Class 6 Mathematics - Ch 3: Mastery Guide: Ratios, Rates, Unitary Method, Continued Ratios & Percentage Applications (FBISE)
Instructional Guide: Unit 3 Ratio, Rate and Percentage
- Understand ratio as a quantitative comparison of two quantities of the same kind ($a:b$).
- Simplify ratios to their lowest terms using HCF and convert mixed units ($1.5\text{ h} : 30\text{ min}$).
- Compute rates and apply the Unitary Method to find unit costs, speeds, and fuel consumptions.
- Construct and simplify continued ratios ($x:y:z$) using common elements and solve proportional sharing problems.
- Interpret percentage as a special fraction with denominator $100$ using $100$-grid models.
- Calculate percentage increases, percentage decreases, and compare ratios/percentages in real life.
- Grade 5 Fractions: Equivalent fractions, reducing fractions to simplest form.
- Grade 5 Unitary Method: Finding single unit cost by division, total by multiplication.
- Decimals & Metric Units: Unit conversion ($1\text{ km} = 1000\text{ m}$, $1\text{ kg} = 1000\text{ g}$, $1\text{ h} = 60\text{ min}$).
- Misconception 1 (Units): Trying to write ratios with units (e.g. $3\text{ kg} : 4\text{ kg}$). Ratios are unitless pure numbers!
- Misconception 2 (Unit Matching): Forgetting to convert different units before forming ratios ($1.5\text{ h} : 30\text{ min} \neq 1.5 : 30$).
- Misconception 3 (Order): Inverting antecedent and consequent ($a:b \neq b:a$).
- Misconception 4 (Percentage Base): Dividing by the new quantity instead of the original base quantity for percentage increase/decrease.
💡 Kid-Friendly Rhymes & Visual Memory Anchors
"First comes first, second comes last,
Make the units match super fast!
Drop the units, divide by HCF,
The simplest ratio is all that's left!"
"Step 1: Divide down to find just ONE.
Step 2: Multiply up till the job is done!"
$\text{Many} \xrightarrow{\div} 1 \xrightarrow{\times} \text{Target}$
"Find the middle friend who's in both teams,
Multiply to make their value match your dreams!"
$x:y$ and $y:z \implies$ Make $y$ equal!
"Percent means Per Hundred, clear and bright,
Put $\frac{\text{Part}}{\text{Whole}} \times 100$ and get it right!"
🌍 Real-World Connections: Why Ratios & Percentages Matter
- Cooking & Baking: Mixing rice and water in the ratio $1 : 2$, or flour to sugar in $3 : 1$.
- Travel & Automobile Mileage: Fuel consumption rates like $18\text{ km/litre}$ or highway speed $100\text{ km/h}$.
- Finance & Shopping: $20\text{%}$ off discount sales, $12\text{%}$ annual salary raises, bank interest rates.
- Technology & Telecom: Mobile phone internet data usage ($40\text{%}$ of $35\text{ GB}$ used) and call minutes.
Comprehensive Conceptual Foundations
1. Understanding Ratios & Simplest Form
A Ratio is a mathematical comparison of two quantities of the same kind and measured in the same units by division.
- Notation: The ratio of $a$ to $b$ is written as $a : b$ or as a fraction $\frac{a}{b}$ (where $b \neq 0$).
- Terms: In $a : b$, the first term $a$ is called the antecedent, and the second term $b$ is called the consequent.
- Unitless: A ratio has no units because the units in the numerator and denominator cancel out. For example, $3\text{ kg} : 4\text{ kg} = 3 : 4$.
- Simplest Form: A ratio $a : b$ is in lowest/simplest form when $\text{HCF}(a, b) = 1$.
2. Rates & The Unitary Method
A Rate compares two quantities having different units. For example, speed is measured in $\text{km/hour}$, earnings in $\text{Rs./day}$, and typing speed in $\text{words/minute}$.
The Unitary Method:
- Step 1: Find the value of one single unit by dividing the total given value by the number of units: $\text{Unit Rate} = \frac{\text{Total Value}}{\text{Given Units}}$.
