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BNU1501 Assignment 2 Semester 1 2022
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BNU1501 Assignment 2 Semester 1 2022 - All Answers are Correct
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BNU1501/202/1/2022Tutorial Letter 202/1/2022
Basic Numeracy
BNU1501
Semester 1
Department of Decision Sciences
Important information:
Solutions to Assignment 02
Bar code
, BNU1501/202/2022
ASSIGNMENT 02: Solutions
Answers
1. [3] 6. [3] 11. [4] 16. [2]
2. [2] 7. [2] 12. [1] 17. [3]
3. [4] 8. [1] 13. [1] 18. [2]
4. [1] 9. [3] 14. [2] 19. [1]
5. [4] 10. [2] 15. [3] 20. [4]
Explanations
Question 1
6 c
1 2 c
B
1 8 c m
The perimeter of the shaded figure is
2πr
12 + 8,1 + 16,1 + 12 + cm
2
= (48,200 + π(9)) cm
= 76,4743 cm
≈ 76,47 cm
The correct option is [3].
2
, BNU1501/202/2022
Question 2
The area of the shaded part of the figure in question 1, is
area of triangle EAD + area of rectangle ABCD − area of semi-circle
base × height πr2
= + (length × width) −
2 2
18 × 6 π.92
= + (18 × 12) −
2 2
254,4690
= 54 + 216 −
2
= 142,7655
≈ 142,77
Thus the shaded area is 142,77 cm2.
The correct option is [2].
Question 3
1 0 c
The volume of the rectangular box = ℓ × b × h = (40 × 30 × 10) cm3 = 12 000 cm3
The volume of one tin = πr2h = (π × 5 × 5 × 10) cm3 = 785,3982 cm3
The amount of sand necessary is therefore
(12 000 − 785,3982 × 12) cm3
= 2 575,22 cm3
= 2,58 litres because 1ℓ = 1 000 cm3
The correct option is [4].
3
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