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Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

Q1: Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

Sol: Let actual marks be x
∴ 9 × [ Actual marks + 10] = [Square of actual marks]
or 9 × (x + 10) = x2
⇒ 9x + 90 = x2
⇒ x2 – 9x – 90 = 0
⇒ x2 – 15x + 6x – 90 = 0
⇒ x(x – 15) + 6(x – 15) = 0
⇒(x + 6) (x – 15) = 0
Either  x + 6 =  0
⇒ x = – 6 or x – 15 = 0
⇒ x = 15
But marks cannot be less than 0.
∴ x = –6 is not desired.

Thus, Ravita got 15 marks in her Mathematics test.

Q2: A motor boat 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.

Sol: Let the speed of the stream = x km/hr
∴ speed of the motor boat:
upstream = (18 − x) km/hr
downstream = (18 + x) km/hr
⇒ Time taken by the motor boat in going:

Class 10 Maths Chapter 4 HOTS Questions - Quadratic EquationsClass 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

24 km upstream = 20/18-x hours

According to the condition:

Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

⇒ 24 × (18 + x) − 2(18 − x) = 1 (18 − x) (18 + x)
⇒ 24 [18 + x − 18 + x] = 182 − x2
⇒ 24 [2x] = 324 − x2
⇒ 48x = 324 − x2
⇒ x2 + 48x − 324 = 0
⇒ x2 + 54x − 6x − 324 = 0
⇒ x (x + 54) − 6 (x + 54) = 0
⇒ (x − 6) (x + 54) = 0    
Either x − 6 = 0 ⇒ x = 6  
or x + 54  = 0 ⇒ x = − 54

But speed cannot be negative
∴ Rejecting x =− 54, we have

x = 6 ⇒ Speed of the boat = 6 km/hr.

Q3: In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately.

Sol: Let marks obtained by P in Maths be ‘x’.
∴ His marks in Science = (28 − x)
According to the condition,

(x + 3) (28 − x − 4) = 180
⇒ (x + 3) (− x + 24) = 180
⇒ 24x − x2 + 72 − 3x = 180
⇒ − x2 + 21x + 72 − 180 = 0
⇒ − x2 + 21x − 108 = 0
⇒ x2 − 21x + 108 = 0
⇒ x2 − 12x − 9x + 108 = 0
⇒ x (x − 12x) − 9(x − 12) = 0
⇒ (x − 9) (x − 12) = 0
⇒ (x − 9) (x − 12) = 0

Either  x − 9 = 0
⇒ x = 9 or  x − 12 = 0  
⇒ x = 12 When x = 9 then 28 − x = 28 − 9 = 19
When x = 12 then 28 − x = 28 − 12 = 16
Thus P’s marks in Maths = 9 and Science = 19

Or

P’s marks in Maths = 12 and Science = 16.

Q4: At ‘t’ minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than  Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations minutes. Find ‘t’.

Sol: For a minute-hand time needed to show 2 pm to 3 pm is ‘60’ minutes. It has already covered ‘t’ minutes. ∴ Time required by the minute-hand to reach to 12 (at 3 pm) = (60 – t) minutes.
Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations
Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations
⇒ t2+ 4t – 252 = 0

Solving, we get,  t = 14 or – 18
But t = – 18 is not desirable (being negative)
Thus, t = 14 minutes.

Q5: A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/ hr more. Find the original speed of the train.

Sol: Let the original speed be x km/hr

∴ Original time taken = 360/x.

New speed = (x + 5) km/hr

Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

According to the condition,

Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

Solving for x, we get x = – 50 or 45
Speed cannot be negative
∴ Rejecting x = – 50, we have x = 45
Thus, the original speed of the train = 45 km/hr.

The document Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 4 HOTS Questions - Quadratic Equations

1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of degree 2, which can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.
2. How do you solve a quadratic equation by factoring?
Ans. To solve a quadratic equation by factoring, we need to set the equation equal to zero and then factor it into two binomials. By setting each binomial equal to zero, we can find the values of x that satisfy the equation.
3. What are the different methods to solve a quadratic equation?
Ans. The different methods to solve a quadratic equation are: 1. Factoring method 2. Square root method 3. Completing the square method 4. Quadratic formula
4. When does a quadratic equation have one solution?
Ans. A quadratic equation has one solution when its discriminant (b^2 - 4ac) is equal to zero. This means that the equation has a repeated root, and the graph of the equation will touch the x-axis at only one point.
5. How can we use the quadratic formula to solve a quadratic equation?
Ans. The quadratic formula is used to solve any quadratic equation of the form ax^2 + bx + c = 0. The formula is x = (-b ± √(b^2 - 4ac)) / (2a). By substituting the values of a, b, and c into the formula and simplifying, we can find the solutions for x.
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