Class 10 Maths Full Syllabus Practice Paper

Q1. If a = (2² × 3³ × 5⁴) and b = (2³ × 3² × 5) then HCF (a, b) = ?
a) 540 b) 360
c) 180 d) 90 –

Answer: (c) 180
Explanation:
It is given that: a = (2² × 3³ × 5⁴) and b = (2³ × 3² × 5)
∴ HCF (a, b) = Product of smallest power of each common prime factor in the numbers = 2² × 3² × 5 = 180

Q2. 2 - √3 is
a) an integer b) a whole number
c) an irrational number d) a rational number

Answer: (c) an irrational number
Explanation: –
Let 2 - √3 be rational number
– p
2 - √3 = q where p and q are composite numbers
– p
√3 = q + 2
– (p+2q)
√3 = q
(p+2q) since p, q are integers, so q is rational –
∴√3 is an irrational number
it shows our supposition was wrong –
hence 2-√3 is an irrational number.

Q3. If the prime factorisation of 2520 is 2³ × 3a × b × 7, then the value of a + 2b is:
a) 9 b) 10
c) 7 d) 12

Answer: (d) 12
Explanation:
2520 = 2³ × 3² × 5 × 7
on comparing
a = 2, b = 5
So,
a + 2b = 2 + 2 × 5
= 12

Q4. If 3825 = 3x × 5y × 17z, then the value of x + y - 2z is:
a) 1 b) 3
c) 0 d) 2

Answer: (d) 2
Explanation:
3825 = 3² × 5² × 17
On comparing
x = 2, y = 2, z = 1
x + y - 2z = 2 + 2 - 2 × 1
= 4 - 2
= 2

Q5. The prime factorisation of 1728 is
a) 2⁵ × 3⁴ b) 2⁶ × 3²
c) 2⁵ × 3³ d) 2⁶ × 3³

Answer: (d) 2⁶ × 3³
Explanation: 2⁶ × 3³

Q6. If p and p are two odd prime numbers such that p > p , then p² − p² is 1 2 1 2 1 2
a) an odd prime number b) a prime number
c) an odd number d) an even number

Answer: (d) an even number
Explanation:
Let p₁ and p₂ be 5 two odd primes.
Then,
p² − p² = (p − p )(p + p )
1 2 1 2 1 2 We know that sum and difference of two odd numbers is even
∴ (p₁ − p₂) and (p₁ + p₂) are even numbers.
Also, we know that product of even numbers is an even number, therefore
p² − p² = (p₁ − p₂)(p₁ + p₂), is an even number.
1 2

Q7. Prime factorisation of 424 is:
a) 2³ × 53 b) 2⁴ × 53
c) 2 × 53 × 2 d) 2 × 53 × 4

Answer: (a) 2³ × 53
Explanation: 2 424 2 212 2 106 53 53 1
424 = 2³ × 53

Q8. If HCF (26,169) = 13, then LCM (26,169) =
a) 338 b) 52
c) 13 d) 26

Answer: (a) 338
Explanation:
HCF (26, 169) = 13
We have to find the value for LCM (26, 169) We know that the product of numbers is equal to the product of their HCF and LCM.
Therefore,
13(LCM) = 26(169)
26(169)
LCM = ₁₃
LCM = 338

Q9. If the HCF of 72 and 234 is 18, then the LCM (72, 234) is:
a) 936 b) 836
c) 324 d) 234

Answer: (a) 936
Explanation:
(72×234)
LCM (72, 234) = ₁₈ = 936
Therefore, the LCM of (72, 234) is 936.

Q10. Four different electronic devices make a beep after every 30 minutes, 1 hour,1 ¹ hour and 1 hour 45 minutes 2 respectively. All the devices beeped together at 12 noon. They will again beep together at ________.
a) 3 a.m. b) 12 midnight
c) 9 a.m. d) 6 a.m.

Answer: (c) 9 a.m.
Explanation:
L.C.M. (30, 60, 90, 105) = 2²× 3² × 5 × 7
= 1260 mins = 21 hours

Q11. The LCM of two numbers is 1200. Which of the following cannot be their HCF?
a) 500 b) 200
c) 400 d) 600

Answer: (a) 500
Explanation: It is given that the LCM of two numbers is 1200 . We know that the HCF of two numbers is always the factor of LCM. 500 is not the factor of 1200. So this cannot be the HCF.

Q12. 7 × 11 × 13 + 13 is a/an:
a) odd number but not composite b) composite number
c) prime number d) square number

Answer: (b) composite number
Explanation:
We have 7 × 11 × 13 + 13 = 13 (77 + 1) = 13 × 78. Since the given number has 2 more factors other than 1 and itself,
therefore it is a composite number.

