Chapter 6: Solving Problems on Quadratic Equations
Word problems test your conceptual clarity, variable modeling skills, and logical deduction. In the ICSE
Class 10 Board Examination, this chapter accounts for 4 to 6 marks in Section B (and
frequently 3 marks in Section A). Success requires converting verbal constraints into a quadratic equation
$ax^2 + bx + c = 0$, solving it correctly, and rigorously applying physical Root Rejection
Criteria.
Step 1: Variable Declaration & Scope: Let the principal unknown quantity be $x$
(or $y$). Explicitly state its physical unit (e.g., km/h, hours, cm, rupees, years, natural number).
Step 2: Equation Formulation & Algebraic Reduction: Express all given conditions
in terms of $x$. Simplify fractions or expand terms to obtain the standard quadratic form:
$$\mathbf{ax^2 + bx + c = 0} \quad (a \ne 0)$$
Step 3: Solution & Inadmissible Root Selection: Solve using Factorisation or the
Quadratic Formula ($x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$). Inspect both mathematical roots against
real-world constraints and explicitly write the rejection statement for extraneous roots.
Exhaustive Root Rejection Rules
Physical Quantity
Mathematical Constraint
Extraneous Root Rejection Justification
Speed / Velocity
$x > 0$
Speed cannot be negative ($\text{Speed} \ne -v$).
Distance / Length / Breadth
$x > 0$
Geometrical dimensions must be strictly positive.
Time Duration / Hours / Days
$x > 0$
Time elapsed cannot be negative.
Number of Persons / Items / Rows
$x \in \mathbb{N}$ (Positive Integer)
Count of discrete objects cannot be negative or fractional.
Natural Numbers / Whole Numbers
$x \in \mathbb{N}$ or $x \in \mathbb{W}$
Negative numbers and non-integers are inadmissible.
Age
$x > 0$
Age must be a positive real number.
Fraction Denominator
$\text{Denominator} \ne 0$
Division by zero is undefined in real numbers.
ICSE Rule: Rounding Off Irrational Roots
When a quadratic word problem does not factorise easily, ICSE exam instructions often state:
"Give your answer correct to 2 decimal places" or "correct to 3 significant figures".
Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Evaluate $\sqrt{D}$ up to 3 decimal places before final division.
Example: If $x = \frac{10 \pm \sqrt{68}}{2} = \frac{10 \pm 8.246}{2} = 9.123 \text{ or } 0.877$, round
to $9.12$ or $0.88$.
2. Category 1: Problems Based on Numbers, Digits & Fractions
This category deals with relations between integers, natural numbers, two-digit numbers, reciprocals, and
fractions.
Key Algebraic Formulations
Consecutive Integers: $x$ and $x+1$; Consecutive Even/Odd: $x$ and
$x+2$.
Sum of Reciprocals: If sum of numbers is $S$, then $\frac{1}{x} + \frac{1}{S-x} =
k$.
Difference of Reciprocals: $\frac{1}{x} - \frac{1}{x+d} = \frac{d}{x(x+d)}$. (Note:
$\frac{1}{x} > \frac{1}{x+d}$ for $x > 0$).
Two-Digit Number: Let tens digit = $x$, units digit = $y$. Original Number = $10x +
y$. Reversed Number = $10y + x$.
Fractions: Let fraction = $\frac{x}{y}$. If numerator is modified: $\frac{x \pm
a}{y \pm b}$.
ICSE Solved Board Questions — Numbers & Digits
Q1. Find two natural numbers which differ by 3 and the sum of whose squares is 117.
ICSE Benchmark
Step-by-Step Solution:
Let the smaller natural number = $x$. Since they differ by 3, the larger natural number = $x +
3$.
According to the given condition:
$$x^2 + (x+3)^2 = 117$$
$$x^2 + x^2 + 6x + 9 = 117 \implies 2x^2 + 6x - 108 = 0$$
Dividing the entire equation by 2:
$$x^2 + 3x - 54 = 0$$
Factorising by splitting the middle term ($9 \times (-6) = -54$ and $9 - 6 = 3$):
$$x^2 + 9x - 6x - 54 = 0 \implies x(x+9) - 6(x+9) = 0$$
$$(x+9)(x-6) = 0 \implies x = -9 \quad \text{or} \quad x = 6$$
Root Rejection: Since $x$ must be a natural number ($x \in \mathbb{N}$), $x = -9$
is rejected.
$\therefore x = 6$. The larger number = $6 + 3 = 9$. Final Answer: The two natural numbers are 6 and 9.
