**Class 10 Maths MCQ Questions of Quadratic Equations with Solutions for Term-2 CBSE Board 2021-22**

Class 10 Maths MCQ Questions of Quadratic Equations with Solutions for Term-2 CBSE Board 2021-22 are created here for the purpose of helping the class 10 students in CBSE Board exam 2021-22-Term-2.It should be awared to all students of CBSE Board that in term 2 CBSE board exam of mathematics question paper there would be 50 percent MCQ questions and 50 percent descriptive questions therefore now you are needed little more hard work in maths comparatively to term 1 exam of CBSE board.

**Three Ways of Solving Quadratic Equations**

**CBSE mathematics class X solutions of important questions chapter 4 โQuadratic equationsโ**

## Class 10 Maths MCQ Questions of Quadratic Equations with Solutions for Term-2 CBSE Board 2021-22

**Click for online shopping**

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

**Q1.The roots of quadratic equation 15xยฒ – 7x -2=0 areย **

**(a) (2/3,-3/2)ย ย ย (b) (-2/3,3/2)ย ย (c) (-1/5,2/3)ย ย (d) (1/5,-2/3)ย **

Ans. (c) (-1/5,2/3)

Factoring the LHS part of the quadratic equation

15xยฒ – 7x -2=0

15xยฒ – 10x +3x-2=0

5x(3x – 2) +1(3x-2) =0

(3x-2)(5x +1) =0

x = 2/3,x =-1/5

**Q2.The quadratic equation whose one rational root is 5 – โ3**

**(a) xยฒ +10x +22=0ย ย ย (b) xยฒ -10x +22=0ย ย ย (c) xยฒ +10x -22 =0ย ย (d) -xยฒ +10x +22=0**

Ans.ย (b) xยฒ -10x +22=0

The given rational root of quadratic equation is 5 – โ3 then other root must be its cojugate i.e 5 + โ3

The sum ofย of the roots = (5 – โ3) +(5 + โ3)=10

The produtofย of the roots = (5 – โ3) (5 + โ3)=25-3=22

Using the following formula for creating the quadratic equation

xยฒ -(sum ofย of the roots )x + produtofย of the roots =0

xยฒ -10x +22 =0

**Q3. The quadratic equation 8xยฒ +kx +2 =0 has two equal roots then the value of the k is**

**(a) 8ย ย ย (b) -8ย ย (c) ยฑ8ย ย (d)ยฑโ8**

Ans. ย (c) ยฑ8

The roots of theย quadratic equation axยฒ+bx +c=0 are equal when

bยฒ – 4ac =0

Comparing the given equation 8xยฒ +kx +2 =0 with standard equation axยฒ+bx +c=0

kยฒ -4ร8ร2 =0

kยฒ -64=0 โ kยฒ =ยฑ64โk=ยฑ8

**Q4.The difference between the square of two numbers is 20, if the square of smaller number is 4 more than twice of larger number, then the numbers are .**

**(a) 4,7ย ย ย (b) 7,4ย ย (c) 4,6ย (d)5,6**

Ans.(c) 4,6

Let the larger number is = x

Square of larger number = xยฒ

The square of smaller number = xยฒ-20

A t Q

The square of smaller number=ย twice of larger number +4

xยฒ -20= 2x +4

xยฒ – 2x -20 -4 =0

xยฒ – 2x -24 =0

xยฒ -6x + 4x -24 =0

x(x -6) +4(x – 6) =0

(x -6)(x +4) =0

x =6, x =-4(neglecting it because here number means natural number)

Square of smaller number is = 6ยฒ-20 =36 -20 =16

Smaller number =โ16 =4

**Q5. 5 years ago a man was 22 times as old as his son, now his age is equal to the square of the age of his son. Their present ages are.**

**(a) 7,49ย (b) 8,64ย (c) 6,36ย (d)5,25**

Ans.Let the present age of son is =x

5 years ago the age of son is =x -5

The age of man AtQ = 22(x -5)

The present age of man =22(x -5) +5

AtQ

Present age of man = square of the present age of son

22(x -5) +5 = xยฒ

xยฒ -22x +110 -5 =0

xยฒ -22x +105 =0

xยฒ -15x -7x+105 =0

x(x -15) -7(x -15) =0

(x -15)(x -7) =0

x = 15, 7

x =15 (impossible)

