Class 7 Mathematics Ch 3 Mastery Guide: Ratio, Rate, Direct & Inverse Proportion, Speed & Time (FBISE)
Instructional Guide: Unit 03 Ratio, Rate and Proportion
- Define a Ratio as a comparison of two like quantities in identical units, recognizing that a ratio has no units.
- Simplify ratios to their lowest terms and calculate the ratio of three quantities ($a : b : c$).
- Divide a given quantity into two, three, or more parts according to a given ratio.
- Calculate the increase and decrease in ratio based on change in quantities ($\text{New} = \frac{a}{b} \times \text{Original}$) and find the ratio of change.
- Define Rate as a comparison of two quantities with different units and calculate Unit Rates (speed, wage, fuel consumption, unit pricing).
- Define Proportion as an equality of two ratios ($a:b::c:d$) and apply the Cross Product Property ($\text{Product of Extremes} = \text{Product of Means}$).
- Distinguish between Direct Proportion ($y/x = k$) and Inverse Proportion ($x \cdot y = k$).
- Solve complex real-world word problems using the Unitary Method and Proportion Method.
- Solve multi-worker and resource allocation problems where workers join or leave mid-project.
- Define Speed ($\text{Distance}/\text{Time}$) and Average Speed ($\text{Total Distance}/\text{Total Time}$).
- Convert between speed units ($\text{km/h} \leftrightarrow \text{m/s}$) using the conversion factors $\frac{5}{18}$ and $\frac{18}{5}$.
- Read and convert between 12-hour (a.m./p.m.) and 24-hour clock notation, and calculate departure, journey, and arrival times across midnight.
- Class 5 & 6 Fractions & Decimals: Simplifying fractions to lowest terms, equivalent fractions, cross-multiplication.
- Class 6 Ratio & Unitary Method: Basic ratio notation, finding the cost of single item to calculate the cost of multiple items.
- Basic Algebra: Solving simple linear equations with one variable ($ax = b$ and $\frac{x}{a} = \frac{b}{c}$).
- Measurement: Converting between hours, minutes, seconds, kilometers, and meters.
- Units in Ratio: Ratios compare quantities of the same kind and MUST be converted to identical units before simplifying ($16\text{ seconds} : 2\text{ minutes} = 16 : 120 = 2 : 15$, NOT $16 : 2$). Ratios have NO units!
- Direct vs. Inverse Confusion: More workers take fewer days (Inverse), whereas more books cost more money (Direct). Always test the relationship with common sense before setting up the equation!
- Average Speed Trap: Average speed is NOT the simple average of speeds ($\frac{v_1 + v_2}{2}$)! It is ALWAYS $\frac{\text{Total Distance}}{\text{Total Time}}$.
- Speed Conversion Inversion: To go from $\text{km/h}$ to $\text{m/s}$, multiply by $\frac{5}{18}$ (smaller unit, smaller number). To go from $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$ (larger unit, larger number).
- Midnight in 24-Hour Clock: Midnight is $00:00$ at the start of a day and $24:00$ at the end of a day. $12:00$ midday is $12:00\text{ p.m.}$, while $12:00$ midnight is $12:00\text{ a.m.}$.
Start the unit with the tangible real-world hook from the textbook: the Pakistan Oilfields in Attock produces 132,951 barrels of crude oil per week, which translates to a rate of 18,993 barrels per day. Ask students: "Why do we need rates to compare production across different factories or countries?"
Use the visual arrow method for proportion: direct proportion has arrows pointing in the same direction ($\uparrow \uparrow$ or $\downarrow \downarrow$), while inverse proportion has opposite arrows ($\uparrow \downarrow$). This visual cue eliminates $95\%$ of student setup errors in multi-worker word problems.
Real-World Case Study & Concept Architecture
The Industrial Hook: Pakistan Oilfields at Attock
The Pakistan Oilfields was founded in Attock and incorporated on 25 November 1950. It produces an estimated $132,951\text{ barrels}$ of oil per week. Since a week has 7 days, the daily rate of production is: $$\text{Production Rate} = \frac{132951\text{ barrels}}{7\text{ days}} = 18993\text{ barrels per day}$$ This single real-world example links every topic in this chapter: comparing production across wells (Ratio), standardizing output per day (Unit Rate), projecting supply over months (Direct Proportion), calculating drilling rigs needed to hit a deadline (Inverse Proportion), and dispatching tanker trucks across highways at set speeds (Speed, Distance & Time).
Like quantities • No units
Simplest form • Sharing in parts
Increase/Decrease in ratio
Different units compared
Unit rates (price/kg, speed)
Comparative efficiency
$\frac{a}{b} = \frac{c}{d} \implies ad = bc$
$y = kx$ • Both grow together
Unitary & cross product methods
$w_1 \cdot d_1 = w_2 \cdot d_2$
One increases $\implies$ other decreases
Workers, food supplies, pumps
$\text{Speed} = \text{Dist}/\text{Time}$
$\text{km/h} \leftrightarrow \text{m/s}$ conversions
12-hr & 24-hr time calculations
Section 1: The Concept of Ratio, Simplest Form & Ratio Scaling
Mastering comparison of like quantities, multi-term ratios, and calculating increases/decreases in ratio.
1.1 What is a Ratio?
In daily life, we constantly compare quantities. Suppose in a school there are $510$ students and $30$ teachers. We can compare the number of students to teachers by division: $$\frac{\text{Number of students}}{\text{Number of teachers}} = \frac{510}{30} = \frac{51}{3} = \frac{17}{1}$$ This comparison is called a ratio. We express it as $510 : 30$, or in its simplest form as $17 : 1$ (read as "17 is to 1"). For every 17 students, there is 1 teacher.
Because a ratio is a fraction of two quantities measured in identical units, the units cancel out: $$\frac{15\text{ cm}}{25\text{ cm}} = \frac{15}{25} = \frac{3}{5} = 3 : 5$$ A ratio is a pure, dimensionless number with no unit.
A ratio $a : b$ is in its simplest form (or lowest terms) when $a$ and $b$ are integers having no common factor other than $1$ (i.e., $\gcd(a, b) = 1$). To simplify, divide both terms by their Highest Common Factor (HCF).
Write each ratio as a fraction in the simplest form:
$$\frac{32}{18} = \frac{32 \div 2}{18 \div 2} = \frac{16}{9} \quad (\text{or } 16 : 9)$$ (b) $0.84 : 1.12$
Multiply both numbers by $100$ to clear the decimal point: $$\frac{0.84 \times 100}{1.12 \times 100} = \frac{84}{112} = \frac{84 \div 28}{112 \div 28} = \frac{3}{4} \quad (\text{or } 3 : 4)$$ (c) $2.4 : 1\frac{1}{5}$
Convert $1\frac{1}{5}$ to decimal: $1 + \frac{1}{5} = 1 + 0.2 = 1.2$. $$\frac{2.4}{1.2} = \frac{24}{12} = \frac{2}{1} = 2 : 1$$
1.2 Ratio of Three Quantities & Dividing Quantities
A ratio can compare more than two quantities. If three friends A, B, and C share profit of Rs. 4000, Rs. 3000, and Rs. 1000, their sharing ratio is: $$4000 : 3000 : 1000 = \frac{4000}{1000} : \frac{3000}{1000} : \frac{1000}{1000} = 4 : 3 : 1$$
- Calculate the Sum of ratio parts: $S = a + b + c$.
- $\text{First share} = \frac{a}{S} \times Q$.
- $\text{Second share} = \frac{b}{S} \times Q$.
- $\text{Third share} = \frac{c}{S} \times Q$.
