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

ATQ,  9 × [ Actual marks + 10] = [Square of actual marks]
∴ 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 be x km/hr

Given that, the speed boat in still water is 18 km/hr.

Sspeed of the boat in upstream = (18 - x) km/hr

Speed of the boat in downstream = (18 + x) km/hr

It is mentioned that the boat takes 1 hour more to go 24 km upstream than to return downstream to the same spot

Therefore, One-way Distance traveled by boat (d) = 24 km 

Hence, Time in hour 

T_upstream = T_downstream  + 1

[distance / upstream speed ] = [distance / downstream speed]   + 1

[ 24/ (18 - x) ] = [ 24/ (18 + x) ] + 1 

[ 24/ (18 - x) - 24/ (18 + x) ] = 1 

24 [1/ (18 - x) - 1/(18 + x) ] = 1

24 [ {18 + x - (18 - x) } / {324 - x²} ] = 1

24 [ {18 + x - 18 + x) } / {324 - x²} ] = 1

⇒ 24 [ {2}x / {324 - x²} ] = 1

⇒ 48x = 324 - x²

⇒ x² + 48x - 324 = 0

⇒ x² + 54x - 6x - 324 = 0   ----------> (by splitting the middle-term)

⇒ x(x + 54) - 6(x + 54) = 0

⇒ (x + 54)(x - 6) = 0

⇒ x = -54  or 6

As speed to stream can never be negative, we consider the speed of the stream (x) as 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.

60 - t = (t²/4 - 3)

60 - t = (t² - 12)/4

240 - 4t = t² - 12

t² + 4t - 12 - 240 = 0

t² + 4t - 252 = 0

On factorizing, 

t² - 14t + 18t - 252 = 0

t(t - 14) + 18(t - 14) = 0

(t + 18)(t - 14) = 0

Now, t - 14 = 0

t = 14

Also, t + 18 = 0

t = -18

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.