Chapter 2.3 — Exercise 2.2 — Applications

Geometry Background — The Angle Facts Behind Question 1

Question 1's five parts all rely on one of these geometric facts, and recognising which one applies is the first step in each part:

• Angle sum of a triangle: the three interior angles of any triangle always add up to 180°.
• Exterior angle theorem: an exterior angle of a triangle equals the sum of the two interior angles that aren't next to it — for example ∠ACD = ∠A + ∠B.
• Isosceles triangle: if two sides of a triangle are equal, the angles opposite those sides are equal too.
• Vertically opposite angles: when two straight lines cross, the angles directly across from each other are always equal.

Working through Question 1's five parts with these facts gives: (i) x = 67, (ii) x = 17, (iii) x = 125, (iv) x = 19, (v) x = 20.

Turning a Sentence Into an Equation

Every word problem in the rest of this exercise follows the same sequence: read the situation carefully, choose a variable for the one unknown quantity that everything else can be described in terms of, write every other unknown quantity in terms of that variable, translate the stated condition into an equation, and solve. The final answer should always be stated in the context of the question — the numbers, ages, or lengths asked for — not just the bare value of x.

Word Problems, Question by Question

• Q2 — Two numbers, difference 8: bigger = x, smaller = x − 8; condition x + 2 = 3(x − 8) gives x = 13. Numbers are 13 and 5.
• Q3 — Sum 58, difference 28: bigger = x, smaller = 58 − x; condition x − (58 − x) = 28 gives x = 43. Numbers are 43 and 15.
• Q4 — Two consecutive odd numbers, sum 56: numbers are 2x − 1 and 2x + 1; 4x = 56 gives x = 14. The odd numbers are 27 and 29.
• Q5 — Three consecutive multiples of 7, sum 777: numbers are x, x + 7, x + 14; 3x + 21 = 777 gives x = 252. The multiples are 252, 259 and 266.
• Q6 — A 70 km journey by walk, train and bus: walk = 10 km, train = x km, bus = 2x km; 10 + 3x = 70 gives x = 20. The train leg is 20 km.
• Q7 — A cake cut into three pieces, total 300 g: first = x, second = x + 7, third = x − 4; 3x + 3 = 300 gives x = 99. The pieces weigh 99 g, 106 g and 95 g.
• Q8 — Rectangular field, perimeter 400 m, length 26 m more than breadth: breadth = x, length = x + 26; using l + b = 200, 2x + 26 = 200 gives x = 87. Breadth = 87 m, length = 113 m.
• Q9 — Perimeter 56 m, length 8 m less than twice the breadth: breadth = x, length = 2x − 8; using l + b = 28, 3x − 8 = 28 gives x = 12. Breadth = 12 m, length = 16 m.
• Q10 — Isosceles triangle, perimeter 55 m: third side = x, each equal side = 2x − 5; 5x − 10 = 55 gives x = 13. Sides are 21 m, 21 m and 13 m.
• Q11 — Two complementary angles differing by 12°: angles x and x + 12, summing to 90; 2x + 12 = 90 gives x = 39. The angles are 39° and 51°.
• Q12 — Ages in ratio 5:7, sum 56 four years from now: present ages 5x and 7x; (5x+4)+(7x+4) = 56 gives x = 4. Present ages are 20 and 28 years.
• Q13 — 180 questions, +4 per correct answer, −1 per wrong or unattempted, score 450: correct = x, the rest = 180 − x; 4x − (180 − x) = 450 gives x = 126 correct answers.
• Q14 — ₹500 in ₹5 and ₹10 notes, 90 notes total: ₹5 notes = x, ₹10 notes = 90 − x; 5x + 10(90 − x) = 500 gives x = 80. That's 80 notes of ₹5 and 10 notes of ₹10.
• Q15 — ₹564 spent on pens (₹7) and pencils (₹3), 108 items total: pens = x, pencils = 108 − x; 7x + 3(108 − x) = 564 gives x = 60. That's 60 pens and 48 pencils.
• Q16 — Volleyball court, perimeter 177 ft, length twice the width: width = x, length = 2x; 2(3x) = 177 gives x = 29.5. Width = 29.5 ft, length = 59 ft.
• Q17 — Facing page numbers summing to 373: pages x and x + 1; 2x + 1 = 373 gives x = 186. The pages are 186 and 187.

Choosing the Variable Wisely

Several of these problems get noticeably easier depending on which quantity you let x represent. In Q12's ratio problem, letting the ages be x and x + something forces awkward fractions, while letting them be 5x and 7x directly captures the 5:7 ratio and keeps every step in whole numbers. Similarly, in the two-category problems (Q14, Q15), expressing the second quantity as (total − x) avoids introducing a second unknown that would need a second equation. Spending a moment deciding what x should represent, before writing anything else down, is often what separates a two-line solution from a tangled one.

Where Setups Go Wrong Before the Algebra Even Starts

• In the angle questions, identify which property applies — exterior angle, angle sum, or isosceles triangle — before writing any equation; using the wrong property produces a plausible-looking but wrong answer.
• In ratio problems, represent the two quantities as kx and mx matching the given ratio, not as x and x + (something) — the ratio itself carries the structure of the problem.
• When two categories share a fixed total (notes, pens and pencils, correct and incorrect answers), always write the second quantity as (total − x) and expand any resulting brackets carefully before collecting like terms.
• Complementary angles sum to 90°; supplementary angles sum to 180° — confusing the two changes the entire equation.

A Skill That Reappears Constantly From Here On

The equation-formation skill practised across these seventeen questions reappears throughout the rest of Class 8 — in ratio and proportion and direct and inverse proportion — and again in Class 10 when forming and solving quadratic equations. For the transposition technique this exercise builds on, revisit Exercise 2.1.