ICSE Class 10 Maths quadratic problems: direct answer
ICSE Class 10 Maths problems based on quadratic equations teach you to convert a word problem into ax^2+bx+c=0, solve the equation, and then keep only the root that fits the condition in the question. In Concise Mathematics Selina Class 10 Chapter 6, the main difficulty is not factorisation alone; it is choosing the correct variable and translating the sentence correctly.
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 6 method
In Concise Mathematics Selina Solutions Class 10 ICSE Chapter 6 Solving (simple) Problems (Based on Quadratic Equations), write the model before solving. A good answer states the unknown, forms the equation, simplifies it to a quadratic equation, solves it, and rejects any root that violates the condition.
Concept snapshot: A word problem is a translation task. “Two parts of 20” becomes x and 20-x. “Two consecutive odd numbers” becomes x and x+2. Once this translation is correct, the algebra usually becomes routine.
- For two parts of N, use x and N-x.
- For consecutive integers, use x and x+1.
- For consecutive odd or even numbers, use x and x+2.
- For numbers differing by d, use x and x+d.
Formula and modelling reference
| Item | Formula or model | Check |
|---|---|---|
| Quadratic equation | ax^2+bx+c=0,\ a\ne0 | Bring all terms to one side. |
| Quadratic formula | x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} | Use when factors are not clear. |
| Reciprocal sum | \dfrac{1}{x}+\dfrac{1}{y} | Keep denominators non-zero. |
| Square sum | x^2+y^2 | Check whether negative integer roots are allowed. |
Worked examples for quadratic word problems
Worked example 1: Consecutive positive integers
Find two consecutive positive integers whose product is 182.
Step 1: Let the integers be x and x+1.
x(x+1)=182
x^2+x-182=0
(x+14)(x-13)=0
Step 2: Since the integers are positive, x=13.
Final answer: The integers are 13 and 14.
Worked example 2: Rectangle dimensions
The area of a rectangle is 84\ \text{cm}^2. Its length is 5\ \text{cm} more than its breadth. Find its dimensions.
Step 1: Let the breadth be x\ \text{cm}. The length is \((x+5)\ \text{cm}\).
x(x+5)=84
x^2+5x-84=0
(x+12)(x-7)=0
Step 2: Breadth cannot be negative, so x=7.
Final answer: Breadth =7\ \text{cm}, length =12\ \text{cm}.
Worked example 3: Speed and time
A car covers 180\ \text{km}. If its speed is increased by 9\ \text{km/h}, it takes 1 hour less. Find the original speed.
Step 1: Let the original speed be x\ \text{km/h}.
\frac{180}{x}-\frac{180}{x+9}=1
\frac{1620}{x(x+9)}=1
x^2+9x-1620=0
(x+45)(x-36)=0
Final answer: The original speed is 36\ \text{km/h}.
Exercise 6(A) step-by-step solutions
Question 1(a): Sum of natural numbers and reciprocals
The sum of two natural numbers is 5, and the sum of their reciprocals is \dfrac{5}{6}.
Step 1: Let the numbers be x and 5-x.
\frac{1}{x}+\frac{1}{5-x}=\frac{5}{6}
\frac{5}{x(5-x)}=\frac{5}{6}
6=x(5-x)
x^2-5x+6=0
(x-2)(x-3)=0
Final answer: The numbers are 2 and 3.
Question 1(b): Consecutive even whole numbers
The product of two consecutive even whole numbers is 24.
Step 1: Let the numbers be x and x+2.
x(x+2)=24
x^2+2x-24=0
(x-4)(x+6)=0
Step 2: Reject x=-6, since a whole number is not negative.
Final answer: The numbers are 4 and 6.
Question 1(c): Consecutive integers with square sum 41
The sum of the squares of two consecutive integers is 41.
Step 1: Let the integers be x and x+1.
x^2+(x+1)^2=41
2x^2+2x-40=0
x^2+x-20=0
(x+5)(x-4)=0
Final answer: The integers are 4,5 or -5,-4.
Question 2: Divide 15 into two parts
Divide 15 into two parts such that the sum of their reciprocals is \dfrac{3}{10}.
Step 1: Let the parts be x and 15-x.
