Ratio and Proportion

What you will learn

A ratio tells us how two quantities compare by multiplication. You will learn to preserve their order and units, simplify their comparison, find a missing proportional value, and recover actual quantities from a total, a difference or a known part. You will also connect two ratios through a shared quantity.

This lesson uses the exact fraction and decimal calculation from Numbers and Calculation. We do not need square-root methods here. The sharing problems use positive quantities. In a:b interpreted as a/b, b cannot be zero. A ratio gives a relationship, not usually the actual sizes: 2:3 could describe 2 and 3, 20 and 30, or 0.2 and 0.3.

1 Retrieve the arithmetic tools

Attempt these before reading the feedback.

1. Simplify 18/30.
2. Calculate (3/5) × 140.
3. Using 1 m = 100 cm, convert 1.25 m to centimetres.
4. Calculate 72 ÷ 0.6.

Feedback:

1. Divide numerator and denominator by 6: 18/30 = 3/5. Dividing both by the same nonzero number preserves the fraction.
2. One fifth of 140 is 28, so three fifths is 3 × 28 = 84.
3. 1.25 × 100 = 125 cm. A measurement changes its number when the unit changes; its length stays the same.
4. Multiply both dividend and divisor by 10: 72 ÷ 0.6 = 720 ÷ 6 = 120. Dividing by a positive number below 1 can give a larger answer.

If the fraction or decimal steps were difficult, revisit Fractions Decimals and Exact Calculation. Keep the units written while working here.

2 Name the comparison before writing the ratio

A difference answers “how much more?” A ratio answers “how many times as much?” For two quantities A and B measured in the same unit, A:B represents A/B, with B nonzero. Changing the order usually changes the ratio.

When counting categories, name both categories. “Red to blue” and “red to all” have different second quantities. A part-to-part ratio a:b does not make the first part a/b of the whole. If these are the only two parts, its fraction of the whole is a/(a+b).

Worked example 1 Part to part, then part to whole

A bag contains 15 red tokens and 25 blue tokens, and no other tokens. Find red:blue and red:all.

Red:blue = 15:25 = 3:5 after dividing both terms by 5. The total is 15 + 25 = 40, so red:all = 15:40 = 3:8. In the first comparison the reference is blue; in the second it is the whole bag.

Check: 15/25 = 3/5, but 15/40 = 3/8. Red is 3 of the 8 equal ratio parts of the whole. Blue:red would be 5:3, not 3:5. The ratio does not mean that the bag contains only 3 red and 5 blue tokens; those are its simplest proportional terms.

Worked example 2 Match units first

Find 84 cm:1.4 m in that order.

Convert the second length: 1.4 m = 140 cm. Then 84:140 = 3:5, dividing both terms by 28. We compare the same kind of measurement in the same unit, so the units cancel in 84 cm/140 cm.

Check: 84 cm is shorter than 1.4 m, and 3/5 is below 1. Dividing 84 by 1.4 without matching units would compare bare numbers from different scales. Do not apply this cancellation to different kinds of quantities: ₹44 per kg is a rate with units, not a unitless length-to-length ratio. Rates are used in the next lesson.

3 Equivalent ratios use one common scale

Multiplying both terms by the same positive number, or dividing both by the same positive number, preserves their quotient. For positive integer terms, dividing by their HCF gives simplest integer terms. For decimal or fractional terms, first clear the decimal places or denominators with one common multiplier; then reduce.

We do not separately round the two terms. Rounding may change the relationship.

Worked example 3 A ratio containing fractions

Simplify (3/4):(5/6) exactly.

The denominators 4 and 6 have common multiple 12. Multiply both terms by 12: (3/4) × 12 = 9 and (5/6) × 12 = 10. Thus the ratio is 9:10. There is no need to turn 5/6 into an approximate decimal.

Check the quotient: (3/4) ÷ (5/6) = (3/4) × (6/5) = 18/20 = 9/10. Using only the numerators, 3:5, would lose the unequal denominators.

4 Proportion means two ratios are equal

A:B = C:D states an equality of two ordered comparisons. For nonzero B and D, write A/B = C/D. Multiplying both sides by B × D gives A × D = B × C. This explains cross multiplication; it is not a rule for multiplying any four numbers that happen to appear in a question.

