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.
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
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 |
Work in Progress
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 |
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress
Work in Progress