Ratios, shares and proportional models

What this lesson helps you decide

A ratio tells us relative size. It does not usually tell us the actual amounts. A proportional model goes further: it says what stays unchanged when a quantity varies. Confusing these two ideas can give a neat calculation with an unjustified answer.

This lesson develops four connected skills: comparing like quantities in common units; recovering actual shares from a total or a difference; combining ratios through a common quantity; and deciding whether a constant quotient or a constant product describes a situation. You will also learn why a straight graph with a fixed offset is not a direct-proportion graph.

Before you begin

You need multiplication, division, equivalent fractions, decimal place value and simple unit conversion. Here is the small amount of algebra used throughout.

• Multiplying both terms of 2:5 by 3 gives 6:15, because 6/15 = 2/5. Multiplying just one term changes the comparison.
• To find 3/8 of 72, first find one eighth: 72 ÷ 8 = 9. Then take three such parts: 3 × 9 = 27.
• If 6k = 54, divide both sides by 6 to obtain k = 9. Substituting back gives 6 × 9 = 54.
• Since 1 km = 1000 m, 0.9 km = 900 m. Since 1 kg = 1000 g, 0.45 kg = 450 g. Conversion changes the numerical description, not the amount.
• A 25% increase means adding 25/100 of the original value: new = old × (1 + 25/100) = 1.25 × old. Percent change always needs its comparison base.

Check yourself: simplify 18:30; find 5/6 of 42; convert 0.32 L to mL. Answers: 3:5 after division by 6; 35 because 42 ÷ 6 × 5 = 35; 320 mL because 1 L = 1000 mL. If any step is unfamiliar, work it through before using a proportion shortcut.

A ratio compares quantities in a stated order

For positive amounts A and B, A:B = a:b means A/B = a/b. The order matters: A:B and B:A reverse the comparison. In this lesson share counts and scale factors are positive; any denominator used in division must be nonzero.

When A and B measure the same kind of quantity, convert them to the same unit first. Only then do the units cancel. For example, a length-to-length ratio is dimensionless after both lengths are expressed in metres. A length-to-time comparison is different: distance/time is a rate, so a unit such as m/s must remain. Do not convert unlike quantities into an imaginary common unit.

An equivalent ratio comes from multiplying or dividing every term by the same nonzero factor. It does not come from adding the same number. Thus 4:6 = 2:3, whereas adding 2 gives 6:8 = 3:4, a different relationship. With decimals, multiplying both terms by 10, 100 or another convenient common factor can first remove decimal places.

To test A/B = a/b without rounding, multiply both sides by Bb: Ab = Ba. This cross-product test is valid because B and b are nonzero. Keep the compared quantities in the same order; cross multiplication cannot repair a wrongly chosen model.

One shared scale factor turns a ratio into amounts

If A:B = a:b, write A = ak and B = bk. The same k represents the amount in one ratio part. The ratio cancels k, so many pairs share the same ratio. For 3:5, both 6:10 and 30:50 work. Without another piece of information, neither absolute amount is determined.

If A + B = T, then ak + bk = (a + b)k = T. Therefore k = T/(a + b), A = aT/(a + b) and B = bT/(a + b). The fractions of the whole are a/(a + b) and b/(a + b), not a/b and b/a.

A known difference can also set the scale. If B − A = D and b is greater than a, then (b − a)k = D and k = D/(b − a). If a = b, the amounts are equal and a nonzero stated difference is inconsistent. If only their ratio is given, do not invent a total or a difference.

For counts, the final amounts must also be possible whole numbers. A mathematically correct fractional share of money or length may be acceptable, but 3.5 indivisible components is not. Check the context after finding k.

Eight equal budget parts

Allocation	Number of equal parts	Fraction of the whole
Printing	3	3/8
Transport	5	5/8
Together	8	1

Caption: The eight parts have the same value. “3:5” compares the two allocations; “3/8” compares printing with the whole budget. The original diagram supplied with this panel shows three labelled P parts followed by five labelled T parts; the table carries the same information without colour.