- Step 2: Find the value of the required quantity by multiplying the unit rate by the desired number of units: $\text{Required Value} = \text{Unit Rate} \times \text{Required Units}$.
3. Continued Ratios & Proportional Sharing
When we compare three or more quantities simultaneously, the ratio is called a Continued Ratio, written as $x : y : z$.
Method to Combine Two Ratios ($x : y$ and $y : z$):
- Identify the common term (usually $y$).
- Find the $\text{LCM}$ of the values of the common term in both ratios.
- Multiply both ratios by appropriate factors so the middle term becomes identical.
- Write the combined ratio: $x : y : z$.
- Dividing a Total Amount into Given Ratio ($a : b : c$): $$\text{Sum of Ratio Terms} = a + b + c$$ $$\text{Share of First Person} = \frac{a}{\text{Sum of Ratios}} \times \text{Total Amount}$$
4. Percentage ($100$-Grid Models & Conversions)
The word Percent comes from the Latin per centum, meaning "out of one hundred". It is denoted by the symbol $\%$.
- $x\%$ means $\frac{x}{100}$ or $x$ parts out of $100$.
- Converting Fraction to Percentage: Multiply by $100\%$: $\frac{a}{b} \times 100\%$.
- Converting Decimal to Percentage: Shift decimal point $2$ places to the right: $0.75 = 75\%$.
- Expressing One Quantity as Percentage of Another: $$\text{Percentage} = \frac{\text{Given Part}}{\text{Total Whole}} \times 100\%$$
- Percentage Increase / Decrease: $$\text{Percentage Change} = \frac{\text{Actual Increase or Decrease}}{\text{Original Amount}} \times 100\%$$
Complete Textbook Exercises & Detailed Step-by-Step Solutions
Q1. Express each of the following in the lowest form:
(i) $12 : 72$
Divide both terms by $\text{HCF}(12, 72) = 12$: $\frac{12 \div 12}{72 \div 12} = \frac{1}{6} = \mathbf{1 : 6}$.
(ii) $0.4 : 20$
Multiply both terms by $10$ to remove decimal: $4 : 200$. Divide by $4$: $\frac{4 \div 4}{200 \div 4} = \mathbf{1 : 50}$.
(iii) $\frac{3}{2} : \frac{1}{3}$
Multiply both terms by $\text{LCM}(2, 3) = 6$: $\left(\frac{3}{2} \times 6\right) : \left(\frac{1}{3} \times 6\right) = 9 : 2 = \mathbf{9 : 2}$.
(iv) $3 : 2\frac{1}{2}$
Convert mixed fraction: $2\frac{1}{2} = \frac{5}{2}$. Multiply both by $2$: $(3 \times 2) : 5 = \mathbf{6 : 5}$.
(v) $0.7 : 70$
Multiply by $10$: $7 : 700$. Divide by $7$: $\mathbf{1 : 100}$.
(vi) $15 : 125$
Divide both by $\text{HCF}(15, 125) = 5$: $\frac{15 \div 5}{125 \div 5} = \mathbf{3 : 25}$.
Q2. Express each as a ratio of the first quantity to the second in its lowest form:
(i) $\text{Rs. } 350, \text{Rs. } 425$
Ratio = $350 : 425$. Divide by $25$: $\frac{350 \div 25}{425 \div 25} = \mathbf{14 : 17}$.
(ii) $210^\circ, 360^\circ$
Ratio = $210 : 360$. Divide by $30$: $\frac{210 \div 30}{360 \div 30} = \mathbf{7 : 12}$.
(iii) $3\text{ kg}, 2000\text{ g}$
Convert to same unit ($1\text{ kg} = 1000\text{ g}$): $3000\text{ g} : 2000\text{ g} = 3000 : 2000 = \mathbf{3 : 2}$.
(iv) $1200\text{ m}, 2\text{ km}$
Convert to same unit ($2\text{ km} = 2000\text{ m}$): $1200\text{ m} : 2000\text{ m} = 1200 : 2000 = \mathbf{3 : 5}$.
(v) $1.5\text{ hour}, 30\text{ min}$
Convert to minutes ($1.5\text{ h} = 1.5 \times 60 = 90\text{ min}$): $90 : 30 = \mathbf{3 : 1}$.
