Class 10 Maths Full Syllabus Practice Paper 2

Q1. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number is:
a) 36 b) 45
c) 81 d) 9

Answer: (c) 81
Explanation:
Let the two numbers be x and y.
It is given that:
x = 54
HCF = 27
LCM = 162
We know,
x × y = HCF × LCM
⇒ 54 × y = 27 × 162
⇒ 54y = 4374
4374
⇒ ∴ y = ₅₄ = 81

Q2. If 1080 = 2x × 3y × 5, then (x - y) is equal to:
a) 1 b) 0
c) 6 d) -1

Answer: (b) 0
Explanation:
1080 = 2³ × 3³ × 5
On comparing
x = 3, y = 3
x - y = 3 - 3 = 0

Q3. If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
a) 1 b) 3
c) 2 d) 4

Answer: (d) 4
Explanation:
LCM (a, 18) = 36
HCF (a, 18) = 2
We know that the product of numbers is equal to the product of their HCF and LCM.
Therefore,
18a = 2(36)
2(36)
a = ₁₈
a = 4

Q4. The ratio of HCF to LCM of the least composite number and the least prime number is:
a) 2 : 1 b) 1 : 1
c) 1 : 2 d) 1 : 3

Answer: (c) 1 : 2
Explanation: Least composite number is 4 and the least prime number is 2.
LCM (4, 2) = 4
HCF (4, 2) = 2
The ratio of HCF to LCM = 2 : 4 or 1 : 2.

Q5. If two positive integers m and n can be expressed as m = x²y⁵ and n = x³y², where x and y are prime numbers,
then HCF(m, n) =
a) x³y² b) x²y²
c) x³y³ d) x²y³

Answer: (b) x²y²
Explanation:
x²y⁵ = y³(x²y²)
x³y³ = x(x²y²)
Therefore HCF (m, n) is x²y²

Q6. If p and q are co-prime numbers, then p² and q² are
a) not coprime b) odd
c) coprime d) even

Answer: (c) coprime
Explanation: We know that the co-prime numbers have no factor in common, or, their HCF is 1. Thus, p² and q² have the same factor with exponent 2 each. which again will not have any common factor. Thus we can conclude that p² and q² are co-prime numbers.

Q7. LCM of (2³ × 3 × 5) and (2⁴ × 5 × 7) is
a) 1680 b) 40
c) 1120 d) 560

Answer: (a) 1680
Explanation:
LCM = Product of greatest power of each prime factor involved in the numbers
= 2⁴ × 3 × 5 × 7
= 16 × 3 × 5 × 7
= 1680

Q8. Two tanks contain 504 and 735 litres of milk respectively. Find the maximum capacity of a container which can measure the milk of either tank in exact number of times.
a) 21 litres b) 7 litres
c) 6 liters d) 42 litres

Answer: (a) 21 litres
Explanation:
We have, 504 = 2³ × 3² × 7 and 735 = 3 × 5 × 7².
∴ H.C.F. (504, 735) = (3 × 7) = 21
∴ Capacity of the container = 21 litres.

Q9. ________ is neither prime nor composite.
a) 3 b) 2
c) 1 d) 4

Answer: (c) 1
Explanation: 1 is neither prime nor composite. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself e.g. 5 is prime because 1 and 5 are its only positive integers factors but 6 is composite because it has divisors 2 and 3 in addition to 1 and 6.

Q10. A charitable trust donates 28 different books of Maths, 16 different books of Science and 12 different books of Social Science to poor students. Each student is given maximum number of books of only one subject of their interest and each student got equal number of books.
i. Find the number of books each student got.
ii. Find the total number of students who got books.
a) (i) - (3), (ii) - (10) b) (i) - (4), (ii) - (14)
c) (i) - (4), (ii) - (10) d) (i) - (3), (ii) - (15)

Answer: (b) (i) - (4), (ii) - (14)
Explanation:
i. H.C.F. (28, 16, 12) = 2 × 2 = 4
∴ Number of books each student got = 4
28
ii. Number of students who got Maths books = ₄ = 7
16
Number of students who got Science books = ₄ = 4
Number of students who got Social Science books = ¹² = 3
4
∴ Total number of students who got books = 7 + 4 + 3 = 14.

Q11. 2.35 is
a) a rational number b) a natural number
c) an integer d) an irrational number – – ₂

Answer: (a) a rational number
Explanation: p It can be expressed in q form
2.35 = ²³⁵
100 so, 2.35 is a rational number

Q12. The number (√3 + √5) is
a) an integer b) an irrational number
c) not a real number d) a rational number

Answer: (b) an irrational number
Explanation: – – 2 – 2 – 2 – –
(√3 + √5) = (√3) + (√5) + 2 × √3 × √5
−−
= 3 + 5 + 2√15
−−
= 8 + 2√15
−− – –
Here, √15 = √3 × √5
– – – – 2
Since √3 and √5 both are an irrational number. Therefore, (√3 + √5) is an irrational number.

