CBSE Class 10 Mathematics Quadratic Equation Worksheet Set E

Read and download free pdf of CBSE Class 10 Mathematics Quadratic Equation Worksheet Set E. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 4 Quadratic Equation in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 10 Mathematics Chapter 4 Quadratic Equation

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 4 Quadratic Equation in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

Class 10 Mathematics Worksheet for Chapter 4 Quadratic Equation

 

QUADRATIC EQUATIONS

 

Q.- A two digit number is such that the product of the digits is 35. When 18 is added to this number the digits interchange their places.
Determine the number.
 
Sol. Let ten’s digit of the numbers = x and its unit digit = y. 
∴The two digit number is 10x + y. 
Given : x . y = 35 and 10x + y + 18 = 10y + x 
=> y = 35/x and 9x + 18 = 9y
i.e., x + 2 = y
On substituting y =  35/x in x + 2 = y ; we get:
x + 2 =35/x
=> x2 + 2x = 35
and x2 + 2x – 35 = 0
On factorising, we get : (x + 7)(x – 5) = 0
i.e., x = – 7 or x = 5
Since, x is digit, therefore x = 5 and y = 35/x =7

∴ The required two digit number = 10x + y

= 10 × 5 + 7 = 57 

Q.- The sides (in cm) of a right triangle are x – 1, x and x + 1. Find the sides of triangle.
 
Sol. It is clear that the largest side x + 1 is hypotenuse of the right triangle.
Quadratic equations notes 3
According to Pythagoras Theorem, we have :
x2 + (x – 1)2 = (x + 1)2
=> x2 + x2 – 2x + 1 = x2 + 2x + 1
This gives x2 – 4x = 0
=> x(x – 4) = 0 i.e., x = 0 or x = 4
Since, with x = 0 the triangle is not possible;
hence x = 4.
∴ Sides, are x – 1, x and x + 1 = 4 – 1
i.e., 3 cm, 4 cm and 5 cm
 
Q.- The hypotenuse of a right triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
 
Sol. Let the shortest side be x m.

Quadratic equations notes 4

 

∴ Hypotensuse = (2x – 1) m and the third side = (x + 1) m
Applying Pythagoras theorem, we get ;
(2x – 1)2 = x2 + (x + 1)2
=> 4x2 – 4x + 1 = x2 + x2 + 2x + 1
i.e., 2x2 – 6x = 0
=> x2 – 3x = 0
i.e., x(x – 3) = 0
=> x = 0 or x = 3
Since, x = 0 makes the triangle impossible.
therefore, x = 3
And, sides of the triangle are = x, 2x – 1 and x + 1 = 3, 2 × 3 –1 and 3 + 1 = 3m, 5m and 4m
 
Q.-  If the perimeter of a rectangular plot is 68 m and its diagonal is 26 m. Find its area.
Sol. Let the length of plot = x m

 Quadratic equations notes 5

2(length + breadth) = perimeter 
=> 2(x + breadth) = 68
=> x + breadth = 68/2 and breadth = (34 – x) m
Given its diagonal = 26 m and we know each angle of the rectangle = 90º. 
∴  x2 + (34 – x)2 = 262
[Applying Pythagoras Theorem]
=> x2 + 1156 – 68x + x2 – 676 = 0
=> 2x2 – 68x + 480 = 0
=> x2 – 34x + 240 = 0
i.e., x2 – 34x + 240 = 0
 
On factorising, we get : (x – 24) (x – 10) = 0
i.e., x = 24 or x = 10
x = 24
=> length = 24 m and breadth
= (34 – 24) m = 10 m and, x = 10
=> length = 10 m and breadth
= (34 – 10) m = 24 m 
∴ Dimensions of the given rectangular plot are 24 m and 10 m.
Hence, its area = length × breadth
= 24 m × 10m = 240 m2
 
Q.- In an auditorium, the number of rows was equal to the number of seats in each row. If the number of roots is doubled and the number of seats in each row is reduced by5, then the total number of seats is increased by 375. How many rows were there ?
 
