This MCQ module is based on: Trairasika and Proportional Sharing
TOPIC 21 OF 22
Trairasika and Proportional Sharing
🎓 Class 8
Mathematics
CBSE
Theory
Ch 7 — Playing with Polynomials
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Trairasika and Proportional Sharing
Targeting Class 8 level in Algebra, with Basic difficulty.
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7.4 More Worked Examples
Example 4: Teacher-to-Student Ratio
In my school, there are 5 teachers and 170 students. The ratio of teachers to students in my school is \(5 : 170\). Count the number of teachers and students in your school. What is the ratio of teachers to students in your school? Write it below.
Is the teacher-to-student ratio in your school proportional to the one in my school?
Reduce \(5 : 170\) to simplest form: divide by 5 → 1 : 34. To compare, reduce your school's ratio and see if it is also 1 : 34. If it is, the ratios are proportional; otherwise they are not.
Example 5: Blackboard Dimensions
Measure the width and height (to the nearest cm) of the blackboard in your classroom. What is the ratio of width to height of the blackboard?
Q. Can you draw a rectangle in your notebook whose width and height are proportional to the ratio of the blackboard?
Reduce the measured ratio to its simplest form \(a : b\). Then draw a rectangle with width = \(a\) cm and height = \(b\) cm (or any scaled version like \(2a:2b, 3a:3b,\dots\)). It will look similar to the blackboard.
Reduce the measured ratio to its simplest form \(a : b\). Then draw a rectangle with width = \(a\) cm and height = \(b\) cm (or any scaled version like \(2a:2b, 3a:3b,\dots\)). It will look similar to the blackboard.
Example 6: Neelima's Age Ratio
When Neelima was 3 years old, her mother's age was 10 times hers. What is the ratio of Neelima's age to her mother's age? What would be the ratio of their ages when Neelima is 12 years old? Would it remain the same?
Solution:
When Neelima is 3, her mother is 30 years old. Ratio = 3 : 30 = 1 : 10.
When Neelima is 12 (that is, 9 years later), her mother is \(30+9=39\) years old. Ratio = 12 : 39 = 4 : 13 (after dividing by HCF 3).
No, the ratio changes because adding (or subtracting) the same constant to both terms is not a proportional change.
When Neelima is 12 (that is, 9 years later), her mother is \(30+9=39\) years old. Ratio = 12 : 39 = 4 : 13 (after dividing by HCF 3).
No, the ratio changes because adding (or subtracting) the same constant to both terms is not a proportional change.
Example 7: Missing Terms in Proportional Ratios
Fill in the missing numbers for the following ratios that are proportional to 14 : 21.
___ : 42 6 : ___ 2 : ___
Solution:
___ : 42. Second term goes from 21 → 42 (×2). So first term = 14 × 2 = 28.
6 : ___ . First term goes from 14 → 6. Factor = \(\frac{6}{14}=\frac{3}{7}\). Second term = \(21 \times \frac{3}{7} = 9\). So the ratio is 6 : 9.
2 : ___ . Divide both terms of \(14:21\) by 7 to get \(2:3\). So the missing term is 3.
6 : ___ . First term goes from 14 → 6. Factor = \(\frac{6}{14}=\frac{3}{7}\). Second term = \(21 \times \frac{3}{7} = 9\). So the ratio is 6 : 9.
2 : ___ . Divide both terms of \(14:21\) by 7 to get \(2:3\). So the missing term is 3.
Filter Coffee — A Flavour Problem
Filter coffee is a beverage made by mixing coffee decoction with milk. Manjunath usually mixes 15 mL of coffee decoction with 35 mL of milk to make one cup. So, the ratio of coffee decoction to milk is \(15 : 35 = 3 : 7\).
If customers want stronger filter coffee, Manjunath mixes 20 mL of decoction with 30 mL of milk. The ratio here is \(20 : 30 = 2 : 3\).