Q13. The sum of two irrational numbers is always
a) a rational number or an irrational number b) a rational number
c) an integer d) an irrational number

Answer: (a) a rational number or an irrational number
Explanation: The sum of two irrational numbers can be either a rational number or an irrational number. – – – –
e.g 5√3 + 3√2 = 5√3 + 3√2 sum is irrational
– –
(2 + 6√7) + ( - 6√7) = 2 sum is rational
Hence sum can be either rational or irrational

Q14. If p is a prime number, then –p is
√
a) Integer b) Rational
c) Prime number d) Irrational
√5+√2

Answer: (d) Irrational
Explanation: – –
√p is an irrational number because the square root of every prime number is an irrational number. (for example √3 is an
irrational number)

Q15. The number is
√5−√2
a) an integer b) not a real number
c) a rational number d) an irrational number

Answer: (d) an irrational number
Explanation:
√5+√2
√5−√2
√5+√2 √5+√2
= ×
√5−√2 √5+√2
(√5+√2)²
= 2 2
(√5) −(√2)
(√5)²+(√2)²+2×√5×√2
= ₅−₂
5+2+2√10
= ₃
7+2√10
= ₃
−− – –
Here √10 = √2 × √5
– –
Since √2 and √5 both are an irrational number
√5+√2
Therefore, is an irrational number.
√5−√2

Q16. If a is a non-zero rational and √b is irrational, then a√b is:
a) a natural number b) a rational number
c) an integer d) an irrational number –

Answer: (d) an irrational number
Explanation:
If possible let a√b be rational.
p
Then a√b = q , where p and q are non-zero integers, having no common factor other than 1.
p
Now, a√b = q
p
⇒ √b = aq ... (i)
But, p and aq are both rational and aq ≠ 0
p ∵ aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational.

Q17. 2√3 is
a) a whole number b) an irrational number
c) an integer d) a rational number

Answer: (b) an irrational number
Explanation: an irrational number

Q18. The number 1.732 is
a) an integer b) a whole number
c) a rational number d) an irrational number – –

Answer: (c) a rational number
Explanation: Clearly, 1.732 is a terminating decimal.
Hence, it is a rational number.

Q19. The number (5 - 3√5 + √5) is:
a) a whole number b) a rational number
c) an integer d) an irrational number –

Answer: (d) an irrational number
Explanation: an irrational number

Q20. (2 + √2) is
a) an integer b) a rational number
c) A real number d) an irrational number

Answer: (d) an irrational number
Explanation: –
(2 + √2) is an irrational number.
If it is rational, then the difference of two rational is rational. – –
∴ (2 + √2) − 2 = √2 = irrational, which is a contradiction.
–
Hence, (2 + √2), is an irrational number.

Q21. The graph of y = f(x) is shown in the figure for some polynomial f(x).
The number of zeroes of f(x) is
a) 6 b) 4
c) 8 d) 5 ₂ –

Answer: (d) 5
Explanation: Graph of f(x) intersect the x-axis at 5 times.
hence, No. of zeroes of f(x) = 5

Q22. The zeros of the polynomial x − √2x − 12 are
a) 3, -1 b) 3, 1 – – – –
c) 3√2, −2√2 d) √2, −√2

Answer: – –
(c) 3√2, −2√2
Explanation: ₂ – ₂ – –
x − √2x − 12 = x − 3√2x + 2√2x − 12
– – – – –
= x(x − 3√2) + 2√2(x − 3√2) = (x − 3√2)(x + 2√2)
– –
∴ x = 3√2 or x = −2√2

Q23. The graph of y = p(x) is shown in the figure for some polynomial p(x). The number of zeroes of p(x) is/are:
a) 3 b) 0
c) 1 d) 2

Answer: (b) 0
Explanation: 0

Q24. A polynomial of the form ax⁵ + bx³ + cx² + dx + e has atmost ________ zeroes.
a) 5 b) 11
c) 3 d) 7

Answer: (a) 5
Explanation: Since, degree of given polynomial is 5, so ax⁵ + bx³ + cx² + dx + e has atmost 5 zeroes.

Q25. Which of the following is not the graph of a quadratic polynomial?
a) b)
c) d)

Answer: (d)
Explanation: The shape of a quadratic polynomial is either upward or downward U - shaped curve i.e., an upward or downward parabola. Also, the graph of the quadratic equation cuts the X - axis at the most at two points, but in fig it cuts the X - axis at three points.
∴ fig is not the graph of a quadratic polynomial.

Q26. Which of the following graph has more than three distinct real roots?
a) b)
c) d)

Answer: (b)
Explanation: For more than three distinct real roots the graph must cut x-axis at least four times.

