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

mcq quadratic equation class 10

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.

mcq quadratic equation class 10

Three Ways of Solving Quadratic Equations

CBSE mathematics class X solutions of important questions chapter 4 ‘Quadratic equations’

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 15x² – 7x -2=0 are 

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

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

Factoring the LHS part of the quadratic equation

15x² – 7x -2=0

15x² – 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 – √3

(a) x² +10x +22=0     (b) x² -10x +22=0     (c) x² +10x -22 =0    (d) -x² +10x +22=0

Ans. (b) x² -10x +22=0

The given rational root of quadratic equation is 5 – √3 then other root must be its cojugate i.e 5 + √3

The sum of  of the roots = (5 – √3) +(5 + √3)=10

The produtof  of the roots = (5 – √3) (5 + √3)=25-3=22

Using the following formula for creating the quadratic equation

x² -(sum of  of the roots )x + produtof  of the roots =0

x² -10x +22 =0

Q3. The quadratic equation 8x² +kx +2 =0 has two equal roots then the value of the k is

(a) 8      (b) -8    (c) ±8    (d)±√8

Ans.   (c) ±8

The roots of the  quadratic equation ax²+bx +c=0 are equal when

b² – 4ac =0

Comparing the given equation 8x² +kx +2 =0 with standard equation ax²+bx +c=0

k² -4×8×2 =0

k² -64=0 ⇒ k² =±64⇒k=±8

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) 4,7      (b) 7,4   (c) 4,6  (d)5,6

Ans.(c) 4,6

Let the larger number is = x

Square of larger number = x²

The square of smaller number = x²-20

A t Q

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

x² -20= 2x +4

x² – 2x -20 -4 =0

x² – 2x -24 =0

x² -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 = 6²-20 =36 -20 =16

Smaller number =√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) 7,49  (b) 8,64  (c) 6,36  (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 = x²

x² -22x +110 -5 =0

x² -22x +105 =0

x² -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 7²=49

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

(a) 13,14     (b)15,16   (c) 11,12  (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)²+ x² =365

x² +2x + 1+ x²-365=0

2x² +2x – 364 =0

x² + x – 182 =0

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

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

Ans.  (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

x² + 12x =160

x² + 12x -160 =0

x² + 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 ax² + bx +c =0 are in the ratio of 2:3 ,then

(a) 6a² = 5 c²       (b) 6b² = 5ac     (c) 6 b² = 25ac        (d) 6b² = ac

Ans.(c) 6 b² = 25ac

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

Sum of the roots =2α+ 3α=5α=-b/a⇒α=-b/5a

Product of the roots = 2α× 3α=6α²=c/a=α²=c/6a

(-b/5a)² = c/6a

b²/25a² = c/6a

b².6a = c.25a²

6ab² = 25a²c

6b² = 25ac

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

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

Ans.  (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

∴ 6x×6(x+1) = 1080

x(x +1) =30

x² + x -30 =0

x² + 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 6×5=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) 30      (b)50     (c) 40      (d)20

Ans.(d)20

Let the actual marks of Shreya be x

According to question

x² = 10(x +20)

x² – 10x – 200 =0

x² – 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 5x² – 4x + 5 are

(a) Real and equal    (b)Real and unequal  (c) Not real      (d)Not real but equal

Ans.(c) Not real

The given quadratic equation is 5x² – 4x + 5

The discriminant,D = b² – 4ac

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

D = (-4)² – 4×5×5 = 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 – √3x + 5 = 0 by the method of completing the square?

(a) 1/4      (b)3/4    (c) √3/4     (d)3/16

Ans.(d)3/16

The given quadratic equation is 4x2 – √3x + 5 = 0

Rewriting it

(2x)2 – √3x + 5 = 0

Comparing it (a +b)² =a² +b² + 2ab

2ab =√3x

Replacing a =2x

2×2x×b = √3x

b = √3/4

Therefore we have to add and subtract (√3/4)²= 3/16

Q13. One of the root of the quadratic equation 3x²- 2k +5 =0 is 1/3,find the value of k.

(a) 4      (b)3    (c) 5    (d)8

Ans.(d)8

The given quadratic equation is 3x²- 2kx +5 =0

The given root is 1/3

Therefore putting x = 1/3 in given quadratic equation

3(1/3)²- 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) 4      (b)3    (c) 7    (d)8

Ans.(c) 7

Let the natural number is x

As per the question

x + 1/ x = 50/7

x² + 1 =50x/7

7x² +7 =50x

7x² – 50x +7 = 0

Factorizing it

7x² – 50x +7 = 0

7x² – 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) 40,48      (b)72,80    (c) 56,64   (d)80,88

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

As per the question

8x × 8(x +1) =3584

64 x² + 64x – 3584 =0

x² + x – 56 =0

x² + 8x – 7x-56 =0

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

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

x = -8, 7

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

Therefore  the successive multiple of 8 are 8x=8×7=56 and 8(7 +1)=8×8 =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) 8,11      (b) 10,13       (a) 15,18       (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 – 5) = 40

(x – 5)(x -2) = 40

X2 – 7x + 10 = 40

X2 – 7x + 10-40=0

X2 – 7x -30=0

X2 – 10x + 3x – 30 =0

X(x -10) + 3(x – 10) =0

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

X = 10,-3

Therefore present age of Sonu’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 kg  onion is Rs 5 more than the  twice of the cost of 1 kg potato ,then the cost of 1 kg potato and 1 kg onion is

(a) Rs 20,Rs 45      (b) Rs 30,Rs 65    (a) Rs 25,Rs 55    (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 – 5400 =0

2x² + 5x – 900 =0

x = [-5 ±√(5² -4×2×-900]/(2×2) = [-5 ±√(25+7200)]/(2×2)= [-5 ±√(7225)]/4 =(-5 ±85)/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 =2×20 +5 = Rs 45

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