Q13. The LCM of smallest 2-digit number and smallest composite number is
a) 20 b) 4
c) 12 d) 40

Answer: (a) 20
Explanation: As we know, the smallest two-digit number is 10 and the smallest composite number is 4. By prime factorisation, we get;
4 = 2 × 2
10 = 2 × 5
Now, LCM of 4, 10 = 2 × 2 × 5 = 20
Therefore, the LCM of the smalles two-digit number and the smallest composite number is 20.
– ₂

Q14. Which of the followings is an irrational number? – ₂ – 2
a) (√2 − 1) b) (2√3 − 1 )
√3
– – (√2+5√2)
c) √2 − (2 + √2) d)
√2

Answer: (a) (√2 − 1)
Explanation: – ₂
(√2 − 1)

Q15. The prime factorisation of the number 5488 is
a) 2⁴ × 7⁴ b) 2⁴ × 7³
c) 2³ × 7⁴ d) 2³ × 7³

Answer: (b) 2⁴ × 7³
Explanation: 2⁴ × 7³

Q16. A quadratic polynomial having zeroes -6 and 0 is:
a) 6(x² - x) b) x(x² + 6)
c) 6x2 - 1 d) 6x (x + 6)

Answer: (d) 6x (x + 6)
Explanation: 6x (x + 6)

Q17. If -2 and 3 are the zeros of the quadratic polynomial x² + (a + 1)x + b then
a) a = -2, b = 6 b) a = -2, b = -6
c) a = 2, b = -6 d) a = 2, b = 6
– ₂

Answer: (b) a = -2, b = -6
Explanation:
α + β = 3 + (−2) = 1 and αβ = 3 × (−2) = −6
∴ -(a + 1) = 1
⇒ a + 1 = -1 ⇒ a = -2
Also, b = -6

Q18. The sum of zeroes of the polynomial √2 x - 17 are given as:
a) 0 b) ¹⁷√²
2
c) 1 d) ¹⁷√²
− ₂

Answer: (a) 0
Explanation: – ₂
Given; P(x) = √2 x - 17
coeff of x
Sum of zeroes = ₂
coeff of x
= 0

Q19. If one root of the polynomial f(x) = 3x² + 11x + p is reciprocal of the other, then the value of p is
a) -3 b) 0
c) 3 d) ¹ 3

Answer: (c) 3
Explanation:
Let one root be q.
∴ Other root = ¹
q 1 p p
⇒ q × q = ₃ ⇒ 1 = ₃ ⇒ p = 3

Q20. If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having ¹ and ¹ is its zero is
α β
a) qx² + px + 1 b) x² − px + q
c) d) x² + qx + p px² + qx + 1

Answer: (a) qx² + px + 1
Explanation:
Let α and β be the zeros of the polynomial f(x) = x² + px + q .Then,
− Coefficient of x p
α + β = ₂ = − ₁ = −p
Coefficient of x Constant term q
And αβ = ₂ = ₁ = q
Coefficient of x
Let S and R denote respectively the sum and product of the zeros of a polynomial whose zeros are ¹ and ¹ , then
α β 1 1 α+β −p
S = α + β = αβ = q
R = ¹ × ¹ = ¹ = ¹
α β αβ q
Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by
x² − Sx + R = 0
x² + P x + ¹ = 0
q q qx²+Px+1
q = 0
⇒ qx² + px + 1
So g(x) = qx² + px + 1

Q21. If α, β are the zeros of polynomial f(x) = x² − p (x + 1) − c, then (α + 1) (β + 1) =
a) c − 1 b) c
c) 1 − c d) 1 + c

Answer: (c) 1 − c
Explanation:
Since α and β are the zeros of quadratic polynomial f(x) = x² − p(x + 1) − c
= x² − px − p − c
− Coefficient of x
α + β = ₂
Coefficient of x −p
= − ( ₁ ) = p
Constant term
α × β = ₂
Coefficient of x −p−c
= ₁ = −p − c
We have (α + 1)(β + 1)
= αβ + β + α + 1
= αβ + (α + β) + 1
= −p − c + (p) + 1
= −c + 1
= 1 - c
The value of (α + 1)(β + 1) is 1 - c.

Q22. The number of polynomials having zeros 1 and -2 is
a) more than 3 b) 2
c) 3 d) 1

Answer: (a) more than 3
Explanation: Since, 1 and -2
Sum of Roots = 1+(-2) = -1
Product of roots = (1) (-2) = -2
Therefore, the polynomial (p(x)) is: [p(x) =K[x² - (sum of roots)x + product of roots]
p(x) = K [ x² - (-1)x+ (-2)]
Therefore, There are infinitely many polynomials that can have (1) and (-2) as their zeros. We can multiply or divide the
polynomial by any nonzero constant(k), and the zeros will remain the same. So, the required number of polynomials is infinite!

