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

# 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 โQuadratic equationsโ

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

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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 – โ3

(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 – โ3 then other root must be its cojugate i.e 5 + โ3

The sum ofย  of the roots = (5 – โ3) +(5 + โ3)=10

The produtofย  of the roots = (5 – โ3) (5 + โ3)=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)ยฑโ8

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 -4ร8ร2 =0

kยฒ -64=0 โ kยฒ =ยฑ64โk=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 =โ16 =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/aโฮฑ=-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

โด 6xร6(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 6ร5=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 – 4ร5ร5 = 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 4x2 – โ3x + 5 = 0 by the method of completing the square?

(a) 1/4ย  ย  ย  (b)3/4ย  ย  (c) โ3/4ย  ย  ย (d)3/16

Ans.(d)3/16

The given quadratic equation is 4x2 – โ3x + 5 = 0

Rewriting it

(2x)2 – โ3x + 5 = 0

Comparing it (a +b)ยฒ =aยฒ +bยฒ + 2ab

2ab =โ3x

Replacing a =2x

2ร2xรb = โ3x

b = โ3/4

Therefore we have to add and subtract (โ3/4)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=8ร7=56 and 8(7 +1)=8ร8 =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

X2 โ 7x + 10 = 40

X2 โ 7x + 10-40=0

X2 โ 7x -30=0

X2 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 Sonuโs 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

12x2 + 30x โ 5400 =0

2xยฒ + 5x – 900 =0

x = [-5 ยฑโ(5ยฒ -4ร2ร-900]/(2ร2) = [-5 ยฑโ(25+7200)]/(2ร2)= [-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 =2ร20 +5 = Rs 45

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