Q27. The zeroes of the quadratic polynomial x² + 99x + 127 are
a) both positive b) both equal
c) one positive and one negative d) both negative

Answer: (d) both negative
Explanation: As the Discriminant of the given quadratic polynomial x² + 99x + 127 is more than Zero.
∴ Both the zeros are negative.

Q28. If one root of the polynomial f(x ) = 5x² + 13x + k is reciprocal of the other, then the value of k is
a) ¹ b) 0 6
c) 5 d) 6

Answer: (c) 5
Explanation:
The Given polynomial is f(x) = 5x² + 13x + k.
Product of roots = k/5
k
1 = ₅
⇒ k = 5

Q29. The zeroes of the polynomial p(x) = 25x² - 49 are:
7 7 7 7
a) ₅ , − ₅ b) ₅ , ₅ 49 49 49 49
c) − ₂₅ , + ₂₅ d) ₂₅ , ₂₅

Answer: (a) ⁷ , − ⁷ 5 5
Explanation:
p(x) = 25x² - 49 = 0
= (5x - 7)(5x + 7) = 0
7 −7
∴ x = ₅ and ₅

Q30. For what value of k, the product of zeroes of the polynomial kx² - 4x - 7 is 2? 7 2
a) ₂ b) − ₇ 1 7
c) − ₁₄ d) − ₂

Answer: (d) − ⁷ 2
Explanation:
Product of zeros = c
a (−7)
2 = k
−7
k = ( ₂ )

Q31. If α and β are the zeroes of the polynomial 3x² + 4x - 3, then value of αβ is
a) − ⁴ b) ⁴ 3 3
c) -1 d) 1

Answer: (c) -1
Explanation: -1

Q32. If one zero of the polynomial p(x) = (a² + 9)x² + 45x + 6a is reciprocal of the other, then the value of a is
a) 1 b) 0
c) 2 d) 3

Answer: (d) 3
Explanation:
Let one zero be β then the other zero will be ¹
α c 1 6a
Since αβ = a ⇒α × α = ₂
a +9
⇒ 1 = ⁶a
a²+9
⇒ 6a = a² + 9
⇒ a² - 6a + 9 = 0
⇒ (a - 3)(a - 3) = 0
a - 3 = 0 and a - 3 = 0
⇒ a = 3 and a = 3

Q33. If p(x) = x² + 5x + 6, then p(-2) is:
a) 8 b) 20
c) 0 d) -8

Answer: (c) 0
Explanation:
p(-2) = (-2)² + 5(-2) + 6
p(-2) = 4 - 10 + 6
= 0

Q34. If the sum of the zeroes of the quadratic polynomial kx² + 2x + 3k is equal to their product, then k equals.
a) ² b) − ² 3 3
c) − ¹ d) ¹ 3 3

Answer: (b) − ² 3
Explanation: −2 3k −2 −2
α + β = αβ ⇒ k = k ⇒ k = 3 ⇒ k = ₃

Q35. If am = bl and bn ≠ cm, then the system of equations
ax + by = c
Ix + my = n
a) Has no solution. b) Has a unique solution.
c) Has infinitely many solutions. d) May or may not have a solution.

Answer: (a) Has no solution.
Explanation:
We have, ax + by - c and lx + my = n
Now, a = b ≠ c (given)
l m n
∴The given system of equations has no solution.

Q36. When L₁ and L₂ are coincident, then the graphical solution of system of linear equation have
a) infinite number of solutions b) no solution
c) unique solution d) one solution

Answer: (a) infinite number of solutions
Explanation: When L₁ & L₂ are co-incident, a₁ b₁ c₁
⇒ a = b = c
2 2 2
⇒ infinite number solution.

Q37. If a pair of linear equations has infinitely many solutions, then the lines representing them will be
a) parallel b) always coincident
c) intersecting or coincident d) always intersecting

Answer: (b) always coincident
Explanation: equation has infinite many solutions if a₁ b₁ c₁
a = b = c
2 2 2 i.e. always co-incident.

Q38. For what value of k, do the equations
3x – y + 8 = 0
and 6x – ky = –16
represent coincident lines?
a) ¹ b) –2 2
c) − ¹ d) 2 2

Answer: (d) 2
Explanation: Condition for coincident lines is -
a₁/a₂ = b₁/b₂ = c₁/c₂ …(i)
Given lines are,
3x - y + 8 = 0
and 6x - ky + 16 = 0;
Comparing with the standard form, gives
a₁ = 3, b₁ = - 1, c₁ = 8;
a₂ = 6, b₂ = - k, c₂ = 16;
and, from Eq. (i), ³ = ¹ = ⁸
6 k 16
1 = 1
k 2
So, k = 2

Q39. Graphically, the pair of equations 6x - 3y + 10 = 0, 2x - y + 9 = 0 represents two lines which are
a) coincident b) Intersect at two points
c) parallel d) intersect at a point

Answer: (c) parallel
Explanation:
Given: a₁ = 6, a₂ = 2, b₁ = -3, b₂ = -1, c₁ = 10 and c₂ =9
a₁ = 6, a₂ = 2, b₁ = −3, b₂ = −1, c₁ = 10 and c₂ = 9
a1 6 3 b1 −3 3 c1 10
Here a = ₂ = ₁ , b = −₁ = ₁ , c = ₉
2 2 2 c₁ 10
but c = ₉
2 a₁ b₁ c₁
∵ a = b ≠ c
2 2 2
Therefore, the lines are parallel.