Q23. The zeroes of the polynomial 3x² + 11x - 4 are: −1 1
a) , -4 b) , -4 3 3 −1 1
c) , 4 d) , 4 3 3

Answer: (b) ¹ , -4 3
Explanation:
Let f(x) = 3x² + 11x - 4
f(x) = 3x² + 12x - x - 4
f(x) = 3x(x + 4) - 1 (x + 4)
f(x) = (x + 4)(3x - 1)
Put both the factors equal to zero.
x + 4 = 0, x = -4
1
3x - 1 = 0, x = ₃
The zeroes of the polynomial 3x² + 11x - 4 are ¹ and - 4. 3

Q24. A quadratic polynomial with sum and product of its zeros as 8 and -9 respectively is
a) x² + 8x - 9 b) x² - 8x - 9
c) x² - 8x + 9 d) x² + 8x + 9 ₁ –

Answer: (b) x² - 8x - 9
Explanation:
Given,
α + β = 8
αβ = -9
p(x) = k(x² - (α + β)x + αβ)
= k(x² - (8)x + (-9))
= k(x² - 8x - 9)
for k = 1,
p(x) = x² - 8x - 9

Q25. A quadratic polynomial whose product and sum of zeroes are and √2 respectively is
3
a) ₂ – b) ₂ –
3x - x + 3√2x 3x + x - 3√2x
c) ₂ – d) ₂ –
3x + 3√2x + 1 3x - 3√2x + 1

Answer: ₂ –
(d) 3x - 3√2x + 1
Explanation:
√2 −(−√2) −(−3√2)
Given: α + β = ₁ = ₁ = ₃
c 1 –
And αβ = a = ₃ On comparing, we get, a = 3, b = −3√2, c = 1
Putting these values in the general form of a quadratic polynomial ax² + bx + c, ₂ –
we have 3x - 3√2 + 1

Q26. If one root of the polynomial p(y) = 5y² + 13y + m is reciprocal of other, then the value of m is
a) 5 b) 6
c) ¹ d) 0 5 2 α β

Answer: (a) 5
Explanation:
p(y) = 5y² + 13y + m.
Given one root of p(x) is reciprocal of other
1
i.e. If α = a then β = a
−b
sum of roots (α + β) = a
a + ¹ = − ¹³
a 5
Product of roots (α ⋅ β) = c
a 1 m
a ⋅ a = ₅ .
m
1 = ₅ .
m = 5

Q27. If α and β are the zeroes of the polynomial ax + bx + c, then the value of + α is β b² c²
a) ac b) ab b²−2ac a²
c) d) ac bc

Answer: b²−2ac
(c) ac
Explanation: Since α²+β²
= αβ
(α+β)²−2αβ
=
αβ −b 2 c ( a ) −2× a
= c
a b2 2c 2 − a
= a
c a b²−2ac a
= ₂ × c
a b²−2ac
= ac

Q28. If α and β are the zeroes of the quadratic polynomial p(x) = x² - ax - b, then the value of α² + β² is:
a) b² + 2a b) a² - 2b
c) b² - 2a d) a² + 2b

Answer: (d) a² + 2b
Explanation:
Given, P(x) = x² - ax - b
α + β = a, αβ = -b
(α + β)² = α²+β² + 2αβ
a² = α²+β² - 2b
α²+β² = a² + 2b

Q29. The sum and product of the zeroes of the polynomial x² - 6x + 8 are respectively −3
a) 6 and 8 b) ₂ and – 1 −3 3
c) ₂ and 1 d) ₂ and 1

Answer: (a) 6 and 8
Explanation: −b 6
Sum of the zeroes of the polynomial = a = ₁ = 6
And Product of the zeroes of the polynomial = c = ⁸ = 8
a 1

Q30. The polynomial having zeroes -3 and 4 is:
a) x² - x - 12 b) x² - 2x + 1
c) x² + 2x - 1 d) x² + 2x + 1

Answer: (a) x² - x - 12
Explanation: A quadratic polynomial is always in the form of x² - (sum of zeros)x + (product of Zeros) hence the required polynomial is x² - (1)x + (-12)
= x² - x - 12

Q31. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of ‘k’ is
a) – 10 b) 10
c) 5 d) – 5 3 −1

Answer: (a) – 10
Explanation:
Given Polynomial is p(x) = x² + 3x + k
According to question, p(x) = 0 (Put x = 2)
p(2) = 0
2
⇒(2) + 3 × 2 + k = 0
⇒4 + 6 + k = 0
⇒k = −10

