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Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Case Study - 1

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.
Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Q1: Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be
(a) 20 km/hr
(b) (20 + x) km/hr
(c) (20 – x) km/hr
(d) 2 km/hr
Ans: 
(c)
Explanation: The speed of the motorboat in still water is given as 20 km/hr. When moving upstream (against the current), the speed of the motorboat is reduced by the speed of the stream because it is moving against the direction of the stream.
Let's denote the speed of the stream as 'x' km/hr. Therefore, the speed of the motorboat while moving upstream will be the speed of the motorboat in still water minus the speed of the stream.
In mathematical terms, this can be represented as (20 - x) km/hr.
Step-by-step process:
1) Identify the speed of the motorboat in still water, which is given as 20 km/hr.
2) Understand that when moving upstream, the speed of the motorboat is reduced by the speed of the stream.
3) Denote the speed of the stream as 'x' km/hr.
4) Subtract the speed of the stream from the speed of the motorboat in still water to find the speed of the motorboat upstream.
5) Represent this as (20 - x) km/hr.
Therefore, the answer is (c) (20 – x) km/hr. 

Q2: What is the relation between speed, distance and time?
(a) speed = (distance )/time
(b) distance = speed x time
(c) time = speed x distance
(d) speed = distance x time

Ans: (b)
Explanation: The relation between speed, distance, and time is given by the formula: distance = (speed )/time.
Here's how it works:
Speed is defined as the rate at which something or someone is able to move or operate. In simpler terms, it is how fast an object is moving.
Distance, on the other hand, is a scalar quantity that refers to "how much ground an object has covered" during its motion.
Time is simply the duration during which an event occurs.
In physics, we can connect these three quantities using the formula: Speed = Distance/Time, which is rearranged to get Distance = Speed x Time.
So, if we know the speed at which an object is moving and the time for which it moves, we can calculate the distance it has covered.
Therefore, option (b) is correct - distance = speed x time.
To illustrate, let's take the given case. If a motor boat is moving at a speed of 20 km/hr and it travels for, let's say, 1 hour, then the distance it will cover is Distance = 20 km/hr x 1 hr = 20 km. 

Q3: Which is the correct quadratic equation for the speed of the current?
(a) x2 + 30x − 200 = 0
(b) x2 + 20x − 400 = 0
(c) x2 + 30x − 400 = 0
(d) x2 − 20x − 400 = 0
Ans:
(c)
Explanation: The speed of the motor boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'x' km/hr.
When the boat is moving downstream (i.e., along the direction of the current), the effective speed of the boat becomes (20 + x) km/hr, while upstream (i.e., against the direction of the current) the effective speed becomes (20 - x) km/hr.
Given that the distance covered by the boat is the same both times (15 km), we can set up the following equation based on the concept that time = distance / speed:
Time taken downstream = 15 / (20 + x)
Time taken upstream = 15 / (20 - x)
The problem states that the boat took 1 hour more for upstream than downstream, therefore:
15 / (20 - x) = 15 / (20 + x) + 1
We can simplify this equation further to get the quadratic equation:
(x2) - 30x - 400 = 0
Therefore, option (c) is the correct quadratic equation for the speed of the current. 


Q4: What is the speed of current?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour

Ans: (b)
Explanation: The speed of a boat in still water is given as 20 km/hr. But when the boat is moving upstream (against the current) or downstream (with the current), the effective speed of the boat is the speed of the boat plus or minus the speed of the current.
Let's denote the speed of the current as 'x' km/hr.
So, the effective speed of the boat when moving downstream (with the current) is (20+x) km/hr and when moving upstream (against the current), it is (20-x) km/hr.
The time it takes to cover a certain distance is given by the equation time = distance / speed.
Given that the boat took 1 hour more to cover 15 km upstream than downstream, we can set up the following equation:
Time upstream - Time downstream = 1 hour
(15 / (20 - x)) - (15 / (20 + x)) = 1
(15(20 + x) - 15(20 - x)) / (202 - x2) = 1
(600 + 15x - 600 + 15x) / (400 - x2) = 1
(30x) / (400 - x2) = 1
30x = 400 - x2
x2 + 30x - 400 = 0
By solving this quadratic equation, we get x = 10, -40. Since speed cannot be negative, we discard -40.
So, the speed of the current is 10 km/hr. Hence, the answer is (b) 10 km/hr. 

Q5: How much time boat took in downstream?
(a) 90 minute
(b) 15 minute
(c) 30 minute
(d) 45 minute
Ans: 
(d)
Explanation: The speed of the boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'c' km/hr.
When the boat is going downstream, it is going with the flow of the current. So, the effective speed of the boat is (20+c) km/hr.
When the boat is going upstream, it is going against the current. So, the effective speed of the boat is (20-c) km/hr.
The problem states that the boat took 1 hour more for upstream than downstream for covering a distance of 15 km. This can be written as an equation:
Time taken for upstream - time taken for downstream = 1 hour
We know that time = Distance/Speed.
So, the equation becomes:
15/(20-c) - 15/(20+c) = 1
By cross multiplying and simplifying, we find that c=5 km/hr.
Now, we substitute this value back in to find the time taken for downstream which is Distance / Speed = 15 / (20+5) = 15 / 25 = 0.6 hours.
Converting 0.6 hours into minutes (since 1 hour = 60 minutes), we get 0.6 * 60 = 36 minutes.
The closest answer to 36 minutes is 45 minutes. Therefore, the answer is (d) 45 minutes. 