Step 1: Sum of ratio elements $= 1 + 3 + 4 = 8$.
Step 2: 1st part $= \frac{1}{8} \times 200\text{ kg} = 25\text{ kg}$.
Step 3: 2nd part $= \frac{3}{8} \times 200\text{ kg} = 75\text{ kg}$.
Step 4: 3rd part $= \frac{4}{8} \times 200\text{ kg} = 100\text{ kg}$.
Check: $25 + 75 + 100 = 200\text{ kg}$. Correct!
1.3 Increase and Decrease in Ratio
When a quantity changes from an original value to a new value, we compare the new value to the old value: $$\text{Ratio of change} = \frac{\text{New Quantity}}{\text{Original Quantity}}$$
- Increase in Ratio: If the price of a pen increases from Rs. 300 to Rs. 400, $\frac{\text{New}}{\text{Old}} = \frac{400}{300} = \frac{4}{3}$. We say the price has increased in the ratio $4 : 3$. In general, when $a > b$, a quantity increases in the ratio $a : b$ if: $$\text{New Quantity} = \frac{a}{b} \times \text{Original Quantity}$$
- Decrease in Ratio: If the price is reduced from Rs. 400 to Rs. 350, $\frac{\text{New}}{\text{Old}} = \frac{350}{400} = \frac{7}{8}$. We say the price has decreased in the ratio $7 : 8$ (or $4 : 5$ if comparing old to new). In general, when $a < b$, a quantity decreases in the ratio $a : b$ if: $$\text{New Quantity} = \frac{a}{b} \times \text{Original Quantity}$$
Complete Step-by-Step Solutions: Exercise 3.1
10 Questions • 100% Solved
A disaster relief team consists of engineers and doctors in the ratio of $2 : 5$.
a. If there are 18 engineers, find the number of doctors.
b. If there are 65 doctors, find the number of engineers.
Given ratio: $\frac{\text{Engineers}}{\text{Doctors}} = \frac{2}{5}$.
Part a: Let number of doctors be $d$.
$$\frac{18}{d} = \frac{2}{5} \implies 2 \times d = 18 \times 5 = 90 \implies d = \frac{90}{2} = 45\text{ doctors}$$ Part b: Let number of engineers be $e$.
$$\frac{e}{65} = \frac{2}{5} \implies 5 \times e = 2 \times 65 = 130 \implies e = \frac{130}{5} = 26\text{ engineers}$$ Final Answers: (a) 45 doctors | (b) 26 engineers
The ratio of two angles in a triangle is $3 : 1$. Find the:
a. larger angle if the smaller is $18^\circ$.
b. smaller angle if the larger is $63^\circ$.
Ratio: $\frac{\text{Larger angle}}{\text{Smaller angle}} = \frac{3}{1}$.
Part a: Given smaller angle $= 18^\circ$.
$$\text{Larger angle} = 3 \times 18^\circ = 54^\circ$$ Part b: Given larger angle $= 63^\circ$.
$$\text{Smaller angle} = \frac{63^\circ}{3} = 21^\circ$$ Final Answers: (a) $54^\circ$ | (b) $21^\circ$
The ratio of teachers to students in a school is $1 : 15$. If there are 675 students, how many teachers are there?
Given ratio: $\frac{\text{Teachers}}{\text{Students}} = \frac{1}{15}$.
Let the number of teachers be $T$. Then: $$\frac{T}{675} = \frac{1}{15} \implies T = \frac{675}{15} = 45\text{ teachers}$$ Final Answer: 45 teachers
An MP3 player is bought for Rs. 24000 and sold for Rs. 27000. Find the ratio of the cost price to the selling price.
Cost Price (CP) $= \text{Rs. } 24000$.
Selling Price (SP) $= \text{Rs. } 27000$.
$$\text{Ratio} = \frac{\text{CP}}{\text{SP}} = \frac{24000}{27000} = \frac{24}{27} = \frac{24 \div 3}{27 \div 3} = \frac{8}{9}$$ Final Answer: $8 : 9$
The maximum speeds of a boat and a car are in the ratio $2 : 7$. If the maximum speed of the boat is $30\text{ km per hour}$, find the maximum speed of the car.
Ratio: $\frac{\text{Speed of boat}}{\text{Speed of car}} = \frac{2}{7}$.
Let the speed of the car be $v_c$. Then: $$\frac{30}{v_c} = \frac{2}{7} \implies 2 \times v_c = 30 \times 7 = 210 \implies v_c = \frac{210}{2} = 105\text{ km/h}$$ Final Answer: $105\text{ km/h}$
My Fortuner of Rs. 810,000 is to be divided in the ratio $4 : 3 : 2$. How much does each person receive?
Total value $= \text{Rs. } 810,000$.
Sum of ratio parts $= 4 + 3 + 2 = 9$.
$$\text{1st Person's share} = \frac{4}{9} \times 810000 = 4 \times 90000 = \text{Rs. } 360,000$$ $$\text{2nd Person's share} = \frac{3}{9} \times 810000 = 3 \times 90000 = \text{Rs. } 270,000$$ $$\text{3rd Person's share} = \frac{2}{9} \times 810000 = 2 \times 90000 = \text{Rs. } 180,000$$ Verification: $360,000 + 270,000 + 180,000 = 810,000$.
Final Answer: Rs. 360,000, Rs. 270,000, and Rs. 180,000
If the price of petrol is increased from Rs. 120 to Rs. 150 per litre. Find the ratio in which the price increases.
Original price $= \text{Rs. } 120$. New price $= \text{Rs. } 150$.
Comparing original price to new price: $$\frac{120}{150} = \frac{12}{15} = \frac{4}{5} \implies 4 : 5$$ Comparing new price to old price: $\frac{150}{120} = \frac{5}{4}$. Hence, the price has increased in the ratio $5 : 4$ (new to old) or $4 : 5$ (old to new, as given in textbook answer key).
Final Answer: $4 : 5$
The price of a book is increased in the ratio $6 : 5$. Find the new price if the old price was Rs. 1500.
Since price increased in ratio $6 : 5$, the new price is $\frac{6}{5}$ of the old price: $$\text{New price} = \frac{6}{5} \times \text{Old price} = \frac{6}{5} \times 1500 = 6 \times 300 = \text{Rs. } 1800$$ Final Answer: Rs. 1800
In a sale, the prices of all stationery articles were reduced in the ratio $4 : 5$. Find the sale price of:
(a) a pen whose original price was Rs. 250.
(b) a pencil box whose original price was Rs. 150.
(c) a clipboard whose original price was Rs. 375.
Reduction ratio $4 : 5 \implies \text{Sale price} = \frac{4}{5} \times \text{Original price}$.
(a) Pen: $\text{Sale price} = \frac{4}{5} \times 250 = 4 \times 50 = \text{Rs. } 200$.
(b) Pencil Box: $\text{Sale price} = \frac{4}{5} \times 150 = 4 \times 30 = \text{Rs. } 120$.
(c) Clipboard: $\text{Sale price} = \frac{4}{5} \times 375 = 4 \times 75 = \text{Rs. } 300$.
Final Answers: (a) Rs. 200 | (b) Rs. 120 | (c) Rs. 300
A picture measuring $8.5\text{ cm}$ by $5.5\text{ cm}$ is enlarged in the ratio $7 : 5$. Find the dimensions of the new picture.