Q32. A quadratic polynomial whose zeros are ₅ and ₂ , is
a) 10x² - x -3 b) 10x² - x + 3
c) 10x² + x + 3 d) 10x² + x - 3

Answer: (a) 10x² - x -3
Explanation: 3 1 1 3 −1 −3
α + β = ( ₅ − ₂ ) = ₁₀ , αβ = ₅ × ( ₂ ) = ₁₀
2 1 3 2 Required polynomial is x − ₁₀ x − ₁₀ , i.e., 10x - x - 3

Q33. If one zero of the polynomial f(x) = (k² + 4)x² + 13x + 4k is reciprocal of the other, then k =
a) 1 b) −2
c) 2 d) −1

Answer: (c) 2
Explanation:
We are given f(x) = (k² + 4)x² + 13x + 4k then
− Coefficient of x
α + β = ₂
Coefficient of x −13
= ₂
k +4 Constant term
α × β = ₂
Coefficient of x
= 4k
k²+4 One root of the polynomial is reciprocal of the other. Then, we have
α × β = 1
⇒ ⁴k = 1
k²+4
⇒ (k − 2)² = 0
⇒ k² − 4k + 4 = 0
⇒ k = 2

Q34. The zeroes of the polynomial p(x) = x² + 3x + 2 are given as.
a) -2, 1 b) 2, -1
c) 1, 2 d) -2, -1

Answer: (d) -2, -1
Explanation:
P(x) = x² + 3x + 2= 0
x² + 2x + x + 2 = 0
x(x + 2) + 1(x + 2) = 0
(x + 1)(x + 2) = 0
x = -1, -2
hence, -1 & -2 are the zero of P(x)

Q35. The sum and product of zeroes of the polynomial p(x) = 3x² - 5x + 2 are
2 −5 −2
a) 1, ₃ b) ₃ , ₃ −5 2 5 2
c) ₃ , ₃ d) ₃ , ₃

Answer: (d) ⁵ , ² 3 3
Explanation:
Let α, β be the zero of Polynomial P(x)
P(x) = 3x2 - 5x + 2
−b −(−5) 5
α + β = a = ₃ = ₃
c 2
αβ = a = ₃

Q36. At the end of the year 2002, Sam was half as old as his grandfather. The sum of the years in which they were born is 3854. Age of Sam at the end of year 2003 is ________.
a) 36 years b) 51 years
c) 50 years d) 35 years

Answer: (b) 51 years
Explanation:
Let the year in which Sam was born be x and the year in which Sam's grandfather was born be y.
2002−y
Then, according to question, 2002 - x = ₂
⇒ 2x - y = 2002 ...(i)
and x + y = 3854 ....(ii)
Solving (i) and (ii), we get ⇒ x = 1952
Thus in 2003, Sam's age would be 2003 - 1952 = 51 yrs

Q37. Graphically, the pair of linear equations 3x - y + 8 = 0 and 3x - y = 24 represents two lines which are:
a) intersecting exactly at two points b) coincident
c) parallel d) intersecting exactly at one point 2x y 1 x 2y

Answer: (c) parallel
Explanation: parallel

Q38. If − + = 0 and + = 3 then
3 2 6 2 3
a) x = -2, y = 3 b) x = - 2, y = -3
c) x = 2, y = 3 d) x = 2, y = -3

Answer: (c) x = 2, y = 3
Explanation: We have, 2x y 1
₃ − ₂ = − ₆ …(i)
x 2y
₂ + ₃ = 3 …(ii)
Now, multiplying (i) and (ii) by 6 we get:
4x - 3y = - 1 …(iii)
3x + 4y = 18 …(iv)
Now, multiplying (iii) by 4 and (iv) by 3 and adding them we get:
16x + 9x = - 4 + 54
x = ⁵⁰ = 2
25 Putting the value of x in (iv) we get:
3 × 2 + 4y = 18
18−6
y = ₄
y = 3

Q39. The value of k, if (6, k) lies on the line represented by x - 3y + 6 = 0, is
a) 4 b) -12
c) 12 d) -4

Answer: (a) 4
Explanation:
x - 3y + 6 = 0
6 - 3k + 6 = 0
⇒ k = 4

Q40. The value of k for which the pair of equations kx = y + 2 and 6x = 2y + 3 has infinitely many solutions,
a) is k = -3 b) does not exist
c) is k = 3 d) is k = 4

Answer: (b) does not exist
Explanation: does not exist

Q41. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction 1 becomes ₂ , then the fraction is 8 5
a) ₇ b) ₇ 6 4
c) ₇ d) ₇