If they want lighter filter coffee, he mixes 10 mL of decoction with 40 mL of milk. The ratio is \(10 : 40 = 1 : 4\).
Fig 7.4 — Coffee-to-milk ratios for three strengths of filter coffee
| Coffee Decoction (in mL) | Milk (in mL) | Regular / Strong / Light |
|---|---|---|
| 300 | 600 | Stronger (1 : 2) |
| 150 | 400 | Lighter (3 : 8) |
| 200 | 400 | Stronger (1 : 2) |
| 24 | 56 | Regular (3 : 7) |
| 100 | 300 | Lighter (1 : 3) |
Trairāśika — The Rule of Three
Example 8: Rice for 150 Students
For the mid-day meal in a school with 120 students, the cook usually makes 15 kg of rice. On a rainy day, only 80 students came to school. How many kilograms of rice should the cook make so that the food is not wasted?
The ratio of the number of students to the amount of rice needs to be proportional. So, \(120 : 15 :: 80 : ?\)
The factor of change in the first term is \(\frac{80}{120}=\frac{2}{3}\). Reducing the second term by the same factor: \(15 \times \frac{2}{3} = 10\). So the cook should make 10 kg of rice.
The situation above is a typical example of a problem where we need to use proportional reasoning to find a solution. Four quantities are linked proportionally, out of which three are known and we must find the fourth unknown quantity.
Rule of Three (Trairāśika)
To solve such problems, we can model two proportional ratios using algebraic notation as \(a : b :: c : d\).For these two ratios to be proportional, term \(c\) should be a multiple of term \(a\) by a factor, say \(f\), and term \(d\) should be a multiple of term \(b\) by the same factor \(f\). So \(c = fa\) and \(d = fb\).
Dividing, \(f = \frac{c}{a} = \frac{d}{b}\). Hence \(ab = bc\), that is, \(\boxed{ad = bc}\). This is known as cross multiplication of terms.
Since \(ad = bc\), we obtain \(\displaystyle d = \frac{bc}{a}\). The fourth quantity can be found using cross multiplication.
Historical Note — Āryabhaṭa (c. 499 CE)
In ancient India, Āryabhaṭa (499 CE) and others called such problems proportionally like or Rule of Three problems. There were 3 numbers given — the pramāṇa (measure — '\(a\)' in our case), the phala (fruit — '\(b\)'), and the icchā (requisition — '\(c\)' in our case). To find the icchāphala (yield — '\(d\)' in our case), Āryabhaṭa says, "Multiply the phala by the icchā and divide the resulting product by the pramāṇa." In other words:\[ \text{icchāphala} = \frac{phala \times icch\bar{a}}{pram\bar{a}na} \] i.e., \(d = \frac{bc}{a}\).
Example 9: Travel Distance
A car travels 90 km in 150 minutes. If it continues at the same speed, what distance will it cover in 4 hours?
Solution using Cross Multiplication:
4 hours = 240 minutes. Proportion: \(150 : 90 :: 240 : x\).
Cross multiplication: \(150 \times x = 240 \times 90\).
\(x = \frac{240 \times 90}{150} = \frac{21600}{150} = \mathbf{144}\) km.
The car covers 144 km in 4 hours.
Cross multiplication: \(150 \times x = 240 \times 90\).
\(x = \frac{240 \times 90}{150} = \frac{21600}{150} = \mathbf{144}\) km.
The car covers 144 km in 4 hours.
Example 10: Tea Per-100 g Price
A small farmer in Himachal Pradesh sells each 200 g packet of tea for ₹200. A large estate in Meghalaya sells each 1 kg packet of tea for ₹800. Are the weight-to-price ratios in both places proportional? Which tea is more expensive?
Solution:
Himachal: \(200:200 = 1:1\). Meghalaya: \(1000:800 = 5:4\). These simplest forms differ, so the ratios are not proportional.
To decide which tea is more expensive, find cost of 1 kg of each.
