ORIGINAL PRACTICE · 1
Solve 3(x−2)=18.
Reveal answer & reasoning
x=8.
Divide both sides by the nonzero constant 3 to get x−2=6, then add 2. Substitute:3(8−2)=18. The parentheses require subtracting 2 before multiplying.
UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON
Equations and inequalities
Translate quantities into equations, preserve inequality direction and check algebra against geometric meaning.
BY THE END, YOU CAN
- Solve and verify linear systems and bounded inequalities.
- Reject algebraic roots outside the stated domain.
- Apply elementary geometry and capacity formulas with correct units.
BEFORE YOU START
- Use signed arithmetic, fractions and square roots of perfect squares.
Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.
LEARN THROUGH ORIGINAL PRACTICE
Original UPSC-style worked applications and practice
All teaching examples, cases, prompts and model answers are original. Official exam questions and answer keys are not reproduced. This common-core lesson does not imply complete historical paper coverage or an official score.
01 / UNDERSTAND & APPLY
Define the quantity before the symbol
An expression represents a quantity, such as 3 x+2. An equation asserts equality between quantities, such as 3 x+2=14. Define x with its unit and domain: a ticket count is a nonnegative integer, while a measured length may be a positive real number. These restrictions can eliminate otherwise algebraic solutions.
Translate each independent statement separately. A total number of tickets and their total revenue yield different equations. “Three more than the width” is width+3, not 3×width. Include consistent units before combining terms.
Algebra does not supply the model’s assumptions. If a problem says only two ticket prices exist and every ticket is sold at one of them, use that. If discounts or unsold tickets were possible but unspecified, the model would need clarification.
02 / UNDERSTAND & APPLY
Independent, dependent and inconsistent equations
Adding the same quantity to both sides or multiplying by the same nonzero quantity preserves an equation’s solution set. Dividing by a variable requires checking it is nonzero, or you may discard a valid zero case. For simple systems, substitution or elimination combines constraints without losing their original meaning.
Two equations do not automatically determine two unknowns. One may merely repeat the other, leaving infinitely many real pairs. They may contradict one another, leaving none. A unique solution needs sufficiently independent and consistent constraints, then still must meet the original domain.
ORIGINAL WORKED EXAMPLE
Two ticket prices
A total of 16 tickets is sold, each at either 40 or 70 units, with total revenue 880. How many tickets were sold at each price?
- Let x be the 40-unit ticket count and y the 70-unit count, both nonnegative integers. Then x+y=16 and 40 x+70 y=880.
- Multiply the count equation by 40:40 x+40 y=640. Subtract it from the revenue equation:30 y=240.
- Thus y=8 and x=16−8=8.
- Check both constraints:8+8=16 and 8×40+8×70=320+560=880.
Answer Eight tickets at each price.
ORIGINAL PRACTICE · 1
Consider x+y=5 and 2 x+2 y=10. How many real solutions exist? How many if x,y must be nonnegative integers?
Reveal answer & reasoning
Infinitely many real pairs; six nonnegative integer pairs.
The second equation doubles the first and adds no new constraint. Real x can vary with y=5−x. In the nonnegative integer domain, x=0,1,2,3,4,5 gives y=5,4,3,2,1,0.
ORIGINAL PRACTICE · 2
Can x+y=5 and x+y=7 hold simultaneously?
Reveal answer & reasoning
No; the system is inconsistent.
The same quantity cannot equal two distinct numbers. Subtracting the equations gives 0=2, exposing the contradiction.
03 / UNDERSTAND & APPLY
Direction and endpoints matter
Adding or subtracting the same quantity preserves inequality direction. Multiplying or dividing by a positive value also preserves it; using a negative value reverses it because order on the number line reverses. Multiplication by zero destroys information and is not an equivalent solving step.
An open endpoint excludes equality and a closed endpoint includes it. A chained inequality imposes both bounds simultaneously. If the domain is integers, list or count only the integers inside the interval rather than treating the entire real interval as the answer.