Answer: (b) ⁵ 7
Explanation:
Let the fraction be x
y Where x is numerator and y be denominator.
ATQ. x + y = 12 ...(i)
again New denominator is y + 3
ATQ. x = ¹
y+3 2
⇒ 2x = y + 3
using 2x - y = 3 ...(ii)
By, Elimination method Add eq (i) & (ii) we get
x = ¹⁵
3
x = 5
put the value of x in eq. (i) we get
5 + y = 12
y = 12 - 5 = 7.
y = 7
Hence Numerator = 5 denominator = 7.
fraction is ⁵ . 7

Q42. The solution of the pair of equations x + y = a + b and ax - by = a² - b² is:
a) x = -a, y = b b) x = b, y = a
c) x = a, y = b d) x = a, y = -b

Answer: (c) x = a, y = b
Explanation: The given equations are
x + y = a + b ...(i)
ax - by = a² - b² ...(ii)
From (i)
y = a + b - x
Substituting y = a + b - x in (ii), we get
ax - b(a + b - x) = a² - b²
⇒ ax - ab - b² + bx = a² - b²
a²+ab
⇒ x = a₊b = a
Now, substitute x = a in (i) to get
a + y = a + b
⇒ y = b
Hence, x = a and y = b.

Q43. The area of the triangle formed by the lines 2x + 3y = 12 with the co – ordinate axis is
a) 12 sq. units b) 20 sq. units
c) 10 sq. unit d) 16 sq. units

Answer: (a) 12 sq. units
Explanation:
The triangle formed by the lines 2x + 3y = 12 with co-ordinate axes is shaded.
The area of the shaded region, i.e., 2x + 3y = 12
Triangle OAB = ¹ × OA × AB
2
= ¹ × 6 × 4 = 12 sq. units
2 x 0 3 6 y 4 2 0

Q44. Which of the following graphs represent the lines 2x + 4y = 8 and 3x - 4y = 12?
a) b)
c) d)

Answer: (d)
Explanation: 15

Q45. In a given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes ¹ . If 7 is 2 subtracted from the numerator and 2 is subtracted from the denominator, it becomes ¹ . The fraction is 3 15 13
a) ₂₆ b) ₂₄ 16 16
c) ₂₁ d) ₂₇

Answer: (a) ₂₆
Explanation: x
Let the fraction be y
According to the question, (x−1) 1
(y+2) = 2
2x - 2 = y + 2
y = 2x - 4 …(i)
And, (x−7) 1
= ₂
(y−2)
3x - 21 = y - 2
3x = y + 19 …(ii)
Using (i) in (ii)
3x = 2x - 4 + 19
x = 15
Using value of x in (i), we get
y = 2 (15) - 4
y = 30 - 4
y = 26
15
Therefore, required fraction = ₂₆

Q46. If (-3, 2) is a solution of the linear equation 5x + 3 ky = 3, then the value of k is ________.
a) 6 b) 3
c) 5 d) 2

Answer: (b) 3
Explanation:
Since, (-3,2) is the solution of 5x + 3/cy = 3. So (-3, 2) satisfies it.
∴ 5 x (-3) + 3
18
⇒ −15 + 6k = 3 ⇒ k = ₆ = 3

Q47. The area of the triangle formed by the lines x = 3, y = 4 and x = y is
a) 3sq. unit b) 1/2 sq. unit
c) 2sq. unit d) 1 sq. unit

Answer: (b) 1/2 sq. unit
Explanation:
Given x = 3, y = 4 and x = y
We have plotting points as (3,4), (3,3), (4,4) when x = y
Therefore, area of △ABC = ¹ (Base × Height) = ¹ (AB × AC) = ¹ (1 × 1) = ¹
2 2 2 2 Area of triangle ABC is ¹ square units. 2

Q48. If a pair of linear equation is consistent, then the lines will be
a) parallel b) intersecting or coincident
c) always coincident d) always intersecting

Answer: (b) intersecting or coincident
Explanation: If a consistent system has an infinite number of solutions, it is dependent. When you graph; the equations, both equations represent the same line. So for consistent line it has to be parallel or even they intersect at one point. If a system has no solution, it is said to be inconsistent. The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

Q49. Graphically, the pair of equations -6x - 2y = 21 and 2x - 3y + 7 = 0 represents two lines which are:
a) intersecting exactly at two points b) intersecting exactly at one point
c) coincident d) parallel

Answer: (b) intersecting exactly at one point
Explanation: intersecting exactly at one point

Q50. The pair of linear equations x + 2y + 5 = 0 and -3x - 6y + 1 = 0 has:
a) a unique solution b) exactly two solutions
c) infinitely many solutions d) no solution

Answer: (d) no solution
Explanation:
Here, a₁ = 1, b₁ = 2, c₁ = 5
a₂ = -3, b₂ = -6, c₂ = 1
a1 1 1
So, a = −₃ = -( ₃ )
2 b1 2 1
b = −₆ = -( ₃ )
c² 1 5
c₂ = 1
a₁ b₁ c₁
a = b ≠ c
2 2 2
Therefore, the pair of equations has no solution.