Join two ratios by matching their shared quantity

Suppose A:B = a:b and B:C = c:d. The B in both statements is the same actual amount, even though one ratio assigns it b parts and the other assigns it c parts. Rescale the first ratio by c and the second by b. Both now give bc parts for B, so A:B:C = ac:bc:bd. A smaller common multiple may simplify the arithmetic; the shared amount must match either way.

Each pair must check after combination. Do not copy the first and last terms into a three-term string while leaving the common quantity on incompatible scales.

Direct proportion needs a constant rate

Suppose x is an input and y is the corresponding output. A direct-proportion model y = kx states that y/x = k for every allowed nonzero x. The constant k is the output per unit input. If x changes by a positive factor s, then the new output is k(sx) = s(kx), so y changes by the same factor. “Both increase” is weaker information and does not establish direct proportion.

For printing, let x be total productive machine-minutes and y be the modelled page output. One machine operating for one minute contributes one machine-minute. If n identical machines each operate productively for t minutes, x = nt. At a constant 12 pages per machine-minute, y = 12x. No setup time, idle time, shared bottleneck or change of rate is included. Completed-page counts are discrete; the line is an idealised constant-rate model, and the marked points give whole-page outputs.

Constant rate through the origin

Productive machine-minutes x	Modelled pages y = 12x	y/x for nonzero x
0	0	Undefined at zero
1	12	12
2	24	12
4	48	12

Caption: On the original graph, x runs from 0 to 4 machine-minutes and y from 0 to 48 pages. The labelled points (0,0), (1,12), (2,24) and (4,48) lie on one straight line. For this cumulative-output model, zero productive time means zero modelled output, so zero is included. The quotient is checked only where x is nonzero; y/x at (0,0) is not 12 or 0, but undefined. If a different physical task permits only positive x, a line may be extended to the origin mathematically without claiming that zero is an allowed operating state.

A graph is a picture of ordered pairs: the first coordinate is read on the horizontal x-axis, the second on the vertical y-axis. A straight line means a constant change in y for equal changes in x. Direct proportion additionally requires that its mathematical line pass through (0,0). A constant slope alone is not enough.

A fixed starting amount changes the model

Compare a usage-only charge C = 12h with a charge F = 30 + 12h, where h is nonnegative hours and amounts are in rupees. The second model includes a ₹30 fee even at zero usage. Both add ₹12 for one extra hour, but only the first is directly proportional to h.

h in hours	C in ₹	F in ₹
0	0	30
1	12	42
2	24	54
4	48	78

Caption: The original comparison graph shows two parallel straight lines. C starts at (0,0); F starts at (0,30). For positive h, C/h = 12 while F/h = 12 + 30/h varies. Doubling h from 1 to 2 doubles C from 12 to 24, but F changes from 42 to 54, not 84. The form y = mx + b is called affine; with a nonzero fixed offset b, it is not direct proportion.

A few observed pairs may fit a proposed relationship without proving that it holds for every input. A stated constant rate or a justified physical assumption is what lets us use the model beyond those pairs. Keep the allowed range and any setup fee explicit.

Inverse proportion comes from a fixed product

For a fixed positive distance d, constant positive speed v and uninterrupted travel time t satisfy d = vt. Dividing by v gives t = d/v. The product vt stays fixed, so doubling v halves t; multiplying v by any positive factor s divides t by s. This is inverse proportion, not a rule based merely on one quantity rising while another falls.

More generally, y = K/x with positive x and fixed positive K means xy = K. The quotient y/x is not constant. Neither x = 0 nor y = 0 belongs to this positive-product model. The units of K are the product of the units of x and y; in travel they reduce from (km/h) × h to km.

Fixed total machine-time

Suppose a divisible job needs exactly 48 machine-hours at identical, constant, independent machine rates. There is no setup delay or shared bottleneck, and work can be assigned so that all machines remain productive. With x machines and y elapsed hours, xy = 48 and y = 48/x.