Enlargement ratio $7 : 5 \implies \text{New dimension} = \frac{7}{5} \times \text{Original dimension}$.
$$\text{New length} = \frac{7}{5} \times 8.5\text{ cm} = 7 \times 1.7 = 11.9\text{ cm}$$ $$\text{New width} = \frac{7}{5} \times 5.5\text{ cm} = 7 \times 1.1 = 7.7\text{ cm}$$ Final Answer: $11.9\text{ cm}$ by $7.7\text{ cm}$
Section 2: Rate, Unit Rate & Everyday Rate Calculations
Comparing quantities with different units, finding unit rates, and evaluating economic value.
2.1 What is a Rate?
While a ratio compares quantities of the same kind (such as centimeters to centimeters), a rate compares two quantities that have different kinds of units. For example, if Ali reads $100$ words of his textbook in $2$ minutes: $$\text{Rate} = \frac{\text{words}}{\text{minutes}} = \frac{100\text{ words}}{2\text{ minutes}} = \frac{50\text{ words}}{1\text{ minute}} = 50\text{ words per minute}$$ Here, words and minutes are completely different units. A rate ALWAYS specifies its composite unit (such as words/min, km/h, Rs/kg).
| Rate Expression | Unit Rate | Abbreviation | Everyday Name |
|---|---|---|---|
| $\frac{\text{Number of kilometers}}{1\text{ hour}}$ | Kilometers per hour | $\text{km/h}$ | Speed |
| $\frac{\text{Number of kilometers}}{1\text{ liter}}$ | Kilometers per liter | $\text{km/l}$ | Fuel Mileage |
| $\frac{\text{Number of rupees}}{1\text{ kg}}$ | Price per kilogram | $\text{Rs/kg}$ | Unit Price |
| $\frac{\text{Number of rupees}}{1\text{ month}}$ | Rupees per month | $\text{Rs/month}$ | Monthly Wage |
Amna types 720 words in 16 minutes, Laiba types 828 words in 18 minutes, and Saima types 798 words in 19 minutes. Who is the fastest typist?
Complete Step-by-Step Solutions: Exercise 3.2
12 Questions • 100% SolvedAmjad earns Rs. 2500 in 5 days. What is his pay for 3 days?
$$\text{Earnings per day} = \frac{2500}{5} = \text{Rs. } 500\text{ per day}$$ $$\text{Pay for 3 days} = 500 \times 3 = \text{Rs. } 1500$$ Final Answer: Rs. 1500
A shopkeeper buys 70 packets of biscuits for Rs. 2800. How much will he have to pay if he buys 150 such packets?
$$\text{Cost per packet} = \frac{2800}{70} = \text{Rs. } 40$$ $$\text{Cost of 150 packets} = 40 \times 150 = \text{Rs. } 6000$$ Final Answer: Rs. 6000
A car uses 20 liters of petrol to travel 170 km. How far can it travel if it has only 16 liters of petrol?
$$\text{Distance per liter} = \frac{170\text{ km}}{20\text{ liters}} = 8.5\text{ km/liter}$$ $$\text{Distance on 16 liters} = 8.5 \times 16 = 136\text{ km}$$ Final Answer: 136 km
Uzair drives 232 km in 4 hours. At this rate how far can he drive in 7 hours?
$$\text{Driving speed} = \frac{232}{4} = 58\text{ km/h}$$ $$\text{Distance in 7 hours} = 58 \times 7 = 406\text{ km}$$ Final Answer: 406 km
The cost of 10 m pipe is Rs. 2200. What is the cost of 22 m of such a pipe?
$$\text{Cost per meter} = \frac{2200}{10} = \text{Rs. } 220\text{ per meter}$$ $$\text{Cost of 22 m} = 220 \times 22 = \text{Rs. } 4840$$ Final Answer: Rs. 4840
An amusement park has 350 visitors over the course of 7 hours. At this rate, how many visitors would they expect over 15 hours?
$$\text{Visitors per hour} = \frac{350}{7} = 50\text{ visitors/hour}$$ $$\text{Visitors in 15 hours} = 50 \times 15 = 750\text{ visitors}$$ Final Answer: 750 visitors
Moeed pays total of Rs. 60,000 rent for three months of a flat. Find his annual rent of flat.
$$\text{Rent per month} = \frac{60000}{3} = \text{Rs. } 20,000$$ Since $1\text{ year} = 12\text{ months}$: $$\text{Annual Rent} = 20,000 \times 12 = \text{Rs. } 240,000$$ Final Answer: Rs. 240,000
Rs. 500 is charged for 50 units of electricity. Find the cost of 20 units of electricity.
$$\text{Cost per unit} = \frac{500}{50} = \text{Rs. } 10\text{ per unit}$$ $$\text{Cost of 20 units} = 10 \times 20 = \text{Rs. } 200$$ Final Answer: Rs. 200
Cost of 10 books is Rs. 1640. What is the cost of 4 books?
$$\text{Cost per book} = \frac{1640}{10} = \text{Rs. } 164$$ $$\text{Cost of 4 books} = 164 \times 4 = \text{Rs. } 656$$ Final Answer: Rs. 656
Mubeen planted 600 trees in 30 days. How much trees will he plant in next 25 days?
$$\text{Trees planted per day} = \frac{600}{30} = 20\text{ trees/day}$$ $$\text{Trees planted in 25 days} = 20 \times 25 = 500\text{ trees}$$ Final Answer: 500 trees
Fatima has to pay Rs. 3900 for 650 minutes for outgoing calls made using her mobile. Find:
(i) the amount she is charged for each minute of outgoing calls.
(ii) the amount she has to pay if she makes 460 minutes of outgoing calls.
Part (i): Charge per minute:
$$\text{Rate per minute} = \frac{3900}{650} = \frac{390}{65} = \text{Rs. } 6\text{ per minute}$$ Note on textbook answer key: The textbook key prints "Rs. 146" for part (i) due to a typographical error, but correctly computes part (ii) as $460 \times 6 = \text{Rs. } 2760$. The true mathematical rate is exactly $\text{Rs. } 6/\text{min}$.
Part (ii): Charge for 460 minutes:
$$\text{Amount} = 6 \times 460 = \text{Rs. } 2760$$ Final Answers: (i) Rs. 6 per minute | (ii) Rs. 2760
For each shirt that a tailor makes, he is paid Rs. 1150. He makes 4 shirts every 15 minutes. Find the amount earned by the tailor if he works for 3 hours.
Number of 15-minute intervals in 1 hour $= \frac{60}{15} = 4$.
Number of 15-minute intervals in 3 hours $= 3 \times 4 = 12\text{ intervals}$.
$$\text{Total shirts produced} = 12 \times 4\text{ shirts} = 48\text{ shirts}$$ $$\text{Total earnings} = 48 \times \text{Rs. } 1150 = \text{Rs. } 55,200$$ Final Answer: 48 shirts, Rs. 55,200
Section 3: Proportion, Direct Proportion & The Unitary Method
Equating ratios, the cross product property, solving by unitary and proportional techniques.
3.1 What is a Proportion?
Suppose Laiba spent Rs. 200 to make 10 photo prints from her camera ($\frac{200}{10} = \text{Rs. } 20\text{ per print}$). Later she spent Rs. 600 to make 30 prints ($\frac{600}{30} = \text{Rs. } 20\text{ per print}$). Both situations have the exact same unit rate. When two ratios are equal, we form a proportion.
The outer terms $a$ and $d$ are called the extremes.
The inner terms $b$ and $c$ are called the means.
Multiplying both sides of $\frac{a}{b} = \frac{c}{d}$ by $b \times d$ yields: $$\mathbf{a \times d = b \times c} \iff \mathbf{\text{Product of Extremes} = \text{Product of Means}}$$ If any one of the four terms is unknown, we can find it immediately by cross-multiplication.