Q2. The sum of two natural numbers is 8 and the difference of their reciprocals is $\frac{2}{15}$.
Find the numbers.
ICSE 2015
Step-by-Step Solution:
Let the smaller natural number = $x$. Since their sum is 8, the larger number = $8 - x$.
Reciprocal of smaller number = $\frac{1}{x}$, Reciprocal of larger number = $\frac{1}{8-x}$.
Since $x < 8-x$, we have $\frac{1}{x} > \frac{1}{8-x}$. Given difference of reciprocals is
$\frac{2}{15}$: $$\frac{1}{x} - \frac{1}{8-x} = \frac{2}{15}$$
$$\frac{(8-x) - x}{x(8-x)} = \frac{2}{15} \implies \frac{8-2x}{8x-x^2} = \frac{2}{15}$$
Divide numerator of left-hand side by 2:
$$\frac{4-x}{8x-x^2} = \frac{1}{15}$$
Cross-multiplying:
$$15(4-x) = 8x - x^2 \implies 60 - 15x = 8x - x^2$$
$$x^2 - 23x + 60 = 0$$
Factorising ($(-20) \times (-3) = 60$ and $-20 - 3 = -23$):
$$(x-20)(x-3) = 0 \implies x = 20 \quad \text{or} \quad x = 3$$
Root Rejection: Since the sum of two natural numbers is 8, individual number
cannot exceed 8. Hence $x = 20$ is rejected.
$\therefore x = 3$. The other number = $8 - 3 = 5$. Final Answer: The numbers are 3 and 5.
Q3. A two-digit number is such that the product of its digits is 12. When 36 is added to the number,
the digits interchange their places. Find the number.
ICSE 2019
Step-by-Step Solution:
Let tens digit = $x$. Since product of digits is 12, units digit = $\frac{12}{x}$.
Original Number = $10(x) + \frac{12}{x}$.
Reversed Number = $10\left(\frac{12}{x}\right) + x = \frac{120}{x} + x$.
According to the problem statement:
$$\text{Original Number} + 36 = \text{Reversed Number}$$
$$\left(10x + \frac{12}{x}\right) + 36 = \frac{120}{x} + x$$
Subtract $x$ and $\frac{12}{x}$ from both sides:
$$9x + 36 = \frac{108}{x}$$
Divide the entire equation by 9:
$$x + 4 = \frac{12}{x} \implies x(x+4) = 12 \implies x^2 + 4x - 12 = 0$$
Factorising:
$$(x+6)(x-2) = 0 \implies x = -6 \quad \text{or} \quad x = 2$$
Root Rejection: A digit in a number cannot be negative ($x > 0$). Thus $x = -6$
is rejected.
$\therefore$ Tens digit $x = 2$, Units digit = $\frac{12}{2} = 6$. Final Answer: Original Number = 26.
Q4. The denominator of a fraction is 1 more than twice its numerator. If the sum of the fraction
and its reciprocal is $2\frac{16}{21}$, find the fraction.
ICSE 2017
Let speed of boat in still water = $y$ km/h, and speed of river stream = $x$ km/h ($x < y$).
Downstream Speed (with current): $v_{\text{down}} = (y + x) \text{ km/h}$. Time taken =
$\frac{D}{y+x}$.
Upstream Speed (against current): $v_{\text{up}} = (y - x) \text{ km/h}$. Time taken =
$\frac{D}{y-x}$.
Key Invariant: Upstream time is ALWAYS greater than downstream time for the same
distance. Therefore:
$$\frac{D}{y-x} - \frac{D}{y+x} = \Delta t \quad (\text{Total time difference})$$
ICSE Solved Board Questions — Speed, Distance & Time
Q1. A car covers a distance of 400 km at a certain average speed. If the speed were increased by 12
km/h, the journey would take 1 hour 40 minutes less. Find the original speed of the car.
ICSE 2024 / 2018
Step-by-Step Solution:
Let original speed of the car = $x$ km/h.
Distance = 400 km.
Original time taken ($t_1$) = $\frac{400}{x}$ hours.
Increased speed = $(x + 12)$ km/h.
New time taken ($t_2$) = $\frac{400}{x+12}$ hours.
Time saved = 1 hour 40 minutes = $1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}$ hours.
