JS2 Mathematics 3rd Term

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)

NEW! Exam Questions. Click on
SS2 Biology 2nd Term to get started.

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.
110112114115116120120120115116  
111112117118119120119117114113  
113110111112119110111118119120  
(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)
 
 1103 
 1113 
 1123 
 1132 
 1142 
 1152 
 1162 
 1172 
 1182 
 1194 
 1205 
 Total Freq30 
    
 (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)
 
 Food40,000.0080 
 Rent10,000.0020 
 Transport60,000.00120 
 Savings50,000.00100 
 Others20,000.0040 
 Total180,000.00360 

STATISTICS PIE CHART

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

Books & References

error: Content is protected !!