Official Class 8 Mathematics Worksheets: Ganita Prakash Part 2 Chapter 03 Proportional Reasoning 2
Access comprehensive chapter-wise worksheets for Ganita Prakash Part 2 Chapter 03 Proportional Reasoning 2 using the CBSE Class 8 Mathematics Proportional Reasoning 2 Long Answer and Case Study Questions with Answers. Designed to align with the 2026-27 academic syllabus for Class 8 Mathematics, these printable practice sets help students reinforce key concepts and improve their overall exam readiness.
Solved Practice Worksheets for Mathematics
Access the complete worksheet PDF for Class 8 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school examinations.
Long Answer Questions
Question 1. Idli batter, spice powders and paints are all made by mixing ingredients in a fixed proportion.
(i) State the condition under which two ratios a : b and c : d are proportional, and use the cross-multiplication method to test whether 14 : 21 and 20 : 30 are proportional.
(ii) A spice powder is ground from spoonfuls of coriander seeds, cumin seeds, toor dal and fenugreek seeds in the ratio 6 : 3 : 2 : 1. A cook has 9 spoons of coriander seeds and wants exactly the same taste. Find the quantity of each of the other three ingredients.
(iii) Write the complete four-term proportionality statement for the two spice powders and explain why both powders taste the same.
Answer:
(i) a : b and c : d are proportional when a × d = b × c, that is, \( \frac{a}{b} \) = \( \frac{c}{d} \).
For 14 : 21 and 20 : 30: 14 × 30 = 420 and 21 × 20 = 420. The products are equal, so the ratios are proportional.
(ii) Coriander : cumin : toor dal : fenugreek = 6 : 3 : 2 : 1. Coriander is 6 parts = 9 spoons, so one part = 9 ÷ 6 = 1.5 spoons.
Cumin = 3 × 1.5 = 4.5 spoons, toor dal = 2 × 1.5 = 3 spoons, fenugreek = 1 × 1.5 = 1.5 spoons.
(iii) 6 : 3 : 2 : 1 :: 9 : 4.5 : 3 : 1.5.
Dividing each term of the second ratio by the matching term of the first: \( \frac{9}{6} \) = \( \frac{4.5}{3} \) = \( \frac{3}{2} \) = \( \frac{1.5}{1} \) = 1.5.
Every ingredient has been multiplied by the same factor 1.5, so the share of each spice in the mix is unchanged and both powders taste the same.
In simple words: 14 : 21 and 20 : 30 are proportional; the cook uses 4.5, 3 and 1.5 spoons, scaling every spice by 1.5 so the taste is the same.
Teacher's Note:
1. Both 14 : 21 and 20 : 30 reduce to 2 : 3.
2. The same factor on every term keeps a many-term ratio proportional.
3. Decimal amounts like 4.5 spoons are fine in recipes.
Question 2. An atlas page of South India carries the Representative Fraction RF 1 : 40,00,000 in its lower right corner.
(i) Explain in your own words what a Representative Fraction is and why a map must show one.
(ii) Show that 1 cm on this map stands for 40 km on the ground.
(iii) Two cities are 7.5 cm apart on the page. Find the geographical distance between them.
(iv) A third city is 180 km from the first. How far from it should it be marked on this page?
Answer:
(i) A Representative Fraction is a ratio that compares a distance measured on the map with the matching distance on the ground. A map is a shrunk picture of the land; without the RF we could measure lengths on the paper but could not turn them into real distances.
(ii) RF 1 : 40,00,000 means 1 cm stands for 40,00,000 cm. Since 1 km = 1,00,000 cm, 40,00,000 cm = \( \frac{40,00,000}{1,00,000} \) km = 40 km.
(iii) Distance = 7.5 × 40 = 300 km.
(iv) Map distance = 180 ÷ 40 = 4.5 cm. Check: 4.5 × 40 = 180 km.
These are geographical (straight-line) distances, not road distances.