According to the problem:
$$t_1 - t_2 = \frac{5}{3} \implies \frac{400}{x} - \frac{400}{x+12} = \frac{5}{3}$$
Take 400 common on left hand side:
$$400 \left( \frac{(x+12) - x}{x(x+12)} \right) = \frac{5}{3} \implies 400 \left( \frac{12}{x^2 +
12x} \right) = \frac{5}{3}$$
$$\frac{4800}{x^2 + 12x} = \frac{5}{3}$$
Divide numerator by 5:
$$\frac{960}{x^2 + 12x} = \frac{1}{3} \implies x^2 + 12x = 2880$$
$$x^2 + 12x - 2880 = 0$$
Factorising ($60 \times (-48) = -2880$ and $60 - 48 = 12$):
$$x^2 + 60x - 48x - 2880 = 0 \implies x(x+60) - 48(x+60) = 0$$
$$(x+60)(x-48) = 0 \implies x = -60 \quad \text{or} \quad x = 48$$
Root Rejection: Speed cannot be negative ($x > 0$). Hence $x = -60$ is
rejected. Final Answer: Original speed of the car = 48 km/h.
Q2. A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than
to return downstream to the same spot. Find the speed of the stream.
ICSE 2020 / 2014
Step-by-Step Solution:
Let speed of stream = $x$ km/h.
Speed of boat in still water = 18 km/h.
Upstream speed = $(18 - x)$ km/h; Downstream speed = $(18 + x)$ km/h.
Upstream time = $\frac{24}{18-x}$ hours; Downstream time = $\frac{24}{18+x}$ hours.
Given: Upstream time $-$ Downstream time = 1 hour.
$$\frac{24}{18-x} - \frac{24}{18+x} = 1$$
$$24 \left( \frac{(18+x) - (18-x)}{(18-x)(18+x)} \right) = 1 \implies 24 \left( \frac{2x}{324 - x^2}
\right) = 1$$
$$\frac{48x}{324 - x^2} = 1 \implies 48x = 324 - x^2$$
$$x^2 + 48x - 324 = 0$$
Factorising ($54 \times (-6) = -324$ and $54 - 6 = 48$):
$$(x+54)(x-6) = 0 \implies x = -54 \quad \text{or} \quad x = 6$$
Root Rejection: Stream speed cannot be negative ($x > 0$). Reject $x = -54$. Final Answer: Speed of the stream = 6 km/h.
Q3. An aeroplane left 30 minutes later than the scheduled time and in order to reach its
destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed.
Find its usual speed.
ICSE 2016
Click to view step-by-step ICSE solution
Let usual speed = $x$ km/h. Increased speed = $(x + 250)$ km/h.
Delay time = 30 mins = $\frac{1}{2}$ hour.
$$\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$$
$$1500 \left( \frac{250}{x^2 + 250x} \right) = \frac{1}{2} \implies \frac{375000}{x^2 + 250x} =
\frac{1}{2}$$
$$x^2 + 250x - 750000 = 0$$
Factorising ($1000 \times (-750) = -750000$):
$$(x+1000)(x-750) = 0 \implies x = -1000 \text{ (rejected)} \quad \text{or} \quad x = 750$$
Final Answer: Usual speed of aeroplane = 750 km/h.
4. Category 3: Problems Based on Time & Work / Pipes & Cisterns
Work efficiency is inversely proportional to time taken. If a person completes a job in $x$ hours, their
1-hour work rate is $\frac{1}{x}$.
Q1. A takes 6 days less than B to do a certain piece of work. If both A and B together can do it in
4 days, find the time taken by B to finish the work alone.
ICSE 2013
Step-by-Step Solution:
Let time taken by B alone = $x$ days.
Time taken by A alone = $(x - 6)$ days.
1 day's work of B = $\frac{1}{x}$, 1 day's work of A = $\frac{1}{x-6}$.
Combined 1 day's work of A and B = $\frac{1}{4}$.
$$\frac{1}{x-6} + \frac{1}{x} = \frac{1}{4}$$
$$\frac{x + (x-6)}{x(x-6)} = \frac{1}{4} \implies \frac{2x-6}{x^2 - 6x} = \frac{1}{4}$$
Cross-multiplying:
$$4(2x - 6) = x^2 - 6x \implies 8x - 24 = x^2 - 6x$$
$$x^2 - 14x + 24 = 0$$
Factorising ($(-12) \times (-2) = 24$ and $-12 - 2 = -14$):
$$(x-12)(x-2) = 0 \implies x = 12 \quad \text{or} \quad x = 2$$
Root Rejection: If $x = 2$, then A's time = $2 - 6 = -4$ days, which is physically
impossible ($t > 0$). Thus $x = 2$ is rejected.
$\therefore x = 12$. Final Answer: B takes 12 days to finish the work alone.