Before calculating, align the same quantities in corresponding positions and match the units. Then choose a simple scale factor if one is visible. Cross multiplication is another way to express the same equality.

Worked example 4 Find a missing proportional term

Find x in 7:12 = x:60.

The second term changes from 12 to 60 by multiplication by 5. The first term must change by the same factor: x = 7 × 5 = 35.

Alternatively, 7/12 = x/60 gives 7 × 60 = 12x, so x = 420 ÷ 12 = 35.

Check: 35:60 reduces to 7:12. Adding 48 to the first term because 48 was added to the second would give 55:60, a different ratio.

Checkpoint A

1. Simplify 45 seconds:2 minutes.
2. Find t in 4:7 = t:35.
3. Is 2:5 equal to 6:15? Give a reason.

Feedback: 2 minutes = 120 seconds, so 45:120 = 3:8. In the second question the common factor is 5, giving t = 20. In the third, both terms were multiplied by 3; 2/5 = 6/15. If you instead changed only one term, the ratio would change.

5 Recover actual quantities from equal ratio parts

If A:B = a:b, write A = a × k and B = b × k. The positive number k is the size of one equal ratio part. It is not necessarily 1.

Choose the equation from the information actually supplied:

• Total T: there are a+b parts, so k = T/(a+b)
• Known B: b parts equal B, so k = B/b
• Positive difference Δ, with b>a: b−a parts equal Δ, so k = Δ/(b−a)

The numbers a and b describe the relationship; k gives its actual scale. If you are dividing indivisible objects, the resulting counts must be whole numbers. A fractional count signals incompatible data or a wrong model, not permission to cut a person or count half a token.

Worked example 5 A total, then a known part

Share ₹630 between funds A and B in the ratio 2:5.

There are 2+5 = 7 equal parts. One part is ₹630 ÷ 7 = ₹90. Therefore A = 2 × ₹90 = ₹180 and B = 5 × ₹90 = ₹450.

Check both conditions: ₹180 + ₹450 = ₹630, and 180:450 = 2:5.

Now hide the total. Suppose you know only A:B = 2:5 and B = ₹450. B represents five parts, so one part = ₹450 ÷ 5 = ₹90. A is still 2 × ₹90 = ₹180; the total is 7 × ₹90 = ₹630. Dividing ₹450 by 7 would incorrectly treat a known part as the whole.

Worked example 6 A difference gives the difference of parts

Two ribbon lengths are in the ratio 4:7 and differ by 81 cm. Find both lengths.

The longer ribbon has 7−4 = 3 more equal parts. These three parts measure 81 cm, so one part is 81 ÷ 3 = 27 cm. The shorter length is 4 × 27 = 108 cm, and the longer is 7 × 27 = 189 cm.

Check: 189−108 = 81 cm and 108:189 = 4:7. The sum 4+7 is useful if a total is given, but 81 cm here is a difference. Dividing 81 by 11 answers the wrong model.

6 Connect ratios through the shared quantity

To combine A:B and B:C, B must represent the same quantity in both statements. Make its ratio terms equal using equivalent ratios; then read A:B:C in order. Do not simply place four unmatched numbers next to each other.

Worked example 7 Join ratios, then share among three parts

Given A:B = 3:4 and B:C = 6:5, find A:B:C.

The shared B terms are 4 and 6. A convenient common value is 12. Multiply 3:4 by 3 to get 9:12. Multiply 6:5 by 2 to get 12:10. Therefore A:B:C = 9:12:10.

Check each original comparison: 9:12 = 3:4 and 12:10 = 6:5.

Coached extension: if A+B+C = 620, find the three quantities. There are 9+12+10 = 31 equal parts, so one part is 620 ÷ 31 = 20. Thus A = 180, B = 240 and C = 200. Their sum is 620, while 180:240 = 3:4 and 240:200 = 6:5. The same whole-sharing mechanism works with three terms; no new formula needs memorising.

Worked example 8 Equal additions need not preserve a ratio

A box has 6 white and 10 black counters. Four of each colour are added. Does white:black stay the same?