The age of son is 7 years and of his father is 7ยฒ=49

**Q6.The difference between square of two consecutive numbers is 27, the numbers are**

**(a) 13,14ย ย ย (b)15,16ย ย (c) 11,12ย (d)9,10**

Ans.Let x is one of the consecutive number then other is x +1

Difference between square of two consecutive numbers is 365

(x +1)ยฒ+ xยฒ =365

xยฒ +2x + 1+ย xยฒ-365=0

2xยฒ +2x – 364 =0

xยฒ + x – 182 =0

**Q7.A natural number when icreased by 12 equals 160 times its reciprocal .Find the number.**

**(a) 3ย ย ย ย ย (b) 8ย ย ย ย (c) 4ย ย ย ย (c) 7**

Ans. ย (b) 8

Let the number is x

When number is increased by 12,it will become =x+12

The reciprocal of the number =1/x

According to question

x+12 = (1/x)160

xยฒ + 12x =160

xยฒ + 12x -160 =0

xยฒ + 20x-8x -160 =0

x(x + 20) -8(x +20) =0

(x + 20)(x -8) =0

x =-20(neglecting), 8

**Q8. If the roots of the quadratic equation axยฒ + bx +c =0 are in the ratio of 2:3 ,then**

**(a) 6aยฒ = 5 cยฒย ย ย ย (b) 6bยฒ = 5acย ย ย (c) 6 bยฒ = 25acย ย ย ย (d) 6bยฒ = ac**

Ans.(c) 6 bยฒ = 25ac

Let the roots of the equation are 2ฮฑ +and 3ฮฑ

Sum of the roots =2ฮฑ+ 3ฮฑ=5ฮฑ=-b/aโฮฑ=-b/5a

Product of the roots = 2ฮฑร 3ฮฑ=6ฮฑยฒ=c/a=ฮฑยฒ=c/6a

(-b/5a)ยฒ = c/6a

bยฒ/25aยฒ = c/6a

bยฒ.6a = c.25aยฒ

6abยฒ = 25aยฒc

6bยฒ = 25ac

**Q9.The product of two successive integral multiples of 6 is 1080,then numbers are**

**(a) 36,42ย ย ย (b)54,60ย ย (c) 42,48ย (d)30,36**

Ans.ย (d)30,36

Let the two successive multiples of 6 are 6x and 6(x +1)

The product of both successive multiples is given 1080

โด 6xร6(x+1) = 1080

x(x +1) =30

xยฒ + x -30 =0

xยฒ + 6x – 5x – 30 =0

x(x +6) -5(x +6) =0

(x +6)(x – 5) =0

x =-6,5

Neglecting the value x =-6,taking positive integral value 5

Therefore successive integral multiples are 6ร5=30 and 6(x +1)=6(5+1)=36

**Q10.Shreya scored 20 more marks in her test out of 50 marks, 10 times these marks would have been the square of her actual marks. How many marks did she get in test.**

**(a) 30ย ย ย (b)50ย ย ย (c) 40ย ย ย (d)20**

Ans.(d)20

Let the actual marks of Shreya be x

According to question

xยฒ = 10(x +20)

xยฒ – 10x – 200 =0

xยฒ – 20x +10x- 200 =0

x(x -20) + 10(x – 20) =0

(x -20) (x +10) =0

x =20,-10

Neglecting the value x =-10,taking positive integral value 20

Therefore actual marks of Shreya are 20

**Q11. The roots of quadratic equation 5xยฒ – 4x + 5 are**

**(a) Real and equalย ย (b)Real and unequal ย (c) Not realย ย ย (d)Not real but equal**

Ans.(c) Not real

The given quadratic equation is 5xยฒ – 4x + 5

The discriminant,D = bยฒ – 4ac

a = 5, b = -4, c = 5

D = (-4)ยฒ – 4ร5ร5 = 16 – 100 = -84

Since D < 0

Therefore roots of given quadratic equation are not real

**Q12.Which constant should be added and subtracted to solve the quadratic equation 4x ^{2} – โ3x + 5 = 0 by the method of completing the square?**

