NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4
Understanding the Question 🧐
Welcome to this detailed NCERT Solutions guide! In this post, we solve Exercise 2.4 Question 4 from Class 9 Maths — step by step, in the simplest possible way. This question involves forming and solving quadratic equations from real-life word problems. No prior expertise needed — just follow along!
Question 4 (iii): Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Question 4 (iv): A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Our Expert’s Approach
Our experts tackle both parts using the Formulate → Simplify → Factorise → Validate method. For each word problem, we first define a variable, translate the English sentence into a mathematical equation, reduce it to standard quadratic form &&ax^2 + bx + c = 0&&, and then apply the factorisation method. Finally, we reject any root that is not physically meaningful (e.g., negative age or negative speed). This structured approach is trusted by thousands of Class 9 students across India.
Part (iii): Rohan’s Age Problem 📝
Step 1: Define the variable. Let Rohan’s present age be &&x&& years.
Then, his mother’s present age &&= x + 26&& years.
Step 2: Express ages 3 years from now.
| Person | Present Age | Age 3 Years from Now |
|---|---|---|
| Rohan | &&x&& | &&x + 3&& |
| Rohan’s Mother | &&x + 26&& | &&x + 29&& |
Step 3: Form the equation. According to the problem, the product of their ages 3 years from now is 360:
&&(x + 3)(x + 29) = 360&&
Step 4: Expand the left-hand side.
&&x^2 + 29x + 3x + 87 = 360&&
&&x^2 + 32x + 87 = 360&&
Step 5: Bring all terms to one side to get standard quadratic form.
&&x^2 + 32x + 87 – 360 = 0&&
Standard Quadratic Form: &&x^2 + 32x – 273 = 0&&
Step 6: Factorise the quadratic. We need two numbers whose product is &&-273&& and whose sum is &&+32&&.
Those numbers are &&+39&& and &&-7&&, because &&39 \times (-7) = -273&& and &&39 + (-7) = 32&&.
&&x^2 + 39x – 7x – 273 = 0&&
&&x(x + 39) – 7(x + 39) = 0&&
&&(x – 7)(x + 39) = 0&&
Step 7: Apply the Zero Product Property.
Either &&x – 7 = 0 \Rightarrow x = 7&&
Or &&x + 39 = 0 \Rightarrow x = -39&&
⚠️ Caution: Age cannot be negative. So we reject &&x = -39&&. Always validate your roots against the real-world context of the problem.
Therefore, Rohan’s present age &&= 7&& years, and his mother’s present age &&= 7 + 26 = 33&& years.
Verification: Ages 3 years from now: Rohan &&= 10&&, Mother &&= 36&&. Product &&= 10 \times 36 = 360&& ✅
✅ Final Answer (iii): Rohan’s present age is &&\boxed{7}&& years.
Part (iv): Train Speed Problem 📝
Step 1: Define the variable. Let the uniform speed of the train be &&x&& km/h.
Step 2: Express time taken in both cases. We use the formula: &&\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}&&
| Case | Speed (km/h) | Distance (km) | Time (hours) |
|---|---|---|---|
| Original | &&x&& | 480 | &&\dfrac{480}{x}&& |
| Reduced Speed | &&x – 8&& | 480 | &&\dfrac{480}{x – 8}&& |
Step 3: Form the equation. At the reduced speed, the train takes 3 hours more:
&&\dfrac{480}{x – 8} – \dfrac{480}{x} = 3&&
Step 4: Simplify by taking LCM. The LCM of &&(x – 8)&& and &&x&& is &&x(x – 8)&&:
&&\dfrac{480x – 480(x – 8)}{x(x – 8)} = 3&&
&&\dfrac{480x – 480x + 3840}{x(x-8)} = 3&&
&&\dfrac{3840}{x(x – 8)} = 3&&
Step 5: Cross-multiply and simplify.
&&3840 = 3 \cdot x(x – 8)&&
&&3840 = 3x^2 – 24x&&
&&3x^2 – 24x – 3840 = 0&&
Divide throughout by &&3&&:
Standard Quadratic Form: &&x^2 – 8x – 1280 = 0&&
Step 6: Factorise the quadratic. We need two numbers whose product is &&-1280&& and whose sum is &&-8&&.
