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QMI1500 ASSIGNMENT FEEDBACK

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QMI1500 Assignment 1 & 2 Questions and Answers 2017

Institution
Module

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QMI1500
Assignment 1 & 2
Questions and Answers 2017

, QMI1500/201/1


Solutions to compulsory Assignment 01

Question 1

2 2
x3 × x5 2 2 2

− 24
= x3+5+4
x
2×5×4+2×3×4+2×3×5
= x 60
40+24+30
= x 60
94
= x 60
17
= x1 30


Answer: Option 1

Question 2
The easiest way to get the solution for this question is to apply the concept of worker-days. 8 workers
require 24 days to complete the job, therefore (8 × 24), i.e. 192 worker-days are required to complete
the job.
If 18 workers are involved to complete a 192 worker-days job, then the number of days required to
complete the paint job, will be
192 2
= 10
18 3
Answer: Option 2

Question 3
There are 4 parts to the meal, each with several choices.
Sandwich: 4 choices
Soup: 3 choices
Dessert: 2 choices
Drink: 5 choices
Thus the number of possible meals is

4 × 3 × 2 × 5 = 120

Answer: Option 3

Question 4
You can get 0, 1, 2, 3, 4 or 5 toppings from the 11 choices. Therefore the number of possible choices
for the first pizza is as follows:

11 C0 + 11 C1 + 11 C2 + 11 C3 + 11 C4 + 11 C5 = 1 + 11 + 55 + 165 + 330 + 462 = 1 024



2

, QMI1500/201/1


The same applies for the second pizza. Thus the number of choices available for two pizzas is
1 024 × 1 024 = 1 048 576

Answer: Option 4

Question 5

Z A B Y




D P X C

In order to answer this question, we first find the area of the trapezium ABCD in terms of x. To
find the area, label the diagram as above.
Find the areas of ZYCD, ZAD and BYC.
But Area of ZAD=Area of APD and Area of BYC=Area of BCX


Area of ABCD = Area of ZY CD − Area of ZAD − Area of BY C
= Area of ZY CD − Area of AP D − Area of BY X
Hence
1
66 = [(3x + 1) × 6] − [6 × (3x + 1) − (x + 5)]
2
66 = 18x + 6 − 9x − 3 + 3x + 15
66 = 12x + 18
48 = 12x
4 = x

Alternative

A B




D P X C


3

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