When the sign of a variable multiplier is unknown, split into positive, negative and zero cases as applicable. Do not multiply through an inequality as though the sign were known.
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ORIGINAL WORKED EXAMPLE
A bounded inequality
Solve−3 x+5<14 together with x≤2 for real x, then list integer solutions.
- Subtract 5:−3 x<9.
- Divide by−3 and reverse the sign: x>−3.
- Intersect with x≤2 to obtain−3<x≤2, written(−3,2].
- For integer x, include−2,−1,0,1,2; exclude−3 because the left bound is strict.
Answer Real solutions form(−3,2]; integer solutions are−2,−1,0,1,2.
ORIGINAL PRACTICE · 1
Solve 2≤3 x−1<8 for real x.
Reveal answer & reasoning
1≤x<3.
Add 1 throughout to get 3≤3 x<9, then divide by positive 3. The left endpoint remains included and the right remains excluded.
04 / UNDERSTAND & APPLY
Candidate roots need a domain check
The zero-product rule says that if a product of real factors is zero, at least one factor is zero. Factoring a quadratic can therefore reveal more than one candidate root. Do not keep only a positive-looking root unless the domain requires it; equally, do not retain a negative length in a physical model.
Squaring or multiplying by expressions can introduce or conceal restrictions in more complicated equations. Substitute candidates into the original equation and check any excluded denominator. The final answer is the valid solution set, not merely the factors you found.
ORIGINAL WORKED EXAMPLE
A rectangle from its area
A rectangle has positive width x centimetres and length x+3 centimetres. Its area is 40 square centimetres. Find its dimensions.
- Area gives x(x+3)=40, hence x²+3 x−40=0.
- Factor as(x+8)(x−5)=0, giving candidate roots−8 and 5.
- Width must be positive, so reject−8. With x=5, length=8.
- Check area 5×8=40 and the length exceeds width by 3.
Answer Width 5 cm and length 8 cm. The negative algebraic root is outside the physical domain.
ORIGINAL PRACTICE · 1
Solve x²=49 over the real numbers, then state the answer if x is required to be positive.
Reveal answer & reasoning
Real roots are−7 and 7; the positive-domain answer is 7.
Both signs square to 49. Positivity is an additional restriction, not a consequence of the equation alone.
05 / UNDERSTAND & APPLY
Use stated geometric conditions
In a Euclidean triangle, interior angles sum to 180 degrees and the sum of any two side lengths exceeds the third. Equality would collapse the shape into a straight segment rather than a nondegenerate triangle. These checks can reject impossible dimensions before further calculation.
For a right triangle, the square of the hypotenuse equals the sum of the squares of the perpendicular legs. The hypotenuse is opposite the right angle and is the longest side. Do not use the formula on an arbitrary triangle without a stated or established right angle.
Similar triangles have equal corresponding angles and proportional corresponding lengths. Similarity may be established by appropriate angle information or proportional sides; visual resemblance alone is insufficient. A scale factor k multiplies corresponding lengths by k and areas by k².
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ORIGINAL WORKED EXAMPLE
A right triangle and a similar copy
A right triangle has perpendicular legs 6 cm and 8 cm. Find its hypotenuse, area and perimeter. A similar copy scales every length by 3/2; find its dimensions and area.
- Hypotenuse c satisfies c²=6²+8²=100, so c=10 cm because length is positive.
- Area=(1/2)×6×8=24 cm²; perimeter=6+8+10=24 cm. Same number, different units and meanings.
- Scaled sides are 9,12 and 15 cm. Area multiplies by(3/2)²=9/4, giving 24×9/4=54 cm².
- Check directly:(1/2)×9×12=54 cm².
Answer Original: hypotenuse 10 cm, area 24 cm², perimeter 24 cm. Similar copy: sides 9,12,15 cm and area 54 cm².
ORIGINAL PRACTICE · 1
Can 2 cm,3 cm and 6 cm be the sides of a nondegenerate triangle?