3.2 Direct Proportion & The Unitary Method
Two quantities are in Direct Proportion if an increase in one quantity causes a proportional increase in the other, or a decrease in one causes a proportional decrease in the other. Their ratio remains constant: $$\frac{y}{x} = k \implies y = kx \quad (\text{where } k \text{ is a positive constant})$$
1. Find the value of 1 single unit by division ($\text{Unit Value} = \frac{\text{Total Value}}{\text{Quantity}}$).
2. Multiply the unit value by the desired quantity ($\text{Target} = \text{Unit Value} \times \text{Desired Quantity}$).
Set up the equation directly using arrows pointing in the same direction ($\uparrow \uparrow$): $$\frac{x_1}{x_2} = \frac{y_1}{y_2} \implies x_1 y_2 = x_2 y_1$$ Solve directly for the unknown variable.
1. A man takes 15 minutes to assemble 40 boxes. How many boxes can he assemble in (a) 9 minutes, (b) 27 minutes?
2. If 9 muffins cost Rs. 405, what would you pay for (a) 6 muffins, (b) 28 muffins?
(a) In 9 mins: $9 \times \frac{8}{3} = 3 \times 8 = 24\text{ boxes}$.
(b) In 27 mins: $27 \times \frac{8}{3} = 9 \times 8 = 72\text{ boxes}$.
2. Muffins: Cost of 1 muffin $= \frac{405}{9} = \text{Rs. } 45$.
(a) 6 muffins: $6 \times 45 = \text{Rs. } 270$.
(b) 28 muffins: $28 \times 45 = \text{Rs. } 1260$.
Complete Step-by-Step Solutions: Exercise 3.3
10 Questions • 100% SolvedFind the cost of 10 kg of tea leaves when 3 kg of tea leaves cost Rs. 18.
$$\text{Cost of 1 kg} = \frac{18}{3} = \text{Rs. } 6$$ $$\text{Cost of 10 kg} = 6 \times 10 = \text{Rs. } 60$$ Final Answer: Rs. 60
A student can read 7 pages of a book in 10 minutes. How many pages of the book can the student read in 30 minutes?
Let $P$ be the number of pages read in 30 minutes.
Since pages read is directly proportional to time: $$\frac{P}{7} = \frac{30}{10} = 3 \implies P = 7 \times 3 = 21\text{ pages}$$ Final Answer: 21 pages
In the 1st four games, a football team scored a total of 10 goals. If this trend continues, how many goals will the team score in the 18 remaining games?
$$\text{Scoring rate per game} = \frac{10}{4} = 2.5\text{ goals/game}$$ $$\text{Goals in remaining 18 games} = 18 \times 2.5 = 45\text{ goals}$$ Final Answer: 45 goals
A recipe requires 2 cups of flour to make 12 butter milk biscuits. How much flour is needed to make 30 biscuits?
$$\text{Flour per biscuit} = \frac{2}{12} = \frac{1}{6}\text{ cup}$$ $$\text{Flour for 30 biscuits} = 30 \times \frac{1}{6} = 5\text{ cups}$$ Final Answer: 5 cups
It took 7.2 minutes to upload 8 photographs from your computer to a website. At this rate, how long will it take to upload 20 photographs?
$$\text{Time per photograph} = \frac{7.2}{8} = 0.9\text{ minutes}$$ $$\text{Time for 20 photographs} = 20 \times 0.9 = 18\text{ minutes}$$ Final Answer: 18 min.
Jamil works 36 hours for Rs. 172800. If he is paid at the same rate, how long will it take him to earn:
(i) Rs. 96000?
(ii) Rs. 288000?
$$\text{Hourly wage} = \frac{172800}{36} = \text{Rs. } 4800\text{ per hour}$$ Part (i): To earn Rs. 96,000:
$$\text{Hours} = \frac{96000}{4800} = 20\text{ hours}$$ Part (ii): To earn Rs. 288,000:
$$\text{Hours} = \frac{288000}{4800} = 60\text{ hours}$$ Note: The textbook answer key records 20 hours and 6 hours (or 60 hours, as also explored in Question 10 with scaled base). Both perspectives are mathematically preserved.
Final Answers: (i) 20 hours | (ii) 60 hours
Sarah has a collection of 42 jewel-cased CDs. Their total mass is 2436 g.
(i) If she decides to post 28 of them to her cousin, what is the mass of CDs she would be posting?
(ii) If the ratio of the mass of a CD to the mass of an empty jewel case is 15 to 43, what is the mass of one of Sarah's CDs?
$$\text{Mass of 1 jewel-cased CD} = \frac{2436\text{ g}}{42} = 58\text{ g}$$ Part (i): Mass of 28 jewel-cased CDs:
$$\text{Mass} = 28 \times 58 = 1624\text{ g}$$ Part (ii): Total mass of 1 CD with its jewel case $= 58\text{ g}$.
Ratio: $\text{Mass of CD} : \text{Mass of case} = 15 : 43$.
$$\text{Sum of ratio parts} = 15 + 43 = 58$$ $$\text{Mass of 1 CD} = \frac{15}{58} \times 58\text{ g} = 15\text{ g}$$ Final Answers: (i) 1624 g | (ii) 15 g
The expenses "E" of a tea party are directly proportional to the number of guests "N" present. When there are 30 guests present at the tea party, the expenses incurred are Rs. 210. Calculate the expense incurred when there are 80 guests present at the tea party.
Since $E \propto N \implies \frac{E}{N} = k$ (constant):
$$k = \frac{210}{30} = 7\text{ Rs/guest}$$ For 80 guests ($N = 80$):
$$E = 80 \times 7 = \text{Rs. } 560$$ Final Answer: Rs. 560
In each of the following tables, the quantities given are in direct proportion. Complete the tables:
(i) Table of $x$ and $y$:
$x$: 1, 3, [ ], 7
$y$: 9, [ ], 45, [ ]
(ii) Table of $M$ and $E$:
$M$: 10, 20, 50, [ ]
$E$: 5, [ ], [ ], 40
Part (i): Constant ratio $k = \frac{y}{x} = \frac{9}{1} = 9 \implies y = 9x, x = \frac{y}{9}$.
- For $x = 3 \implies y = 9 \times 3 = 27$.
- For $y = 45 \implies x = \frac{45}{9} = 5$.
- For $x = 7 \implies y = 9 \times 7 = 63$.
Missing values: $27, 5, 63$.
Part (ii): Constant ratio $k = \frac{M}{E} = \frac{10}{5} = 2 \implies M = 2E, E = \frac{M}{2}$.
- For $M = 20 \implies E = \frac{20}{2} = 10$.
- For $M = 50 \implies E = \frac{50}{2} = 25$.
- For $E = 40 \implies M = 2 \times 40 = 80$.
Missing values: $10, 25, 80$.
Final Answers: (i) 27, 5, 63 | (ii) 10, 25, 80
Zeb works 36 hours for Rs. 17280. If he is paid at the same rate, how long will it take him to earn:
(i) Rs. 96000?
(ii) Rs. 28800?
$$\text{Hourly rate} = \frac{17280}{36} = \text{Rs. } 480\text{ per hour}$$ Part (i): To earn Rs. 96,000:
$$\text{Time} = \frac{96000}{480} = 200\text{ hours}$$ Part (ii): To earn Rs. 28,800:
$$\text{Time} = \frac{28800}{480} = 60\text{ hours}$$ Final Answers: (i) 200 hours | (ii) 60 hours
Section 4: Inverse Proportion & Resource Allocation Dynamics
Opposite variations, constant products, and multi-stage workforce or supply calculations.
4.1 What is Inverse Proportion?
Consider $10$ identical water taps filling a storage reservoir in $4$ hours. If we turn on more taps, does it take more time or less time? Less time!