Sol. Let the number of rows be x 
=> No. of seats in each rows = x
∴ The total number of seats in the auditorium
= x × x = x2
Now, the new no. of rows = 2x and the new no. of seats in each row = x – 5
∴ The new no. of total seats in the auditorium = 2x(x – 5).
Given : 2x (x – 5) – x2 = 375
=> 2x2 – 10x – x2 = 375 and x2 – 10x – 375 = 0
On factorising, we get : (x – 25) (x + 15) = 0
i.e., x = 25 or x = – 15
Neglecting x = – 15, we get : no. of rows = 25
 
Q.- Find the roots of the equation x2 – 2x – 8 = 0.
 
Sol. Quadratic Equation x2 – 2x – 8 = 0
After factorization (x – 4) (x + 2) = 0
=> x = 4, – 2
 

More question-

 SECTION A: (1 MARK)

1. Find the value of  (-2,3)

2.A polygon of n sides has diagonals. How many sides has a polygon with 54 diagonals? (12)

3. Find the roots of the equation ax2 + a = a2x + x (a, )

SECTION B: (2 MARKS)

4. Solve for x : (CBSE BOARD 2012) (-4,)

5. Find the value of p for which x2 + 5px + 16 = 0 has no real roots. ()

6.One day, I asked the son of my close friend about his age. The child replied in a different way. He said, “One year ago, my dad was 8 times as old as me and now his age is equal to square of my age.” Represent this situation in the form of a quadratic equation. (CBSE BOARD 2007) (x2-8x+7=0)

7. Find the value of p for which the quadratic equation 4x2 – 3px + 9 = 0 has real roots. ( p )

8. If y =1 is a common root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0, find ab. (CBSE BOARD 2012) ( 3 )

SECTION C: (3 MARKS)

9. Solve for x : 2 2x + 3 = 65(2x – 2) + 122 (CBSE BOARD 2012) (-3,3)

10. Solve for x : x2 + 5x – (a2 + a – 6) = 0. [-(a+3),(a-2)]

11. Solve for x: 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0.

12. Solve for x: , x (CBSE BOARD 2004) [ (a+b), ]

13. Solve for x : (x (CBSE BOARD 2005) ( -a , -b)

14. Solve for x: + = (x ≠ 0,1) (CBSE BOARD 2000) ( )

SECTION D: (4 MARKS)

15. The numerator of a fraction is 1 less than the denominator. If 3 is added to each of the numerator and denominator, the fraction is increased by 3/28. Find the fraction. (CBSE BOARD 2016) (3/4)

16. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples. (CBSE BOARD 2016) (14,21)

17. The total cost of a certain length of a piece of cloth is ₹200. If the piece was 5m longer and each meter of cloth costs ₹2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per meter? (CBSE BOARD 2012) (20m,₹10)

18. If the roots of the quadratic equation x2 + 2px + mn = 0 are real and equal, show that the roots of the quadratic equation x2 – 2(m + n)x + (m2 + n2 + 2p2) = 0 are also equal. (CBSE BOARD 2008)

19. A tank can be filled by one pipe in x minutes and emptied by another pipe in (x + 5) minutes. Both the pipes when opened together can fill the empty tank in 16.8 minutes. Find x. (EXEMPLAR PROBLEM) (7 hours)

20. Solve for x: 5 x + 1 + 5 2 – x = 126 (2,-1)

21. If the roots of the equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c. (EXEMPLAR PROBLEM)

22. Solve for x : (-1)

23. Students of class X collected ₹18000. They wanted to divide it equally among a certain number of students residing in slum area. When they started distributing the amount, 20 more students from the nearby slums also joined. Now each student get ₹240 less. (a) Find the number of students living in the slum. (b) Which value is depicted by the students? (CBSE BOARD 2013) (30)

24. Out of a number of saras birds , one fourth of the number are moving about in lots, 1/9 th coupled with ¼ th as well as 7 times the square root of the number move on a hill, 56 birds in vakula trees. What is the total number of birds? (CBSE BOARD 2004) (576)

25. At t minutes past 2 p.m. the time needed by the minutes hand of a clock to show 3 p.m. was found to be 3 minutes less than t2/4 minutes. Find t.

 

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