Machines x	Elapsed hours y	Product xy
1	48	48
2	24	48
4	12	48
8	6	48

Caption: As x doubles, y halves. The plotted positive-branch curve falls and bends; it is not the straight line in the constant-rate graph and has no point at x = 0. In the machine context only positive whole-number machine counts are feasible; the smooth curve is a mathematical guide between them.

The distinction between fixed output and fixed time is essential. For output Q = rnt, with per-machine rate r fixed, output is directly proportional to n when t is fixed. For Q also fixed, solving gives t = Q/(rn), which is inversely proportional to n. The same physical relationship produces different comparisons depending on what is held constant.

If speed rises by a fraction p of its old value, new speed is (1 + p)v. At fixed distance, new time is t/(1 + p). Its fractional decrease relative to old time is [t − t/(1 + p)]/t = p/(1 + p). Thus a 25% speed increase uses p = 0.25 and gives a time decrease of 0.25/1.25 = 0.20 = 20%. For a positive speed reduction fraction q smaller than 1, the corresponding time-increase fraction is q/(1 − q). These results assume the same distance and no additional stopped time.

Decide whether there is enough information

“Six workers finished a job” does not tell us how a larger team will perform. Perhaps all can work in parallel at the same constant rate. Perhaps a single tool restricts useful work to six people, or the extra workers have different skills. To calculate an inverse-proportion completion time, state the equal-rate, additive-work and fixed-work assumptions. Otherwise name the missing model instead of manufacturing a number.

Use this decision sequence before calculating:

1. Name each quantity, its unit and the requested comparison order.
2. Identify the shared scale, fixed quotient, fixed product, or fixed offset stated in the problem.
3. Find the extra datum that determines the scale: a total, difference, one amount, rate, or fixed workload.
4. Write the relationship with units, then solve. Keep fractions exact until the final step.
5. Substitute back into every given condition. Check direction of change, feasible counts and whether rounding was requested.

Six worked examples

Worked example 1 — Normalize the units first

Simplify the ratio 750 m : 1.2 km.

Both quantities are lengths, but their written units differ. Convert 1.2 km to 1.2 × 1000 = 1200 m. Then 750 m : 1200 m = 750:1200. Dividing both terms by 150 gives 5:8.

Check: 5 × 150 m = 750 m and 8 × 150 m = 1200 m. The first length is smaller, so the first ratio term should also be smaller. Keeping 750:1.2 and simplifying the displayed numbers mixes metres with kilometres; reversing the terms would instead answer 1.2 km : 750 m.

Answer: 5:8

Worked example 2 — Allocate a whole by equal parts

A budget of ₹640 is split between printing and transport in the ratio 3:5. Find both allocations.

Write printing = 3k rupees and transport = 5k rupees. Their total is 8k = 640, so one part k = 640/8 = ₹80. Printing receives ₹240 and transport ₹400.

Two separate checks are needed: ₹240 + ₹400 = ₹640, and 240:400 = 3:5. The printing share of the whole is 3/8, not 3/5. Using (3/5) × 640 = 384 treats the whole budget as if it were the transport amount. A ratio compares its named components; it is not automatically a fraction of the total.

Answer: Printing ₹240; transport ₹400

Worked example 3 — Reconcile two scales for one quantity

Positive quantities satisfy a:b=2:3 and b:c=4:5. Find a:b:c.

In the first statement b has 3 parts; in the second it has 4. Use 12 parts for the same b. Multiply 2:3 by 4 to obtain 8:12. Multiply 4:5 by 3 to obtain 12:15. Therefore a:b:c = 8:12:15.

Check the supplied pairs: 8:12 reduces to 2:3; 12:15 reduces to 4:5. Writing 2:3:5 would give b:c = 3:5, contradicting the second statement. These ratios still do not give absolute amounts: 8,12,15 and 16,24,30 both fit. A total or one known amount would fix their common multiplier.

Answer: 8:12:15

Worked example 4 — Count productive machine-minutes

Five identical printers produce 420 pages in 7 minutes. With the same constant per-printer rate and no delays, how many pages do eight printers produce in 10 minutes?

