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# Hots Questions - Quadratic Equations Class 10 Notes | EduRev

## Mathematics (Maths) Class 10

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## Class 10 : Hots Questions - Quadratic Equations Class 10 Notes | EduRev

The document Hots Questions - Quadratic Equations Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

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:  24 km upstream = 20/18-x hours

According to the condition: ⇒ 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 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.  ⇒ 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 According to the condition, 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.

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