Rate, Ratio, Proportion and Percentages Comprehensive Revision

Fundamentals of Rates and Consumptions

• Hourly Wage Calculation: Rate of payment is determined by dividing total wages by total time. For example, if Otieno's wage is $sh.\,240$ per eight-hour working day, the hourly rate is calculation is expressed as:   $\text{Rate} = \frac{240}{8} = sh.\,30\text{ per hour}$

• Daily Consumption Rate: Calculating the rate of consumption per unit of time. If twelve bags of beans are consumed over a duration of $120\,days$, the consumption rate is:   $\text{Rate} = \frac{12}{120} = 0.1\text{ bags per day}$

• Fuel Consumption and Distance: Distance and fuel usage follow a direct variation. If a car travels $40\,km$ on $5\,litres$ of petrol, the distance covered using $12\,litres$ is calculated using the rate per litre:   $\text{Distance} = \frac{40}{5} \times 12 = 8 \times 12 = 96\,km$

Ratio Operations and Comparisons

• Consolidating Ratios: To find the ratio between two endpoints ($a:c$) when given intermediate ratios ($a:b$ and $b:c$):   - Given: $a:b = 3:4$ and $b:c = 5:7$   - To find $a:c$, normalize $b$ to a common multiple or multiply the ratios: $\frac{a}{b} \times \frac{b}{c} = \frac{3}{4} \times \frac{5}{7} = \frac{15}{28}$

• Increasing/Decreasing Quantities by Ratios:   - To increase $20$ in the ratio $5:4$, calculate: $20 \times \frac{5}{4} = 25$   - To decrease $45$ in the ratio $7:9$, calculate: $45 \times \frac{7}{9} = 35$

• Compound Geometric Reductions: If a photograph is reduced in size multiple times:   - Newspaper reduction: $3:5$   - Textbook reduction: $4:5$   - The final ratio of newspaper size to textbook size is found by taking the ratio of the reduction factors:     $\text{Ratio} = \frac{3}{5} : \frac{3}{5} \times \frac{4}{5} \rightarrow 1 : \frac{4}{5} \text{ (or } 5:4)$

• Comparing Ratios: To determine which ratio ($2:3$ or $4:5$) is greater, convert them to percentages or common denominators:   - $2:3 = \frac{2}{3} \approx 66.67\%$   - $4:5 = \frac{4}{5} = 80\%$   - Therefore, $4:5$ is greater.

Proportional Sharing and Division

• Dividing a Total Amount into Ratios:   - For a $72\text{-hectare}$ farm shared among three sons in the ratio $2:3:4$:     - Total units: $2 + 3 + 4 = 9$     - Share 1: $\frac{2}{9} \times 72 = 16\,hectares$     - Share 2: $\frac{3}{9} \times 72 = 24\,hectares$     - Share 3: $\frac{4}{9} \times 72 = 32\,hectares$

• Ratio Division with Fractions: Dividing $100\,cm^3$ in the ratio $\frac{1}{4} : \frac{1}{2} : \frac{1}{5}$:   - Convert fractions to a common denominator (e.g., $20$): $5:10:4$   - Total parts: $5 + 10 + 4 = 19$   - Parts: $\frac{5}{19} \times 100 \approx 26$, $\frac{10}{19} \times 100 \approx 53$, and $\frac{4}{19} \times 100 \approx 21$

Percentages and Financial Mathematics

• Conversions and Increases:   - To convert a decimal like $0.67$ into a percentage: $0.67 \times 100 = 67\%$   - Calculating a pay rise: If a man earns $sh.\,4800$ and receives a $25\%$ rise:     $\text{New Salary} = 4800 \times (1 + 0.25) = 4800 \times 1.25 = sh.\,6000$

• Harvest Statistics: If a farmer harvests $250$ bags of maize and sells $200$ bags, the percentage sold is:   $\frac{200}{250} \times 100 = 80\%$

• Geometry and Percentages: If length increases by $20\%$ and width decreases by $10\%$   - New Length = $1.2L$   - New Width = $0.9W$   - New Area = $1.2 \times 0.9 = 1.08\,LW$   - Percentage change in area = $(1.08 - 1) \times 100 = 8\%\text{ increase}$

• Investment Returns: A man invested $sh.\,36,000$ in Company P ($11.25\%$ dividend) and Company Q ($10.5\%$ dividend), yielding a total return of $10.75\%$. Using alligation or simultaneous equations:   - Let amount in P be $x$: $0.1125x + 0.105(36000 - x) = 0.1075(36000)$

Labor, Productivity, and Time Rates

• Direct and Inverse Labor Proportions:   - Cottage Construction: If $5\,men$ erect $2\,cottages$ in $21\,days$, to construct $6\,cottages$ in the same time:     - Cottages increase by a factor of 3 ($6 \div 2 = 3$).     - Men required: $5 \times 3 = 15\,men$.     - Additional men needed: $15 - 5 = 10\,men$.

• Food Rationing and Soldiers: Inverse relationship between consumers and time.   - Initial scenario: $10\,soldiers$, $7\,days$ of food. After $4\,soldiers$ desert, $6\,soldiers$ remain.   - If the desertion happened at the start: $Days = \frac{10 \times 7}{6}$. If it happens after certain days, one must calculate the remaining "soldier-days" of food.