Q40. The ratio of a 2-digit number to the sum of digits of that number is 4 : 1. If the digit in the units place is 3 more than the digit in the tens place, then what is the number?
a) 63 b) 36
c) 24 d) 40

Answer: (b) 36
Explanation:
Let the digit at units place be x and the digit at tens place e be y, then the number = 10y + x
10y+x 4
Now, according to the question, y+x = 1
⇒10y + x = 4y + 4x
⇒6y = 3x⇒x = 2y ...(i)
Also, x = 3 + y⇒2y = 3 + y [From (i)]
⇒y = 3 and x = 6
∴Required number = 36

Q41. Which equation satisfies the data given in the table? x -1 0 1 2
y -3 -1 1 3
a) y = 2x - 1 b) y = x - 2
c) y = x + 1 d) y = 3x - 3
x₊y x₋y –

Answer: (a) y = 2x - 1
Explanation:
y = 2x - 1

Q42. If 2 = 2 = √8 then the value of y is
a) ¹ b) 0 2 3
c) ₂ d) 1

Answer: (b) 0
Explanation: x+y x-y 3/2 3 3
2 = 2 = 2 ⇒ x + y = ₂ and x - y = ₂ . So, by adding above two equations we get and x= y = 0

Q43. In a cyclic quadrilateral ABCD, if ∠A = (2x - 1)o, ∠B = (y + 5)o, ∠C = (2y + 15)o and ∠D = (4x - 7)o, then the
value of ∠C is
a) 65o b) 55o
c) 115o d) 125o

Answer: (c) 115o
Explanation: Since the sum of the opposite angles of a cyclic quadrilateral is 180o
∴ ∠A + ∠C = 180o
⇒ 2x - 1 + 2y + 15 = 180o
⇒ x + y = 83o ... (i)
And ∠B + ∠D = 180o
⇒ y + 5 + 4x - 7 = 180o
⇒ 4x + y = 182o ... (ii)
Subtracting eq. (ii) from eq. (i),
we get -3x = -99o
⇒ x = 33o
Putting the value of x in eq. (i),
we get 33o + y = 83o
⇒ y = 50o
∴ ∠C = (2y + 15)o = (2 × 50 + 15)o = 115o

Q44. The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is
a) 87 b) 96
c) 69 d) 78

Answer: (d) 78
Explanation:
Let us assume the tens and the unit digits of the required number be x and y respectively
∴ Required number = (10x + y)
According to the given condition in the question, we have
x + y = 15 .....(i)
By reversing the digits, we obtain the number = (10y + x)
∴ (10y + x) = (10x + y) + 9
10y + x - 10x - y = 9
9y - 9x = 9
y - x = 1 .....(ii)
Now, on adding (i) and (ii) we get:
2y = 16
∴ y = ¹⁶ = 8
8 Putting the value of y in (i), we get:
x + 8 = 15
x = 15 - 8
x = 7
∴ Required number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78

Q45. In △ABC, if ∠C = 3∠B = 2(∠A + ∠B), then ∠C =
a) 120o b) 60o
c) 150o d) 90o

Answer: (a) 120o
Explanation:
Since ∠A + ∠B + ∠C = 180o ... (i)
∠C = 3∠B = 2(∠A+∠B)
3∠B = 2(∠A+∠B)
3∠B - 2∠B = 2∠A
∠B = 2∠A
∠A = ∠ B
2 from (i),
∠ B + ∠B + 3∠B = 180o
2
9∠ B = 180o
2
∠B = 40o
∠C = 3∠B
∠C = 3 × 40 = 120o

Q46. If 29x + 37y = 103 and 37x + 29y = 95 then
a) x = 2, y = 1 b) x = 3, y = 2
c) x = 1, y = 2 d) x = 2, y = 3

Answer: (c) x = 1, y = 2
Explanation:
29x + 37y=103 .......(i)
37x+29y=95 .........(ii)
Adding (i) and (ii), we get 66 (x + y) = 198 ⇒ x + y = 3.
Subtracting (ii) from (i), we get 8 (y - x) = 8 ⇒ y - x = 1.
Solve above equations we get
x = 1, y = 2

Q47. The sum of the digits of a two-digit number is 12. The number obtained by interchanging the two digits exceeds the given number by 18. Find the number.
a) 58 b) 57
c) 75 d) 85

Answer: (b) 57
Explanation:
Let the units and tens digits in the number be y and x respectively.
So, the number be 10x + y.
According to the question,
x + y = 12 ...(i)
Also, 10x + y + 18 = 10y + x
⇒ 9x - 9y = -18 ⇒ x - y = -2 ...(ii)
Solving (i) and (ii), we get x = 5 and y = 7
∴ Required number is 57.