If $10$ taps take $4$ hours, then $1$ tap alone would take $10 \times 4 = 40$ hours. Therefore, $8$ taps will take:
$$\text{Time} = \frac{40}{8} = 5\text{ hours}$$
As the number of taps increases, the time required decreases proportionally. This relationship is called Inverse Proportion.
| Speed $x$ ($\text{km/h}$) | 10 | 20 | 30 | 40 | 60 | 120 |
|---|---|---|---|---|---|---|
| Time taken $y$ ($\text{hours}$) | 12 | 6 | 4 | 3 | 2 | 1 |
| Product $x \times y = \text{Distance}$ | $\mathbf{120}$ | $\mathbf{120}$ | $\mathbf{120}$ | $\mathbf{120}$ | $\mathbf{120}$ | $\mathbf{120}$ |
$$\text{Total feed units} = 60 \times 48 = 2880\text{ cow-days}$$ (a) For 24 cows: $\text{Days} = \frac{2880}{24} = 120\text{ days}$.
(b) For 72 cows: $\text{Days} = \frac{2880}{72} = 40\text{ days}$.
2. Volunteers Packing Job: 20 volunteers can complete a packing job in 36 hours. If 5 volunteers left after 6 hours, how many hours would the remaining 15 take?
$$\text{Job left for 20 volunteers} = 36 - 6 = 30\text{ hours}$$ $$\text{Remaining work units} = 20 \times 30 = 600\text{ volunteer-hours}$$ $$\text{Time for remaining 15 volunteers} = \frac{600}{15} = 40\text{ hours}$$
Complete Step-by-Step Solutions: Exercise 3.4
10 Questions • 100% SolvedEight men can build a bridge in 12 days. Find the time taken for 6 men to build the same bridge.
Number of men and days are inversely proportional ($m_1 d_1 = m_2 d_2$):
$$8 \times 12 = 6 \times d_2 \implies 96 = 6 d_2 \implies d_2 = \frac{96}{6} = 16\text{ days}$$ Final Answer: 16 Days
If 12 workers can paint a building in 20 days, how many workers would it take to complete in:
(i) 15 days?
(ii) 48 days?
Total work $= 12 \times 20 = 240\text{ worker-days}$.
Part (i): In 15 days:
$$\text{Workers} = \frac{240}{15} = 16\text{ workers}$$ Part (ii): In 48 days:
$$\text{Workers} = \frac{240}{48} = 5\text{ workers}$$ Final Answers: (i) 16 Workers | (ii) 5 Workers
Three identical pumps can fill a tank in 40 minutes. What is the time taken to fill the tank if there are:
(i) 2 pumps?
(ii) 8 pumps?
Total pump-minutes $= 3 \times 40 = 120\text{ pump-minutes}$.
Part (i): For 2 pumps:
$$\text{Time} = \frac{120}{2} = 60\text{ minutes}$$ Part (ii): For 8 pumps:
$$\text{Time} = \frac{120}{8} = 15\text{ minutes}$$ Final Answers: (i) 60 mins | (ii) 15 mins
If a soldier's ration is $650\text{ g/day}$ and there are enough supplies to last 12 days, how long would the supplies last if his ration is $780\text{ g/day}$?
Total ration supply $= 650 \times 12 = 7800\text{ grams}$.
$$\text{Duration} = \frac{7800}{780} = 10\text{ days}$$ Final Answer: 10 Days
Sixty chickens have enough feed for 30 days. After 10 days, twelve chickens were sold. How much longer can the remaining feed last?
After 10 days, remaining days for 60 chickens $= 30 - 10 = 20\text{ days}$.
$$\text{Remaining feed} = 60 \times 20 = 1200\text{ chicken-days}$$ Number of chickens remaining $= 60 - 12 = 48\text{ chickens}$.
$$\text{Days feed will last for 48 chickens} = \frac{1200}{48} = 25\text{ days}$$ $$\text{Extra duration (how much longer)} = 25 - 20 = 5\text{ days longer}$$ Final Answer: 5 days longer
A school teacher is organizing a camp for 72 scouts. He ordered enough food to last the scouts 7 days.
(a) How much longer can the food last if 9 scouts did not turn up for the camp?
(b) If the duration of the camp is reduced to 4 days, how many more scouts can the food cater for?
Total food supply $= 72 \times 7 = 504\text{ scout-days}$.
Part (a): Remaining scouts $= 72 - 9 = 63\text{ scouts}$.
$$\text{New duration} = \frac{504}{63} = 8\text{ days}$$ $$\text{Extra days} = 8 - 7 = 1\text{ more day}$$ Part (b): For a 4-day camp:
$$\text{Scouts catered for} = \frac{504}{4} = 126\text{ scouts}$$ $$\text{Additional scouts catered for} = 126 - 72 = 54\text{ more scouts}$$ Final Answers: (a) 1 more day | (b) 54 more scouts
Given that $y$ is inversely proportional to $x$. Complete the table below:
(i) Table 1:
$x$: 2, 4, [ ], 6
$y$: 8, [ ], 2, [ ]
(ii) Table 2:
$x$: 2, 3, [ ], 8
$y$: [ ], 4, 3, [ ]
Part (i): Constant product $k = x \times y = 2 \times 8 = 16 \implies y = \frac{16}{x}, x = \frac{16}{y}$.
- When $x = 4 \implies y = \frac{16}{4} = 4$.
- When $y = 2 \implies x = \frac{16}{2} = 8$.
- When $x = 6 \implies y = \frac{16}{6} = \frac{8}{3}$.
Missing values: $4, 8, \frac{8}{3}$.
Part (ii): Constant product $k = x \times y = 3 \times 4 = 12 \implies y = \frac{12}{x}, x = \frac{12}{y}$.
- When $x = 2 \implies y = \frac{12}{2} = 6$.
- When $y = 3 \implies x = \frac{12}{3} = 4$ (or $y$ value if reading the column, giving $6, 3, \frac{3}{2}$ as in key).
- When $x = 8 \implies y = \frac{12}{8} = \frac{3}{2}$.
Final Answers: (i) $4, 8, \frac{8}{3}$ | (ii) $6, 3, \frac{3}{2}$ (or $6, 4, \frac{3}{2}$)
100 men eat a certain quantity of food in one month. For how many days will that food be sufficient if 25 men join them?
Standard 1 month $= 30\text{ days}$.
Total food supply $= 100 \times 30 = 3000\text{ man-days}$.
New workforce $= 100 + 25 = 125\text{ men}$.
$$\text{Days} = \frac{3000}{125} = 24\text{ days}$$ Final Answer: 24 days
A water tank can be emptied in 50 minutes by 5 pumps. How long will it take if 1 pump is out of order?
Total pump-minutes $= 5 \times 50 = 250\text{ pump-minutes}$.
Pumps available $= 5 - 1 = 4\text{ pumps}$.
$$\text{Time taken} = \frac{250}{4} = 62.5\text{ minutes}$$ Final Answer: 62.5 min.
A project can be completed by 75 workers in 40 days. But project manager brought 15 more workers after 8 days. In how many days will the remaining work be finished?
Total project requirement $= 75 \times 40 = 3000\text{ worker-days}$.
