CBSE Class 10 Maths Chapter-4 Quadratic Equations – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 4 – Quadratic Equations prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-4 Quadratic Equations Important Questions
CBSE Class 10 Maths Important Questions Chapter 4 – Quadratic Equations
4 Mark Questions
1. Find the roots of the following Quadratic Equations by applying quadratic formula.
(i) 2x2 − 7x+ 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 
(iv) 2x2 + x + 4 = 0
Ans . (i) 2x2 − 7x+ 3 = 0
Comparing quadratic equation 2x2 − 7x+ 3 = 0 with general form ax2+bx+c=0, we geta=2, b=-7 and c=3
Putting these values in quadratic formula 
⇒ 
⇒ 
⇒ 
⇒ x=3, ½
(ii) 2x2 + x – 4 = 0
Comparing quadratic equation 2x2 + x – 4 = 0 with the general form ax2+bx+c=0, we get a=2, b=1 and c=−4
Putting these values in quadratic formula
⇒ 
⇒ 
(iii)
Comparing quadratic equation with the general form ax2+bx+c=0, we geta=4,b=
and c=3
Putting these values in quadratic formula
⇒ 
⇒ 
A quadratic equation has two roots. Here, both the roots are equal.
Therefore, 
(iv) 2x2 + x + 4 = 0
Comparing quadratic equation 2x2 + x + 4 = 0 with the general form ax2+bx+c=0, we geta=2,b=1 and c= 4
Putting these values in quadratic formula
⇒ 
But, square root of negative number is not defined.
Therefore, Quadratic Equation 2x2 + x + 4 = 0 has no solution.
2. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of two trains.
Ans. Let average speed of passenger train = x km/h
Let average speed of express train = (x+11) km/h
Time taken by passenger train to cover 132 km = 
hours
Time taken by express train to cover 132 km = 
hours
According to the given condition,
⇒ 
⇒ 
⇒ 132(11)=x(x+11)
⇒ 1452=x2+11x
⇒ x2+11x– 1452=0
Comparing equation x2+11x– 1452=0 with general quadratic equation ax2+bx+c=0, we get a=1,b=11 and c=−1452
Applying Quadratic Formula
⇒ 
⇒ 
⇒ 
⇒ 
⇒ x=33,−44
As speed cannot be in negative.Therefore, speed of passenger train = 33 km/h
And, speed of express train = x+11=33+11=44 km/h
3. Sum of areas of two squares is 468 m2. If, the difference of their perimeters is 24 metres, find the sides of the two squares.
Ans. Let perimeter of first square = x metres
Let perimeter of second square = (x+24) metres
Length of side of first square = 
metres {Perimeter of square = 4 × length of side}
Length of side of second square = 
metres
Area of first square = side × side = 
Area of second square = 
According to given condition:
⇒ 
⇒ 
⇒ 2x2+576+48x=468×16
⇒ 2x2+48x+576=7488
⇒ 2x2+48x– 6912=0
⇒ x2+24x– 3456=0
Comparing equation x2+24x– 3456=0 with standard form ax2+bx+c=0,
We get a=1,b=24 andc= −3456
Applying Quadratic Formula
⇒ 
⇒ 
⇒ 
⇒ x=48,−72
Perimeter of square cannot be in negative. Therefore, we discard x=−72.
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x+24=48+24=72 metres
⇒ Side of First square = 
And, Side of second Square = 
4. Is it possible to design a rectangular park of perimeter 80 metres and area 400 m2. If so, find its length and breadth.
Ans. Let length of park = x metres
We are given area of rectangular park = 400 m2
Therefore, breadth of park = 
metres {Area of rectangle = length ×breadth}
Perimeter of rectangular park = 2(length+breath)=
metres
We are given perimeter of rectangle = 80 metres
According to condition:
⇒ 
⇒ 2x2+800=80x
⇒ 2x2−80x+800=0
⇒ x2−40x+400=0
Comparing equation, x2−40x+400=0 with general quadratic equation ax2+bx+c=0, we geta=1,b=−40 and c=400
Discriminant = b2−4ac=(−40)2 – 4(1)(400)=1600 – 1600=0
Discriminant is equal to 0.
Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area 400 m2.