In simple words: On this map each cm is 40 km, so 7.5 cm is 300 km and 180 km is drawn as 4.5 cm.
Teacher's Note:
1. Map to ground: multiply; ground to map: divide.
2. Always state that scale distances are straight-line distances.
Question 3. Dividing a whole in a given ratio is used in trade as well as in geometry.
(i) Write the general rule for dividing a quantity x in the ratio a : b : c.
(ii) A trader blends 48 kg of tea from three varieties A, B and C in the ratio 3 : 2 : 1. Find the mass of each variety used.
(iii) The angles of a triangle are in the ratio 1 : 2 : 3. Find the three angles and name the type of triangle obtained.
Answer:
(i) Add the terms and give each share its fraction of x: the parts are \( x \times \frac{a}{a+b+c} \), \( x \times \frac{b}{a+b+c} \) and \( x \times \frac{c}{a+b+c} \).
(ii) Here x = 48 kg and the terms add to 3 + 2 + 1 = 6, so one part = 48 ÷ 6 = 8 kg.
Variety A = 3 × 8 = 24 kg; variety B = 2 × 8 = 16 kg; variety C = 1 × 8 = 8 kg. Check: 24 + 16 + 8 = 48 kg.
(iii) The angle sum 180° is divided in the ratio 1 : 2 : 3. The terms add to 6, so one part = 180° ÷ 6 = 30°.
The angles are 30°, 60° and 90°. One angle is 90°, so it is a right-angled triangle.
In simple words: Each share is its fraction of the whole: the tea is 24, 16 and 8 kg, and the triangle has angles 30°, 60°, 90°.
Teacher's Note:
1. The same rule works for money, mass or angles.
2. In (iii) the whole is 180°, the angle sum of a triangle.
| Sport | Cricket | Football | Kabaddi | Hockey |
|---|---|---|---|---|
| Number of students | 20 | 15 | 15 | 10 |
(ii) Draw a pie chart for this data, writing down the steps of construction you follow.
(iii) Name the two sports whose slices are of equal size and give the fraction of the circle taken by cricket.
Answer:
(i) 360° stands for 60 students, so one student stands for 6°.
Cricket = \( \frac{20}{60} \) × 360° = 120°; football = \( \frac{15}{60} \) × 360° = 90°; kabaddi = 90°; hockey = \( \frac{10}{60} \) × 360° = 60°.
Check: 120° + 90° + 90° + 60° = 360°.
(ii) Step 1: Draw a circle with centre A and draw one radius AB.
Step 2: With the protractor at A, measure 120° from AB and draw radius AC. Sector BAC is cricket.
Step 3: Measure 90° from AC and draw AD, giving the football sector.
Step 4: Measure 90° from AD and draw AE, giving the kabaddi sector.
Step 5: The remaining sector should measure 60° (hockey); this checks the drawing.
Step 6: Shade and label each slice with the name of the sport.
(iii) Football and kabaddi have equal slices of 90° each, because 15 students chose each. Cricket takes \( \frac{120}{360} \) = \( \frac{1}{3} \) of the circle.
In simple words: The slices are 120°, 90°, 90° and 60°; football and kabaddi are equal, and cricket is one-third of the circle.
Teacher's Note:
1. Find the angle for one student first.
2. Measure each new angle from the last radius drawn.
3. The last sector is a free check of the drawing.
Question 5. The pie chart below shows how a family of monthly income ₹18,000 spends the whole of it. The central angles are marked on the chart.
(i) Write the amount spent under each of the four heads.
(ii) Verify that the four amounts add up to the monthly income.
(iii) Write the ratio food : rent : travel : savings in its simplest form and check that the same ratio is obtained from the four angles.
Answer:
(i) 360° stands for ₹18,000, so 1° stands for ₹50.
Food = \( \frac{144}{360} \) × 18000 = ₹7200; rent = \( \frac{90}{360} \) × 18000 = ₹4500.