Q51. If the lines represented by equations 3x + 2my = 2 and 2x + 5y + 1 = 0 are parallel, then the value of m is:
3 15
a) ₂ b) ₄ 5 2
c) − ₄ d) ₅

Answer: 15
(b) ₄
Explanation: Condition for the lines to be parallel is a₁ b₁ c₁
a = ≠ c
2 b2 2 Here the equations are
3x + 2my = 2 and 2x + 5y + 1 = 0
So, a₁ = 3, b₁ = 2m, c₁ = -2 and a₂ = 2, b₂ = 5, c₂ = 1
a₁ 3 b₁ 2m c₁ −2
∴ a = ₂ , = ₅ and c = ₁ = -2
2 b2 2 2m 3
∴ ₅ = ₂
15
∴ m = ₄

Q52. If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
a) always coincident b) intersecting or coincident
c) always intersecting d) parallel

Answer: (b) intersecting or coincident
Explanation: If a pair of linear equations in two variables is consistent, then its solution exists.
∴ The lines represented by the equations are either intersecting or coincident.

Q53. If the pair of equations 3x - y + 8 = 0 and 6x - ry + 16 = 0 represent coincident lines, then the value of r is:
a) − ¹ b) ¹ 2 2
c) -2 d) 2

Answer: (d) 2
Explanation: a₁ b₁ c₁
a = b = c
2 2 2 3 −1 8
⇒ 6 = −k = 16
Taking, 3 −1
6 = −k
⇒ ¹ = ¹
2 k
⇒ k = 2
−1 8
−k = 16
⇒ ¹ = ¹
k 2
⇒ k = 2
So, the answer is k = 2

Q54. The value of k for which the pair of linear equations 5x + 2y - 7 = 0 and 2x + ky + 1 = 0 don't have a solution, is:
5
a) 5 b) ₄ 4 5
c) ₅ d) ₂

Answer: (c) ⁴ 5
Explanation: For no solution a₁ b₁ c₁
a = b ≠ c
2 2 2
5 = 2
2 k
k = ⁴
5

Q55. If the system 6x – 2y = 3, kx – y = 2 has a unique solution, then
a) k = 3 b) k ≠ 3
c) k ≠ 4 d) k = 4

Answer: (b) k ≠ 3
Explanation: a₁ b₁
If the system has a unique solution, then a ≠ b
2 2
Here a₁ = 6, a₂ = k, b₁ = −2
and b₂ = −1
6 −2
∴ k ≠ −₁ ⇒3k ≠ 6 ⇒k ≠ 3
2k ≠ 6
k ≠ 3

Q56. Which of the given is a quadratic equation?
a) x + ¹ = x² b) 2x² − 5x = (x − 1)²
x ₂ −− 1
c) x − 3√x + 2 = 0 d) x + ₂ = 5
x

Answer: (b) 2x² − 5x = (x − 1)²
Explanation:
2x² − 5x = (x − 1)² using (a − b)² = a² + b² − 2ab
2x² − 5x = x² − 2x + 1
2x² − 5x − x² + 2x − 1 = 0
x² − 3x − 1 = 0
a = 1, b = -3 and c = -1
This is of the form ax² + bx + c = 0 i.e. of degree 2(a ≠ 0, a, b, c are real numbers)
Hence this is a quadratic equation.

Q57. If p and q are the roots of the equation x² + px + q = 0, then
a) p = - 2, q = 0 b) b = 0, 9 = 1
c) p = 1, q = - 2 d) p = - 2, q = l

Answer: (c) p = 1, q = - 2
Explanation:
Given sum of roots, S = p + q = – p and product pq = q
⇒ q(p – 1) = 0 i.e. q = 0 or p = 1
Now If q = 0 then p = 0, this implies p = q
If p = 1, then p + q = – p
q = – 2p
q = – 2(1)
q = – 2

Q58. The roots of the quadratic equation x² - 4 = 0 is/are:
a) 2 only b) -2, 2
c) -4, 4 d) 4 only