Q48. In△ABC, if ∠C = 50° and ∠A exceeds ∠B by 44o, then ∠A =
a) 40o b) 43o
c) 67o d) 87o

Answer: (d) 87o
Explanation:
Let x and y be the measures of ∠A and ∠B respectively.
Now,∠A +∠B +∠C = 18o[By angle sum property]
⇒x + y + 50o = 180o[Given,∠C = 50o]
⇒x + y = 130o ...(i)
Also,∠A -∠B = 44o⇒x - y = 44o ...(ii)
Adding (i) and (ii), we get
2x = 174o⇒x = 87o⇒∠A = 87o

Q49. If ∠A and ∠B are complementary angles and ∠A is x, then which equation can be used to find ∠B which is denoted by y?
a) y = (180° - x) b) y = (90° - x)
c) y = (x + 180°) d) y = (90° + x)

Answer: (b) y = (90° - x)
Explanation:
We have given, ∠A + ∠B = 90°
⇒ x + y = 90° ⇒ y = (90° - x)

Q50. 5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is
a) 47 years b) 50 years
c) 40 years d) 45 years

Answer: (c) 40 years
Explanation:
Let us assume the present age of men be x years
Also, the present age of his son be y years According to question, after 5 years:
(x + 5) = 3 (y + 5)
x + 5 = 3y + 15
x - 3y = 10 …(i)
Also, five years ago:
(x - 5) = 7 (y - 5)
x - 5 = 7y - 35
x - 7y = - 30 …(ii)
Now, on subtracting (i) from (ii) we get:
- 4y = - 40
y = 10
Putting the value of y in (i), we get
x - 3 × 10 = 10
x - 30 = 10
x = 10 + 30
x = 40
∴ The present age of men is 40 years

Q51. A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, then find the present ages of A, B and C respectively.
a) 40 years, 20 years, 15 years b) 20 years, 15 years, 10 years
c) 25 years, 20 years, 10 years d) 50 years, 25 years, 20 years

Answer: (c) 25 years, 20 years, 10 years
Explanation:
Let the present ages of A, B, C and D are x, y, z and t respectively.
Since, present age of D = t = 48 years.
According to question,
x = y + 5
1
z = ₂ y
f = 2y + 8
From (iii), 48 = 2y + 8
⇒ From (iii), 48 = 2y + 8
From (ii), z = ¹ x 20 = 10 years
2
From (i), x = 20 + 5 = 25 years
So, present ages of A, B and C are 25 years, 20 years and 10 years respectively.

Q52. The cost of a notebook is twice the cost of a pen. If the cost of a notebook is ₹ x and that of a pen is ₹ y, then a linear equation in two variables to represent the given condition is ________.
a) x - 2y = 0 b) 2x - y = 0
c) 2x + y = 0 d) x + 2y = 0

Answer: (a) x - 2y = 0
Explanation: According to question,
2 x Cost of pen = Cost of notebook
⇒ 2y = x ⇒ x - 2y = 0

Q53. Sum of two numbers is 80 and their difference is 36. Find the numbers.
a) 44, 36 b) 40, 40
c) 58, 22 d) 52, 28

Answer: (c) 58, 22
Explanation:
Let the two numbers be x and y.
Then, x + y = 80 ...(i)
Also, x - y = 36 ...(ii)
or y - x = 36 ...(iii)
a. If x - y = 36, from (i) and (ii), we have 2x = 116
⇒x = 58 and y = 22
b. If y - x = 36, from (i) and (iii), we have
2y = 116⇒y = 58 and x = 22
∴Numbers are 58 and 22.