Initially, 6:10 = 3:5. Afterwards, 10:14 = 5:7. These ratios differ: 3 × 7 = 21, while 5 × 5 = 25. Equal additions preserve the difference between counts, but generally do not preserve their quotient.

This is a counterexample to the claim that equal additions always preserve a ratio. It does not mean they can never do so: equal counts, such as 6:6, stay in the ratio 1:1 after equal additions. Check the actual relationship.

Checkpoint B

1. A:B = 3:5 and the total is 96. Find A and B.
2. In the same ratio, suppose B alone is 65. Find the total.
3. Two positive lengths are in ratio 2:7 and differ by 40 cm. Find the shorter.

Feedback: (1) Eight parts make 96, so one part is 12; A=36 and B=60. (2) Five parts make 65, so one part is 13; the total is 8×13=104. (3) Five parts make the 40 cm difference, so one part is 8 cm and the shorter is 16 cm. Name the given total, part or difference before dividing.

7 Independent practice

Attempt all twelve questions before the key. Choose one option and write the named comparison or equal-parts model that supports it. This is untimed learning practice: +1 for correct, 0 otherwise, with no negative marking or pass cutoff. It is not a full RRB CBT simulation.

1. There are 12 green and 18 yellow counters. What is green:yellow in simplest form?

A. 3:2 B. 2:3 C. 2:5 D. 3:5

2. A bag contains only 21 red and 35 blue tokens. What is red:all in simplest form?

A. 3:5 B. 5:8 C. 3:8 D. 8:3

3. Which is the simplest integer ratio equivalent to 0.6:1.5?

A. 2:5 B. 5:2 C. 6:5 D. 2:3

4. Find x in 5:9=x:45.

A. 41 B. 9 C. 45 D. 25

5. Find 72 cm:1.2 m in simplest form, preserving this order.

A. 60:1 B. 3:5 C. 5:3 D. 6:1

6. Simplify (2/3):(5/9) exactly.

A. 2:5 B. 5:6 C. 6:5 D. 3:5

7. ₹840 is divided into two shares in the ratio 3:4. What is the smaller share?

A. ₹360 B. ₹480 C. ₹630 D. ₹120

8. Two positive amounts are in the ratio 5:8 and differ by ₹96. What is the larger amount?

A. ₹96 B. ₹160 C. ₹416 D. ₹256

9. Positive quantities A:B:C=2:3:7 and A+B+C=360. What is C?

A. 90 B. 210 C. 360 D. 30

10. Let A, B and C be positive quantities satisfying A:B=4:5 and B:C=10:3. What is A:B:C?

A. 4:5:3 B. 4:10:3 C. 8:10:3 D. 8:5:3

11. A learner says adding 3 to both terms of 2:5 preserves the ratio. Which response is correct?

A. False: the new ratio is 5:8, not 2:5 B. True: equal addition always preserves 2:5 C. False: the new ratio is 5:5 D. False: the new ratio is 2:8

12. A:B=7:11 and B=154. What is A+B?

A. 98 B. 154 C. 168 D. 252

8 Explained answer key

1. B. Divide 12:18 by 6 to get 2:3. Check 12/18=2/3. A reverses the categories. C is green:all because there are 30 counters; D is yellow:all. The question compares green with yellow, not either colour with the whole.

2. C. Total=21+35=56. Red:all=21:56=3:8 after dividing by 7. A uses blue rather than all as the second quantity. B gives blue:all. D reverses the required comparison. A part cannot exceed this positive whole, which also rejects 8:3.

3. A. Multiply both terms by 10 to obtain 6:15, then divide both by 3 to get 2:5. Check 0.6/1.5=0.4=2/5. B reverses the ratio. C fails to reduce the first term with the same factor as the second. D replaces the second quantity by the difference: 0.6:(1.5−0.6)=0.6:0.9=2:3. That is a different comparison.

4. D. Since 45=9×5, x=5×5=25. Equivalently, 9x=5×45=225, so x=25. Check 25:45=5:9. A adds the change 36 instead of scaling. B repeats a known term without satisfying the equality. C would give 45:45=1:1.