(vi) $6\text{ m}, 80\text{ cm}$
Convert to cm ($6\text{ m} = 600\text{ cm}$): $600 : 80 = 60 : 8 = \mathbf{15 : 2}$.
Q3. In a class of $25$ students, there are $11$ boys.
(i) How many girls are in the class?
$$\text{Number of girls} = \text{Total students} - \text{Boys} = 25 - 11 = \mathbf{14\text{ girls}}$$
(ii) What is the ratio of girls to boys?
$$\text{Ratio of girls : boys} = \mathbf{14 : 11}$$
(iii) What is the ratio of boys to girls?
$$\text{Ratio of boys : girls} = \mathbf{11 : 14}$$
Q4. Daily income of a labourer and carpenter is $\text{Rs. } 600$ and $\text{Rs. } 800$ respectively. Find the ratio between their income in lowest form.
Q5. Ayesha earned $\text{Rs. } 1800$ and spent $\text{Rs. } 1200$ in a day. Find the ratio of her income to expenditure.
Q1. Amjad earns $\text{Rs. } 2500$ in $5$ days. What is his pay for $3$ days?
$$\text{Pay for } 1\text{ day} = \frac{\text{Rs. } 2500}{5} = \text{Rs. } 500$$
$$\text{Pay for } 3\text{ days} = \text{Rs. } 500 \times 3 = \mathbf{\text{Rs. } 1500}$$
Q2. A shopkeeper buys $70$ packets of biscuits for $\text{Rs. } 2800$. How much will he have to pay if he buys $150$ such packets?
$$\text{Cost of } 1\text{ packet} = \frac{\text{Rs. } 2800}{70} = \text{Rs. } 40$$
$$\text{Cost of } 150\text{ packets} = \text{Rs. } 40 \times 150 = \mathbf{\text{Rs. } 6000}$$
Q3. An amusement park has $350$ visitors over the course of $7$ hours. At this rate, how many visitors would they expect over $15$ hours?
$$\text{Visitors per hour} = \frac{350}{7} = 50\text{ visitors/hour}$$
$$\text{Visitors in } 15\text{ hours} = 50 \times 15 = \mathbf{750\text{ visitors}}$$
Q4. A car uses $20\text{ litres}$ of petrol to travel $170\text{ km}$. How far can it travel if it has only $16\text{ litres}$ of petrol?
$$\text{Distance on } 1\text{ litre} = \frac{170}{20} = 8.5\text{ km/litre}$$
$$\text{Distance on } 16\text{ litres} = 8.5 \times 16 = \mathbf{136\text{ km}}$$
Q5. Uzair drives $232\text{ km}$ in $4$ hours. At this rate how far can he drive in $7$ hours?
$$\text{Speed (Rate)} = \frac{232}{4} = 58\text{ km/hour}$$
$$\text{Distance in } 7\text{ hours} = 58 \times 7 = \mathbf{406\text{ km}}$$
Q6. The cost of $10\text{ m}$ pipe is $\text{Rs. } 2200$. What is the cost of $22\text{ m}$ of such a pipe?
$$\text{Cost of } 1\text{ m pipe} = \frac{2200}{10} = \text{Rs. } 220$$
$$\text{Cost of } 22\text{ m pipe} = 220 \times 22 = \mathbf{\text{Rs. } 4840}$$
Q7. Cost of $10$ books is $\text{Rs. } 1640$. What is the cost of $4$ books?
$$\text{Cost of } 1\text{ book} = \frac{1640}{10} = \text{Rs. } 164$$
$$\text{Cost of } 4\text{ books} = 164 \times 4 = \mathbf{\text{Rs. } 656}$$
Q8. $\text{Rs. } 500$ is charged for $50$ units of electricity. Find the cost of $20$ units of electricity.
$$\text{Cost of } 1\text{ unit} = \frac{500}{50} = \text{Rs. } 10$$
$$\text{Cost of } 20\text{ units} = 10 \times 20 = \mathbf{\text{Rs. } 200}$$
Q9. Moeed pays total of $\text{Rs. } 60,000$ rent for three months of a flat. Find his annual rent of flat.
