๐Ÿ“˜ BOARD LEVEL SAMPLE PAPER-Class 10 Mathematics โ€“ Chapter: Quadratic Equations

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prepared strictly as per CBSE board pattern, with proper marks distribution, variety of questions, and exam difficulty level.

๐Ÿ“„ CLASS 10 โ€“ MATHEMATICS

CHAPTER 4: QUADRATIC EQUATIONS

๐Ÿ•’ Time: 1ยฝ Hoursโ€ƒโ€ƒ๐Ÿ“Š Maximum Marks: 40

๐Ÿ”น GENERAL INSTRUCTIONS:

  1. All questions are compulsory.
  2. Use of calculator is not permitted.
  3. Draw neat figures wherever required.
  4. Show all necessary steps for full marks.

๐ŸŸข SECTION A โ€“ VERY SHORT ANSWER

(1 ร— 5 = 5 Marks)

  1. Write the standard form of a quadratic equation.
  2. Find the value of discriminant for the equation
    x2โˆ’6x+9=0x^2 โ€“ 6x + 9 = 0x2โˆ’6x+9=0.
  3. How many roots does a quadratic equation have?
  4. Is 5xโˆ’3=05x โ€“ 3 = 05xโˆ’3=0 a quadratic equation? Give reason.
  5. If the discriminant of a quadratic equation is negative, what can you say about its roots?

๐ŸŸข SECTION B โ€“ SHORT ANSWER (I)

(2 ร— 5 = 10 Marks)

  1. Find the roots of the equation
    x2โˆ’7x+10=0x^2 โ€“ 7x + 10 = 0x2โˆ’7x+10=0 by factorisation.
  2. Find the nature of roots of the quadratic equation
    2x2โˆ’4x+1=02x^2 โ€“ 4x + 1 = 02ร—2โˆ’4x+1=0.
  3. Find the value of kkk for which the equation
    x2+kx+9=0x^2 + kx + 9 = 0x2+kx+9=0 has equal roots.
  4. If the sum and product of roots of a quadratic equation are 1 and โ€“6 respectively, form the quadratic equation.
  5. Find the discriminant of the quadratic equation
    3x2โˆ’2x+5=03x^2 โ€“ 2x + 5 = 03ร—2โˆ’2x+5=0 and state the nature of its roots.

๐ŸŸข SECTION C โ€“ SHORT ANSWER (II)

(3 ร— 5 = 15 Marks)

  1. Solve the quadratic equation
    2x2โˆ’5xโˆ’3=02x^2 โ€“ 5x โ€“ 3 = 02ร—2โˆ’5xโˆ’3=0 using the quadratic formula.
  2. Solve the equation
    x(x+4)=12x(x + 4) = 12x(x+4)=12.
  3. Find two consecutive positive integers whose product is 306.
  4. Solve for x:
    1x+1+1xโˆ’1=13\frac{1}{x+1} + \frac{1}{x-1} = \frac{1}{3}x+11โ€‹+xโˆ’11โ€‹=31โ€‹
  5. If the roots of the equation
    x2โˆ’5x+k=0x^2 โ€“ 5x + k = 0x2โˆ’5x+k=0 are equal, find the value of kkk.

๐ŸŸข SECTION D โ€“ LONG ANSWER

(5 ร— 2 = 10 Marks)

  1. Solve the quadratic equation by completing the square method:

x2+6xโˆ’7=0x^2 + 6x โ€“ 7 = 0x2+6xโˆ’7=0

  1. The product of two positive consecutive integers is 1320.
    Find the integers.

โญ INTERNAL CHOICE (For Board Pattern)

  • Question 11 OR Solve: 3x2โˆ’7x+2=03x^2 โ€“ 7x + 2 = 03ร—2โˆ’7x+2=0
  • Question 17 OR Find two numbers whose sum is 27 and product is 182.

โœ… MARKING SCHEME HINT (Teacher Use)

  • Correct formula โ†’ 1 mark
  • Substitution โ†’ 1 mark
  • Calculation โ†’ 2 marks
  • Final answer โ†’ 1 mark

โœ… ANSWER KEY / SOLUTIONS

๐ŸŸข SECTION A โ€“ VERY SHORT ANSWER

1. Standard form of a quadratic equation isax2+bx+c=0,ย aโ‰ 0ax^2 + bx + c = 0,\ a \neq 0ax2+bx+c=0,ย a๎€ =0

2. Given equation: x2โˆ’6x+9=0x^2 โ€“ 6x + 9 = 0x2โˆ’6x+9=0D=b2โˆ’4ac=(โˆ’6)2โˆ’4(1)(9)=36โˆ’36=0D = b^2 โ€“ 4ac = (-6)^2 โ€“ 4(1)(9) = 36 โ€“ 36 = 0D=b2โˆ’4ac=(โˆ’6)2โˆ’4(1)(9)=36โˆ’36=0

3. A quadratic equation has two roots.

4. No, 5xโˆ’3=05x โ€“ 3 = 05xโˆ’3=0 is not quadratic because the highest power of x is 1.

5. If discriminant is negative, the equation has no real roots.

๐ŸŸข SECTION B โ€“ SHORT ANSWER (I)

6. Solve: x2โˆ’7x+10=0x^2 โ€“ 7x + 10 = 0x2โˆ’7x+10=0x2โˆ’5xโˆ’2x+10=0x^2 โ€“ 5x โ€“ 2x + 10 = 0x2โˆ’5xโˆ’2x+10=0 (xโˆ’5)(xโˆ’2)=0(x โ€“ 5)(x โ€“ 2) = 0(xโˆ’5)(xโˆ’2)=0 x=5,ย 2x = 5,\ 2x=5,ย 2

7. Equation: 2x2โˆ’4x+1=02x^2 โ€“ 4x + 1 = 02ร—2โˆ’4x+1=0D=(โˆ’4)2โˆ’4(2)(1)=16โˆ’8=8>0D = (-4)^2 โ€“ 4(2)(1) = 16 โ€“ 8 = 8 > 0D=(โˆ’4)2โˆ’4(2)(1)=16โˆ’8=8>0

Roots are real and distinct.