Assume each printer works independently and their outputs add, with enough supplies and no shared bottleneck. The first run uses 5 × 7 = 35 printer-minutes. Hence the per-printer rate is 420/35 = 12 pages/(printer·minute). This is the y = 12x model in the constant-rate graph with x measured in printer-minutes.

The second run uses 8 × 10 = 80 printer-minutes, giving 80 × 12 = 960 pages. Unit check: printer·minute × pages/(printer·minute) = pages. A second calculation gives 420 × (8/5) × (10/7) = 960.

Both printer count and productive time increase, so output should exceed 420. Using an inverse rule here would wrongly hold total output fixed. If all printers had to share a device that limited throughput, the independent-rate assumption would fail and this calculation would need a different model.

Answer: 960 pages

Worked example 5 — Keep the rate unit visible

A coating process uses 12 L for 80 m² at a fixed consumption per unit area. How much is needed for 50 m²?

Let A be coated area and V the coating volume. Fixed consumption per unit area means V/A = k, so V = kA. Find k = 12/80 = 0.15 L/m². Then V = 0.15 × 50 = 7.5 L.

Check by scaling: 50/80 = 5/8, so the new volume is 12 × 5/8 = 7.5 L. The smaller area uses less liquid under this model. The product m² × L/m² leaves litres. A fixed cleaning loss, changed layer thickness or different absorption could break the constant-consumption assumption; none is included here. Do not assume every painting or coating situation follows this model without such a condition.

Answer: 7.5 L

Worked example 6 — Derive the inverse percentage change

A vehicle travels a fixed distance. Its constant speed is increased by 25%. By what percentage does its travel time decrease?

Let the original positive speed and uninterrupted travel time be v and t. Distance is d = vt. The new speed is 1.25v = (5/4)v. Holding distance fixed gives new time = d/[(5/4)v] = (4/5)t = 0.8t.

The decrease is t − 0.8t = 0.2t. Divide by the old time t and multiply by 100 to get 20%. A numerical check: travelling 150 km at 50 km/h takes 3 h; at 62.5 km/h it takes 2.4 h. The 0.6 h decrease is 0.6/3 = 20%.

A 25% time reduction would give 0.75t, whose product with 1.25v is 0.9375vt, not the fixed distance. The percentages do not cancel because the relation is reciprocal. Distance must remain unchanged and no separate stop duration may be added.

Answer: 20% decrease

Ten practice questions

Use the relationship and its assumptions before calculating. For a choice question, select one answer; for a numerical question, give the requested value and unit. Write your reasoning before reading the solutions.

Practice question 1

What is the simplest ratio of 0.84 kg to 630 g, in that order?

• A. 4:3
• B. 3:4
• C. 4:30
• D. 840:63

Practice question 2

Positive quantities u and v, both measured in litres, satisfy u:v = 5:8. If u = 35 L, find v in litres.

Practice question 3

₹990 is divided between accounts A and B in the ratio 4:7. How much goes to B?

• A. ₹360
• B. ₹630
• C. ₹1732.50
• D. ₹90

Practice question 4

Two rods have lengths in the ratio shorter:longer = 3:8. The longer rod exceeds the shorter by 45 cm. Find the shorter length in centimetres.

Practice question 5

Three boxes P, Q and R contain a total of 124 beads. Their counts satisfy P:Q = 3:4 and Q:R = 6:5. How many beads are in R?

• A. 20
• B. 48
• C. 40
• D. 50

Practice question 6

For nonnegative usage x, two charges in rupees are P = 9x and Q = 15 + 9x. Which statement is correct?

• A. Only P is directly proportional to x.
• B. Only Q is directly proportional to x.
• C. Both are directly proportional because both graphs are straight.
• D. Neither is directly proportional because division by x is undefined at zero.

Practice question 7

Three identical packing machines make 252 packs in 7 minutes. They have constant independent additive rates, no setup delay and enough supplies. At the same rates, how many packs will five machines make in 9 minutes?

