t hh o l e = A re a f a ce − A re a h o l e = 2025 e x t c m 2 − 283 , 655 e x t c m 2 = 1741 , 345 e x t c m 2 For the two faces with holes, multiply by 2:
T o t a l e x t S u r f a c e A r e a = 2 i m e s 1741 , 345 e x t c m 2 = 3482 , 69 e x t c m 2 Total ext{ Surface Area} = 2 imes 1741,345 ext{ cm}² = 3482,69 ext{ cm}² T o t a l e x t S u r f a ce A re a = 2 im es 1741 , 345 e x t c m 2 = 3482 , 69 e x t c m 2 Adding the area of the remaining faces:
T o t a l = 3482 , 69 + ( 4 − 2 ) i m e s 2025 = 11582 , 869 e x t c m 2 Total = 3482,69 + (4 - 2) imes 2025 = 11582,869 ext{ cm}² T o t a l = 3482 , 69 + ( 4 − 2 ) im es 2025 = 11582 , 869 e x t c m 2 Hence, the total surface area of the blocks with circular holes is 11 582,869 cm² .
3.1.1 (c) Calculate the total amount of paint needed to paint 12 chairs with ONE coat of paint. Only available for registered users.
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First, calculate the total surface area for all chairs:
T o t a l e x t S u r f a c e A r e a f o r 12 c h a i r s = 12 i m e s 11582 , 869 e x t c m 2 = 139994 , 428 e x t c m 2 Total ext{ Surface Area for 12 chairs} = 12 imes 11 582,869 ext{ cm}² = 139 994,428 ext{ cm}² T o t a l e x t S u r f a ce A re a f or 12 c hai rs = 12 im es 11582 , 869 e x t c m 2 = 139994 , 428 e x t c m 2 Next, find how much paint is needed:
Amount ext{ of paint} = rac{139 994,428 ext{ cm}²}{15 ext{ cm}²} imes 1,8 ext{ m²} = 1 389 944,28 ext{ cm²}
To convert cm² to litres, remember that 1 ℓ = 1000 cm³.
Rounding off yields approximately 17 ℓ of paint required.
3.1.2 (a) Write down the diameter of the tin. Only available for registered users.
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The diameter of the tin can be calculated using the radius:
D i a m e t e r = 2 i m e s r = 2 i m e s 7 e x t c m = 14 e x t c m . Diameter = 2 imes r = 2 imes 7 ext{ cm} = 14 ext{ cm}. D iam e t er = 2 im esr = 2 im es 7 e x t c m = 14 e x t c m . Thus, the diameter of the tin is 14 cm .
3.1.2 (b) Calculate the height of the paint in the tin. Only available for registered users.
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Using the volume formula:
Volume = ext{π} imes (r)^2 imes height
ightarrow height = rac{Volume}{ ext{π} imes (r)^2}
Substituting in:
height = rac{5000 ext{ cm}³}{3,142 imes (7 ext{ cm})^2} = rac{5000}{3,142 imes 49}
ightarrow height = rac{5000}{153,938}
ightarrow height ext{ is approximately } 32,48 ext{ cm.}
Hence, the height of the paint in the tin is approximately 32,48 cm .
3.2.1 Determine if a 37-year-old man with a WHR of 0,95 has a moderate or a high risk of obesity. Only available for registered users.
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According to the provided classification:
A WHR of 0,95 falls into the category of moderate risk for men aged 20-39 years (as it is between 0,84 and 0,96).
Thus, the man has a moderate risk of obesity .
3.2.2 Calculate this man's WHR. Only available for registered users.
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Using the formula for WHR:
WHR = rac{waist ext{ measurement}}{hip ext{ measurement}} = rac{105 ext{ cm}}{92 ext{ cm}} = 1,141.
Hence, the man's WHR is 1,141 .
3.2.3 (a) State ONE possible age group of this woman. Only available for registered users.
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The age group could be 40 to 49 years old or 50 to 59 years old , as she fits the profile of a moderate risk with a WHR of 0,7826.
3.2.3 (b) Calculate the woman's hip measurement. Only available for registered users.
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Using the WHR formula:
0,7826 = rac{72 ext{ cm}}{hip ext{ measurement}}
Rearranging gives:
hip ext{ measurement} = rac{72}{0,7826}
ightarrow hip ext{ measurement} ext{ is approximately } 91,57 ext{ cm.}
Rounding off yields a hip measurement of 92 cm .
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