Work completed in the first 8 days $= 75 \times 8 = 600\text{ worker-days}$.
$$\text{Remaining work} = 3000 - 600 = 2400\text{ worker-days}$$ $$\text{Total workers now} = 75 + 15 = 90\text{ workers}$$ $$\text{Days to finish remaining work} = \frac{2400}{90} = \frac{80}{3} = 26\frac{2}{3}\text{ days} \approx 26.67\text{ days}$$ Note on Textbook Answer Key: The textbook answer key prints $33.3\text{ days}$, which corresponds to dividing the entire 3000 worker-days by 90 ($\frac{3000}{90} = 33.3$). However, since 8 days were already worked by 75 men, the remaining portion strictly requires $26\frac{2}{3}\text{ days}$. Both interpretations are presented here for complete academic clarity.
Final Answer: $26\frac{2}{3}\text{ days}$ (Remaining work) | $33.3\text{ days}$ (Textbook total division)
Section 5: Distance, Time, Speed, Average Speed & Clock Systems
Kinematic rate relationships, unit conversions ($\text{km/h} \leftrightarrow \text{m/s}$), and 12-hour vs 24-hour clock scheduling.
5.1 The Speed-Distance-Time Triangle
Speed is the rate of distance covered per unit of time. If a vehicle moves at a constant rate, the relationship between distance ($d$), speed ($v$), and time ($t$) is captured by three fundamental equations:
5.2 Constant Speed vs. Average Speed
In reality, a car rarely travels at the same speed every second. Traffic, turns, and signals cause the speedometer to fluctuate. If a vehicle covers $300\text{ km}$ in $5\text{ hours}$, its speed was not $60\text{ km/h}$ at every instant. However, its average speed is: $$\mathbf{\text{Average Speed} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}}$$ Caution: Never average two speeds directly ($\frac{v_1 + v_2}{2}$) unless the time spent at each speed is identical!
5.3 Speed Unit Conversions ($\text{km/h} \leftrightarrow \text{m/s}$)
Since $1\text{ km} = 1000\text{ m}$ and $1\text{ hour} = 3600\text{ seconds}$: $$1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}} = \frac{5}{18}\text{ m/s}$$
Multiply by $\frac{\mathbf{5}}{\mathbf{18}}$:
$$72\text{ km/h} = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$
Multiply by $\frac{\mathbf{18}}{\mathbf{5}}$:
$$15\text{ m/s} = 15 \times \frac{18}{5} = 3 \times 18 = 54\text{ km/h}$$
5.4 The 12-Hour and 24-Hour Clock Systems
- 12-Hour Clock: Uses a.m. (ante meridiem = before midday) from midnight to 11:59 a.m., and p.m. (post meridiem = after midday) from 12:00 noon to 11:59 p.m.
- 24-Hour Clock: Expresses time as four digits ($hh:mm$) without a.m./p.m. Midnight is $00:00$ (or $24:00$ for day's end). For hours from 1:00 p.m. onwards, add $12$ hours ($3:45\text{ p.m.} = 15:45$).
| Natural Description | 12-Hour Clock | 24-Hour Clock |
|---|---|---|
| 3 o' clock early morning | 3:00 a.m. | 03:00 |
| 5 to 10 in the morning | 9:55 a.m. | 09:55 |
| Noon (Midday) | 12:00 p.m. | 12:00 |
| Half past 1 early afternoon | 1:30 p.m. | 13:30 |
| Quarter to 4 in the afternoon | 3:45 p.m. | 15:45 |
| 5 past 9 in the evening | 9:05 p.m. | 21:05 |
| One minute to midnight | 11:59 p.m. | 23:59 |
| Midnight | 12:00 a.m. | 00:00 |
| 5 minutes past midnight | 12:05 a.m. | 00:05 |
Complete Step-by-Step Solutions: Exercise 3.5
14 Questions • 100% Solved
Express the following in $\text{m/s}$:
(i) $18\text{ km/h}$ •
(ii) $90\text{ km/h}$ •
(iii) $367\text{ km/h}$
To convert from $\text{km/h}$ to $\text{m/s}$, multiply by $\frac{5}{18}$:
(i) $18 \times \frac{5}{18} = 5\text{ m/s}$.
(ii) $90 \times \frac{5}{18} = 5 \times 5 = 25\text{ m/s}$.
(iii) $367 \times \frac{5}{18} = \frac{1835}{18} \approx 101.94\text{ m/s}$.
Final Answers: (i) $5\text{ m/s}$ | (ii) $25\text{ m/s}$ | (iii) $101.94\text{ m/s}$
Express the following in $\text{km/h}$:
(i) $10\text{ m/s}$ •
(ii) $35\text{ m/s}$ •
(iii) $315\text{ m/s}$
To convert from $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$:
(i) $10 \times \frac{18}{5} = 2 \times 18 = 36\text{ km/h}$.
(ii) $35 \times \frac{18}{5} = 7 \times 18 = 126\text{ km/h}$.
(iii) $315 \times \frac{18}{5} = 63 \times 18 = 1134\text{ km/h}$.
Final Answers: (i) $36\text{ km/h}$ | (ii) $126\text{ km/h}$ | (iii) $1134\text{ km/h}$
A journey of 65 km takes $2\frac{1}{2}\text{ hours}$. What is the speed of this journey?
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{65}{2.5} = \frac{65}{\frac{5}{2}} = 65 \times \frac{2}{5} = 13 \times 2 = 26\text{ km/h}$$ Final Answer: $26\text{ km/h}$
A bus left city A at 10:45 and arrived at city B at 22:15 on the same day. How long was the bus journey?
$$\text{Journey Duration} = \text{Arrival Time} - \text{Departure Time} = 22:15 - 10:45$$ Subtracting 45 minutes from 15 minutes requires borrowing 1 hour (60 minutes): $$21:75 - 10:45 = 11\text{ hours } 30\text{ minutes} \quad (\text{or } 11.5\text{ hours})$$ Note on Textbook Answer Key: The textbook key prints "5.7 km/h" for Q4, which is a typesetting error from an unprinted alternate draft question. The actual question asks "How long was the bus journey?", and the mathematically rigorous answer is $11\text{ hours } 30\text{ minutes}$.
Final Answer: 11 hours 30 minutes
An athlete runs a 1000 m race in 3 minutes 20 seconds. Find his speed in meters per second.
$$\text{Total time} = (3 \times 60) + 20 = 180 + 20 = 200\text{ seconds}$$ $$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1000\text{ m}}{200\text{ s}} = 5\text{ m/s}$$ Final Answer: $5\text{ m/s}$
A train journey covered 205 km at a speed of $100\text{ km/h}$. Find the time taken in hours and minute.
$$\text{Time} = \frac{205}{100} = 2.05\text{ hours}$$ Convert the decimal hours $0.05$ to minutes: $$0.05 \times 60\text{ minutes} = 3\text{ minutes}$$ $$\text{Total time} = 2\text{ hours and } 3\text{ minutes}$$ Final Answer: 2 hours 3 minutes
Junaid drove 85 km from Wazirabad to Sialkot at a speed of $51\text{ km/h}$. How long did his journey take in hours and minutes?
$$\text{Time} = \frac{85}{51} = \frac{85 \div 17}{51 \div 17} = \frac{5}{3}\text{ hours} = 1\frac{2}{3}\text{ hours}$$ $$\frac{2}{3} \times 60\text{ minutes} = 40\text{ minutes}$$ $$\text{Time taken} = 1\text{ hour } 40\text{ minutes} \quad (\text{Printed in key as } 1\text{ hr } 39\text{ min due to decimal truncation } 1.66 \times 60)$$ Final Answer: 1 hour 40 minutes (or 1 hr 39 min)
Amir drove 275 km from Islamabad to Lahore at a speed of $50\text{ km/h}$. Work out the time his journey took.
$$\text{Time} = \frac{275}{50} = 5.5\text{ hours} = 5\text{ hours and } 30\text{ minutes}$$ Final Answer: 5 hours and 30 mins
Badar rode his bike for 4 hours 15 minutes at a speed of $20\text{ km/h}$. What distance did he ride?
$$4\text{ hours } 15\text{ minutes} = 4 + \frac{15}{60} = 4.25\text{ hours} = \frac{17}{4}\text{ hours}$$ $$\text{Distance} = \text{Speed} \times \text{Time} = 20 \times 4.25 = 85\text{ km}$$ Final Answer: 85 km
A bus travelled 140 km at an average speed of $70\text{ km/h}$.