Travel = \( \frac{72}{360} \) × 18000 = ₹3600; savings = \( \frac{54}{360} \) × 18000 = ₹2700.
(ii) ₹7200 + ₹4500 + ₹3600 + ₹2700 = ₹18,000, the whole income. Also 144° + 90° + 72° + 54° = 360°.
(iii) From the amounts: 7200 : 4500 : 3600 : 2700 = 8 : 5 : 4 : 3 (dividing each term by 900).
From the angles: 144 : 90 : 72 : 54 = 8 : 5 : 4 : 3 (dividing each term by 18).
The two agree, because each angle is proportional to the amount it stands for.
In simple words: Each degree is ₹50: food ₹7200, rent ₹4500, travel ₹3600, savings ₹2700, in the ratio 8 : 5 : 4 : 3.
Teacher's Note:
1. ₹50 per degree makes every amount a quick multiplication.
2. Angles and amounts always give the same simplest ratio.
(i) Explain, with the table below, why the number of workers and the number of days are in inverse proportion.
| Number of workers | 36 | 45 | 50 |
|---|---|---|---|
| Number of days | 25 | ? | ? |
(iii) The wall must now be finished in 15 days. How many workers are needed?
(iv) State two assumptions you have made while solving this problem.
Answer:
(i) The wall is of a fixed size, so the total work is fixed. More workers build more each day and need fewer days; one quantity grows by the same factor as the other shrinks.
Their product (workers × days) is the total work and stays constant: 36 × 25 = 900 worker-days.
(ii) 45 workers: days = \( \frac{900}{45} \) = 20. 50 workers: days = \( \frac{900}{50} \) = 18. Check: 45 × 20 = 900 and 50 × 18 = 900.
(iii) Workers needed = \( \frac{900}{15} \) = 60 workers.
The days fell from 25 to 15 (factor \( \frac{3}{5} \)) and the workers rose from 36 to 60 (factor \( \frac{5}{3} \)), the inverse factor.
(iv) Every worker works at the same rate for the same hours each day; and all of them can work on the wall at once without getting in each other's way, with no stoppage for material or weather.
In simple words: The work is 900 worker-days, so 45 workers take 20 days, 50 take 18 days, and 15 days need 60 workers.
Teacher's Note:
1. Write the constant with its unit, worker-days.
2. Inverse factors: 3/5 for days, 5/3 for workers.
3. Assumptions are what make the product truly constant.
Question 7. Concrete is mixed as cement : sand : gravel :: 1 : 1.5 : 3.
(i) A slab needs 132 units of concrete. Find the units of cement, sand and gravel required, and verify that they add up to 132.
(ii) On another day only 18 units of sand are available and the builder wants to use all of it in the same ratio. Find the units of cement and gravel to be mixed with it and the total concrete obtained.
(iii) Explain why the answer in part (ii) cannot be increased by simply adding more cement.
Answer:
(i) The terms add to 1 + 1.5 + 3 = 5.5, so one part = 132 ÷ 5.5 = 24 units.
Cement = 1 × 24 = 24 units; sand = 1.5 × 24 = 36 units; gravel = 3 × 24 = 72 units.
Check: 24 + 36 + 72 = 132, and 24 : 36 : 72 = 1 : 1.5 : 3 on dividing by 24.
(ii) Sand is 1.5 parts = 18 units, so one part = 18 ÷ 1.5 = 12 units.
Cement = 12 units, gravel = 3 × 12 = 36 units; total concrete = 12 + 18 + 36 = 66 units.
(iii) The three ingredients must stay in the ratio 1 : 1.5 : 3. Extra cement has no sand and gravel to go with it, so the ratio would change and the mix would not be the concrete planned, nor have its strength. With only 18 units of sand, 66 units is the most that can be made.
In simple words: 132 units need 24, 36 and 72 units; 18 units of sand allow only 66 units of concrete, and extra cement would spoil the ratio.
