QMI1500
Elementary Quantitative Methods
Department of Decision Sciences
Exam Paper 2 May/June 2018
Question 1
Simplify the following fraction:
8
9
7
15
57
[1] 1
63
19
[2] 1
21
56
[3] 1
135
23
[4] 1
16
Answer:
8 7
÷
9 15
,Apply the division rule:
8 15
×
9 7
Reduce the numbers with the greatest common factor 3:
8 5
×
3 7
Multiply The fractions:
40
21
Convert the improper fraction to a mixed fraction.
40 divided by 21 is 1 with a remainder of (40 − 21 = 19).
19
1
21
NB: Things to remember:
• If you don’t find your answer as one of the options, this means you need to convert your answer
to a manner in which it will be one of the option. Or simplify it further.
• You are allowed to use a programmable pocket calculator (e.g. Casio scientific calculator) which
can save a lot of time on numerical questions such as these.
Question 2
Tumelo bought a cellphone for R3 000. He also bought airtime costing 1% of the cellphone price. At
the pay point the cashier gave him a 20% discount on the total bill. What is the amount that he had
to pay?
[1] R2 370
[2] R2 376
[3] R2 424
[4] R2 640
Answer:
What is this question trying to test?
Percentages and proportions
Cellphone + airtime
𝑅3 000 + 1% 𝑜𝑓 𝑅3 000
𝑅3 000 + 30
𝑅3 030
Discount = 20% 𝑜𝑓 𝑅3 030
= 𝑅606
Amount paid =R3 030-R606
= 𝑅2 424
, Question 3
If a building has eight entrances and sixteen exits (including emergency exits) in how many ways can
a person enter and leave the building?
[1] 2 ways
[2] 128 ways
[3] 12870 ways
[4] 518 918 400 ways
Answer:
What is this question trying to test?
Permutations and combinations.
Think of it this way:
A person can enter in eight ways AND can exit in sixteen ways. Therefore,
8 × 16 = 128 𝑤𝑎𝑦𝑠
Therefore, a person can enter and exit the building in 128 different ways.
Question 4
Determine the value of 𝑥 that satisfies
3 15
=
𝑥 14
21
[1] 1
24
5
[2]
24
4
[3] 4
5
8
[4]
15
Answer:
What is this question trying to test?
Fractions
3 15
=
𝑥 14
72 = 15𝑥
72
𝑥 = 15
4
= 45
Elementary Quantitative Methods
Department of Decision Sciences
Exam Paper 2 May/June 2018
Question 1
Simplify the following fraction:
8
9
7
15
57
[1] 1
63
19
[2] 1
21
56
[3] 1
135
23
[4] 1
16
Answer:
8 7
÷
9 15
,Apply the division rule:
8 15
×
9 7
Reduce the numbers with the greatest common factor 3:
8 5
×
3 7
Multiply The fractions:
40
21
Convert the improper fraction to a mixed fraction.
40 divided by 21 is 1 with a remainder of (40 − 21 = 19).
19
1
21
NB: Things to remember:
• If you don’t find your answer as one of the options, this means you need to convert your answer
to a manner in which it will be one of the option. Or simplify it further.
• You are allowed to use a programmable pocket calculator (e.g. Casio scientific calculator) which
can save a lot of time on numerical questions such as these.
Question 2
Tumelo bought a cellphone for R3 000. He also bought airtime costing 1% of the cellphone price. At
the pay point the cashier gave him a 20% discount on the total bill. What is the amount that he had
to pay?
[1] R2 370
[2] R2 376
[3] R2 424
[4] R2 640
Answer:
What is this question trying to test?
Percentages and proportions
Cellphone + airtime
𝑅3 000 + 1% 𝑜𝑓 𝑅3 000
𝑅3 000 + 30
𝑅3 030
Discount = 20% 𝑜𝑓 𝑅3 030
= 𝑅606
Amount paid =R3 030-R606
= 𝑅2 424
, Question 3
If a building has eight entrances and sixteen exits (including emergency exits) in how many ways can
a person enter and leave the building?
[1] 2 ways
[2] 128 ways
[3] 12870 ways
[4] 518 918 400 ways
Answer:
What is this question trying to test?
Permutations and combinations.
Think of it this way:
A person can enter in eight ways AND can exit in sixteen ways. Therefore,
8 × 16 = 128 𝑤𝑎𝑦𝑠
Therefore, a person can enter and exit the building in 128 different ways.
Question 4
Determine the value of 𝑥 that satisfies
3 15
=
𝑥 14
21
[1] 1
24
5
[2]
24
4
[3] 4
5
8
[4]
15
Answer:
What is this question trying to test?
Fractions
3 15
=
𝑥 14
72 = 15𝑥
72
𝑥 = 15
4
= 45