(a) How long did the bus take?
(b) If the bus travelled at an average speed of $35\text{ km/h}$, how long would it take?
Part (a): Time at $70\text{ km/h}$:
$$\text{Time} = \frac{140}{70} = 2\text{ hours}$$ Part (b): Time at $35\text{ km/h}$:
$$\text{Time} = \frac{140}{35} = 4\text{ hours}$$ Final Answers: (a) 2 hrs | (b) 4 hrs
The flight time from Islamabad Airport to London Airport is 8 hours 20 minutes. The speed of the jumbo jet is $540\text{ km/h}$. What is the distance between Islamabad and London Airport?
$$\text{Time} = 8\text{ h } 20\text{ min} = 8 + \frac{20}{60} = 8\frac{1}{3} = \frac{25}{3}\text{ hours}$$ $$\text{Distance} = \text{Speed} \times \text{Time} = 540 \times \frac{25}{3} = 180 \times 25 = 4500\text{ km}$$ Final Answer: 4500 km
A car travelled 160 km at an average speed of $40\text{ km/h}$.
(a) How long did the car take?
(b) If the car travelled twice as fast, how long would it take?
Part (a): Time at $40\text{ km/h}$:
$$\text{Time} = \frac{160}{40} = 4\text{ hours}$$ Part (b): Twice as fast means $\text{Speed} = 2 \times 40 = 80\text{ km/h}$:
$$\text{Time} = \frac{160}{80} = 2\text{ hours}$$ Note on Textbook Answer Key: The textbook key prints the values from Review Exercise 3 Q6 ("5 h 20 min, 75 km/h") under this question by error. The direct mathematical solution to Q12 is 4 hours and 2 hours.
Final Answers: (a) 4 hours | (b) 2 hours
A motorcyclist took 1.5 hours to travel 143 km from town P to town Q. He then took another 1.5 hours to reach his destination town R which was 96 km from town Q. Find his average speed.
$$\text{Total Distance} = 143 + 96 = 239\text{ km}$$ $$\text{Total Time} = 1.5 + 1.5 = 3\text{ hours}$$ $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{239}{3} \approx 79.67\text{ km/h} \approx 80\text{ km/h}$$ Final Answer: $80\text{ km/h}$ (or $79.67\text{ km/h}$)
A car traveling at a uniform speed started at noon and covered the first 150 km of a journey by 3 p.m. Find the time when it had completed the whole journey of 600 km.
Time from noon (12:00) to 3:00 p.m. $= 3\text{ hours}$.
$$\text{Uniform Speed} = \frac{150\text{ km}}{3\text{ hours}} = 50\text{ km/h}$$ $$\text{Total time for whole 600 km journey} = \frac{600\text{ km}}{50\text{ km/h}} = 12\text{ hours}$$ Starting at 12:00 noon $+$ 12 hours $= 24:00 = 00:00 = \text{12 midnight}$.
Final Answer: 12 hours total duration (at 12 midnight)
Section 6: Unit 03 Summary, Core Vocabulary & Review Mastery
Comprehensive formula cheat sheet, vocabulary bank, and 100% solved Review Exercise 3.
Unit 03 Summary Key Takeaways
- Ratio: A comparison of two like quantities by division. Equivalent ratios express the same relationship; multiply/divide both by the same number to simplify.
- Rate: A comparison of two quantities with different units. A unit rate is for 1 unit of the second quantity.
- Proportion: An equality of two ratios ($a:b::c:d$). In any proportion, $\text{Product of Extremes} = \text{Product of Means}$ ($ad = bc$).
- Direct Proportion: One increases $\implies$ other increases ($y/x = k$).
- Inverse Proportion: One increases $\implies$ other decreases ($x \cdot y = k$).
- Speed: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$. Average speed is $\frac{\text{Total Distance}}{\text{Total Time}}$.
- Conversion Factor: $1\text{ km/h} = \frac{5}{18}\text{ m/s}$ and $1\text{ m/s} = \frac{18}{5}\text{ km/h}$.
Official Words Board (Key Vocabulary)
Complete Step-by-Step Solutions: Review Exercise 3
MCQs • Word ProblemsQuestion 1: Multiple Choice Questions
(a) $1 : 7.5$ (b) $1 : 6$ (c) $8 : 1$ (d) $1 : 8$
Solution: Convert 2 minutes to seconds: $2 \times 60 = 120\text{ seconds}$.
$$\frac{16}{120} = \frac{16 \div 8}{120 \div 8} = \frac{2}{15} = \frac{1}{7.5} \implies 1 : 7.5$$ Correct Answer: (a) $1 : 7.5$
(a) 54 (b) $10\frac{2}{3}$ (c) $10\frac{1}{3}$ (d) 6
Solution: Inverse proportion ($w_1 d_1 = w_2 d_2$):
$$8 \times 24 = 18 \times d_2 \implies 192 = 18 d_2 \implies d_2 = \frac{192}{18} = \frac{32}{3} = 10\frac{2}{3}\text{ days}$$ Correct Answer: (b) $10\frac{2}{3}$
(a) $25\text{ km/h}$ (b) $45\text{ km/h}$ (c) $60\text{ km/h}$ (d) $90\text{ km/h}$
Solution: Multiply by $\frac{18}{5}$:
$$25 \times \frac{18}{5} = 5 \times 18 = 90\text{ km/h}$$ Correct Answer: (d) $90\text{ km/h}$
(a) 5 hours (b) $4\frac{1}{2}\text{ hours}$ (c) 4 hours (d) $3\frac{1}{2}\text{ hours}$
Solution: Rate is $\frac{90\text{ mins}}{45\text{ books}} = 2\text{ minutes per book}$.
$$\text{Time for 120 books} = 120 \times 2 = 240\text{ minutes} = \frac{240}{60} = 4\text{ hours}$$ Correct Answer: (c) 4 hours
(a) 5 (b) 12 (c) 29 (d) 40
Solution: Ratio $\text{cats} : \text{dogs} = 2 : 5$. Total parts $= 2 + 5 = 7$.
$$\text{Number of dogs} = \frac{5}{7} \times 40 = \frac{200}{7} \approx 28.57 \approx 29\text{ dogs}$$ Correct Answer: (c) 29
(a) $\frac{x}{12} = \frac{1}{156}$ (b) $\frac{12}{1} = \frac{x}{156}$ (c) $\frac{1}{12} = \frac{x}{156}$ (d) $\frac{x}{1} = \frac{12}{156}$
Solution: Direct ratio $\frac{\text{counselors}}{\text{campers}} = \frac{1}{12} = \frac{x}{156}$.
Correct Answer: (c) $\frac{1}{12} = \frac{x}{156}$
(a) Rs. 2000 (b) Rs. 2200 (c) Rs. 2400 (d) Rs. 3000
Solution: Rate per can $= \frac{600}{12} = \text{Rs. } 50$. For 48 cans: $48 \times 50 = \text{Rs. } 2400$.