Teacher's Note:
1. The scarcest ingredient decides how much mix can be made.
2. Check every answer by reducing back to 1 : 1.5 : 3.
Question 8. Two maps of the same town are printed with different scales: Map P has RF 1 : 30,000 and Map Q has RF 1 : 1,50,000.
(i) Find the ground distance, in kilometres, represented by 1 cm on each map.
(ii) A park is 2.4 km long. Find the length of the park as drawn on Map P and on Map Q.
(iii) A canal measures 6 cm on Map Q. Find its real length, and the length it would measure on Map P.
(iv) Which of the two maps shows more detail of the town? Give a reason.
Answer:
(i) Map P: 1 cm stands for 30,000 cm = 0.3 km. Map Q: 1 cm stands for 1,50,000 cm = 1.5 km.
(ii) 2.4 km = 2,40,000 cm. On Map P: \( \frac{2,40,000}{30,000} \) = 8 cm. On Map Q: \( \frac{2,40,000}{1,50,000} \) = 1.6 cm.
(iii) On Map Q, 6 cm stands for 6 × 1,50,000 = 9,00,000 cm = 9 km. On Map P it would be \( \frac{9,00,000}{30,000} \) = 30 cm long.
(iv) Map P. Its RF 1 : 30,000 shrinks the ground less than Map Q's 1 : 1,50,000, so the same land is drawn five times longer on Map P, and small features like lanes and parks can be shown. Map Q covers more area on the same sheet but with less detail.
In simple words: Map P shows 0.3 km per cm and Map Q 1.5 km per cm, so Map P draws everything five times longer and shows more detail.
Teacher's Note:
1. The smaller the second term of the RF, the larger the drawing.
2. 1,50,000 ÷ 30,000 = 5 explains the factor of five.
Question 9. The pie chart below shows the modes of transport used by the children of a school. The central angles are marked on the chart. It is known that 30 children come by two-wheeler.
(i) Find the total number of children surveyed.
(ii) Find the number of children using each of the other four modes of transport, and verify your five numbers against the total.
(iii) Write the ratio walk : bus : cycle : two-wheeler : car in its simplest form, obtaining it once from the angles and once from the numbers of children, and comment on what you notice.
Answer:
(i) The two-wheeler sector is 54° and stands for 30 children, so 1° stands for \( \frac{30}{54} \) = \( \frac{5}{9} \) child. Total = 360 × \( \frac{5}{9} \) = 200 children.
(ii) Each number = \( \frac{\text{angle}}{360} \) × 200.
Walk = \( \frac{108}{360} \) × 200 = 60; bus = \( \frac{90}{360} \) × 200 = 50; cycle = \( \frac{72}{360} \) × 200 = 40; car = \( \frac{36}{360} \) × 200 = 20.
Check: 60 + 50 + 40 + 30 + 20 = 200, the total found in (i).
(iii) From the angles: 108 : 90 : 72 : 54 : 36 = 6 : 5 : 4 : 3 : 2 (HCF 18).
From the children: 60 : 50 : 40 : 30 : 20 = 6 : 5 : 4 : 3 : 2 (HCF 10).
The two are the same, because every angle is \( \frac{360}{200} \) = 1.8 times the number of children it stands for. So a pie chart can be read as a ratio straight from its angles, even without knowing the total.
In simple words: 200 children were surveyed: 60 walk, 50 bus, 40 cycle, 30 two-wheeler and 20 car, in the ratio 6 : 5 : 4 : 3 : 2.
Teacher's Note:
1. Use the one known slice to find the whole.
2. Angles and counts differ only by a constant factor, here 1.8.
Question 10. Pipes, pumps and machines give everyday examples of inverse proportion.
(i) Pipe A alone fills a tank in 6 hours and pipe B alone fills the same tank in 3 hours. Working out the part of the tank each pipe fills in one hour, find the time the two pipes together take to fill the tank.