Answer: (b) -2, 2
Explanation:
x² - 4 = 0
x² = 4
x = ±2
roots are +2, -2

Q59. If α and β are the roots of ax² + bx + c = 0, then the wrong statement is
c b
a) αβ = a b) α + β = a
2 2 b²−2ac 1 1 −b
c) α + β = ₂ d) α + β = c
a

Answer: (b) α + β = b
a
Explanation:
If α and β are the roots of ax² + bx + c = 0,
−b
then α + β = a

Q60. If x² + 5kx + 16 = 0, has equal roots, then the value of k is
25 64
a) ± ₆₄ b) ± ₂₅ 8 5
c) ± ₅ d) ± ₈

Answer: (c) ± ⁸ 5
Explanation:
Here, a = 1, b = 5k, c = 16
If x² + 5kx + 16 = 0 has equal roots,
then, b² - 4ac = 0
⇒ (5k)² - 4 × 1 × 16 = 0
⇒ 25k² - 64 = 0
⇒ 25k² = 64
2 64
⇒ k = ₂₅
8
⇒ k = ± ₅

Q61. If the roots of 5x² -kx + 1 = 0 are real and distinct then
– – –
a) −2√5 < k < 2√5 b) k < −2√5 only
– – –
c) either k > 2√5 or k < −2√5 d) k > 2√5 only

Answer: – –
(c) either k > 2√5 or k < −2√5
Explanation:
The roots of 5x² - kx + 1 = 0 are real and distinct.
∴ (k² − 4 × 16) > 0 ⇒ k² − 20 > 0
– –
This gives; k < −2√5 and k > 2√5

Q62. If the equation x² + 2(k + 2)x + 9k = 0 has equal roots then k = ?
a) -1 or 4 b) 1 or - 4
c) 1 or 4 d) -1 or - 4

Answer: (c) 1 or 4
Explanation:
Since the roots are equal, we have D = 0.
∴ 4(k + 2)² - 36k = 0 ⇒ {k + 2)² - 9k = 0
k² - 5k + 4 = 0 ⇒ k² - 4k - k + 4 = 0
⇒ k(k − 4) − (k − 4) = 0
⇒ (k − 4)(k − 1) = 0 ⇒ k = 4 or k = 1 .

Q63. The perimeter of a right triangle is 70cm and its hypotenuse is 29cm. The area of the triangle is
a) 200 sq.cm b) 180 sq.cm
c) 210 sq.cm d) 250 sq.cm

Answer: (c) 210 sq.cm
Explanation:
Let base of the right triangle be x cm.
Given: Perpendicular = x + 29 = 70 ⇒ Perpendicular = (41 − x) cm
Now, using Pythagoras theorem,
(29)² = x² + (41 − x)²
⇒841 = 1681 + x² − 82x + x²
⇒2x² − 82x + 840 = 0
⇒x² − 41x + 420 = 0
⇒x² − 20x − 21x + 420 = 0
⇒x (x − 20) − 21 (x − 20) = 0
⇒(x − 20) (x − 21) = 0
⇒x − 20 = 0 and x − 21 = 0
⇒x = 20 and x = 21
Therefore, the two sides other than hypotenuse are of 20 cm and 21 cm.
∴ Area of right triangle = ¹ × Base × Perpendicular = ¹ × 20 × 21 = 210 sq. cm
2 2

Q64. If one root of the equation x² + ax + 3 = 0 is 1, then its other root is
a) 2 b) 3
c) -3 d) -2

Answer: (b) 3
Explanation:
The given equation is x² + ax + 3 = 0
One root = 1
and product of roots = c = ³ = 3
a 1
Second root = ³ = 3
1

Q65. Which of the following is not a quadratic equation?
a) 2(x − 1)² = 4x² − 2x + 1 b) x = x² + 3 + 4x²
– – 2 2 2 2 2
c) (√2x + √3) + x = 3x − 5x d) 2x − x = x + 5

Answer: – – 2 2 2
(c) (√2x + √3) + x = 3x − 5x
Explanation: – – 2 2 2
In equation (√2x + √3) + x = 3x − 5x
2 – 2 2
⇒2x + 3 + 2√6x + x = 3x − 5x
2 2 –
⇒3x − 3x + 5x + 2√6x + 3 = 0
–
⇒(5 + 2√6)x + 3 = 0
It is not the quadratic equation because its degree is not 2.

Q66. The perimeter of a rectangle is 82 m and its area is 400 m². The breadth of the rectangle is
a) 20 m b) 9 m
c) 16 m d) 25 m

Answer: (c) 16 m
Explanation:
2(l + b) = 82 ⇒ l + b = 41 ⇒ l = (41 - b).
And, lb = 400 ⇒ (41 - b)b = 400
⇒ b² - 41b + 400 = 0 ⇒ b² - 25b - 16b + 400 = 0
⇒ b(b - 25) - 16(b - 25) = 0
⇒ (b - 25)(b - 16) = 0
∴ b = 25 or b = 16.
But for b = 25 we have l = (41 - 25) = 16 < b.
∴ breadth = 16 m.