Q2. Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter
takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each
tap can separately fill the tank.
ICSE Benchmark
Click to view step-by-step ICSE solution
Let time taken by smaller tap = $x$ hours. Time for larger tap = $(x - 10)$ hours.
Combined time = $9\frac{3}{8} = \frac{75}{8}$ hours $\implies$ Combined 1-hour rate =
$\frac{8}{75}$.
$$\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}$$
$$\frac{(x-10) + x}{x(x-10)} = \frac{8}{75} \implies \frac{2x-10}{x^2-10x} = \frac{8}{75}$$
Divide numerator by 2:
$$\frac{x-5}{x^2-10x} = \frac{4}{75} \implies 75(x-5) = 4(x^2 - 10x)$$
$$75x - 375 = 4x^2 - 40x \implies 4x^2 - 115x + 375 = 0$$
Factorising ($4 \times 375 = 1500$; $-100 \times (-15) = 1500$):
$$4x^2 - 100x - 15x + 375 = 0 \implies 4x(x-25) - 15(x-25) = 0$$
$$(4x-15)(x-25) = 0 \implies x = 25 \quad \text{or} \quad x = \frac{15}{4} = 3.75$$
If $x = 3.75$, larger tap time = $3.75 - 10 = -6.25 < 0$ (impossible). Reject $x = 3.75$. Final Answer: Smaller tap takes 25 hours, Larger tap takes 15 hours.
5. Category 4: Problems Based on Geometrical Figures & Mensuration
Rectangle Area & Perimeter: Area = $L \times B$, Perimeter = $2(L + B)$, Diagonal
$D = \sqrt{L^2 + B^2}$.
Uniform Footpath / Border:
If a path of uniform width $x$ runs inside a field of size $L \times B$:
$$\text{Path Area} = (L \times B) - (L - 2x)(B - 2x)$$
Q1. The hypotenuse of a right-angled triangle is 13 cm. If the difference between the other two
sides is 7 cm, find the sides of the triangle.
ICSE 2011
Step-by-Step Solution:
Let the shorter side = $x$ cm.
Then the longer side = $(x + 7)$ cm.
Hypotenuse = 13 cm.
By Pythagoras Theorem:
$$x^2 + (x+7)^2 = 13^2$$
$$x^2 + x^2 + 14x + 49 = 169 \implies 2x^2 + 14x - 120 = 0$$
Divide by 2:
$$x^2 + 7x - 60 = 0$$
Factorising ($12 \times (-5) = -60$ and $12 - 5 = 7$):
$$(x+12)(x-5) = 0 \implies x = -12 \quad \text{or} \quad x = 5$$
Root Rejection: Length of a side cannot be negative ($x > 0$). Reject $x = -12$.
$\therefore x = 5$ cm, longer side = $5 + 7 = 12$ cm. Final Answer: The sides of the triangle are 5 cm, 12 cm, and 13 cm.
Q2. A footpath of uniform width runs all around the inside of a rectangular park 32 m long and 24 m
wide. If the area of the footpath is 208 m², find the width of the footpath.
ICSE 2012
Step-by-Step Solution:
Outer length = 32 m, Outer breadth = 24 m.
Outer Area = $32 \times 24 = 768 \text{ m}^2$.
Let uniform width of footpath = $x$ m.
Inner length = $(32 - 2x)$ m, Inner breadth = $(24 - 2x)$ m.
Inner Area = $(32 - 2x)(24 - 2x) = 768 - 64x - 48x + 4x^2 = 4x^2 - 112x + 768$.
Given Footpath Area = 208 m²:
$$\text{Outer Area} - \text{Inner Area} = 208$$
$$768 - (4x^2 - 112x + 768) = 208 \implies 112x - 4x^2 = 208$$
Rearranging and dividing by 4:
$$4x^2 - 112x + 208 = 0 \implies x^2 - 28x + 52 = 0$$
Factorising ($(-26) \times (-2) = 52$ and $-26 - 2 = -28$):
$$(x-26)(x-2) = 0 \implies x = 26 \quad \text{or} \quad x = 2$$
Root Rejection: Path width $x = 26$ m is impossible because park breadth is only 24
m ($2x < 24 \implies x < 12$). Thus $x = 26$ is rejected. Final Answer: Width of the footpath = 2 metres.
Commercial word problems involve total budgets shared among individuals, articles purchased for a fixed
cost, and Cost Price vs Profit/Loss percentage relations.