Using quadratic formula to solve equation,
Here, both the roots are equal to 20.
Therefore, length of rectangular park = 20 metres
Breadth of rectangular park = 
5. If I had walked 1 km per hour faster, I would have taken 10 minutes less to walk 2 km. Find the rate of my walking.
Ans. Distance = 2 km
Let speed = 
km/hr
New speed = (x+1) km/hr
Time taken by normal speed = 
Time taken by new speed = 
According to question,
So, speed is
km/hr
6. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
Ans. Let B takesdays to finish the work, then A alone can finish it in 
days
According to question,
(Neglect)
So, B takes 
days.
7. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Ans. Let usual speed 
km/hr
New speed 
km/hr
Total distance = 1500 km
Time taken by usual speed = 
hr
Time taken by new speed = 
hr
According to question,
Therefore, usual speed is 750 km/hr, -1000 is neglected.
8. A motor boat, whose speed is 15 km/hr in still water, goes 30 km downstream and comes back in a total time of 4 hr 30 minutes, find the speed of the stream.
Ans. Speed of motor boat in still water = 15 km/hr
Speed of stream = km/hr
Speed in downward direction = 
Speed in downward direction = 
According to question,
Speed of stream = 5 km/hr
9. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.
Ans. Let be the number of hours required by the second pipe alone to till the pool and first pipe 
hour while third pipe 
hour
10. A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number the digit interchange their places. Find the number.
Ans. Let digit on unit’s place =
Digit on ten’s place = 
(given)
Number = 10.
=
According to question,
Number = 
= 92
11. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years.
Ans. According to question,
2P 
12. Two pipes running together can fill a cistern in 
if one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Ans. Let the faster pipe takes minutes to fill the cistern and the slower pipe will take 
minutes.
According to question,
13. If the roots of the equation 
are equal, prove that 
Ans.
For equal root s,
D = 0
14. Two circles touch internally. The sum of their areas is 116
cm2 and the distance between their centres is 6 cm. Find the radii of the circles.
Ans. Let 
and 
be the radius of two circles
According to question,
(Given)
Put the value of in eq. … (i)
(Neglect) or r1 = 4 cm
15. A piece of cloth costs Rs. 200. If the piece was 5 m longer and each metre of cloth costs Rs. 2 less the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Ans. Let the length of piece = m
Rate per metre = 
New length = (x+5)
New rate per metre = 
According to question,
Rate per metre = 10
16. ax2+bx+x = 0, a ≠≠ 0 solve by quadratic formula.
Ans. ax2+bx+x = 0
17. The length of the hypotenuse of a right-angled triangleexceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
Ans. Let base = x
Altitude = y
Hypotenuse = h
According to question,
Base = 15 cm
Altitude = 
Hypotenuse = 17 cm
18. Find the roots of the following quadratic equations if they exist by the method of completing square.
(i) 2x2−7x+3=0
(ii) 2x2+x– 4=0
(iii) 
(iv) 2x2+x+4=0
Ans. (i) 2x2−7x+3=0
First we divide equation by 2 to make coefficient of x2 equal to 1,
We divide middle term of the equation by 2x, we get 
We add and subtract square of 
from the equation ,
⇒ 
{(a−b)2=a2+b2−2ab}
⇒ 
⇒ 
Taking Square root on both sides,
⇒ 
⇒ 
Therefore, 
(ii) 2x2+x– 4=0
Dividing equation by 2,
Following procedure of completing square,
⇒ 
{(a+b)2=a2+b2+2ab}
⇒ 
⇒ 
Taking square root on both sides,
⇒ 
⇒ 
Therefore, 
(iii)
Dividing equation by 4,
Following the procedure of completing square,
⇒ 
⇒ 
{(a+b)2=a2+b2+2ab}
⇒ 
Taking square root on both sides,
⇒ 
⇒ 
(iv) 2x2+x+4=0
Dividing equation by 2,
Following the procedure of completing square,
⇒ 
⇒ 
{(a+b)2=a2+b2+2ab}
⇒ 
⇒ 
Taking square root on both sides
Right hand side does not exist because square root of negative number does not exist.
Therefore, there is no solution for quadratic equation 2x2+x+4=0