$$\text{Rent for } 1\text{ month} = \frac{60,000}{3} = \text{Rs. } 20,000$$
$$\text{Annual Rent (12 months)} = 20,000 \times 12 = \mathbf{\text{Rs. } 240,000}$$
Q10. Mubeen planted $600$ trees in $30$ days. How many trees will he plant in next $25$ days?
$$\text{Trees planted in } 1\text{ day} = \frac{600}{30} = 20\text{ trees/day}$$
$$\text{Trees in } 25\text{ days} = 20 \times 25 = \mathbf{500\text{ trees}}$$
Q1. Find $x : y : z$ and $x : z$ if:
(i) $x : y = 4 : 5, \quad y : z = 5 : 7$
Since $y$ has the same value ($5$) in both ratios:
$$\mathbf{x : y : z = 4 : 5 : 7}, \quad \mathbf{x : z = 4 : 7}$$
(ii) $x : y = 3 : 2, \quad y : z = 4 : 9$
Make $y$ equal (LCM of $2$ and $4$ is $4$). Multiply first ratio by $2$: $x : y = 6 : 4$.
$$\mathbf{x : y : z = 6 : 4 : 9}, \quad x : z = 6 : 9 = \mathbf{2 : 3}$$
(iii) $x : y = 6 : 7, \quad y : z = 8 : 11$
LCM of $7$ and $8$ is $56$. Multiply first by $8$ ($48 : 56$) and second by $7$ ($56 : 77$):
$$\mathbf{x : y : z = 48 : 56 : 77}, \quad \mathbf{x : z = 48 : 77}$$
(iv) $x : y = \frac{2}{3} : 1, \quad y : z = \frac{3}{4} : \frac{1}{2}$
Simplify first: $x : y = 2 : 3$. Simplify second: $y : z = 3 : 2$.
Since $y = 3$ is already equal:
$$\mathbf{x : y : z = 2 : 3 : 2}, \quad x : z = 2 : 2 = \mathbf{1 : 1}$$
(v) $x : y = \frac{2}{3} : \frac{5}{2}, \quad y : z = 1 : \frac{3}{2}$
Simplify first (multiply by $6$): $x : y = 4 : 15$. Simplify second (multiply by $2$): $y : z = 2 : 3$.
LCM of $15$ and $2$ is $30$. Multiply first by $2$ ($8 : 30$) and second by $15$ ($30 : 45$):
$$\mathbf{x : y : z = 8 : 30 : 45}, \quad x : z = 8 : 45 = \mathbf{8 : 45}$$
Q2. In a car park, ratio of black to white cars is $5 : 6$ and white to silver cars is $3 : 10$. Find ratio of black : white : silver.
$$\text{Black} : \text{White} = 5 : 6$$
$$\text{White} : \text{Silver} = 3 : 10 = 6 : 20 \quad (\text{multiply by } 2)$$
$$\mathbf{\text{Black} : \text{White} : \text{Silver} = 5 : 6 : 20}$$
Q3. The salaries of A and B are in ratio $8 : 3$. The salaries of B and C are in ratio $5 : 12$. Express in continued ratio $A : B : C$.
$$\text{LCM}(3, 5) = 15$$
$$A : B = 8 \times 5 : 3 \times 5 = 40 : 15$$
$$B : C = 5 \times 3 : 12 \times 3 = 15 : 36$$
$$\mathbf{A : B : C = 40 : 15 : 36}$$
Q4. In a farm there are $840$ cattle. Ratio of goats to sheep is $4 : 5$. Ratio of cows to sheep is $9 : 3 = 3 : 1$. Find number of goats, sheep and cows.