(ii) A borewell with 5 pumps of the same kind fills an overhead reservoir in 12 hours. Find the time taken if 8 such pumps are used.
(iii) Using part (ii), explain why the product (number of pumps) × (time taken) stays constant, and state one assumption without which this reasoning would fail.
Answer:
(i) Take filling the tank as 1 unit of work. In one hour pipe A fills \( \frac{1}{6} \) and pipe B fills \( \frac{1}{3} \) of the tank.
Together in one hour: \( \frac{1}{6} \) + \( \frac{1}{3} \) = \( \frac{1 + 2}{6} \) = \( \frac{1}{2} \) of the tank.
So filling the whole tank takes 1 ÷ \( \frac{1}{2} \) = 2 hours.
(ii) Pumps and time are inversely proportional: 5 × 12 = 8 × t, so t = \( \frac{60}{8} \) = 7.5 hours (7 hours 30 minutes). Check: 8 × 7.5 = 60.
(iii) The constant 60 is the work needed to fill the reservoir, in pump-hours. n pumps running for t hours do nt pump-hours, and the job is always the same size, so nt must always be 60.
This fails unless every pump delivers water at the same steady rate (and the pumps do not slow one another down by drawing from the same source).
In simple words: Together the pipes fill half the tank each hour, so 2 hours; 8 pumps need 60 ÷ 8 = 7.5 hours because the job is always 60 pump-hours.
Teacher's Note:
1. For joint work, add the parts done in one hour.
2. The constant product is the size of the job.
3. Equal, steady rates are the hidden assumption.
Case Study Based Questions
The panchayat of Kondapur village prepared a record of how the village land is used. The whole of the village land is put to one of four uses only, and the gram sabha wants the record shown as a pie chart on the notice board so that everyone can read it at a glance. The areas recorded by the panchayat are given in the table below.
| Land use | Farming | Forest | Housing | Pond and canals |
|---|---|---|---|---|
| Area (hectares) | 180 | 90 | 60 | 30 |
(ii) Write the ratio farming : forest : housing : pond and canals in its simplest form.
(iii) Find the central angle of the sector that will represent the forest land in the pie chart.
(iv) What fraction of the village land is used for housing?
Answer:
(i) Total land = 180 + 90 + 60 + 30 = 360 hectares.
(ii) 180 : 90 : 60 : 30. The HCF of the terms is 30, so the simplest form is 6 : 3 : 2 : 1.
(iii) Angle for forest = \( \frac{90}{360} \) × 360° = 90°. (Area in hectares and angle in degrees match here because the total is 360 hectares.)
(iv) Fraction for housing = \( \frac{60}{360} \) = \( \frac{1}{6} \) of the village land.
In simple words: The village has 360 hectares in the ratio 6 : 3 : 2 : 1; forest gets 90° and housing is one-sixth of the land.
Teacher's Note:
1. A total of 360 makes each hectare exactly 1°.
2. Simplify with the HCF of all four terms.
Sumitra earns ₹24,000 a month and plans her spending before the month begins. She divides the whole of her income among food, rent, education and savings in the ratio shown in the table, and she does not change the plan during the month.
| Head | Food | Rent | Education | Savings |
|---|---|---|---|---|
| Parts of the ratio | 5 | 4 | 2 | 1 |
(ii) How much money does one part of the ratio stand for?
(iii) Find the amount kept for food and the amount kept for savings.
(iv) By how much does the rent exceed the amount kept for education?
Answer:
(i) Sum of the terms = 5 + 4 + 2 + 1 = 12.
(ii) One part = 24000 ÷ 12 = ₹2000.
(iii) Food = 5 × 2000 = ₹10,000; savings = 1 × 2000 = ₹2000.
(iv) Rent = 4 × 2000 = ₹8000 and education = 2 × 2000 = ₹4000, so rent exceeds education by ₹4000.
Check: ₹10,000 + ₹8000 + ₹4000 + ₹2000 = ₹24,000.