\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}
\frac{15}{x(15-x)}=\frac{3}{10}
150=3x(15-x)
x^2-15x+50=0
(x-5)(x-10)=0
Final answer: The two parts are 5 and 10.
Question 5: Divide 20 into two parts
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Step 1: Let the squared part be x. The other part is 20-x.
3x^2-(20-x)=10
3x^2+x-30=0
(x-3)(3x+10)=0
Step 2: The positive part is x=3, and the other part is 17.
Final answer: The two parts are 3 and 17.
Question 6: Three consecutive natural numbers
The middle number is x. Its square exceeds the difference of the squares of the other two numbers by 60.
Step 1: The numbers are x-1, x, and x+1.
(x+1)^2-(x-1)^2=4x
x^2-4x=60
x^2-4x-60=0
(x-10)(x+6)=0
Final answer: The equation is x^2-4x-60=0; the numbers are 9,10,11.
Question 9: Difference of natural numbers and reciprocals
The difference between two natural numbers is 5, and the difference of their reciprocals is \dfrac{1}{10}.
Step 1: Let the smaller number be x. The larger number is x+5.
\frac{1}{x}-\frac{1}{x+5}=\frac{1}{10}
\frac{5}{x(x+5)}=\frac{1}{10}
x^2+5x-50=0
(x+10)(x-5)=0
Final answer: The numbers are 5 and 10.
Question 12: Two positive numbers differing by 5
Two positive numbers differ by 5. Three times the square of the larger number exceeds twice the square of the smaller number by 334. Find the larger number.
Step 1: Let the smaller number be x. The larger number is x+5.
3(x+5)^2-2x^2=334
x^2+30x-259=0
(x+37)(x-7)=0
Step 2: Since the smaller number is positive, x=7.
Final answer: The larger number is 12.
Quick answer index
| Question | Answer |
|---|---|
| 1(a) | 2,3 |
| 1(b) | 4,6 |
| 1(c) | 4,5 or -5,-4 |
| 2 | 5,10 |
| 5 | 3,17 |
| 6 | Equation x^2-4x-60=0; numbers 9,10,11 |
| 9 | 5,10 |
| 12 | Larger number =12 |
Examiner’s mindset for quadratic-equation problems
A clear ICSE-style solution shows where the quadratic equation came from. Marks are commonly lost when students write only the roots but do not define the variable or explain why one root is rejected. For word problems, write the final answer in words and include units wherever needed.
Common mistakes students make
- Using x+1 for consecutive odd numbers: use x and x+2.
- Reversing reciprocal differences: the reciprocal of the smaller number is larger.
- Rejecting negative roots without checking: negative roots can be valid for integer questions but not for natural-number, length, age, speed or positive-part questions.
- Answering the wrong quantity: if x is the smaller number and the question asks for the larger number, answer x+5.
Sources and related links
This page follows standard ICSE Class 10 algebra treatment of quadratic-equation word problems. For official syllabus framing, refer to the CISCE official website. For textbook publisher information, refer to Selina Publishers.
For more practice, use Concise Mathematics Selina Class 10 equation of a straight line solutions, Concise Mathematics Selina Class 10 trigonometrical identities solutions, Concise Mathematics Selina Class 10 mixed practice solutions, and ICSE Board Class 10 study resources.
Frequently Asked Questions
How do I form a quadratic equation from a word problem in ICSE Class 10 Maths?
Choose one variable, express the other quantity in terms of it, translate the condition into an equation, and reduce it to ax^2+bx+c=0.
When should I reject a root in Chapter 6 quadratic problems?
Reject a root only when it does not fit the wording, such as a negative length, speed, age, natural number or positive part of a number.
Is factorisation enough for Selina Class 10 Chapter 6?
Factorisation is enough when the quadratic splits neatly; otherwise use the formula x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
What is the common error in reciprocal word problems?
Students often reverse the reciprocal difference. If the smaller number is x and the larger is x+5, then the positive difference is \frac{1}{x}-\frac{1}{x+5}.
How should I check my final answer?
Substitute the answer back into the original sentence, not only into the quadratic equation, and confirm that the units and restrictions are satisfied.