Q54. A part of monthly expenses of a family on milk is fixed which is ₹ 700 and remaining varies with quantity of milk taken extra at the rate of ₹ 25 per litre. Taking quantity of milk required extra as x litres and total expenditure on milk as ₹ y, write a linear equation from the above information.
a) -25x + y = 700 b) 20x + 10y = 300
c) 20x + y = 500 d) x + 25y = 900

Answer: (a) -25x + y = 700
Explanation: Since, x litres is the extra quantity of milk and y be total expenditure on milk.
∴ Required linear equation is,
700 + 25x = y ⇒ y - 25x = 700
or -25x + y = 700

Q55. Two numbers whose sum is 12 and the absolute value of whose difference is 4 are the roots of the equation ________.
a) 2x² - 6x + 7 = 0 b) 2x² - 24x + 43 = 0
c) x² - 12x + 30 = 0 d) x² - 12x + 32 = 0

Answer: (d) x² - 12x + 32 = 0
Explanation:
Let the two roots be a and b, then
a + b = 12 ...(i)
and a - b = 4 ...(ii)
⇒ a = 8 and b = 4 (from (i) and (ii))
∴ Required equation is x² - 12x + 32 = 0

Q56. 5x² + 8x + 4 = 2x² + 4x + 6 is a
a) constant b) cubic equation
c) quadratic equation d) linear equation

Answer: (c) quadratic equation
Explanation:
Given: 5x² + 8x + 4 = 2x² + 4x + 6
⇒ 5x² - 2x² + 8x - 4x + 4 - 6
⇒ 3x² + 4x - 2 = 0
Here, the degree is 2, therefore it is a quadratic equation.

Q57. Value of k for which x = 2 is a solution of the equation 5x² - 4x + (2 + k) = 0, is
a) 10 b) -10
c) -14 d) 14

Answer: (c) -14
Explanation:
x = 2 is solution
p(2) = 5(2)² - 4(2) + (2 + k)
0 = 20 - 8 + 2 + k
k = -14

Q58. If y = 1 is one of the solutions of the quadratic equation py² + py + 3 = 0, then the value of p is:
a) -3 b) -2 3
c) − ₂ d) 2

Answer: 3
(c) − ₂
Explanation:
If y = 1 is solution
p(1) = 0
p(1)² + p(1) + 3 = 0
p + p + 3
2p = -3
−3
p = ₂

Q59. If one root of the equation 2x² + ax + 6 = 0 is 2 then a = ?
7
a) 7 b) ₂ −7
c) ₂ d) -7

Answer: (d) -7
Explanation:
One root of the equation 2 x² + ax + 6 = 0 is 2 i.e. it satisfies the equation
2(2)² + 2a + 6=0
8 + 2 a + 6=0
2a = - 14
a = - 7

Q60. The hypotenuse of a right triangle is 6m more than twice the shortest side. The third side is 2m less than the hypotenuse. The representation of the above situation in the form of a quadratic equation is
a) (2x - 6)² = x² - (2x - 4)² b) (2x + 6)² = x² - (2x + 4)²
c) (2x + 6)² + x² = (2x + 4)² d) (2x + 6)² = x² + (2x + 4)²

Answer: (d) (2x + 6)² = x² + (2x + 4)²
Explanation:
Let the shortest side of a right angled triangle be x meters.
Then according to question, its hypotenuse will be (2x + 6) meters and,
the third side will be (2x + 6 -2) = (2x + 4) meters.
Now, using Pythagoras theorem, (Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (2x + 6)² = x² + (2x + 4)²

Q61. If x = 3 is a solution of the equation 3x² + (k - 1)x + 9 = 0 then k = ?
a) 11 b) -11
c) 13 d) -13

Answer: (b) -11
Explanation:
3x² + (k − 1)x + 9 = 0
x = 3 is a solution of the equation means it satisfies the equation
Put x = 3, we get
3(3)² + (k - 1) 3 + 9 = 0
27 + 3 k - 3 + 9 = 0
27 + 3 k + 6 = 0
3 k = - 33
k = - 11

Q62. If one root of the equation 2x² + kx + 4 = 0 is 2, then the other root is
a) 1 b) -1
c) 6 d) -6

Answer: (a) 1
Explanation:
Let α and β be the roots of quadratic equation 2x² + kx + 4 = 0 in such a way that α = 2
Here, a = 2, b = k and c = 4
Then, according to question sum of the roots −b
α + β = a
−k
2 + β = ₂
−k
β = ₂ - 2
−k−4
β = ₂
And the product of the roots c
α ⋅ β = a
4
= ₂
= 2
−k−4
Putting the value of β = ₂ in above
−k−4
2 × ₂ = 2
(-k - 4) = 2
k = -4 - 2
= -6
−k−4
Putting the value of k in β = ₂
−(6)−4
β = ₂
6−4
= ₂
= ²
2
β = 1
Therefore, value of other root be β = 1

Q63. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal 16 is2 ₂₁ , find then fraction. 7 3
a) ₃ b) ₇ 3 4
c) ₄ d) ₃

Answer: 3
(b) ₇
Explanation:
Let the fraction be x .
y According to the question,
y = 2x + 1 ...(i)
x y 16 58
Also, y + x =2 ₂₁ = ₂₁
x 2x+1 58
⇒ 2x+1 + x = 21 [From (i)]
x²+4x²+1+4x 58
⇒ = ₂₁
x(2x+1)
⇒105x² + 84x + 21 = 116x² + 58x
⇒(x - 3)(11x + 7) = 0
7
⇒x = 3, or x =− ₁₁ (Not possible)
∴y = 7
3
∴Required fraction = ₇

Q64. If one root of 5x² + 13x + k = 0 be the reciprocal of the other root then the value of k is
a) 2 b) 5
c) 1 d) 0

Answer: (b) 5
Explanation: 2 1
Let the roots of the equation (5x + 13x + k = 0) be α and α
Product of the roots = c
a
⇒ α × ¹ = k
α 5
⇒ 1 = k
5
⇒ k = 5.