8. Equation: x2+kx+9=0x^2 + kx + 9 = 0x2+kx+9=0
For equal roots, D=0D = 0D=0k2โˆ’36=0k^2 โ€“ 36 = 0k2โˆ’36=0 k=ยฑ6k = \pm 6k=ยฑ6

9. Sum = 1, Product = โ€“6x2โˆ’(1)xโˆ’6=0x^2 โ€“ (1)x โ€“ 6 = 0x2โˆ’(1)xโˆ’6=0

10. Equation: 3x2โˆ’2x+5=03x^2 โ€“ 2x + 5 = 03ร—2โˆ’2x+5=0D=(โˆ’2)2โˆ’4(3)(5)=4โˆ’60=โˆ’56D = (-2)^2 โ€“ 4(3)(5) = 4 โ€“ 60 = -56D=(โˆ’2)2โˆ’4(3)(5)=4โˆ’60=โˆ’56

Roots are not real.

๐ŸŸข SECTION C โ€“ SHORT ANSWER (II)

11. Solve: 2x2โˆ’5xโˆ’3=02x^2 โ€“ 5x โ€“ 3 = 02ร—2โˆ’5xโˆ’3=0x=5ยฑ25+244x = \frac{5 \pm \sqrt{25 + 24}}{4}x=45ยฑ25+24โ€‹โ€‹ x=5ยฑ74x = \frac{5 \pm 7}{4}x=45ยฑ7โ€‹ x=3,ย โˆ’12x = 3,\ -\frac{1}{2}x=3,ย โˆ’21โ€‹

12. Equation: x(x+4)=12x(x + 4) = 12x(x+4)=12x2+4xโˆ’12=0x^2 + 4x โ€“ 12 = 0x2+4xโˆ’12=0 x2+6xโˆ’2xโˆ’12=0x^2 + 6x โ€“ 2x โ€“ 12 = 0x2+6xโˆ’2xโˆ’12=0 (x+6)(xโˆ’2)=0(x + 6)(x โ€“ 2) = 0(x+6)(xโˆ’2)=0 x=โˆ’6,ย 2x = -6,\ 2x=โˆ’6,ย 2

13. Product of consecutive integers = 306x(x+1)=306x(x+1) = 306x(x+1)=306 x2+xโˆ’306=0x^2 + x โ€“ 306 = 0x2+xโˆ’306=0 x=โˆ’1ยฑ352x = \frac{-1 \pm 35}{2}x=2โˆ’1ยฑ35โ€‹ x=17x = 17x=17

Numbers are 17 and 18

14.1x+1+1xโˆ’1=13\frac{1}{x+1} + \frac{1}{x-1} = \frac{1}{3}x+11โ€‹+xโˆ’11โ€‹=31โ€‹ 2xx2โˆ’1=13\frac{2x}{x^2 โ€“ 1} = \frac{1}{3}x2โˆ’12xโ€‹=31โ€‹ 6x=x2โˆ’16x = x^2 โ€“ 16x=x2โˆ’1 x2โˆ’6xโˆ’1=0x^2 โ€“ 6x โ€“ 1 = 0x2โˆ’6xโˆ’1=0 x=3ยฑ10x = 3 \pm \sqrt{10}x=3ยฑ10โ€‹

15. Equation: x2โˆ’5x+k=0x^2 โ€“ 5x + k = 0x2โˆ’5x+k=0
Equal roots โ‡’ D=0D = 0D=025โˆ’4k=025 โ€“ 4k = 025โˆ’4k=0 k=254k = \frac{25}{4}k=425โ€‹

๐ŸŸข SECTION D โ€“ LONG ANSWER

16. Solve: x2+6xโˆ’7=0x^2 + 6x โ€“ 7 = 0x2+6xโˆ’7=0x2+6x=7x^2 + 6x = 7x2+6x=7

Add 999 to both sides:(x+3)2=16(x + 3)^2 = 16(x+3)2=16 x+3=ยฑ4x + 3 = \pm 4x+3=ยฑ4 x=1,ย โˆ’7x = 1,\ -7x=1,ย โˆ’7

17. Product of consecutive integers = 1320x(x+1)=1320x(x + 1) = 1320x(x+1)=1320 x2+xโˆ’1320=0x^2 + x โ€“ 1320 = 0x2+xโˆ’1320=0 x=โˆ’1ยฑ732x = \frac{-1 \pm 73}{2}x=2โˆ’1ยฑ73โ€‹ x=36x = 36x=36

Numbers are 36 and 37

โญ INTERNAL CHOICE ANSWERS

Alternative to Q11:
3x2โˆ’7x+2=03x^2 โ€“ 7x + 2 = 03ร—2โˆ’7x+2=0x=2,ย 13x = 2,\ \frac{1}{3}x=2,ย 31โ€‹

Alternative to Q17:
Sum = 27, Product = 182x2โˆ’27x+182=0x^2 โ€“ 27x + 182 = 0x2โˆ’27x+182=0 (xโˆ’13)(xโˆ’14)=0(x โ€“ 13)(x โ€“ 14) = 0(xโˆ’13)(xโˆ’14)=0

Numbers are 13 and 14

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