Those numbers are &&-40&& and &&+32&&, because &&(-40) \times 32 = -1280&& and &&-40 + 32 = -8&&.
&&x^2 – 40x + 32x – 1280 = 0&&
&&x(x – 40) + 32(x – 40) = 0&&
&&(x + 32)(x – 40) = 0&&
Step 7: Apply the Zero Product Property.
Either &&x + 32 = 0 \Rightarrow x = -32&&
Or &&x – 40 = 0 \Rightarrow x = 40&&
⚠️ Caution: Speed cannot be negative. So we reject &&x = -32&&. Also, if &&x = -32&&, then &&x – 8 = -40&&, which makes no physical sense for a train’s speed.
Therefore, the speed of the train &&= 40&& km/h.
Verification: Original time &&= \dfrac{480}{40} = 12&& hours. Reduced speed &&= 32&& km/h, time &&= \dfrac{480}{32} = 15&& hours. Difference &&= 15 – 12 = 3&& hours ✅
✅ Final Answer (iv): The speed of the train is &&\boxed{40}&& km/h.
Conclusion and Key Points ✅
In this NCERT Solutions post, we successfully solved both parts of Question 4 from Exercise 2.4 (Class 9 Maths) by converting real-life word problems into quadratic equations and solving them using the factorisation method. The key skill here is accurate equation formation — once the equation is set up correctly, factorisation becomes straightforward.
- Standard form of a quadratic equation: &&ax^2 + bx + c = 0&&, where &&a \neq 0&&.
- Zero Product Property: If &&(x – p)(x – q) = 0&&, then &&x = p&& or &&x = q&&.
- Always check both roots and reject the one that is not valid in the real-world context (e.g., negative age, negative speed).
- For age problems: define the present age as &&x&&, then express future/past ages in terms of &&x&&.
- For speed-distance problems: use &&\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}&& and form a difference equation.
Frequently Asked Questions (FAQ)
Q1. What is Rohan’s present age in Exercise 2.4 Question 4 Part (iii)?
Rohan’s present age is &&7&& years. We form the equation &&(x+3)(x+29) = 360&&, which gives &&x^2 + 32x – 273 = 0&&. Factorising: &&(x-7)(x+39) = 0&&, so &&x = 7&& (rejecting &&x = -39&& as age cannot be negative).
Q2. How do you form the quadratic equation for the age problem in Part (iii)?
Let Rohan’s present age &&= x&&. Mother’s present age &&= x + 26&&. Three years from now: Rohan &&= x + 3&&, Mother &&= x + 29&&. Given: &&(x+3)(x+29) = 360&&. Expanding: &&x^2 + 32x + 87 = 360&&, which gives &&x^2 + 32x – 273 = 0&&.
Q3. What is the speed of the train in Part (iv)?
The speed of the train is &&40&& km/h. We solve &&x^2 – 8x – 1280 = 0&&, which factorises as &&(x – 40)(x + 32) = 0&&. Rejecting &&x = -32&&, we get &&x = 40&& km/h.
Q4. Why do we reject the negative root in both parts of Question 4?
In real-life problems, physical quantities like age and speed must be positive. A negative age (&&x = -39&&) or a negative speed (&&x = -32&&) has no real-world meaning. Therefore, the negative root is always rejected in such word problems.
Q5. How can we verify the answer for the train speed problem?
At &&x = 40&& km/h: Time taken &&= \dfrac{480}{40} = 12&& hours. At reduced speed &&32&& km/h: Time taken &&= \dfrac{480}{32} = 15&& hours. Difference &&= 15 – 12 = 3&& hours, which matches the given condition. ✅
Q6. What method is used to solve quadratic equations in NCERT Class 9 Maths Exercise 2.4?
Exercise 2.4 primarily uses the factorisation method. The quadratic equation is written in standard form &&ax^2 + bx + c = 0&&, split into two linear factors, and the zero product property is applied to find the roots.
Q7. What formula is used in the train speed problem to set up the equation?
The formula used is &&\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}&&. Since the distance is fixed at 480 km, the time at original speed is &&\dfrac{480}{x}&& and at reduced speed is &&\dfrac{480}{x-8}&&. The difference of these two times equals 3 hours, giving the equation &&\dfrac{480}{x-8} – \dfrac{480}{x} = 3&&.
Further Reading
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