$$\text{Goats} : \text{Sheep} = 4 : 5$$
$$\text{Sheep} : \text{Cows} = 3 : 9 = 1 : 3 = 5 : 15 \quad (\text{since } \text{Cows}:\text{Sheep} = 9:3)$$
$$\text{Goats} : \text{Sheep} : \text{Cows} = 4 : 5 : 15$$
$$\text{Sum of ratios} = 4 + 5 + 15 = 24$$
$$\text{Goats} = \frac{4}{24} \times 840 = \mathbf{140}$$
$$\text{Sheep} = \frac{5}{24} \times 840 = \mathbf{175}$$
$$\text{Cows} = \frac{15}{24} \times 840 = \mathbf{525}$$
Q5. Find difference between largest and smallest shares when $\text{Rs. } 160$ is shared in ratio $1 : 6 : 9$.
$$\text{Sum of ratios} = 1 + 6 + 9 = 16$$
$$\text{Largest share} = \frac{9}{16} \times 160 = \text{Rs. } 90$$
$$\text{Smallest share} = \frac{1}{16} \times 160 = \text{Rs. } 10$$
$$\text{Difference} = 90 - 10 = \mathbf{\text{Rs. } 80}$$
Q6. A sum of money is divided among three people in ratio $15 : 18 : 7$. Calculate each share in $\text{Rs. } 1600$.
$$\text{Sum of ratios} = 15 + 18 + 7 = 40$$
$$\text{Share 1} = \frac{15}{40} \times 1600 = \mathbf{\text{Rs. } 600}$$
$$\text{Share 2} = \frac{18}{40} \times 1600 = \mathbf{\text{Rs. } 720}$$
$$\text{Share 3} = \frac{7}{40} \times 1600 = \mathbf{\text{Rs. } 280}$$
Q7. A sum of money is divided in ratio $3 : 5 : 9$. Calculate smallest share given largest share is $\text{Rs. } 369$.
$$9\text{ parts} = \text{Rs. } 369 \implies 1\text{ part} = \frac{369}{9} = \text{Rs. } 41$$
$$\text{Smallest share (3 parts)} = 41 \times 3 = \mathbf{\text{Rs. } 123}$$
Q8. An alloy consists of metals X, Y and Z with $X : Y = 2 : 3$ and $Y : Z = 5 : 4$. Calculate $X : Z$.
$$X : Y = 10 : 15, \quad Y : Z = 15 : 12$$
$$X : Y : Z = 10 : 15 : 12 \implies X : Z = 10 : 12 = \mathbf{5 : 6}$$
Q9. Money divided among Fiza, Hadia, Maha in ratio $13 : 12 : 7$. If Fiza gets $\text{Rs. } 390$, calculate how much Hadia gets.
$$13\text{ parts} = \text{Rs. } 390 \implies 1\text{ part} = \frac{390}{13} = \text{Rs. } 30$$
$$\text{Hadia's share (12 parts)} = 30 \times 12 = \mathbf{\text{Rs. } 360}$$
Q10. Divide $\text{Rs. } 74000$ among Akbar, Abid, Azam where $\text{Akbar} : \text{Abid} = 4 : 5$ and $\text{Abid} : \text{Azam} = 3 : 2$.
$$\text{Akbar} : \text{Abid} = 12 : 15, \quad \text{Abid} : \text{Azam} = 15 : 10$$
$$\text{Continued ratio} = 12 : 15 : 10, \quad \text{Sum of parts} = 12 + 15 + 10 = 37$$
$$\text{Akbar} = \frac{12}{37} \times 74000 = \mathbf{\text{Rs. } 24,000}$$
$$\text{Abid} = \frac{15}{37} \times 74000 = \mathbf{\text{Rs. } 30,000}$$
$$\text{Azam} = \frac{10}{37} \times 74000 = \mathbf{\text{Rs. } 20,000}$$
Q1. Complete the following statements:
(i) $9\%$ means $\mathbf{9}$ out of $100$ or $\mathbf{\frac{9}{100}}$.
(ii) $21\%$ means $\mathbf{21}$ out of $100$ or $\mathbf{\frac{21}{100}}$.
(iii) $31\%$ means $\mathbf{31}$ out of $100$ or $\mathbf{\frac{31}{100}}$.
(iv) $73\%$ means $\mathbf{73}$ out of $100$ or $\mathbf{\frac{73}{100}}$.
(v) $47\%$ means $\mathbf{47}$ out of $100$ or $\mathbf{\frac{47}{100}}$.