**(a) 1/4ย ย ย (b)3/4ย ย (c) โ3/4ย ย ย (d)3/16**

Ans.(d)3/16

The given quadratic equation is 4x^{2} – โ3x + 5 = 0

Rewriting it

(2x)^{2} – โ3x + 5 = 0

Comparing it (a +b)ยฒ =aยฒ +bยฒ + 2ab

2ab =โ3x

Replacing a =2x

2ร2xรb = โ3x

b = โ3/4

Therefore we have to add and subtract (โ3/4)ยฒ= 3/16

**Q13. One of the root of the quadratic equation 3xยฒ- 2k +5 =0 is 1/3,find the value of k.**

**(a) 4ย ย ย (b)3ย ย (c) 5 ย ย (d)8**

Ans.(d)8

The given quadratic equation is 3xยฒ- 2kx +5 =0

The given root is 1/3

Therefore putting x = 1/3 in given quadratic equation

3(1/3)ยฒ- 2k(1/3) +5 =0

3/9 -2k/3 = -5

-2k/3 = -5 – 3/9 =-5 -1/3

-2k/3 =(-15 -1)/3 = -16/3

k = 8

**Q14. A natural number when added to its reciprocal become 50/3,find the number.**

**(a) 4ย ย ย (b)3ย ย (c) 7ย ย (d)8**

Ans.(c) 7

Let the natural number is x

As per the question

x + 1/ x = 50/7

xยฒ + 1 =50x/7

7xยฒ +7 =50x

7xยฒ – 50x +7 = 0

Factorizing it

7xยฒ – 50x +7 = 0

7xยฒ – 49x -x+7 = 0

7x(x – 7) -1(x – 7) =0

(x – 7)(7x – 1) =0

x =7,1

Neglecting x =1,since reciprocal of 1 is the same as the number

Therefore required number is 7

**Q15.The product of two successive integral multiples of 8 is 3584. Then the numbers are:**

**(a) 40,48ย ย ย (b)72,80ย ย (c) 56,64ย ย (d)80,88**

Ans. Let the successive multiple of 8 be 8x and 8(x +1)

As per the question

8x ร 8(x +1) =3584

64 xยฒ + 64x – 3584 =0

xยฒ + x – 56 =0

xยฒ + 8x – 7x-56 =0

x(x +8) -7(x +8) =0

(x +8)(x-7) =0

x = -8, 7

Neglecting x = -8ย because the multiple of 8 can’t be negative

Therefore ย the successive multiple of 8 are 8x=8ร7=56 and 8(7 +1)=8ร8 =64

**Q16.The present age of Sonu is 3 years more than the age of his younger brother,if 5 years back the product of their age is 40 then find their present age.**

**(a) 8,11ย ย ย ย ย (b) 10,13ย ย ย ย ย ย (a) 15,18ย ย ย ย ย ย (a) 4,7**

Ans. (b) 10,13

Let the present age of his brother is x

The age of Sonu is x + 3

According to question

(x -5)(x + 3 โ 5) = 40

(x โ 5)(x -2) = 40

X^{2} โ 7x + 10 = 40

X^{2} โ 7x + 10-40=0

X^{2} โ 7x -30=0

X^{2} โ 10x + 3x โ 30 =0

X(x -10) + 3(x โ 10) =0

(x -10)(x +3) =0

X = 10,-3

Therefore present age of Sonuโs brother is 10 years and the age of Sonu is 10 +3 =13 years

**Q17.The product of the cost of 2 kg potato and 3 kg onion is 5400,if the cost of 1 kgย onion is Rs 5 more than theย twice of the cost of 1 kg potato ,then the cost of 1 kg potato and 1 kg onion is**

**(a) Rs 20,Rs 45ย ย ย (b) Rs 30,Rs 65 ย ย (a) Rs 25,Rs 55 ย ย (a)Rs 15,Rs 35**

Ans.(a) Rs 20,Rs 45

Let the cost of 1 kg potato is x and the cost of 1 kg onion is 2x +5

2x.3(2x +5)= 5400

12x^{2} + 30x โ 5400 =0

2xยฒ + 5x – 900 =0

x = [-5 ยฑโ(5ยฒ -4ร2ร-900]/(2ร2) = [-5 ยฑโ(25+7200)]/(2ร2)= [-5 ยฑโ(7225)]/4 =(-5 ยฑ85)/4

x =(-5 +85)/4 = 80/4 =20

x =(-5 -85)/4 = -90/4 (neglecting it because the cost can’t be negative)

The cost of 1 kg potato is Rs 20 and the cost of 1 kg onion is 2x +5 =2ร20 +5 = Rs 45

**You can compensate us by donating any amount of money for our survival**

**Our Paytm No 9891436286**

**NCERT Solutions ofย Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 scienceย **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |