Overview
This chapter introduces direct and inverse proportions — fundamental ideas that describe how two quantities change with respect to each other. In a direct proportion, one quantity increases or decreases in the same ratio as another (y = kx); in an inverse proportion, one quantity increases while the other decreases so that their product is constant (xy = k). The chapter develops methods such as the unitary method, cross-multiplication, finding the constant of proportionality, and simple graphing for proportions. Through worked examples and real-life problems (cost and quantity, speed and distance, time and work), students learn to set up equations, solve numerical and word problems, and interpret results. Understanding these concepts is important for solving everyday problems, for science and economics applications, and as a foundation for algebra and coordinate geometry in higher classes. The chapter also builds reasoning and problem-solving skills by encouraging students to recognize proportional relationships and choose efficient solution strategies.
Learning Objectives
- Define direct proportion and inverse proportion using mathematical expressions
- Explain the meaning of the constant of proportionality in direct and inverse relations
- Formulate linear equations that represent direct proportion and reciprocal equations for inverse proportion
- Determine the constant of proportionality from a table of values or a given relation
- Solve numerical problems involving direct proportion using ratio and unitary methods
- Solve numerical problems involving inverse proportion using the product-constant method
- Calculate missing terms in a proportion using cross-multiplication and verify results
- Distinguish between direct and inverse proportion from given data, tables, or word problems
Topics in this chapter
7 topics · tap a topic title to jump straight to it.
Proportion and Proportionality
Proportion and Proportionality
Key Point: a : b = c : d ⇔ a/b = c/d ⇔ ad = bc (cross-multiplication)
Proportion means an equality between two ratios. If a : b = c : d (with b and d ≠ 0), we say a, b, c, d are in proportion and write a/b = c/d. A useful test is cross-multiplication: a/b = c/d ⇔ ad = bc. In a proportion, a and d are called the extremes and b and c the means; product of extremes equals product of means.
Proportionality describes a specific relationship between two quantities. Two main types appear in Class 8:
- Direct proportionality: One quantity varies directly as the other. If y is directly proportional to x, write y ∝ x, or y = kx, where k is the constant of proportionality. Graph: a straight line through the origin with slope k.
- Inverse proportionality: One quantity varies inversely as the other. If y is inversely proportional to x, write y ∝ 1/x, or y = k/x (k ≠ 0). Graph: a rectangular hyperbola with asymptotes x = 0 and y = 0.
Key points and properties:
- If a/b = c/d = k (same constant), then (a+c)/(b+d) = k and (a−c)/(b−d) = k (provided denominators ≠ 0).
- To check proportionality from data, compute ratios y/x for every pair; if all equal, they are directly proportional.
- Linear relationship y = mx + c is proportional only when c = 0. Proportionality requires the graph to pass through the origin.
Understanding proportionality lets you scale quantities, solve problems with unit rates, and interpret graphs in physics and everyday life.
- Recipe scaling: A cake recipe needs 2 cups flour for 8 muffins. For 12 muffins, flour = 2 × (12/8) = 3 cups (direct proportion).
- Cost and quantity: If 5 pens cost ₹60, then cost of 8 pens = 60 × (8/5) = ₹96 (direct proportion).
- Map distance: If 1 cm on the map represents 5 km, then 7 cm represents 35 km (direct proportion with constant 5 km/cm).
- Workers and time (fixed work): 6 workers finish a job in 10 days. How long for 12 workers? Time ∝ 1/number of workers, so days = 10 × (6/12) = 5 days (inverse proportion).
- Speed and travel time (fixed distance): For a fixed distance, if speed doubles, time halves — time ∝ 1/speed (inverse proportion).
- \[a : b = c : d ⇔ a/b = c/d ⇔ ad = bc (cross-multiplication)\]
- \[Direct proportion: y ∝ x ⇔ y = kx where k = y/x (constant of proportionality)\]
- \[Inverse proportion: y ∝ 1/x ⇔ y = k/x (k ≠ 0)\]
- \[Scaling rule: If quantity y = kx\]\[then scaling x by factor t scales y by the same factor t (y_new = k·(t·x) = t·y).\]
- \[Proportional check for data: compute y1/x1\]\[y2/x2, ...\]\[if all equal\]\[direct proportionality holds.\]
Direct Proportion (Direct Variation)
Direct Proportion (Direct Variation)
Key Point: Direct proportion: y ∝ x
Definition: Two quantities are in direct proportion (or direct variation) when one quantity increases or decreases exactly in the same ratio as the other. If y is directly proportional to x, we write y ∝ x.
Mathematical form: Direct proportion is expressed as y = kx, where k (a nonzero constant) is called the constant of proportionality or unit rate. For every pair (x, y) that satisfy the relation, the ratio y/x = k is the same.
How to test: Given pairs (x₁, y₁) and (x₂, y₂), they are directly proportional if y₁/x₁ = y₂/x₂. Equivalently, y₁x₂ = y₂x₁ (cross-multiplication).
Properties:
- The graph of a direct proportion is a straight line passing through the origin (0, 0).
- The slope of this line equals the constant of proportionality k.
- If k > 0, both quantities increase together; if k < 0, one increases while the other decreases (less common in basic Class 8 examples).
Solving word problems: Translate the statement into y = kx or use ratios. Find k from known values (k = y/x), then use it to find unknowns, or set up proportion equations like y₁ : x₁ = y₂ : x₂.
Difference from general linear relations: All direct proportions are linear functions that pass through the origin. A linear equation like y = mx + c with c ≠ 0 is not a direct proportion.
- Example 1 — Cost of apples: If 1 kg of apples costs ₹80, cost (C) is directly proportional to weight (w). C = 80w. If you buy 2.5 kg, C = 80 × 2.5 = ₹200.
- Example 2 — Speed and distance: At a constant speed, distance (d) is directly proportional to time (t). If a car goes 60 km in 1 hour, d = 60t. In 3.5 hours it travels d = 60 × 3.5 = 210 km.
- Example 3 — Wages: If a worker earns ₹150 per hour, total wage W = 150h. For 8 hours, W = 150 × 8 = ₹1200.
- Example 4 — Printing pages: A printer prints 30 pages in 2 minutes. Pages printed P are directly proportional to time t, P = k t. Find k = 30/2 = 15 pages/min. In 7 minutes it prints P = 15 × 7 = 105 pages.
- Example 5 — Recipe scaling: A recipe needs 200 g flour for 4 servings. Flour F is proportional to servings s: F = (200/4) s = 50 s. For 7 servings, F = 50 × 7 = 350 g.
- \[Direct proportion: y ∝ x\]
- \[Equation form: y = kx (k = constant of proportionality)\]
- \[Find k: k = y / x\]
- \[Test proportionality (two pairs): y1 / x1 = y2 / x2 (or) y1 * x2 = y2 * x1\]
- \[To scale: if y1 corresponds to x1\]\[then y2 = (y1 / x1) × x2\]
Inverse Proportion (Inverse Variation)
Inverse Proportion (Inverse Variation)
Key Point: y = k / x (k is a nonzero constant)
Definition: Two quantities x and y are in inverse proportion (or inverse variation) if one quantity increases when the other decreases so that their product is constant. We write y \propto 1/x or y = k/x, where k is a nonzero constant.
Mathematical form and key property: If y varies inversely as x, then y = k/x for all allowed x (x ≠ 0). Equivalently xy = k (constant). From one known pair (x1, y1) you get k = x1y1, and for any other x, y = k/x.
How to use it (procedure): 1) Identify the two variables and confirm one decreases as the other increases. 2) Use a known pair to compute the constant k = x·y. 3) Use y = k/x (or x·y = k) to find unknown values. For comparing two situations: x1y1 = x2y2.
Difference from direct proportion: In direct proportion y \propto x (y = m x). In inverse proportion y \propto 1/x (y = k/x) so graphs and behaviour are opposite: doubling x halves y (when k is fixed).
Domain and graph behaviour: x cannot be 0. For k>0 the curve is a rectangular hyperbola in the first and third quadrants (y>0 for x>0, y<0 for x<0). The axes x=0 and y=0 are asymptotes. For practical Class 8 contexts we usually consider positive x and y, so the graph lies in the first quadrant.
- Workers and days: 6 workers finish a job in 10 days. How many days will 15 workers take? k = 6×10 = 60, so days = 60/15 = 4 days.
- Speed and time for a fixed distance: If a 120 km trip takes 3 hours at 40 km/h, then time ∝ 1/speed. For 60 km/h time = 120/60 = 2 hours (half the time when speed doubles).
- Machines producing items: 5 machines produce a batch in 8 hours. How long for 10 machines? k = 5×8 = 40, so time = 40/10 = 4 hours.
- Pipes filling a tank (inverse relation between number of identical taps and time): If 4 taps fill a tank in 6 hours, then 8 taps fill it in 4·? Actually k = 4×6 = 24, so time with 8 taps = 24/8 = 3 hours.
- \[y = k / x (k is a nonzero constant)\]
- \[x y = k (product of the two variables is constant)\]
- \[Given (x1\]\[y1) and (x2\]\[y2): x1·y1 = x2·y2\]
- \[To find k: k = x·y for any known pair\]
- \[If x is multiplied by a factor a\]\[then y is divided by the same factor: x→a·x ⇒ y→y/a\]
Graphical Representation
Graphical Representation
Key Point: Direct: y = kx (k = constant of proportionality, k = y/x)
What it means
Graphical representation shows how two quantities vary with each other by plotting points on a coordinate plane and joining them. For direct and inverse proportions we use simple, characteristic graphs:
Direct proportion (y ∝ x): if y = kx (k is constant), points (x,y) lie on a straight line that passes through the origin (0,0). The slope of the line is the constant k.
Inverse proportion (y ∝ 1/x): if y = k/x (k ≠ 0), points (x,y) form a curve (rectangular hyperbola) that does not pass through the origin. The product xy is constant (= k). The axes x = 0 and y = 0 are asymptotes.
How to draw
1. Make a table of values: choose several x-values (for inverse, avoid x = 0), compute corresponding y-values using the relation.
2. Mark each (x,y) on the coordinate plane, label axes and choose equal scales.
3. For direct proportion, draw a straight line through plotted points and the origin. For inverse, draw a smooth curve through the plotted points approaching the axes.
How to recognise from a graph
- If the graph is a straight line through the origin, quantities are in direct proportion.
- If xy is constant for points on the curve and the graph is a hyperbola with axes as asymptotes, quantities are in inverse proportion.
- Direct proportion — Cost and weight: Price per kg = Rs 40. Table: (kg, price) = (1,40), (2,80), (3,120). Plot these; points lie on a straight line through the origin with slope 40.
- Direct proportion — Distance and time at constant speed: Speed = 60 km/h. Distance = 60 × time. Table: (time h, distance km) = (0.5,30), (1,60), (2,120). Graph is a straight line through origin.
- Inverse proportion — Workers and days to finish a job: 4 workers take 15 days → k = workers × days = 60. Relation: days = 60 / workers. Table: (workers, days) = (2,30), (3,20), (4,15), (5,12). Plot gives a hyperbola in the first quadrant.
- Inverse proportion — Time and speed for fixed distance: Distance = 120 km. Time = 120 / speed. Table: (speed km/h, time h) = (30,4), (40,3), (60,2). Graph is a rectangular hyperbola (first quadrant).
- \[Direct: y = kx (k = constant of proportionality\]\[k = y/x)\]
- \[Slope interpretation (direct): slope = Δy/Δx = k\]
- \[Inverse: y = k/x (k = xy\]\[constant product)\]
- \[Asymptotes (inverse): x = 0 and y = 0\]\[curve approaches axes but never touches them\]
- \[To test from data: if y/x is constant → direct proportionality\]\[if x·y is constant → inverse proportionality\]
Methods and Techniques for Solving Problems
Methods and Techniques for Solving Problems
Key Point: Direct proportion: y ∝ x ⇒ y = kx, where k is constant of proportionality.
Direct and inverse proportion problems require identifying how two quantities change relative to each other and then using an appropriate technique to compute unknown values. Common steps are:
- Identify the two variables and whether the relation is direct (both increase/decrease together) or inverse (one increases while the other decreases when a certain product is constant).
- Choose a method: unitary method, ratio/proportion (cross-multiplication), constant-of-proportionality, or graphs.
- Set up the relation (y = kx for direct, y = k/x for inverse) or a proportion equation, solve for the unknown, and check units and reasonableness.
Key techniques:
- Unitary method — find value for one unit, then multiply to get required units; useful for costs, recipes, and simple scaling.
- Ratio and cross-multiplication — use when two ratios are equal: a/b = c/d ⇒ ad = bc; quick for comparing fractions of quantities.
- Constant of proportionality — find k from known pair(s) then use y = kx (direct) or y = k/x (inverse).
- Inverse-proportion manipulation — if x and y are inversely proportional, xy = k (constant). Use xy = x1y1 = x2y2 to find unknowns.
- Graphical method — plot points and read off values: straight line through origin for direct proportion; rectangular hyperbola for inverse proportion.
Always label variables, units, and state which type of proportion is used. When solving word problems, translate words to a mathematical relation (for example: "more workers → less time" implies inverse proportion if total work is fixed).
- Example 1 (Direct proportion, unitary method): Cost of 7 kg apples is ₹420. Find cost of 3 kg. Method: cost per 1 kg = 420 ÷ 7 = ₹60. Cost of 3 kg = 60 × 3 = ₹180.
- Example 2 (Direct proportion, cross-multiplication): 5 pens cost ₹75. How much do 8 pens cost? Set up proportion 5/8 = 75/x ⇒ 5x = 8×75 ⇒ x = (8×75)/5 = 8×15 = ₹120.
- Example 3 (Inverse proportion, product method): 4 machines complete a job in 12 days. How many days will 6 machines take (assuming same efficiency)? Since machines × days = constant, 4×12 = 6×d ⇒ 48 = 6d ⇒ d = 8 days.
- Example 4 (Inverse proportion with speed/time): A car takes 5 hours to cover a fixed distance at 40 km/h. If speed increases to 50 km/h, how long will it take? Time ∝ 1/speed, so t1/t2 = v2/v1 ⇒ 5/t2 = 50/40 ⇒ 5/t2 = 5/4 ⇒ t2 = 4 hours.
- Example 5 (Recipe scaling): A cake recipe needs 250 g flour for 4 servings. For 10 servings, flour required = (250/4)×10 = 62.5×10 = 625 g (unitary method).
- Example 6 (Using constant k): If y is directly proportional to x and y = 15 when x = 3, find y when x = 8. k = y/x = 15/3 = 5, so y = 5×8 = 40.
- \[Direct proportion: y ∝ x ⇒ y = kx\]\[where k is constant of proportionality.\]
- \[Inverse proportion: y ∝ 1/x ⇒ y = k/x\]\[equivalently x·y = k (constant).\]
- \[Proportion equality: a/b = c/d ⇒ ad = bc (useful for cross-multiplication).\]
- \[Unitary method: value per one unit = given value ÷ number of units\]\[required value = (value per one unit) × desired units.\]
- \[Scaling ratios: For direct proportion x1/x2 = y1/y2\]\[For inverse proportion x1/x2 = y2/y1 (because x1·y1 = x2·y2).\]
Compound Proportion
Compound Proportion
Key Point: General compound form: Y ∝ (A^p * B^q * ...) / (C^r * D^s * ...).
What is Compound Proportion?
Compound proportion deals with a situation where a quantity depends simultaneously on two or more other quantities, some directly proportional and some inversely proportional. Instead of a single direct or inverse relation, the combined effect (compound) is found by multiplying the individual proportional factors.
General idea
If Y depends directly on A and B and inversely on C and D, we write (informally):
Y ∝ (A * B) / (C * D).
In practice, to find a new value Y2 from a known value Y1 when the influencing factors change, multiply Y1 by the product of the ratios of directly proportional factors and divide by the product of the ratios of inversely proportional factors:
Y2 = Y1 * (A2/A1) * (B2/B1) * (C1/C2) * (D1/D2)
How to solve (stepwise):
- Identify which quantities are directly proportional and which are inversely proportional to the required quantity.
- Write the ratio for each factor as new/old for direct factors and old/new for inverse factors.
- Multiply the known value by all these ratios to get the new value.
Key point: Treat each factor separately as a ratio, then multiply all ratios to adjust the original value.
Short worked example (explained)
If 6 workers, working 10 days, produce 240 units, how many units will 8 workers produce in 12 days? Units are directly proportional to both number of workers and number of days. So:
- Units per worker-day = 240 / (6*10) = 240 / 60 = 4 units.
- New worker-days = 8 * 12 = 96.
- Units = 96 * 4 = 384 units.
This demonstrates the compound proportion principle: change in two direct factors (workers and days) changes the result by the product of their ratios (8/6 * 12/10 = 4/3 * 6/5 = 24/15 = 8/5 = 1.6; 240 * 1.6 = 384).
- Example 1: (All direct) If 6 workers in 10 days produce 240 items, how many items will 8 workers produce in 12 days? Solution: Items ∝ workers * days. Items per worker-day = 240/(6*10) = 4. New items = 8*12*4 = 384.
- Example 2: (Direct + Inverse) 5 machines take 12 hours to produce 600 items. How long will 8 machines take to produce 1000 items? Time ∝ items / machines. So time2 = 12 * (1000/600) * (5/8) = 12 * (5/3) * (5/8) = 12 * 25/24 = 12.5 hours.
- Example 3: (Different physical quantities) Fuel used is directly proportional to distance and inversely proportional to fuel-efficiency. If 40 liters cover 480 km at 12 km/l, then liters needed for 600 km at 15 km/l = 40 * (600/480) * (12/15) = 40 * 1.25 * 0.8 = 40 liters.
- \[General compound form: Y ∝ (A^p * B^q * ...) / (C^r * D^s * ...).\]
- \[When moving from a known case (Y1\]\[A1\]\[B1\]\[C1...) to a new case (Y2\]\[A2\]\[B2\]\[C2...): Y2 = Y1 * (A2/A1)^{p} * (B2/B1)^{q} * (C1/C2)^{r} * (D1/D2)^{s}.\]
- \[For simple integer-exponent cases (each exponent = 1): Y2 = Y1 * (A2/A1) * (B2/B1) * (C1/C2) * (D1/D2).\]
- \[Tip: Convert everything to ratios (new/old for direct\]\[old/new for inverse) then multiply all ratios with the known value.\]
Applications and Word Problems
Applications and Word Problems
Key Point: Direct proportion: y = kx (k is the constant of proportionality).
What the topic means
Applications and word problems in Direct and Inverse Proportions involve translating real-life situations into mathematical relations where one quantity changes in proportion to another. You must first decide whether the quantities are directly proportional (both increase or decrease together) or inversely proportional (one increases while the other decreases so that their product remains constant).
How to approach such problems
- Read the problem and identify the two quantities and whether one is directly or inversely proportional to the other.
- Introduce variables and write the proportionality: for direct proportion use y = kx; for inverse proportion use xy = k (or y = k/x).
- Use the given values to find the constant of proportionality k.
- Substitute k to find the unknown value; check units and reasonableness of the result.
Common types of real-life problems: price & quantity, ingredients (recipes), distance & time (at constant speed), work problems (workers & days), wages, machine & time, concentration problems, scaling geometric figures.
Solution techniques: ratio method (unitary method), forming equations y = kx or xy = k and solving, or cross-multiplication for proportions. Always label units and, if useful, make a table of values.
- Example 1 — Direct proportion (cost and quantity): If 5 kg of apples cost Rs. 150, how much do 8 kg cost? Solution: Price ∝ quantity. Price per kg = 150/5 = 30. So 8 kg costs 30 × 8 = Rs. 240. (Or 150 × 8/5 = 240.)
- Example 2 — Inverse proportion (speed and time for a fixed distance): A car covers a fixed distance of 360 km in 6 hours. How long will it take to cover the same distance at 90 km/h? Solution: Time = Distance/Speed. At 60 km/h time = 360/60 = 6 h (given). For 90 km/h: time = 360/90 = 4 h. Using inverse proportion: time ∝ 1/speed, so t1 × v1 = t2 × v2 ⇒ 6 × 60 = t2 × 90 ⇒ t2 = (6×60)/90 = 4 h.
- Example 3 — Inverse proportion (work and workers): 12 workers can finish a job in 20 days. How many days will 8 workers take (same efficiency, same total work)? Solution: Workers × Days = constant. So 12 × 20 = 8 × D ⇒ D = (12×20)/8 = 30 days.
- Example 4 — Direct proportion (recipe scaling): A recipe for 4 servings needs 200 g flour. How much flour for 10 servings? Solution: Flour ∝ servings. Flour per serving = 200/4 = 50 g. For 10 servings = 50 × 10 = 500 g (or 200 × 10/4 = 500 g).
- \[Direct proportion: y = kx (k is the constant of proportionality).\]
- \[From two pairs (x1\]\[y1) and (x2\]\[y2) in direct proportion: y1/x1 = y2/x2 = k or y2 = y1 × (x2/x1).\]
- \[Inverse proportion: xy = k or y = k/x (k is constant).\]
- \[From two pairs (x1\]\[y1) and (x2\]\[y2) in inverse proportion: x1·y1 = x2·y2 = k or y2 = (x1·y1)/x2.\]
- \[Common solution methods: unitary method (find one-unit value then scale)\]\[cross-multiplication for proportion equations.\]
Key Concepts
- Ratio
- A comparison of two quantities by division, written as a:b or a/b.
- Proportion
- An equality of two ratios, written as a:b = c:d, meaning a/b = c/d.
- Direct Proportion (Direct Variation)
- Two quantities are directly proportional if their ratio is constant; doubling one doubles the other.
- Inverse Proportion (Inverse Variation)
- Two quantities are inversely proportional if their product is constant; increasing one decreases the other.
- Constant of Proportionality
- The fixed number k in y = kx (direct) or xy = k (inverse) that relates two proportional quantities.
- Unitary Method
- A method to find the value of one unit and then scale to find the required value.
- Cross-multiplication
- A technique to compare or solve proportions by multiplying diagonally: a/b = c/d ⇒ a·d = b·c.
- Equivalent Ratios
- Different ratio expressions that represent the same relationship between quantities.
- Proportionality Equation
- An equation that expresses proportionality, e.g., y = kx for direct and y = k/x for inverse relationships.
- Directly Proportional Quantities
- Pairs of quantities where one is a constant multiple of the other (y = kx).
- Inversely Proportional Quantities
- Pairs of quantities where their product is constant (xy = k).
- Graph of Direct Proportion
- A straight line through the origin when y is plotted against x for y = kx.
- Graph of Inverse Proportion
- A rectangular hyperbola in the first quadrant for positive x and y when xy = k.
- Scaling (Scale Factor)
- Multiplying all dimensions of an object by the same factor to enlarge or reduce it proportionally.
- Continued Proportion
- A sequence of three or more quantities where consecutive pairs are in the same ratio: a:b = b:c = c:d...
- Mean Proportional (Geometric Mean)
- If a:b = b:c, then b is the mean proportional between a and c; b = √(a·c).
- Proportional Division (Dividing in a Given Ratio)
- Splitting a total quantity into parts proportional to given numbers (ratios).
- Unit Rate
- A rate expressed per one unit of the second quantity, often found using the unitary method.
- Proportionate Increase/Decrease
- Change in a quantity that is directly proportional to change in another quantity; described by scaling factor.
Practice Questions
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If y is directly proportional to x and y = 15 when x = 3, what is y when x = 8? / यदि y, x के सीधे अनुपात में है और x = 3 पर y = 15 है, तो x = 8 पर y क्या होगा? (a) 30 / 30 (b) 40 / 40 (c) 45 / 45 (d) 24 / 24
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(b) — Direct proportion: y = kx. Find k = 15/3 = 5. Then y = 5 × 8 = 40. / सीधा अनुपात: y = kx। k = 15/3 = 5, तो y = 5 × 8 = 40।
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6 workers can finish a job in 12 days. How many days will 9 workers take for the same job? / 6 मजदूर एक काम 12 दिन में पूरा करते हैं। 9 मजदूर उसी काम को कितने दिन में करेंगे? (a) 18 days / 18 दिन (b) 8 days / 8 दिन (c) 6 days / 6 दिन (d) 10 days / 10 दिन
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(b) — Inverse proportion: workers × days = constant. 6 × 12 = 72. Days for 9 workers = 72 ÷ 9 = 8 days. / व्युत्क्रम अनुपात: मजदूर × दिन = स्थिरांक। 6 × 12 = 72। 9 मजदूरों के लिए दिन = 72 ÷ 9 = 8।
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The graph of a direct proportion y = kx is a _____. / सीधे अनुपात y = kx का ग्राफ _____ है। (a) Parabola / परवलय (b) Straight line not through origin / मूल बिंदु से न गुजरने वाली सीधी रेखा (c) Straight line through the origin / मूल बिंदु से गुजरने वाली सीधी रेखा (d) Rectangular hyperbola / आयताकार अतिपरवलय
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(c) — For y = kx, when x = 0, y = 0, so the line always passes through the origin (0, 0) with slope k. / y = kx में x = 0 पर y = 0, इसलिए रेखा सदैव मूल बिंदु (0, 0) से गुजरती है।
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If 5 kg of apples cost ₹200, the cost of 8 kg is ₹_____. / यदि 5 kg सेब का मूल्य ₹200 है, तो 8 kg का मूल्य ₹_____ होगा।
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₹320 / ₹320 — Direct proportion: cost ∝ quantity. Rate = 200/5 = ₹40 per kg. Cost of 8 kg = 40 × 8 = ₹320. / सीधा अनुपात: मूल्य ∝ मात्रा। दर = ₹40/kg। 8 kg का मूल्य = ₹320।
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In an inverse proportion, if one quantity is doubled, the other quantity is _____. / व्युत्क्रम अनुपात में, यदि एक राशि दोगुनी हो जाती है, तो दूसरी राशि _____।
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Halved / आधी हो जाती है — In inverse proportion xy = k (constant). If x doubles to 2x, then y must become k/(2x) = y/2. / व्युत्क्रम अनुपात में xy = k। यदि x दोगुना होकर 2x हो, तो y = k/(2x) = y/2 हो जाता है।
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True or False: The graph of an inverse proportion is a straight line passing through the origin. / सत्य या असत्य: व्युत्क्रम अनुपात का ग्राफ मूल बिंदु से गुजरने वाली सीधी रेखा होती है।
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False / असत्य — The graph of an inverse proportion y = k/x is a rectangular hyperbola, not a straight line. It never passes through the origin (as x = 0 is undefined). / व्युत्क्रम अनुपात y = k/x का ग्राफ आयताकार अतिपरवलय होता है, सीधी रेखा नहीं।
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A recipe for 6 people needs 300 g of flour. How much flour is needed for 10 people? / 6 लोगों के लिए एक रेसिपी में 300 g आटे की आवश्यकता है। 10 लोगों के लिए कितना आटा चाहिए?
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500 g / 500 g — Direct proportion: flour ∝ people. Flour per person = 300/6 = 50 g. For 10 people = 50 × 10 = 500 g. / सीधा अनुपात: आटा ∝ लोग। प्रति व्यक्ति आटा = 50 g। 10 लोगों के लिए = 500 g।
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4 taps can fill a tank in 6 hours. How long will it take 8 taps to fill the same tank (assuming same flow rate)? / 4 नल एक टंकी 6 घंटे में भर सकते हैं। 8 नल उसी टंकी को कितने समय में भरेंगे?
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3 hours / 3 घंटे — Inverse proportion: taps × time = constant. 4 × 6 = 24. Time for 8 taps = 24 ÷ 8 = 3 hours. More taps fill the tank faster, showing inverse relationship. / व्युत्क्रम अनुपात: नल × समय = 24। 8 नलों के लिए समय = 24 ÷ 8 = 3 घंटे।
Related Laws & Principles
Explore allFoundational laws & principles behind this chapter. Each one opens a full page — what it says, why it matters, five practice questions and the mistakes to avoid.