Reveal answer & reasoning
No.
The two shorter lengths sum to 5 cm, which is less than 6 cm. They cannot meet to form the third side under the triangle inequality.
06 / UNDERSTAND & APPLY
Linear and square units
Perimeter adds boundary lengths; area measures a two-dimensional region. A rectangle has perimeter 2(l+w) and area lw. Scaling both dimensions by k scales perimeter by k but area by k². Changing only one dimension follows a different calculation.
For composite regions, divide into nonoverlapping shapes or subtract an excluded part from a whole. For example, a 10-by 8 rectangle with a 2-by 3 corner removed has area 80−6=74 square units. Adding the original area and the removed corner would double count material not present.
An area answer needs square units. Converting centimetres to metres before squaring gives the same result as using 1 m²=10000 cm²; a linear conversion factor 100 is not the area conversion factor.
ORIGINAL PRACTICE · 1
A rectangle measures 4 cm by 7 cm. Find perimeter and area, then both after doubling every length.
Reveal answer & reasoning
Original perimeter 22 cm and area 28 cm²; doubled perimeter 44 cm and area 112 cm².
Perimeter 2(4+7)=22 and area 4×7=28. Doubling both dimensions gives 8×14, so perimeter doubles while area quadruples.
07 / UNDERSTAND & APPLY
Volume is not surface area
A right prism has volume base area multiplied by perpendicular height. A right circular cylinder therefore has volume πr²h. Radius is centre-to-rim distance, half the diameter; substituting diameter as r would multiply the circular area by four.
Capacity models often ignore wall thickness and assume the stated interior shape and fill height. Make that assumption explicit. Surface area instead measures the material covering a boundary and has square units; volume has cubic units.
One litre equals 1000 cm³. A cubic conversion applies the length factor three times:1 m³=(100 cm)³=1000000 cm³=1000 L. Use a stated approximation to π consistently rather than presenting an approximate numerical result as exact for the usual irrational π.
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ORIGINAL WORKED EXAMPLE
A cylinder’s capacity
An ideal cylindrical container has interior radius 7 cm and fill height 20 cm. Ignore wall thickness and use π=22/7 for this arithmetic exercise. Find the filled volume in cubic centimetres and litres.
- Base area=πr²=(22/7)×49=154 cm² under the supplied approximation.
- Volume=154×20=3080 cm³.
- Convert by 1000 cm³ per litre:3080/1000=3.08 L.
- The radius, not diameter, is 7 cm; the volume is not the side surface area.
Answer 3080 cm³, or 3.08 L, using the stipulated π approximation and interior-capacity model.
ORIGINAL PRACTICE · 1
A rectangular box has interior dimensions 10 cm,20 cm and 30 cm. Find its capacity in litres.
Reveal answer & reasoning
6 L.
Volume=10×20×30=6000 cm³, and 6000/1000=6 L. This assumes the full rectangular interior is available for filling.
08 / UNDERSTAND & APPLY
Report the shape of the solution
An answer may be one value, several discrete values, an interval, infinitely many pairs or no feasible solution. State which applies. A calculator output alone does not establish uniqueness or physical feasibility.
Substitute into every original constraint, not only a simplified equation. Check ticket counts are integers, lengths positive, inequality endpoints correctly included and geometry assumptions supplied. An algebraic root can satisfy the equation while failing the context.
The geometry here is an editorial application of elementary numeracy. It is not presented as a separately enumerated official syllabus bullet or a claim that a specific shape must appear. The transferable method is defining quantities, preserving conditions and verifying the result.
KEEP THE BIG IDEAS
Quick revision
- Define variables, units and domains before forming equations.
- Dependent equations can leave many solutions; contradictory equations leave none.
- Reverse inequality direction when dividing by a negative value.
- Check quadratic roots against the original and physical constraints.
- Lengths, areas and volumes scale differently and use different units.
READ WITH CONTEXT
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