Q67. The value(s) of k for which the quadratic equation 3x² - kx + 3 = 0 has equal roots, is (are)
a) -6 b) ±6
c) 6 d) 9

Answer: (b) ±6
Explanation: For equal roots
D = 0
b² - 4ac = 0
(-k)² - 4(3)(3) = 0
k² - 36 = 0
k² = 36
k = ±6

Q68. The values of k for which the quadratic equation 2x² – kx + k = 0 has equal roots is
a) 0, 8 b) 8 only
c) 0 only d) 4

Answer: (a) 0, 8
Explanation:
If a quadratic equation ax² + bx + c = 0, a ≠ 0 has two equal roots, then its discriminant value will be equal to zero i.e., D = b² -
4ac = 0
Given, 2x² – kx + k = 0
For equal roots,
D = b² - 4ac = 0
⇒ (-k)² - 4(2)(k) = 0
⇒ k² - 8k = 0
⇒ k (k - 8) = 0
∴ k = 0,8

Q69. Which of the following equations has 2 as a root?
a) x² + 3x – 12 = 0 b) x² – 4x + 5 = 0
c) 2x² – 7x + 6 = 0 d) 3x² – 6x – 2 = 0

Answer: (c) 2x² – 7x + 6 = 0
Explanation:
Given, 2x² - 7x + 6 = 0
If 2 satisfies the above equation then 2 is a root.
Now, 2(2)² - 7(2) + 6 = 0
∴ 2 is a root of this equation

Q70. 3x² + 2x - 1 = 0 have
a) No Real roots b) Real roots
c) real and equal root d) Real and Distinct roots

Answer: (d) Real and Distinct roots
Explanation:
D = b² - 4ac
D = 2² - 4 × 3 × (-1)
D = 4 + 12
D = 16
D > 0.
Hence Real and distinct roots.

Q71. If p = -7 and q = 12 and x² + px + q = 0, Then the value of x is
a) -3 and 4 b) -3 and -4
c) 3 and 4 d) 3 and -4

Answer: (c) 3 and 4
Explanation: Putting the values of p and q in given equation, we get
x² + (-7)x + 12 = 0
⇒ x² - 7x + 12 = 0
⇒ x² - 4x - 3x + 12 = 0
⇒ x(x - 4) - 3(x - 4) = 0
⇒ (x - 3)(x - 4) = 0
⇒ x - 3 =0 and x - 4 = 0
⇒ x = 3 and x = 4

Q72. (x² + 1)² – x² = 0 has
a) two real roots b) no real roots
c) one real root. d) four real roots

Answer: (b) no real roots
Explanation:
Let, x² = y, then our given equation become
(y + 1)² − y = 0
⇒ y² + y + 1 = 0
D = b² − 4ac = 1² − 4 × 1 × 1 = 1 − 4 = −3 < 0
Hence no real root.

Q73. The ratio of the sum and product of the roots of the quadratic equation 5x² - 6x + 21 = 0 is:
a) 5 : 21 b) 21 : 5
c) 7 : 2 d) 2 : 7 1 2 5

Answer: (d) 2 : 7
Explanation: −b
Sum of roots = a
−(−6) 6
= ₅ = ₅
product of roots = c
a 21
= ₅
ATQ 6 Sum of roots 5 6
= ₂₁ = ₂₁
prod. of roots 5
= ²
7

Q74. If ₂ is a root of the equation x + kx – ₄ = 0, then the value of k is
a) ¹ b) 2 4
c) -2 d) ¹ 2

Answer: (b) 2
Explanation: 1 2 5 1
If ₂ is a root of the equation x + kx - ₄ = 0 then, substituting the value of ₂ in place of x should give us the value of k.
Given, x² + kx - ⁵ = 0 where, x = ¹
4 2 1 2 1 5
⇒ ( ₂ ) + k( ₂ ) − ₄ = 0
⇒ k = ⁵ − ¹
2 4 4
∴ k = 2

Q75. The roots of the quadratic equation ax² + bx + c = 0 are real and distinct, if:
a) b² - 4ac = 0 b) b² - 4ac > 0
c) b² - 4ac ≥ 0 d) b² - 4ac < 0

Answer: (b) b² - 4ac > 0
Explanation:
A quadratic equation ax² + bx + c = 0 has real and distinct roots, if b² - 4ac > 0.