Q2. Study the 100-grid square diagrams and fill the percentage values:
(a) Star Pattern Grid (Total 100 squares):
Black squares = $16 \implies \frac{16}{100} = \mathbf{16\%}$
Blue squares = $36 \implies \frac{36}{100} = \mathbf{36\%}$
White squares = $48 \implies \frac{48}{100} = \mathbf{48\%}$
(b) Cross Pattern Grid:
White squares = $16 \implies \mathbf{16\%}$, Blue squares = $64 \implies \mathbf{64\%}$, Black squares = $20 \implies \mathbf{20\%}$.
Total white + blue = $80 \implies \mathbf{80\%}$.
Q3. Draw a large square divided into 100 equal small squares. Colour $15\%$ black and remaining blue. What percentage is blue?
Q4. A bar is divided into $10$ equal parts with $7$ parts shaded. Find the fraction and percentage shaded.
$$\text{Fraction shaded} = \mathbf{\frac{7}{10}}$$
$$\text{Percentage shaded} = \frac{7}{10} \times 100\% = \mathbf{70\%}$$
Q5. In a university, $220$ out of $1000$ students have science subjects.
(a) What fraction of students selected science? $\frac{220}{1000} = \mathbf{\frac{11}{50}}$
(b) How many per hundred have science subjects? $\frac{220}{10} = \mathbf{22\text{ per hundred}}$
(c) What percentage of students have science subjects? $\mathbf{22\%}$
Q1. Express:
(i) $\text{Rs. } 40$ as a percentage of $\text{Rs. } 80$: $\frac{40}{80} \times 100\% = \mathbf{50\%}$
(ii) $20\text{ km}$ as a percentage of $80\text{ km}$: $\frac{20}{80} \times 100\% = \mathbf{25\%}$
Q2. Which is greater?
(i) $\text{Rs. } 10$ out of $\text{Rs. } 50$ ($20\%$) or $\text{Rs. } 20$ out of $\text{Rs. } 80$ ($25\%)$?
$$\mathbf{\text{Rs. } 20\text{ out of }\text{Rs. } 80}\text{ is greater.}$$
(ii) $70$ out of $80$ marks ($87.5\%$) or $44$ out of $50$ marks ($88\%$)?
$$\mathbf{44\text{ marks out of } 50\text{ marks}}\text{ is greater.}$$
(iii) $16\text{ m}$ out of $60\text{ m}$ ($26.67\%$) or $5\text{ m}$ out of $20\text{ m}$ ($25\%$)?
$$\mathbf{16\text{ m out of } 60\text{ m}}\text{ is greater.}$$
Q3. Which is smaller?
(i) $10\%$ of $50$ ($5$) or $25\%$ of $100$ ($25$)? $\mathbf{10\%\text{ of } 50}$
(ii) $20\%$ of $80$ ($16$) or $12\%$ of $60$ ($7.2$)? $\mathbf{12\%\text{ of } 60}$
Q4. Increase:
(i) $\text{Rs. } 1500$ by $12\%$: $\text{Increase} = \frac{12}{100} \times 1500 = \text{Rs. } 180 \implies \text{New Amount} = 1500 + 180 = \mathbf{\text{Rs. } 1680}$
(ii) $1000\text{ m}$ by $50\%$: $\text{Increase} = 500\text{ m} \implies \text{New Length} = 1000 + 500 = \mathbf{1500\text{ m}}$
Q5. Decrease:
(i) $700\text{ kg}$ by $18\%$: $\text{Decrease} = \frac{18}{100} \times 700 = 126\text{ kg} \implies \text{New Mass} = 700 - 126 = \mathbf{574\text{ kg}}$
(ii) $\text{Rs. } 120$ by $40\%$: $\text{Decrease} = \frac{40}{100} \times 120 = \text{Rs. } 48 \implies \text{New Amount} = 120 - 48 = \mathbf{\text{Rs. } 72}$
Q6. Underground water in Islamabad has fallen from $80\text{ feet}$ to $240\text{ feet}$ in last 10 years. What percentage of water depth has fallen?