Himachal: 200 g costs ₹200 → 1 kg (5 × 200 g) costs \(5 \times 200 = ₹1000\).
Meghalaya: 1 kg costs ₹800.
So Himachal tea is ₹1000/kg, Meghalaya tea is ₹800/kg. Himachal tea is more expensive.
To decide which tea is more expensive, find cost of 1 kg of each.
Himachal: 200 g costs ₹200 → 1 kg (5 × 200 g) costs \(5 \times 200 = ₹1000\).
Meghalaya: 1 kg costs ₹800.
So Himachal tea is ₹1000/kg, Meghalaya tea is ₹800/kg. Himachal tea is more expensive.
Activity 1: Ingredient Ratios in a Recipe
L3 Apply
Materials: Your favourite dish, its recipe (or help from home), notebook
Predict: If you double a recipe, will every ingredient double — or will some be doubled and others kept the same?
- Take your favourite dish and list out all the ingredients and their quantities to make the dish for your family.
- Suppose you are celebrating a festival and you want to invite 15 guests. Find out the quantities of the ingredients required to cook the same dish for them.
- Tabulate the original and scaled quantities and write the scale factor.
Expected outcome: Every ingredient is scaled by the same factor — the factor that expresses the ratio of new servings : original servings. For example, if the family of 5 needs 1 cup rice and you are making for 15 guests, the scale factor is \(\frac{15}{5}=3\), so you need 3 cups of rice, 3× of every other ingredient too.
7.5 Sharing, but Not Equally!
Form a pair. Collect 12 countable objects or counters (kabaddi pebbles). Now, share them between the two of you in different ways.
If you divide them equally, what is the ratio of the number of counters with each of you? Each of you will get 6 counters. So, the ratio is \(6:6 = 1:1\) (in its simplest form).
Q. If your partner gets 5 counters, how many objects will you get? What is the ratio of counters?
You get \(12-5=7\) counters. Ratio = 7 : 5.
You get \(12-5=7\) counters. Ratio = 7 : 5.
Now, if you want to share the counters between the two of you in the ratio 3 : 1, how willingly do it? One way is to share them as follows —
1. Your partner takes 3 counters and you take 1 counter. There are now 8 counters left.
2. Your partner takes 3 more counters and you take 1 more counter. There are now 4 counters left.
3. Your partner takes 3 more counters and you take 1 more counter. There are now no more counters left.
So your partner gets 9 counters in total and you get 3 counters.
When we divide 12 counters in the ratio 3 : 1 between two people, one gets 9 counters and the other gets 3 counters.
Fig 7.5 — Sharing 12 counters in the ratio 3 : 1 gives 9 and 3
Now, if you want to share 42 counters between the two of you in the ratio 4 : 3, how will you do it?
Using the same procedure would take a long time! There is a simpler way. You need to divide 42 into groups such that your partner gets 4 parts and you get 3 parts.
Q. What is the size of each group?
If your partner gets 4 groups and you get 3 groups, the total number of groups is \(4+3=7\). So, the size of each group is \(42 \div 7 = 6\). Multiplying the number of groups by the size of each group gives: partner gets \(4 \times 6 = 24\) counters and you get \(3 \times 6 = 18\) counters. The 'parts' are 6 times the ratio 4 : 3.
If your partner gets 4 groups and you get 3 groups, the total number of groups is \(4+3=7\). So, the size of each group is \(42 \div 7 = 6\). Multiplying the number of groups by the size of each group gives: partner gets \(4 \times 6 = 24\) counters and you get \(3 \times 6 = 18\) counters. The 'parts' are 6 times the ratio 4 : 3.
General Rule — Sharing in Ratio m : n
When we want to divide a quantity, say \(x\), in the ratio \(m : n\), we do the following:1. We need to split it into groups such that the first part has \(m\) groups and the second part has \(n\) groups. This can be found only if the number of groups are \(m+n\).
2. But what is the size of each group? This can be found by dividing the number of groups by \(x\). That is, the size of each group is \(\frac{x}{m+n}\).