• Refugee Food Rations: $10,000\,refugees$ for $35\,days$. After $5\,days$, $2500$ more arrive ($12,500$ total), and all receive half rations.   - Remaining work-days at full ration: $10000 \times 30 = 300,000\,units$   - New burn rate (half ration): $12500 \times 0.5 = 6250\,units/day$   - Additional days: $\frac{300000}{6250} = 48\,days$

• Simultaneous Working (Tap Problems):   - Tap A fills in $6\,hr$, B in $8\,hr$, C empties in $10\,hr$. Combined rate per hour:     $\text{Rate} = \frac{1}{6} + \frac{1}{8} - \frac{1}{10} = \frac{20+15-12}{120} = \frac{23}{120}$   - Time to fill empty tank: $\frac{120}{23} \approx 5.217\,hours$

Mixture, Alligation, and Density

• Blending Commodities (Rice/Tea/Coffee):   - Grade A Rice ($sh.\,80/kg$) mixed with Grade B ($sh.\,60/kg$) to produce a blend at $sh.\,75/kg$:     - Ratio $x:y$ where $80x + 60y = 75(x + y)$     - $5x = 15y \rightarrow x:y = 3:1$

• Density and Mass Mixtures: Liquid mixture of Water (density $1\,g/cm^3$) and Ethanol (density $1.2\,g/cm^3$) in ratio $3:1$ volume.   - For $2.5\,litres$ ($2500\,cm^3$) total volume:     - Water volume: $\frac{3}{4} \times 2500 = 1875\,cm^3$     - Ethanol volume: $\frac{1}{4} \times 2500 = 625\,cm^3$     - Total mass: $(1875 \times 1) + (625 \times 1.2) = 1875 + 750 = 2625\,grams$

• Chemical Concentrations: Mixing $45\%$ concentration acid with $25\%$ to get a $30\%$ solution:   - Ratio $\frac{45-30}{30-25} = \frac{15}{5} = 3:1$ (Inverse of distance from mean).

Construction and Mass Calculations

• Open Market Floor Slabs:   - Area = $800\,m^2$, Thickness = $200\,mm$ ($0.2\,m$).   - Volume of slab: $800 \times 0.2 = 160\,m^3$.   - Mass of dry slab: Volume ($160\,m^3$) $\times$ Density ($200\,kg/m^3$) = $32,000\,kg$.   - Materials in ratio Sand:Cement:Ballast = $3:2:3$.     - Total mass units: $3 + 2 + 3 = 8$     - Mass of cement: $\frac{2}{8} \times 32000 = 8,000\,kg$     - Number of cement bags (at $50\,kg/bag$): $\frac{8000}{50} = 160\,bags$     - Mass of ballast: $\frac{3}{8} \times 32000 = 12,000\,kg$     - Lorry capacity (10 tonnes = $10,000\,kg$): Lorries required = $\frac{12000}{10000} = 1.2$, which implies $2\,lorries$ are necessary.

Profit Sharing and Business Partnerships

• Complex Profit Distribution (Bela, Joan, Trinity):   - Contributions: Bela ($sh.\,112,000$), Joan ($sh.\,128,000$), Trinity ($sh.\,210,000$).   - Total Contribution: $112,000 + 128,000 + 210,000 = sh.\,450,000$.   - Profit: $sh.\,1.35\,million$.     - Retained for business ($40\%$): $0.40 \times 1,350,000 = sh.\,540,000$     - Shared equally ($30\%$ of total): $0.30 \times 1,350,000 = 405,000$. Per person: $405,000 / 3 = sh.\,135,000$.     - Shared by contribution ($30\%$ of total): $0.30 \times 1,350,000 = 405,000$. Shared in ratio $112:128:210$.

• Business Re-investment (Asha, Nangila, Cherop):   - Contributed $60k, 85k, 105k$ respectively. Profit: $225,000$.   - First reduction ($25\%$ back to business): $225,000 \times 0.75 = 168,750$   - Second reduction ($40\%$ for tax/insurance from remaining): $168,750 \times 0.60 = 101,250$   - Final amount shared by contribution ratio.

• Matatu Business Operations (Mutua, Muthoka, Mwikali):   - Contributions: $600,000, 400,000, 800,000$.   - Income: $14\,passengers \times 250\,sh. \times 4\,trips\text{ (2 round trips)} = 14,000\,per\,day$.   - Daily Net Profit: $14,000 - 6,000\text{ (costs)} = sh.\,8,000$.   - Monthly calculation: $8,000 \times 25\,days - 10,000\text{ (service)} = 190,000$.

Average and Mean Shifts

• Impact of Individual Changes on Mean: Form IV class of $30\,students$ with original mean $41$.   - Total original marks: $30 \times 41 = 1230$.   - New mean: $42.5$, so new total: $30 \times 42.5 = 1275$.   - Increase in marks: $1275 - 1230 = 45\,marks$.   - Marks added to Amina, Nduku, and Karimi in ratio $2:3:4$.     - Total ratio parts: $2+3+4 = 9$     - Marks for Karimi: $\frac{4}{9} \times 45 = 20\,marks$     - Marks for Nduku: $\frac{3}{9} \times 45 = 15\,marks$     - Difference: $20 - 15 = 5\,more\,marks$.

Volume and Leakage Rates

• Cylindrical Tank Leakage:   - Diameter: $70\,cm$ ($radius = 35\,cm$).   - Volume change ($V = \pi r^2 h$): For a fall of $h = 20\,cm$:     $V = \frac{22}{7} \times 35^2 \times 20 = 77,000\,cm^3 = 77\,litres$   - Rate of loss: $10\,litres/hour$.   - Time needed: $\frac{77}{10} = 7.7\,hours$.

• Continuous Leakage Problems (Scenario 13/23):   - Rate: $11\,cm^3$ every $5\,seconds$.   - Capacity lost per hour: $\frac{11}{5} \times 3600 = 7920\,cm^3/hour$.   - Draining into a cylindrical tank ($r=30, h=30$):     - Cylinder Volume: $3.142 \times 30^2 \times 30 = 84,834\,cm^3$     - Time to fill: $\frac{84834}{7920} \approx 10.71\,hours$.