In simple words: Each part is ₹2000: food ₹10,000, savings ₹2000, and rent is ₹4000 more than education.
Teacher's Note:
1. The difference is also (4 − 2) parts = ₹4000.
2. Check that all heads add to the income.
A road map of a hilly district is printed with the Representative Fraction 1 : 20,00,000, so that 1 cm on the map stands for a geographical distance of 20 km. A cyclist measures the straight-line map distances between four halting places A, B, C and D with her ruler and notes them down in the table.
| Stretch | A to B | B to C | C to D |
|---|---|---|---|
| Distance on the map (cm) | 3 | 4.5 | 6 |
(ii) Find the geographical distance from A to B.
(iii) Find the geographical distances from B to C and from C to D.
(iv) Find the total geographical distance from A to D along these three stretches.
Answer:
(i) 1 km = 1,00,000 cm, so 20,00,000 cm = \( \frac{20,00,000}{1,00,000} \) km = 20 km.
(ii) A to B = 3 × 20 = 60 km.
(iii) B to C = 4.5 × 20 = 90 km; C to D = 6 × 20 = 120 km.
(iv) Total = 60 + 90 + 120 = 270 km. (Also: map total 3 + 4.5 + 6 = 13.5 cm, and 13.5 × 20 = 270 km.)
These are geographical distances; the actual road on the hills will be a little longer.
In simple words: Each cm is 20 km, so the stretches are 60, 90 and 120 km, 270 km in all.
Teacher's Note:
1. Adding map lengths first gives the same total.
2. Hill roads wind, so road distance is more.
A contractor notes that 12 workers of the same skill complete the plastering of a building in 15 days. He then tries out different team sizes for the same amount of plastering work and prepares the table below, so that he can promise his client a date.
| Number of workers | 12 | 15 | 18 | 20 |
|---|---|---|---|---|
| Number of days | 15 | 12 | ? | 9 |
(ii) What kind of proportion do the two quantities show? Give a reason.
(iii) Fill in the missing entry of the table.
(iv) How many days would 30 workers take for the same plastering work?
Answer:
(i) 12 × 15 = 180 and 15 × 12 = 180 worker-days.
(ii) Inverse proportion: more workers need fewer days, and the product stays the same (180) because the plastering work is fixed.
(iii) For 18 workers, days = \( \frac{180}{18} \) = 10 days.
(iv) For 30 workers, days = \( \frac{180}{30} \) = 6 days. Check: 30 × 6 = 180.
In simple words: The work is 180 worker-days, so 18 workers take 10 days and 30 workers take 6 days.
Teacher's Note:
1. The last column, 20 × 9 = 180, confirms the constant.
2. Divide the constant by the number of workers.
For the foundation of a godown, concrete is prepared by mixing cement, sand and gravel. The ratio chosen depends on the strength needed; for a foundation the site engineer uses the mix given in the table. He must prepare 98 units of concrete in all, and he insists that the ratio must be kept exactly, otherwise the foundation will not carry the load.
| Component | Cement | Sand | Gravel |
|---|---|---|---|
| Parts of the mix | 1 | 2 | 4 |
(ii) Find the units of cement, sand and gravel needed for 98 units of concrete.
(iii) Verify that your three answers add up to 98 units.
(iv) On the next day only 10 units of cement are left in the store. What is the greatest amount of concrete of the same mix that can be prepared?
Answer:
(i) Sum of the parts = 1 + 2 + 4 = 7.
(ii) One part = 98 ÷ 7 = 14 units. Cement = 14 units, sand = 2 × 14 = 28 units, gravel = 4 × 14 = 56 units.
(iii) 14 + 28 + 56 = 98 units. Also 14 : 28 : 56 = 1 : 2 : 4 on dividing by 14.
(iv) Cement is 1 part, so one part = 10 units. Greatest concrete = 7 × 10 = 70 units (with 20 units of sand and 40 units of gravel).