3. So, the first part has \(\displaystyle \frac{m}{m+n}\times x\) objects and the second part has \(\displaystyle \frac{n}{m+n}\times x\) objects.
Example 11: Prashanth and Bhuvan's Fair Share
Prashanth and Bhuvan started a food cart business near their school. Prashanth invested ₹75,000 and Bhuvan invested ₹25,000. At the end of the first month, they gained a profit of ₹4,000. They decided that they would share the profit in the same ratio as that of their investment. What is each person's share of the profit?
Solution:
Ratio of investments = 75000 : 25000 = 3 : 1.
Total groups = \(3+1=4\). Size of each group = \(4000 \div 4 = ₹1000\).
Prashanth's share = \(3 \times 1000 = ₹3000\).
Bhuvan's share = \(1 \times 1000 = ₹1000\).
Total groups = \(3+1=4\). Size of each group = \(4000 \div 4 = ₹1000\).
Prashanth's share = \(3 \times 1000 = ₹3000\).
Bhuvan's share = \(1 \times 1000 = ₹1000\).
Example 12: Adjusting Sand-to-Cement Ratio
A mixture of 40 kg contains sand and cement in the ratio 2 : 3. How much cement should be added to the mixture to make the ratio of sand to cement 1 : 2?
Solution:
Original mixture: ratio 2 : 3, total = 40 kg.
Sand = \(\frac{2}{5}\times 40 = 16\) kg, Cement = \(\frac{3}{5}\times 40 = 24\) kg.
In the new mixture, the ratio 1 : 2 means for each 1 part sand, there are 2 parts cement. Sand is still 16 kg, so new cement = \(16 \times 2 = 32\) kg.
Extra cement required = \(32 - 24 = \mathbf{8}\) kg. (NCERT's worked variant continues with additional adjustment of \(\frac{4}{5}\times 30\); the simplified key step is shown here.)
Sand = \(\frac{2}{5}\times 40 = 16\) kg, Cement = \(\frac{3}{5}\times 40 = 24\) kg.
In the new mixture, the ratio 1 : 2 means for each 1 part sand, there are 2 parts cement. Sand is still 16 kg, so new cement = \(16 \times 2 = 32\) kg.
Extra cement required = \(32 - 24 = \mathbf{8}\) kg. (NCERT's worked variant continues with additional adjustment of \(\frac{4}{5}\times 30\); the simplified key step is shown here.)
Competency-Based Questions
Scenario: A paint supplier sells a pigment-to-base ratio of 2 : 5 to make a soft blue. One litre of this paint covers about 10 m². A school wants to paint a 150 m² wall with the same blue.
Q1. How many litres of paint will the school need, and how much pigment (in litres) will go into the mix? Use the Rule of Three.
L3 ApplyPaint needed. \(10 : 1 :: 150 : x\). Cross multiply: \(10x = 150\), so \(x=15\) L.
Pigment share. Ratio pigment : base = 2 : 5, so pigment share of total = \(\frac{2}{2+5}=\frac{2}{7}\). Pigment in 15 L = \(\frac{2}{7}\times 15 = \frac{30}{7} \approx 4.29\) L.
Pigment share. Ratio pigment : base = 2 : 5, so pigment share of total = \(\frac{2}{2+5}=\frac{2}{7}\). Pigment in 15 L = \(\frac{2}{7}\times 15 = \frac{30}{7} \approx 4.29\) L.
Q2. The school realises the original mix is too dark. They want the new pigment-to-base ratio to be 1 : 4 but keep the total volume at 15 L. Analyse how much extra base is required and how much pigment must be removed.
L4 AnalyseAnswer: For a 15 L mix at 1 : 4, pigment = \(\frac{1}{5}\times 15 = 3\) L; base = \(\frac{4}{5}\times 15 = 12\) L. Original pigment ≈ 4.29 L, original base ≈ 10.71 L. So remove \(4.29 - 3 \approx 1.29\) L pigment and add \(12 - 10.71 \approx 1.29\) L base.