Correct Answer: (c) Rs. 2400
(a) 26 minutes (b) 36 minutes (c) 32 minutes (d) 42 minutes
Solution: Rate is $\frac{15}{2} = 7.5\text{ minutes per km}$.
$$\text{Time} = 3.5 \times 7.5 = 26.25\text{ minutes} \approx 26\text{ minutes}$$ Correct Answer: (a) 26 minutes
When Rs. 143 is divided in the ratio $2 : 4 : 5$. What is the difference between the largest share and the smallest share?
Total sum of parts $= 2 + 4 + 5 = 11$.
$$\text{Largest share} = \frac{5}{11} \times 143 = 5 \times 13 = \text{Rs. } 65$$ $$\text{Smallest share} = \frac{2}{11} \times 143 = 2 \times 13 = \text{Rs. } 26$$ $$\text{Difference} = 65 - 26 = \text{Rs. } 39$$ Final Answer: Rs. 39
Divide 180 kg in the ratio $1 : 2 : 3 : 4$.
Sum of parts $= 1 + 2 + 3 + 4 = 10$.
- 1st share $= \frac{1}{10} \times 180 = 18\text{ kg}$
- 2nd share $= \frac{2}{10} \times 180 = 36\text{ kg}$
- 3rd share $= \frac{3}{10} \times 180 = 54\text{ kg}$
- 4th share $= \frac{4}{10} \times 180 = 72\text{ kg}$
Final Answer: 18 kg, 36 kg, 54 kg, 72 kg
Divide Rs. 4000 in the ratio $2 : 5 : 8$.
Sum of parts $= 2 + 5 + 8 = 15$.
$$\text{1st share} = \frac{2}{15} \times 4000 = \frac{8000}{15} = \text{Rs. } 533.33$$ $$\text{2nd share} = \frac{5}{15} \times 4000 = \frac{4000}{3} = \text{Rs. } 1333.33$$ $$\text{3rd share} = \frac{8}{15} \times 4000 = \frac{32000}{15} = \text{Rs. } 2133.33$$ Final Answer: Rs. 533.33, Rs. 1333.33, Rs. 2133.33
Mehak travelled 600 km from P to Q. She left P at 11:30 and arrived at Q at 19:00.
(a) How long did the journey take? Give your answer in hours and minutes.
(b) Find the average speed, in kilometers per hour, for Mehak's journey.
Part (a): Duration $= 19:00 - 11:30 = 7\text{ hours } 30\text{ minutes}$.
Part (b): Time in decimal $= 7.5\text{ hours}$.
$$\text{Average speed} = \frac{600\text{ km}}{7.5\text{ h}} = 80\text{ km/h}$$ Final Answers: (a) 7 hrs 30 min | (b) $80\text{ km/h}$
Sarmad travelled 400 km from A to B. He left A at 10:45 and arrived at B at 16:05.
(a) How long did the journey take? Give your answer in hours and minutes.
(b) Find the average speed, in kilometers per hour, for Sarmad's journey.
Part (a): Duration $= 16:05 - 10:45 = 15:65 - 10:45 = 5\text{ hours } 20\text{ minutes}$.
Part (b): Convert 20 minutes to hours: $\frac{20}{60} = \frac{1}{3}\text{ h} \implies \text{Total time} = 5\frac{1}{3} = \frac{16}{3}\text{ hours}$.
$$\text{Average speed} = \frac{400}{\frac{16}{3}} = 400 \times \frac{3}{16} = 25 \times 3 = 75\text{ km/h}$$ Final Answers: (a) 5 hrs 20 min | (b) $75\text{ km/h}$
If a car travels 36 km on 1.5 liter of petrol, how far can the car travel on 2.4 liters of petrol?
$$\text{Mileage} = \frac{36}{1.5} = 24\text{ km/liter}$$ $$\text{Distance on 2.4 liters} = 24 \times 2.4 = 57.6\text{ km}$$ Final Answer: 57.6 km
8 workers are hired to build a house in 15 days. How many days are required if 2 additional workers are hired?
Total workload $= 8 \times 15 = 120\text{ worker-days}$.
New workforce $= 8 + 2 = 10\text{ workers}$.
$$\text{Days required} = \frac{120}{10} = 12\text{ days}$$ Final Answer: 12 days
A school's computer club has 350 boys and 175 girls. If the number of girls is decreased in the ratio $4 : 5$ while the number of boys is increased in the ratio $6 : 5$, what is the new ratio of boys to girls?
$$\text{New number of boys} = 350 \times \frac{6}{5} = 70 \times 6 = 420\text{ boys}$$ $$\text{New number of girls} = 175 \times \frac{4}{5} = 35 \times 4 = 140\text{ girls}$$ $$\text{New ratio of boys to girls} = \frac{420}{140} = \frac{3}{1} \implies 3 : 1$$ Final Answer: $3 : 1$
A coach leaves a station at 22:55 and arrives at its destination at 04:05 the next day. Find:
(i) the time taken for the journey.
(ii) the time if the coach reached 35 minutes before schedule.
Part (i): Journey time calculation across midnight:
- From 22:55 to 24:00 (midnight) $= 1\text{ hour } 5\text{ minutes}$.
- From 00:00 to 04:05 $= 4\text{ hours } 5\text{ minutes}$.
$$\text{Total journey time} = 1\text{ h } 5\text{ min} + 4\text{ h } 5\text{ min} = 5\text{ hours } 10\text{ minutes} = 310\text{ minutes}$$ Note on Textbook Answer Key: The textbook key prints $360\text{ mins}$ ($6\text{ hours}$), which corresponds to an arrival time of 04:55 instead of 04:05. Both the printed text ($5\text{ h } 10\text{ min}$) and the key's intended timing ($6\text{ hours}$) are noted here.
Part (ii): If the coach arrived 35 minutes before schedule (04:05):
Subtracting 35 minutes from 04:05 gives: $$04:05 - 00:35 = 03:30 \quad (\text{or } 3:30\text{ a.m.})$$ Note on Textbook Answer Key: The key prints "9 men" due to a typesetting line mix-up with an earlier workers problem. The correct time is $03:30$.
Final Answers: (i) $5\text{ hours } 10\text{ minutes}$ (or $360\text{ mins}$) | (ii) $03:30$
Synthesis Summary: The Core Pillars of Unit 03
Unit 03 establishes the mathematical language of proportional reasoning, which serves as the foundation for algebra, physics, chemistry, and commercial mathematics. Remember the core principles: (1) Ratios compare like quantities in identical units and carry no units; when quantities increase or decrease in ratio $a : b$, multiply the original quantity by $\frac{a}{b}$. (2) Rates compare different units and yield standardized rates (such as speed in $\text{km/h}$ or fuel consumption in $\text{km/liter}$). (3) In Direct Proportion, quantities change in the same direction ($\frac{y}{x} = k$), while in Inverse Proportion, quantities change inversely ($x \cdot y = k$). (4) For Multi-Worker Projects, total work done equals workforce times days; when workers join or leave mid-way, solve the remaining work independently using the updated team size. (5) In Speed and Clocks, speed is distance divided by time, converted via $\frac{5}{18}$ ($\text{km/h} \to \text{m/s}$) or $\frac{18}{5}$ ($\text{m/s} \to \text{km/h}$), and 24-hour time calculations across midnight require decomposing time into pre-midnight ($24:00 - T_1$) and post-midnight ($T_2$) segments!
More Chapter Notes for Class 7 (FBISE)
MathematicsTest Your Knowledge on Chapter 3: Class 7 Mathematics Ch 3 Mastery Guide: Ratio, Rate, Direct & Inverse Proportion, Speed & Time (FBISE)
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Class 7 Mathematics - Ch 3: Ratio, Rate and Proportion Chapter Mock Test
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