In simple words: 98 units need 14, 28 and 56 units; with only 10 units of cement, at most 70 units can be made.
Teacher's Note:
1. The available cement fixes the size of one part.
2. Never add extra of one ingredient alone.
Every one of the 240 students of Class 8 of a school was asked to name one favourite subject. The school wants to display the result as a pie chart in the corridor. The counts are given in the table.
| Subject | Maths | Science | Social Science | Language | Arts |
|---|---|---|---|---|---|
| Number of students | 80 | 60 | 50 | 30 | 20 |
(ii) Find the central angle of the sector for Social Science.
(iii) What percentage of the students chose Science?
(iv) Find the angles for Language and for Arts, and verify that all five angles add up to 360°.
Answer:
(i) Maths = \( \frac{80}{240} \) × 360° = 120°.
(ii) Social Science = \( \frac{50}{240} \) × 360° = 75°.
(iii) Science = \( \frac{60}{240} \) × 100 = 25%.
(iv) Language = \( \frac{30}{240} \) × 360° = 45°; Arts = \( \frac{20}{240} \) × 360° = 30°.
Science = \( \frac{60}{240} \) × 360° = 90°, so the sum = 120° + 90° + 75° + 45° + 30° = 360°; the whole circle is used.
In simple words: Each student is 1.5°: Maths 120°, Social Science 75°, Language 45°, Arts 30°, and Science is 25% (90°).
Teacher's Note:
1. 360° ÷ 240 = 1.5° per student.
2. Percentage uses 100, angle uses 360°.
The overhead tank of a housing block is filled once and then used by the families living there. The water in one filling is enough to supply 40 families for 9 days. The secretary of the block prepares the table below to see how long one filling would last if the number of families changes. Every family is assumed to use the same amount of water each day.
| Number of families | 40 | 45 | 60 | 72 |
|---|---|---|---|---|
| Number of days the water lasts | 9 | ? | 6 | ? |
(ii) Complete the two missing entries of the table.
(iii) If 90 families were to live in the block, for how many days would one filling last?
(iv) State one assumption, other than equal daily use, that is needed for this working.
Answer:
(i) Constant = 40 × 9 = 360 family-days, the water in one filling.
(ii) 45 families: days = \( \frac{360}{45} \) = 8; 72 families: days = \( \frac{360}{72} \) = 5. Check: 60 × 6 = 360 too.
(iii) 90 families: days = \( \frac{360}{90} \) = 4 days.
(iv) The tank must be filled to the same level every time, and no water may be lost by leakage or drawn by anyone else during those days.
In simple words: One filling is 360 family-days, so 45 families get 8 days, 72 get 5 days and 90 get 4 days.
Teacher's Note:
1. The given column 60 → 6 checks the constant.
2. An assumption must keep the total water fixed.
As a mapping activity, Class 8 of a school made a sketch of their own classroom on a chart paper using the scale 1 : 50, and marked the blackboard, the teacher's desk, the fans and the lights on it with suitable symbols. Each length in the sketch was measured with a ruler and entered in the table below. The pupils now want to check their sketch against the real classroom with a measuring tape.
| Item in the sketch | Blackboard | Teacher's desk | Length of the room | Width of the room |
|---|---|---|---|---|
| Length in the sketch (cm) | 8 | 2.4 | 16 | 11 |
(ii) Find the real length of the blackboard and of the teacher's desk.
(iii) Find the real length and the real width of the classroom.
(iv) A cupboard in the room is 1.5 m wide. How wide should it be drawn in the sketch?
Answer:
(i) 1 cm in the sketch stands for 50 cm in the room, that is 0.5 m.
(ii) Blackboard = 8 × 50 = 400 cm = 4 m. Teacher's desk = 2.4 × 50 = 120 cm = 1.2 m.
(iii) Length = 16 × 50 = 800 cm = 8 m. Width = 11 × 50 = 550 cm = 5.5 m.