Q7. Height of mango tree is $20\text{ m}$ and banana tree is $5\text{ m}$. Express mango tree height as percentage of banana tree.
Q8. In a basket of $150$ oranges, $15$ are eaten. What percentage has been eaten? What percentage is left?
$$\text{Percentage Eaten} = \frac{15}{150} \times 100\% = \mathbf{10\%}$$
$$\text{Percentage Left} = 100\% - 10\% = \mathbf{90\%}$$
Q9. Monthly income of Arshad and Asghar is $\text{Rs. } 9500$ and $\text{Rs. } 10500$. Express Arshad's income as percentage of Asghar's income.
Q10. Mr. Farooq used $60$ off-net minutes out of $180$ and $600$ on-net minutes out of $1200$.
(i) Off-net percentage: $\frac{60}{180} \times 100\% = \frac{1}{3} \times 100\% = \mathbf{33.33\%}$
(ii) On-net percentage: $\frac{600}{1200} \times 100\% = \frac{1}{2} \times 100\% = \mathbf{50\%}$
Q11. A piece of elastic $48\text{ cm}$ long is stretched to $64\text{ cm}$. What percentage of original length is increased?
Q12. Factory worker salary is $\text{Rs. } 5000$. He receives $12\%$ increment. Find his new salary.
Q1. Multiple Choice Questions (MCQs):
(i) Ratio is the comparison of different quantities having the same...
(a) meaning (b) units (c) direction (d) length • Answer: (b) units
(ii) Which one cannot be the term of a ratio?
(a) 0 (consequent cannot be 0) / negative • Answer: (a) (or non-positive/0)
(iii) $a : b$ is equivalent to:
(a) $a \div b$ (b) $b \div a$ (c) $a - b$ (d) $b - a$ • Answer: (a) $a \div b$
(iv) Ratio between $12\text{ m}$ and $16\text{ m}$ is written as:
(a) $12\text{m} : 16\text{m}$ (b) $12 : 16$ (c) $3 : 4$ (d) $3\text{m} : 4\text{m}$ • Answer: (c) $3 : 4$
(v) The reduced form of $16 : 20$ is:
(a) $8 : 10$ (b) $4 : 5$ (c) $16 : 20$ (d) $5 : 4$ • Answer: (b) $4 : 5$
(vi) $40\%$ is equal to:
(a) $\frac{2}{5}$ (b) $\frac{3}{10}$ (c) $5$ (d) $\frac{1}{2}$ • Answer: (a) $\frac{2}{5}$
(vii) $2.72$ is equal to:
(a) $1.36\%$ (b) $272\%$ (c) $2.72\%$ (d) $27.2\%$ • Answer: (b) $272\%$
(viii) Of $\text{Rs. } 1000$, $60\%$ is spent. Find the expenditure:
(a) $\text{Rs. } 600$ (b) $\text{Rs. } 700$ (c) $\text{Rs. } 65$ (d) $\text{Rs. } 400$ • Answer: (a) $\text{Rs. } 600$
(ix) What percentage of $\text{Rs. } 25$ is $\text{Rs. } 5$?
(a) $5\%$ (b) $10\%$ (c) $20\%$ (d) $30\%$ • Answer: (c) $20\%$
(x) If $a : b = 2 : 3$ and $c : b = 4 : 3$, then $a : b : c = $
(a) $2 : 3 : 4$ (b) $4 : 3 : 2$ (c) $2 : 4 : 3$ (d) $3 : 2 : 4$ • Answer: (a) $2 : 3 : 4$
(xi) If $b : c = 1 : 2$ and $c : d = 3 : 5$, then $b : c : d = $
(a) $2 : 6 : 5$ (b) $3 : 6 : 10$ (c) $3 : 6 : 5$ (d) $3 : 2 : 10$ • Answer: (b) $3 : 6 : 10$
(xii) The continued ratio among $\text{Rs. } 20, \text{Rs. } 40, \text{Rs. } 60$ is:
(a) $2 : 4 : 6$ (b) $2 : 1 : 3$ (c) $3 : 2 : 1$ (d) $1 : 2 : 3$ • Answer: (d) $1 : 2 : 3$
Q2. Length and width of a rectangle are $6\text{ m}$ and $\frac{9}{5}\text{ m}$ respectively. Find ratio of width to length.