Q65. The least positive value of k, for which the quadratic equation 2x² + kx - 4 = 0 has rational roots, is
–
a) √2 b) 2
–
c) ±2 d) ±2√2

Answer: (b) 2
Explanation: Dividing the equation by the coefficient of x² i.e., 2 we got
x² + kx - 2 = 0
2 k 2 k²
(x + ₄ ) − ₁₆ - 2 = 0
k 2 k²+32
(x + ₄ ) = ₁₆
k²+32
Hence for rational roots, ₁₆ has to be a perfect square.
k²+32 36 6
We get a perfect square at k = ±2 for ( ₁₆ ) i.e., ₁₆ which becomes ₄ upon removing the square
k²+32 81 9
We get a perfect square at k = ±7 for ( ₁₆ ) i.e., ₁₆ which becomes ₄ upon removing the square
Hence the least positive value of k is 2.
c(a−b)

Q66. If one root of the equation a(b - c)x² + b(c - a)x + c(a - b) = 0 is 1, then the other root is ________.
c(a−b) a(b−c)
a) b) a(b−c) b(c−a) b(c−a) a(b−c)
c) d) a(b−c) c(a−b)

Answer: (a) a(b−c)
Explanation:
Given equation is
a(b - c)x² + b(c - a)x + c(a - b) = 0
Let α be the other root, then
c(a−b)
Product of roots =α×1 =
a(b−c) c a−b
⇒α = a ( b−c )

Q67. The equation x² - 8x + k = 0 has real and distinct roots if
a) k = 8 b) k < 16
c) k = 16 d) k > 16

Answer: (b) k < 16
Explanation: D > 0 b² - 4ac > 0 (-8)² - 4(1)(k) > 0
64 - 4k > 0
64 > 4k 64 ( ₄ ) > k 16 > k

Q68. If (a² + b²) x² + 2(ac + bd) x + c² + d² = 0 has no real roots, then
a) ab = cd b) ac = bd
c) ad ≠ bc d) ad = bc

Answer: (c) ad ≠ bc
Explanation:
(a² + b²) x² + 2(ac + bd)x + c² + d² = 0
Here A = a² + b², B = 2(ac + bd), C = c² + d²
D = B² − 4AC = [2(ac + bd)]² − 4(a² + b²) (c² + d²)
= 4[a²c² + b²d² + 2abcd]−4[a²c² + a²d²+ b²c² + b²d²]
= 4a²c² + 4b²d² + 8abcd − 4a²c² −4a²d² - 4b²c² - 4b²d²
= −4a²d² − 4b²c² + 8abcd
= −4(a²d² + b²c² − 2abcd)
= −4(ad − bc)²
∵ Roots are not real
∴ D < 0
∴ −4(ad − bc)² < 0 ⇒ (ad − bc)² < 0
⇒ ad − bc < 0 or ad ≠ bc

Q69. The discriminant of the equation (2a + b) x = x² + 2ab is ________
a) (2a + b)2 b) (2a - b)2
c) (2a + b²) d) (2a - b²)

Answer: (b) (2a - b)²
Explanation:
(2a + b)x = x² + 2ab
x² - (2a + b)x + 2ab = 0
D = b² - 4ac
D = [-(2a + b)]² - 4 × 1 × 2ab
D = 4a² + b² + 4ab - 8ab
D = 4a² + b² - 4ab
D = (2a - b)²

Q70. If 2 is a root of the equation x² + ax + 12 = 0 and the quadratic equation x² + ax + q = 0 has equal roots, then q =
a) 8 b) 16
c) 20 d) 12

Answer: (b) 16
Explanation:
2 is root equation x² + ax + 12 = 0
∴ (2)² + a × 2 + 12 = 0 ⇒ 4 + 2a + 12 = 0
⇒ 2a + 16 = 0
−16
⇒ a = ₂ = −8
and given that roots of x²+ ax + q = 0 are equal.
∴ b² - 4ac = 0
⇒ a² − 4q = 0 ⇒ (−8)² − 4q = 0
⇒ 64 − 4q = 0 ⇒ 4q = 64
64
⇒ q = ₄ = 16
∴ q = 16

Q71. A train travels 360km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey, then the actual speed of the train is
a) 45 km/hr b) 48 km/hr
c) 40 km/hr d) 36 km /hr