5. B. 1.2 m=120 cm, so 72:120=3:5. Check that the first length is shorter and the quotient 3/5 is below 1. A divides 72 by 1.2 without matching units. C reverses the order. D effectively uses an incorrect factor of 10 for metres to centimetres instead of 100.

6. C. Multiply both terms by 9: (2/3)×9=6 and (5/9)×9=5. Thus 6:5 is correct. Check (2/3)÷(5/9)=(2/3)×(9/5)=6/5. A ignores denominators. B reverses the quotient. D does not apply the same scale to both original terms.

7. A. There are 3+4=7 equal parts. One part=₹840÷7=₹120; smaller share=3×₹120=₹360. The other share is ₹480 and the sum is ₹840. B is the larger share. C takes 3/4 of the total, confusing part:part with part:whole. D is just one part, not three.

8. D. The difference is 8−5=3 parts, so one part=₹96÷3=₹32. The larger amount is 8×₹32=₹256; the smaller is ₹160. Check ₹256−₹160=₹96 and 160:256=5:8. A repeats the difference; B gives the smaller amount; C gives the total ₹160+₹256.

9. B. Total parts=2+3+7=12. One part=360÷12=30. Therefore C=7×30=210. Check A=60 and B=90, so 60+90+210=360. Option A gives quantity B, option C gives the whole total, and option D gives one part. All three terms belong in the sum of parts.

10. C. Match B at 10: multiplying both terms of 4:5 by 2 gives A:B=8:10. The other ratio is already B:C=10:3, so A:B:C=8:10:3. Check 8:10=4:5. A joins unmatched B terms; B doubles B but not A; D doubles A but not B. Each violates at least one original ratio.

11. A. The new terms are 2+3=5 and 5+3=8, giving 5:8. It is not equal to 2:5: the cross products are 5×5=25 and 8×2=16. B confuses addition with scaling. C changes only the first term; D changes only the second. Both original terms must receive the stated addition.

12. D. B is 11 parts, so one part=154÷11=14. Then A=7×14=98 and A+B=(7+11)×14=252. Check 98:154=7:11. A gives A alone; B repeats the known quantity. C=154+14 adds only one ratio part to B, but A contains seven parts. The known 154 is not the whole.

9 Sources and scope

These original explanations and questions develop the Ratio and Proportion topic listed in RRB CEN 09/2025 §14.1, printed p.28. This is teaching coverage, not a claim about question frequency or a complete exam course. Mathematical references: NCERT Ganita Prakash, Grade 8, Chapter 7, reprint 2026–27, §§7.1–7.6, printed pp.159–177, especially cross multiplication p.168 and whole-sharing pp.172–175; Part II, Chapter 3, §§3.1–3.4, printed pp.55–60. Linked-ratio joining is an explained application of equivalent ratios. Source pages and textbook questions are not reproduced. No age equations, alligation or capital-time partnership rules are introduced here.

Analogy

Think of a packing pattern that puts 2 red counters with every 5 blue counters. Each repeat adds one complete 2-red/5-blue bundle. Three repeats make 6 red and 15 blue: both quantities were multiplied by 3. The seven positions in one bundle explain why the red share of the whole is 2/7, not 2/5. If only the blue count is known, groups of five reveal the number of repeats.

Limit: the physical bundle is a model for whole counters. Ratios of money, length or fractional quantities can have a non-integer scale; no actual bundle is necessary. Adding a fixed number to each colour is not the same as adding complete bundles, and need not preserve the ratio.

Quick reference

• Label the compared quantities and preserve order: A:B is A/B, B≠0
• Same-kind measurements: convert to the same unit before simplifying
• Equivalent ratio: multiply/divide every term by the same positive scale
• Positive integer terms: divide by HCF; fractional terms: clear denominators with one common multiplier
• Proportion: A/B=C/D implies AD=BC when B,D≠0
• A:B=a:b ⇒ A=ak, B=bk
• Total T: k=T/(a+b); known B: k=B/b
• Difference Δ with b>a: k=Δ/(b−a)
• Three parts a:b:c and total T: one part=T/(a+b+c)
• Join A:B and B:C only after making the B terms equal
• Check original ratio plus supplied total/part/difference; indivisible counts must be whole
• Equal additions generally change a ratio; test instead of assuming