Q3. Hanan is $18$ years old. His elder brother Asim is $6$ years older than Hanan. Find ratio of Hanan's age to Asim's age.
Q4. Ratio of mass of sugar to soap is $5 : 8$. How many times soap is heavier than sugar?
Q5. A and B invested $45\%$ and $55\%$ in joint business. Find share of each in annual profit of $\text{Rs. } 15000$.
$$\text{Share of A} = \frac{45}{100} \times 15000 = \mathbf{\text{Rs. } 6750}$$
$$\text{Share of B} = \frac{55}{100} \times 15000 = \mathbf{\text{Rs. } 8250}$$
Q6. Divide $\text{Rs. } 176$ among Amna, Fatima and Umer such that Amna gets $2$ times Fatima and Umer gets $2\frac{1}{2}$ times Amna.
Let Fatima's share = $1$ part. Then Amna = $2$ parts. Umer = $2.5 \times 2 = 5$ parts.
$$\text{Fatima} : \text{Amna} : \text{Umer} = 1 : 2 : 5, \quad \text{Sum of parts} = 1 + 2 + 5 = 8$$
$$\text{Fatima} = \frac{1}{8} \times 176 = \mathbf{\text{Rs. } 22}$$
$$\text{Amna} = \frac{2}{8} \times 176 = \mathbf{\text{Rs. } 44}$$
$$\text{Umer} = \frac{5}{8} \times 176 = \mathbf{\text{Rs. } 110}$$
Q7. $\frac{1}{5}$ of a water tank is empty. What percentage of the tank is filled with water?
Q8. In a class of $60$ students, $10\%$ students failed. How many students passed?
$$\text{Passed Percentage} = 100\% - 10\% = 90\%$$
$$\text{Passed Students} = \frac{90}{100} \times 60 = \mathbf{54\text{ students}}$$
Q9. $30\%$ of a tower is painted red, $38\%$ is painted green and remaining is white. What percentage is white?
Q10. $5$ students were absent from a class of $45$ students.
(a) What fraction of students was absent? $\frac{5}{45} = \mathbf{\frac{1}{9}}$
(b) What fraction of students was present? $\frac{40}{45} = \mathbf{\frac{8}{9}}$
(c) What percentage of students was present? $\frac{8}{9} \times 100\% = \mathbf{88.89\%}$ (Absent: $\mathbf{11.11\%}$)
📊 Quick Reference Formula Sheet for Unit 3
| Concept | Mathematical Formula / Definition | Key Rule / Example |
|---|---|---|
| Ratio ($a : b$) | $\frac{a}{b}$, $b \neq 0$ | Unitless, same units required ($1.5\text{ h} : 30\text{ min} = 3 : 1$) |
| Unit Rate | $\text{Unit Rate} = \frac{\text{Total Value}}{\text{Number of Units}}$ | $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$, $\text{Cost per item}$ |
| Continued Ratio | $x : y : z$ | Equate the middle term $y$ via LCM |
| Proportional Share | $\text{Share} = \frac{\text{Person's Ratio Term}}{\text{Sum of Ratio Terms}} \times \text{Total}$ | Divide $\text{Rs. } 160$ in $1:6:9$ |
| Percentage Conversion | $\text{Percentage} = \frac{\text{Part}}{\text{Whole}} \times 100\%$ | $\frac{40}{80} \times 100\% = 50\%$ |
| Percentage Change | $\text{Change %} = \frac{\text{Increase / Decrease}}{\text{Original Amount}} \times 100\%$ | $\text{Salary increase from } 5000 \to 5600 = 12\%$ |
More Chapter Notes for Class 6 (FBISE)
MathematicsTest Your Knowledge on Chapter 3: Class 6 Mathematics - Ch 3: Mastery Guide: Ratios, Rates, Unitary Method, Continued Ratios & Percentage Applications (FBISE)
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Class 6 Mathematics - Ch 3: Ratio, Rate and Percentage Chapter Mock Test
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