(iv) 1.5 m = 150 cm, so the cupboard is drawn \( \frac{150}{50} \) = 3 cm wide. Check: 3 × 50 = 150 cm.
In simple words: Each cm of the sketch is 0.5 m: blackboard 4 m, desk 1.2 m, room 8 m by 5.5 m, and the cupboard is drawn 3 cm wide.
Teacher's Note:
1. Sketch to real: multiply by 50; real to sketch: divide by 50.
2. Convert cm to m by dividing by 100.
Question 19. Crops on a farm
A farmer grows exactly four crops on his farm and divides the whole of his land among them. He shows the division as the pie chart given below, with the central angle of each crop marked on it. He remembers that the area under wheat is 25 hectares, but he has lost the paper on which the other areas were written.
(i) Find the total area of the farm.
(ii) Find the area under rice and the area under sugarcane.
(iii) Find the area under vegetables and verify that the four areas add up to the total area.
(iv) Write the ratio rice : wheat : sugarcane : vegetables in its simplest form.
Answer:
(i) Wheat is 90° and stands for 25 hectares. 90° is \( \frac{1}{4} \) of the circle, so the farm = 4 × 25 = 100 hectares.
(ii) Rice = \( \frac{144}{360} \) × 100 = 40 hectares; sugarcane = \( \frac{72}{360} \) × 100 = 20 hectares.
(iii) Vegetables = \( \frac{54}{360} \) × 100 = 15 hectares. Sum = 40 + 25 + 20 + 15 = 100 hectares, the total area. Angles: 144° + 90° + 72° + 54° = 360°.
(iv) Rice : wheat : sugarcane : vegetables = 40 : 25 : 20 : 15 = 8 : 5 : 4 : 3 (dividing by the HCF 5). The angles give the same: 144 : 90 : 72 : 54 = 8 : 5 : 4 : 3.
In simple words: Wheat's quarter of the circle is 25 hectares, so the farm is 100 hectares: rice 40, sugarcane 20, vegetables 15, in the ratio 8 : 5 : 4 : 3.
Teacher's Note:
1. A 90° slice is exactly one quarter of the whole.
2. The angles alone already give the ratio.
A printing press has to print one large order of books. The manager knows that 6 machines of the same kind would finish the whole order in 20 days, and he prepares the table below for other team sizes. All machines print at the same rate and run for the same number of hours each day.
| Number of machines | 6 | 10 | 12 | 15 |
|---|---|---|---|---|
| Number of days | 20 | 12 | ? | 8 |
(ii) In how many days would 8 machines finish the same order?
(iii) Fill in the missing entry of the table.
(iv) The press starts the order with 6 machines, but after 8 days two machines break down and the remaining machines continue alone. Find the number of extra days now needed, and the total number of days the order takes.
Answer:
(i) Constant = 6 × 20 = 120 machine-days. It measures the size of the whole order: one machine alone would need 120 days.
(ii) 8 machines: days = \( \frac{120}{8} \) = 15 days. Check: 8 × 15 = 120.
(iii) 12 machines: days = \( \frac{120}{12} \) = 10, so the missing entry is 10.
(iv) In 8 days, 6 machines do 6 × 8 = 48 machine-days. Work left = 120 − 48 = 72 machine-days.
Now 6 − 2 = 4 machines run, so extra days = \( \frac{72}{4} \) = 18 days. Total time = 8 + 18 = 26 days.
Check: 48 + 4 × 18 = 48 + 72 = 120 machine-days, the whole order.
In simple words: The order is 120 machine-days: 8 machines take 15 days, 12 take 10, and after the breakdown the job takes 18 more days, 26 in all.
Teacher's Note:
1. Subtract the work already done before using the new team size.
2. Machine-days, like worker-days, measure the size of the job.
Free study material for Mathematics
Free CBSE Practice Worksheets: Class 8 Mathematics Ganita Prakash Part 2 Chapter 03 Proportional Reasoning 2
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