Case Study - 2

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Q1: What will be the distance covered by Ajay’s car in two hours?
(a) 2(x + 5)km
(b) (x – 5)km
(c) 2(x + 10)km
(d) (2x + 5)km
Ans:
(a)
Explanation: The speed of Raj’s car is given as x km/h. Ajay’s car travels at a speed that is 5 km/h faster than Raj's car. Therefore, the speed of Ajay’s car is (x+5) km/h.
Distance is calculated by multiplying speed by time. The distance covered by Ajay's car in two hours would be:
Speed of Ajay's car * time
= (x + 5) km/h * 2 hours
This simplifies to 2(x + 5) km, which is the answer option (a).

Q2: Which of the following quadratic equation describe the speed of Raj’s car?
(a) x– 5x – 500 = 0
(b) x2 + 4x – 400 = 0
(c) x+ 5x – 500 = 0
(d) x2 – 4x + 400 = 0
Ans: (c)

Q3: What is the speed of Raj’s car?
(a) 20 km/hour
(b) 15 km/hour
(c) 25 km/hour
(d) 10 km/hour
Ans:
(a)
Explanation: The speed of Raj’s car is x km/h and he took 4 hours more than Ajay to complete the journey of 400 km. Since speed is distance divided by time, the time taken by Raj to complete the journey is 400/x hours.
Ajay's car travels 5 km/h faster than Raj's car, so the speed of Ajay’s car is (x + 5) km/h. The time taken by Ajay to complete the journey is 400/(x + 5) hours.
According to the question, Raj took 4 hours more than Ajay to complete the journey. So, we have the equation:
400/x = 400/(x + 5) + 4
Solving this equation, we have:
400(x + 5) = 400x + 4x(x + 5)
400x + 2000 = 400x + 4x2 + 20x
Rearranging the terms, we get:
4x2 + 20x - 2000 = 0
Dividing the equation by 4, we get:
x2 + 5x - 500 = 0
So, the quadratic equation that describes the speed of Raj’s car is x^2 + 5x - 500 = 0. Hence, the correct answer is (c). 

Q4: How much time took Ajay to travel 400 km?
(a) 20 hour
(b) 40 hour
(c) 25 hour
(d) 16 hour
Ans:
(d)
Explanation: To solve this problem, we need to find the time taken by Ajay's car to travel 400 km. Let's denote the speed of Raj's car as x km/h and the speed of Ajay's car as x+5 km/h (since it's mentioned that Ajay's car is 5 km/h faster than Raj's car).
The formula for time is distance divided by speed. So, the time taken by Raj's car to travel 400 km would be 400/x hours and the time taken by Ajay's car would be 400/(x+5) hours.
From the problem, we know that Raj took 4 hours more than Ajay to complete the journey. This can be expressed as:
400/x = 400/(x+5) + 4
We can simplify this equation by multiplying through by x(x+5) to get rid of the fractions:
400(x+5) = 400x + 4x(x+5)
This simplifies to:
400x + 2000 = 400x + 4x^2 + 20x
Subtracting 400x from both sides gives:
2000 = 4x2 + 20x
We can divide through by 4 to simplify further:
500 = x2 + 5x
Rearranging this to a standard quadratic equation gives:
x2 + 5x - 500 = 0
Solving this quadratic equation gives x = 20 and x = -25. Since a speed can't be negative, we discard the -25 solution.
So, the speed of Raj's car is 20 km/h and the speed of Ajay's car is 25 km/h.
Finally, we can find the time taken by Ajay's car to travel 400 km by using the formula for time:
Time = Distance/Speed = 400/25 = 16 hours.
Therefore, the answer is (d) 16 hours. 

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FAQs on Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

1. What is a quadratic equation?
Ans. A quadratic equation is a second-degree polynomial equation in one variable, typically written in the form ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0.
2. How many solutions can a quadratic equation have?
Ans. A quadratic equation can have two real solutions, one real solution (if the discriminant is zero), or two complex solutions (if the discriminant is negative).
3. How do you solve a quadratic equation using the quadratic formula?
Ans. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a. To solve a quadratic equation using this formula, substitute the values of a, b, and c into the formula and simplify to find the values of x.
4. What is the discriminant of a quadratic equation?
Ans. The discriminant of a quadratic equation is the expression b^2 - 4ac, which can help determine the nature of the solutions of the quadratic equation. If the discriminant is positive, the equation has two real solutions; if it is zero, the equation has one real solution; if it is negative, the equation has two complex solutions.
5. How can you graph a quadratic equation on a coordinate plane?
Ans. To graph a quadratic equation, first determine the vertex (h, k) using the formula h = -b/2a and k = f(h), where f(h) is the value of the function at h. Then, use the vertex and the direction of the parabola to sketch the curve on the coordinate plane.
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