Practice question 8

A vehicle covers the same 180 km without stops, first at 45 km/h and then at 60 km/h, each speed constant. Relative to the first journey, what is the percentage decrease in travel time?

• A. 33⅓%
• B. 15%
• C. 75%
• D. 25%

Practice question 9

A divisible production job needs 432 units. Each identical machine makes 9 units per hour at a constant independent additive rate. There are no delays or shared bottlenecks, and work can keep every machine productive. How many hours do six machines need?

Practice question 10

Six workers completed a job in 18 days. The same job is assigned to nine workers. Nothing is stated about individual productivity, parallel working or shared tools. Which completion time is justified by the information alone?

• A. 12 days
• B. 18 days
• C. 27 days
• D. No unique completion time is determined.

Answers with reasoning

Solution 1

A is correct. Convert 0.84 kg to 840 g; 840:630 = 4:3 after dividing by 210. The first mass exceeds the second, agreeing with 4 being greater than 3. B reverses the requested order. C has a tenfold unit/place-value error and says the first mass is smaller. D changes only the second converted term by a factor of ten; 840:63 = 40:3, not 4:3.

Solution 2

The factor taking 5 parts to 35 L is 35/5 = 7 L per part. Apply it to both terms: v = 8 × 7 = 56 L. Check 35/56 = 5/8. Adding 30 to 8 to get 38 copies an additive change; equivalent ratios require a common multiplier. Computing 35 × 5/8 = 21.875 uses the ratio in reverse. Here one actual amount supplies the scale, so the ratio is sufficient with that extra datum.

Solution 3

B is correct. The total is 11 parts, so one part is 990/11 = ₹90 and B receives 7 × 90 = ₹630. A receives ₹360; the sum is ₹990 and the ratio is 4:7. Option A gives the other account. C is 990 × 7/4, treating the whole as account A, and even exceeds the total. D is only one ratio part, not B's seven parts.

Solution 4

Let the lengths be 3k cm and 8k cm. Their difference is 5k = 45, so k = 9 and the shorter is 27 cm. The longer is 72 cm; 72 − 27 = 45 and 27:72 = 3:8. Dividing 45 by 11 would treat the difference as a total. Giving 9 cm stops at one part. Giving 72 cm answers the longer-length question.

Solution 5

C is correct. Match Q at 12 parts: P:Q = 9:12 and Q:R = 12:10, giving 9:12:10. The total is 31 parts, so k = 124/31 = 4 beads per part and R = 10 × 4 = 40. P = 36 and Q = 48; both given ratios and the total check. A uses the unscaled 5 from the second ratio with the final 4-bead scale, mixing scales. B gives Q instead of R. D gives R:(P+Q) = 50:74 and cannot satisfy the combined ratio with a total of 124.

Solution 6

A is correct. P/x = 9 for positive x, and P = 9x includes P = 0 at x = 0. Q/x = 9 + 15/x varies, and Q begins at (0,15). For x = 2 and 4, P is 18 and 36, but Q is 33 and 51. B selects the fixed-offset model instead of the constant quotient. C mistakes a constant slope for a line through the origin. D confuses testing y/x at positive x with defining y = kx at zero; the quotient need not be defined at zero for P to be a direct-proportion model.

Solution 7

The first run has 3 × 7 = 21 machine-minutes, so the rate is 252/21 = 12 packs/(machine·minute). The new run has 5 × 9 = 45 machine-minutes and produces 45 × 12 = 540 packs. Cross-check: 252 × (5/3) × (9/7) = 540. Multiplying only by 5/3 gives 420 and misses the extra time; multiplying only by 9/7 gives 324 and misses the extra machines. No fixed-total condition justifies an inverse factor here.

Solution 8

D is correct. Original time = 180/45 = 4 h; new time = 180/60 = 3 h. Decrease = 1 h out of the original 4 h, or 25%. A is the percentage increase in speed, or the time difference divided by the new time; neither is the requested old-time-based decrease. B turns the speed difference 15 km/h into a percentage without a base. C is the remaining time as a percentage of the original, 3/4 × 100, not the decrease.