Answer: (c) 40 km/hr
Explanation:
Let the actual speed of the train be x km/hr
360
Time taken to cover 360 km at this speed = x hrs.
360
Time taken to cover 360 km at the increased speed = x₊₅ hrs.
According to condition, ³⁶⁰ − ³⁶⁰ = 1
x x+5
⇒ 360 [ ¹ − ¹ ] = 1
x x+5 x+5−x
⇒ 360 [ ] = 1
x(x+5)
⇒ 360 [ ⁵ ] = 1
x(x+5)
⇒ x² + 5x - 1800 = 0
⇒ x² + 45x - 40x - 1800
⇒ x(x + 45) - 40(x + 45) = 0
⇒ (x - 40)(x + 45) = 0
⇒ x - 40 = 0 and x + 45 = 0
⇒ x = 40 km/hr and x = -45 km/hr [But x = -45 is not possible]
Therefore, the actual speed of the train is 40 km/hr.

Q72. Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360, then Rohan’s present age is
a) 8 years b) 10 years
c) 6 years d) 7 years

Answer: (d) 7 years
Explanation:
Let Rohan’s present age be x years.
Then Rohan’s mother age will be (x + 26) years. And after 3 years their ages will be (x + 3) and (x + 29) years. According to question,
(x + 3)(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x - 273 = 0
⇒ x² + 39x + 7x - 273 = 0
⇒ x(x + 39) -7(x + 39) = 0
⇒ (x - 7)(x + 39) = 0
⇒ (x - 7) = 0 and x + 39 = 0
⇒ x = 7 and x = -39 [x = -39 is not possible]
Therefore, Rohan’s present is 7 years

Q73. If I had walked 1 km per hour faster, I would have taken 10 minutes less to walk 2 km. Then the rate of my walking is
a) 6 km /hr b) 3 km/hr
c) 8 km/hr d) 4 km/hr

Answer: (b) 3 km/hr
Explanation:
Let the rate of my walking be x km/h
∴ Time taken to cover 2 km at the rate of x km/h = ² hrs
x
New rate = (x + 1) km/h
∴ Time taken to cover 2 km at new rate = ² hrs
x+1
According to question, ² − ² = ¹⁰
x x+1 60
⇒ ¹ − ¹ = ¹
x x+1 12
⇒ x+1−x = 1
x(x+1) 12
⇒ ¹ = ¹
x²+x 12
⇒ x² + x - 12 = 0
⇒ x² + 4x - 3x - 12 = 0
⇒ x(x + 4) -3(x + 4) = 0
⇒ (x + 4)(x - 3) = 0
⇒ (x + 4) = 0 and x - 3 = 0
⇒ x = -4 [not possible] and x = 3
Therefore, the rate of my walking is 3 km/h.

Q74. A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, then the time taken by B to finish the work is
a) 20 days b) 25 days
c) 30 days d) 28 days

Answer: (c) 30 days
Explanation:
Let B takes x days to do the work,
then A takes (x - 10) days to do it.
∴ Work done by B in 1 day = ¹ and work done by A in ¹
x x−10 1 1 1
According to question, x + x−10 = 12
x−10+x 1
⇒ x(x−10) = 12
⇒ x² - 10x = 24x - 120
⇒ x² - 34x + 120 = 0
⇒ x² - 30x - 4x + 120 = 0
⇒ x(x - 30) -4(x - 30) = 0
⇒ (x - 30)(x - 4) = 0
⇒ x - 30 = 0 and x - 4 = 0
⇒ x = 30 and x = 4 [x = 4 is not possible]
Therefore, B can finish the work in 30 days.

Q75. The angry Arjun carried some arrows for fighting with Bheeshma. With half the arrows, he cut down the arrows thrown by Bheeshma on him and with six other arrows he killed the rath driver of Bheeshma. With one arrow each, he knocked down respectively the rath, flag and bow of Bheeshma. Finally, with one more than four times the square root of arrows, he laid Bheeshma unconscious on an arrow bed. The total number of arrows that Arjun had, is
a) 100 b) 120
c) 80 d) 96

Answer: (a) 100
Explanation:
Let Arjun had x arrows.
According to question, x −−
₂ + 6 + 3 + 4√x + 1 = x
⇒ 10 + 4 −−x = x
√ ₂
⇒ 20 + 8 −−x = x
√
⇒ 8 −−x = x - 20
√
⇒ 64x = x² - 40x + 400
⇒ x² - 104x + 400 = 0
⇒ x² - 100x - 4x + 400 = 0
⇒ x(x - 100) - 4(x - 100) = 0
⇒ (x - 100)(x - 4) = 0
⇒ x - 100 = 0 and x - 4 = 0
⇒ x = 100 and x = 4 [which is not possible]
Therefore, Arjun had 100 arrows.