Q1. A shopkeeper buys a certain number of books for ₹960. If he had bought 4 more books for the same
amount, each book would have cost him ₹8 less. Find the number of books he bought.
ICSE 2023 / 2017
Step-by-Step Solution:
Let original number of books bought = $x$.
Total expenditure = ₹960.
Original cost price per book = ₹$\frac{960}{x}$.
If he buys $(x + 4)$ books for ₹960, new cost per book = ₹$\frac{960}{x+4}$.
Given price difference per book = ₹8:
$$\frac{960}{x} - \frac{960}{x+4} = 8$$
$$960 \left( \frac{(x+4) - x}{x(x+4)} \right) = 8 \implies 960 \left( \frac{4}{x^2 + 4x} \right) =
8$$
$$\frac{3840}{x^2 + 4x} = 8 \implies x^2 + 4x = \frac{3840}{8} = 480$$
$$x^2 + 4x - 480 = 0$$
Factorising ($24 \times (-20) = -480$):
$$(x+24)(x-20) = 0 \implies x = -24 \quad \text{or} \quad x = 20$$
Root Rejection: Number of books cannot be negative ($x \in \mathbb{N}$). Reject $x
= -24$. Final Answer: Number of books bought = 20.
Q2. By selling an article for ₹24, a trader loses as much percent as the cost price of the article.
Calculate the cost price.
ICSE Classic
Step-by-Step Solution:
Let Cost Price (C.P.) of article = ₹$x$.
Loss Percentage = $x\%$.
Total Loss amount = $x\% \text{ of } x = \frac{x^2}{100}$.
Selling Price (S.P.) = $\text{C.P.} - \text{Loss} = x - \frac{x^2}{100}$.
Given S.P. = ₹24:
$$x - \frac{x^2}{100} = 24 \implies \frac{100x - x^2}{100} = 24$$
$$100x - x^2 = 2400 \implies x^2 - 100x + 2400 = 0$$
Factorising ($(-60) \times (-40) = 2400$):
$$(x-60)(x-40) = 0 \implies x = 60 \quad \text{or} \quad x = 40$$
Valid Roots: Both $x = 60$ and $x = 40$ are positive real prices. If C.P. = ₹60,
Loss = 60%, S.P. = $60 - 36 = 24$. If C.P. = ₹40, Loss = 40%, S.P. = $40 - 16 = 24$. Final Answer: Cost Price is either ₹60 or ₹40.
7. Category 6: Problems Based on Ages
ICSE Solved Board Questions — Age Problems
Q1. Five years ago, a woman's age was the square of her son's age. Ten years hence, her age will be
twice the age of her son. Find the present age of the woman.
ICSE 2007
Step-by-Step Solution:
Let son's age 5 years ago = $x$ years.
Then woman's age 5 years ago = $x^2$ years.
Present age of son = $(x + 5)$ years.
Present age of woman = $(x^2 + 5)$ years.
10 years hence:
Son's age = $(x + 5 + 10) = (x + 15)$ years.
Woman's age = $(x^2 + 5 + 10) = (x^2 + 15)$ years.
Given condition 10 years hence:
$$\text{Woman's age} = 2 \times (\text{Son's age})$$
$$x^2 + 15 = 2(x + 15) \implies x^2 + 15 = 2x + 30$$
$$x^2 - 2x - 15 = 0$$
Factorising ($(-5) \times 3 = -15$):
$$(x-5)(x+3) = 0 \implies x = 5 \quad \text{or} \quad x = -3$$
Root Rejection: Age cannot be negative ($x > 0$). Reject $x = -3$.
$\therefore x = 5$.
Woman's present age = $x^2 + 5 = 5^2 + 5 = 25 + 5 = 30$ years. Final Answer: Present age of the woman is 30 years.
8. Interactive Quadratic Word Problem Solver
⚡ Interactive Word Problem Solver & Discriminant Inspector
Select a problem template, enter given parameters, and view instant algebraic formulation, roots & root
rejection criteria!
9. ICSE Past 15 Years Question Bank (Self-Practice)
Board Exam Practice Set
P1. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their
ages was 124. Find their present ages.
ICSE 2010
Click to view Answer & Key Steps
Equation: $(x-5)(40-x) = 124 \implies x^2 - 45x + 324 = 0 \implies (x-36)(x-9) = 0$.
Father's age = 36 years, Son's age = 9 years.
P2. A trader bought a number of articles for ₹700. Ten articles were damaged. He sold the remaining
at ₹3 more per article than he paid for it, making a total profit of ₹50. Find the number of
articles bought.
ICSE 2009