Solution 9

Six machines together make 6 × 9 = 54 units/h, so time = 432/54 = 8 h. Equivalently the job needs 432/9 = 48 machine-hours; 6t = 48 gives t = 8. Four machines would need 12 h, and both 4 × 12 and 6 × 8 equal 48 machine-hours. Reporting 48 h ignores the six simultaneous machines; reporting 72 h by multiplying 12 by 6 ignores both the old four-machine count and the fixed workload. The answer's unit is hours, not units or machine-hours.

Solution 10

D is correct. Under an added equal, constant, additive-rate model with fully divisible work, 6 × 18 = 108 worker-days and nine workers would need 108/9 = 12 days. Under a different model, a shared resource lets only six workers contribute at once, so the same observed first job can coexist with a second duration of 18 days. These two consistent possibilities already show that the given data do not select one time.

A silently adds the independent equal-rate assumptions. B silently adds a bottleneck or equivalent restriction. C applies a direct multiplier 18 × 9/6 even though fixed work has not been modelled; it is not implied either. Ask for a productivity and concurrency model, not merely another worker count. A conditional calculation is useful only when its added assumptions are disclosed.

References and further reading

The official GATE 2027 General Aptitude syllabus, section 2, includes ratios, percentages and interpretation of graphs and tables. The lesson sequence here is a teaching choice. The official papers and syllabi page states that the test papers are in English; the Hindi explanations here support learning.

For background on equal ratios and solving proportions, see OpenStax Prealgebra 2e, section 6.5. The explanations, situations, questions, data and drawings in this lesson are independently authored; the linked sources are references, not sources of copied exercises.

Analogy

Think of a measuring scoop whose capacity you do not yet know. One container holds three scoops and another five. Their amounts are in the ratio 3:5 regardless of the scoop size. Finding that the two containers together hold 640 mL fixes one scoop at 80 mL, so the amounts are 240 mL and 400 mL. A ratio is the recipe for relative shares; an actual total sets its scale.

For rates, imagine two labels on a job sheet: “same running time” and “same finished job”. With the same running time and independent identical machines, more machines give more output. With the same finished job, more such machines can reduce time. Read the label before choosing direct or inverse proportion. The scoop analogy explains share allocation; it does not by itself establish a real machine's productivity or remove bottlenecks.

Quick reference

Ratios and shares

• Preserve the comparison order. Convert like quantities to the same unit before cancelling units; a rate between unlike quantities keeps a compound unit.
• A:B = a:b means A = ak and B = bk for one common positive k. Multiply or divide all terms by the same nonzero factor; adding the same number need not preserve the ratio.
• Total T: k = T/(a + b). The shares of the whole are a/(a + b) and b/(a + b).
• Positive difference D = B − A with b greater than a: k = D/(b − a). Equal ratio terms cannot have a nonzero difference.
• A:B = a:b and B:C = c:d give A:B:C = ac:bc:bd. Match the shared quantity and check both original ratios.

Choose the model before the arithmetic

• Direct: y = kx, with y/x fixed for allowed nonzero x. Multiplying x by s multiplies y by s. Its mathematical line passes through the origin.
• Fixed offset: y = mx + b with b nonzero is not direct proportion, even though its graph is straight and its slope is constant.
• Inverse: y = K/x for positive x and fixed positive K. The product xy is fixed; multiplying x by s divides y by s. Zero is excluded.
• Constant independent machine rates: Q = rnt. Hold time fixed for output-to-machine direct proportion; hold output fixed for time-to-machine inverse proportion.
• Fixed-distance uninterrupted travel: t = d/v. A speed increase fraction p gives a time-decrease fraction p/(1 + p), measured from the original time.

Final checks

State the units, fixed quantity, positive/nonzero restrictions and productive-rate assumptions. A ratio alone does not fix absolute amounts. A larger team alone does not fix a completion time. Check the total or difference, every supplied ratio, feasible counts and the comparison base. Do not round intermediate fractions unless the task calls for an approximation.