Q3. A shopkeeper claims his 500 mL bottle at ₹180 is a "better deal" than Manjunath's 800 mL bottle at ₹240. Evaluate using price-per-mL ratios.
L5 EvaluateAnswer: Shopkeeper: \(\frac{180}{500}=₹0.36\) per mL. Manjunath: \(\frac{240}{800}=₹0.30\) per mL. Manjunath's bottle is cheaper per mL, so the shopkeeper's "better deal" claim is false.
Q4. Design a "scholarship prize pool" rule: ₹60,000 must be split among 3 winners in the ratio 5 : 3 : 2 (Gold : Silver : Bronze). Compute each prize and propose one additional rule that keeps the proportion fair even if more winners are added later.
L6 CreatePrizes. Total groups = \(5+3+2=10\). Each group = \(60000 \div 10 = ₹6000\). Gold = \(5 \times 6000 = ₹30{,}000\); Silver = \(3 \times 6000 = ₹18{,}000\); Bronze = \(2 \times 6000 = ₹12{,}000\).
Rule proposal: "Each Honourable Mention receives 1 share drawn from the same unit group-size of ₹6000." This keeps the same per-group value and preserves fairness by proportional reasoning.
Rule proposal: "Each Honourable Mention receives 1 share drawn from the same unit group-size of ₹6000." This keeps the same per-group value and preserves fairness by proportional reasoning.
Assertion–Reason Questions
Assertion (A): In the proportion \(6:10::18:30\), cross multiplication gives \(6\times 30 = 10 \times 18\).
Reason (R): If \(a:b::c:d\) then \(ad = bc\).
Reason (R): If \(a:b::c:d\) then \(ad = bc\).
Answer: (a) — \(6 \times 30 = 180\) and \(10\times 18 = 180\). R directly explains why A holds.
Assertion (A): If Neelima is 3 and her mother is 30, the age ratio stays 1 : 10 forever.
Reason (R): Adding the same number to both terms of a ratio preserves the ratio.
Reason (R): Adding the same number to both terms of a ratio preserves the ratio.
Answer: (d) — A is false: After 9 years the ratio becomes 12 : 39 = 4 : 13, not 1 : 10. R is also false. Actually, both parts of the statement fail — this option tests whether students realise both claims are false. (Note: examiners may instead accept "Both A and R are false." Cross-check with teacher.)
Assertion (A): When ₹4000 is split in the ratio 3 : 1, one share is ₹3000 and the other is ₹1000.
Reason (R): The share of the first person is \(\frac{m}{m+n}\times x\) where \(x\) is the total.
Reason (R): The share of the first person is \(\frac{m}{m+n}\times x\) where \(x\) is the total.
Answer: (a) — First share = \(\frac{3}{4}\times 4000 = 3000\), second = \(\frac{1}{4}\times 4000 = 1000\). R is the precise formula used to obtain A.
Frequently Asked Questions
What is the unitary method?
The unitary method finds the value of one unit first, then scales it up to the required quantity. For example, if 5 books cost Rs 200, one book costs Rs 40, and 3 books cost Rs 120.
How does Trairasika work?
If three proportional quantities a, b and c are known and the unknown x corresponds to b in the same way c corresponds to a, then x = (b times c) / a.
What is proportional sharing?
Proportional sharing distributes a total quantity among parties in the ratio of their shares or contributions. Dividing Rs 600 in the ratio 2:3 gives Rs 240 and Rs 360.
Who developed the Trairasika method?
Trairasika appears in Indian mathematical texts by Aryabhata, Brahmagupta and Bhaskara II. It travelled via Arabic mathematics into medieval Europe as the 'golden rule' or rule of three.
When is Trairasika not applicable?
It applies only when the quantities are directly proportional. For inversely proportional quantities, a modified (inverse) rule of three is used.
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