Word Problems on Algebraic Fractions
David and Doyin share #380.00 in such a way that 1/5 of David’s share is equal to 4/5 of Doyin’s. Find the shares. (Source: MM P114 Ex F Q9)
Solution
Let David’s share = V
Let Doyin’s share = Y
i.e. from the question, David’s share plus Doyin’s share = #380.00
i.e. V + Y = #380.00 — 1
Make Y the subject of the formula in equ (1)
i.e. Y = #380.00 – V — 2
1/5 of David’s share = 1/5 x V — 3
4/5 of Doyin’s share = 4/5 x Y — 4
In the question, the fraction of David’s share is equal to the fraction of Doyin’s share.
i.e. 1/5 x V = 4/5 x Y — 5
Cross multiply (CRM)
i.e. 1 x V x 5 = 5 x 4 x Y
i.e. 5V = 20Y — 6
Divide both sides (DBS) by 5
i.e. 5V/5 = 20Y/5
i.e. V = 4Y — 7
Substitute the value of V in equ (7) into equ (2)
i.e. Y = #380.00 – 4Y — 8
Collect like terms (CLT)
i.e. Y + 4Y = #380.00
i.e. 5Y = #380.00
Divide both sides (DBS) by 5
i.e. 5Y/5 = #380.00/5
Therefore, Y = #76.00
Substitute the value of Y into equ (7)
i.e. V = 4 x #76.00
Therefore, V = #304.00
Note that you can also substitute the value of Y into equ (2) to solve for the value of V.
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Data Presentation: Frequency Table
The heights of students in a class in centimetres is given below. | |||||||||||
110 | 112 | 114 | 115 | 116 | 120 | 120 | 120 | 115 | 116 | ||
111 | 112 | 117 | 118 | 119 | 120 | 119 | 117 | 114 | 113 | ||
113 | 110 | 111 | 112 | 119 | 110 | 111 | 118 | 119 | 120 | ||
(a) Prepare a frequency table for this distribution | |||||||||||
(b) Which height occurs most frequently? | |||||||||||
(c) How many students are measured? | |||||||||||
MM Pg 193 Q2 |
(a) Solution | |||
Height (cm) | Freq (f) | ||
110 | 3 | ||
111 | 3 | ||
112 | 3 | ||
113 | 2 | ||
114 | 2 | ||
115 | 2 | ||
116 | 2 | ||
117 | 2 | ||
118 | 2 | ||
119 | 4 | ||
120 | 5 | ||
Total Freq | 30 | ||
(b) Solution | |||
Height 120 cm |
(c) Solution | |||
30 students |
Data Presentation: The Pie Chart
The expenditure pattern of Mr. John working in an oil company per month is as follows. | |||||||||||
MM Page 196 Q6 | |||||||||||
Food | #40,000.00 | ||||||||||
Rent | #10,000.00 | ||||||||||
Transport | #60,000.00 | ||||||||||
Savings | #50,000.00 | ||||||||||
Others | #20,000.00 |
Solution | ||||
Item | Expenditure (N:K) | Angle (⁰) | ||
Food | 40,000.00 | 80 | ||
Rent | 10,000.00 | 20 | ||
Transport | 60,000.00 | 120 | ||
Savings | 50,000.00 | 100 | ||
Others | 20,000.00 | 40 | ||
Total | 180,000.00 | 360 |
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MM P206 Ex B3 Q20 | ||
A box contains 40 oranges, 12 of which are unripe. An orange is picked at random. What is the probability that it is: (a) ripe? (b) unripe? | ||
Total number of possible outcome | = | 40 |
Unripe oranges | = | 12 |
Ripe oranges | = | 40 – 12 |
(a) number of expected oucome (ripe) | = | 28 |
∴ Prob (of ripe) | = | 28/40 |
= | 7/10 | |
(b) number of expected oucome (unripe) | = | 12 |
∴ Prob (of unripe) | = | 12/40 |
= | 3/10 |
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