Mathematics MCQ Quiz - Objective Question with Answer for Mathematics - Download Free PDF

Solution:

• Probability of tie: Occurs when both A and B get equal heads.
  • (A = 0, B = 0): 1/4 × 1/8 = 1/32
  • (A = 1, B = 1): 1/2 × 3/8 = 3/16
  • (A = 2, B = 2): 1/4 × 3/8 = 3/32
  • Total tie probability T = 1/32 + 3/16 + 3/32 = 5/16
• Let E = expected number of rounds until someone wins
• The recurrence is: E = 1 + T × E
• ⇒ E − (5/16)E = 1
• ⇒ (11/16)E = 1
• ⇒ E = 16/11

∴ Final Answer: 11 E = 16 

Solution:

• A tosses 2 coins ⇒ Heads: 0, 1, 2 with probabilities: 1/4, 1/2, 1/4
• B tosses 3 coins ⇒ Heads: 0, 1, 2, 3 with probabilities: 1/8, 3/8, 3/8, 1/8

Favorable outcomes for A:

• (1, 0): 1/2 × 1/8 = 1/16
• (2, 0): 1/4 × 1/8 = 1/32
• (2, 1): 1/4 × 3/8 = 3/32

Total P(A wins) = 1/16 + 1/32 + 3/32 = 3/16

P(B wins): = 1/2

P(Tie): = 1 − (3/16 + 1/2) = 5/16

Let P = final probability that A wins

⇒ P = 3/16 + 5/16 × P

⇒ P − (5/16)P = 3/16

⇒ (11/16)P = 3/16

⇒ P = 3/11

∴ K = 3

Concept:

• Median of a triangle: The line segment joining a vertex to the midpoint of the opposite side.
• Centroid (G): The point of intersection of medians; it divides each median in the ratio 2:1.
• Circle touches triangle side: Use geometrical relationships based on circle properties and distances.
• Important property: If AG:AF = AB², and AG = 2×GD, then algebraic equations help to find required lengths.

Calculation:

Let GD = x, DF = y

⇒ AG = 2x, AF = x + y

⇒ 2x(3x + y) = 36

⇒ xy = 4

⇒ 3x² + 4 = 18

⇒ x² = 14/3

⇒ AD = 3x = √42

Now, AC² + AB² = 2(AD² + BD²)

⇒ AC² + 36 = 2(42 + 4)

⇒ AC² + 36 = 92

⇒ AC² = 56

AC² = 56

Concept:

• Median of a triangle: The line segment joining a vertex to the midpoint of the opposite side.
• Centroid (G): The point of intersection of medians; it divides each median in the ratio 2:1.
• Circle touches triangle side: Use geometrical relationships based on circle properties and distances.
• Important property: If AG:AF = AB², and AG = 2×GD, then algebraic equations help to find required lengths.

Calculation:

Let GD = x, DF = y

⇒ AG = 2x, AF = x + y

⇒ 2x(3x + y) = 36

⇒ xy = 4

⇒ 3x² + 4 = 18

⇒ x² = 14/3

⇒ AD = 3x = √42

AD = √42

AD² = 42

Concept:

• Ordered Triplets: If x, y, z ∈ ℕ and xyz = N, the number of ordered triplets is found using the number of solutions to the product form using prime factorization.
• Multinomial Expansion: The number of terms in the expansion of (x + y + z)ⁿ is equal to the number of non-negative integer solutions to x + y + z = n.
• Inequality Solution: For quadratic inequalities of the form ax² + bx + c ≤ 0, solve the corresponding equation and determine the integer values within the valid interval.
• Number of Solutions of x + y + z = n: For x, y, z ∈ ℕ, the number of solutions is equal to C(n−1, r−1) where r = number of variables.

Calculation:

Given,

(A) xyz = 243

⇒ 243 = 3⁵

⇒ Number of ordered triplets of natural numbers is equal to number of non-negative integer solutions of a + b + c = 5

⇒ Number of solutions = C(5 + 3 - 1, 2) = C(7, 2) = 21

⇒ Triplets are ordered ⇒ 21

(B) Expansion of (x + y + z)⁶

⇒ Number of terms = C(6 + 3 - 1, 2) = C(8, 2) = 28

(C) x² + x - 400 ≤ 0

⇒ Solve equation x² + x - 400 = 0

⇒ x = [-1 ± √(1 + 1600)]/2 = [-1 ± √1601]/2

⇒ √1601 ≈ 40 ⇒ x ranges from 1 to 19

⇒ Valid integer values = 19

(D) x + y + z = 10, with x, y, z ∈ ℕ

⇒ Convert to non-negative solution by setting x' = x−1, y' = y−1, z' = z−1

⇒ x' + y' + z' = 7

⇒ Number of solutions = C(7 + 3 - 1, 2) = C(9, 2) = 36

∴ A-(s), B-(r), C-(p), D-(t) is the correct match

⇒ Option (1)

Concept:

sin (2nπ ± θ) = ±  sin θ

sin (90 + θ) = cos θ

Calculation:

Given: sin (1920°)

⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

\({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + {\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right)} \)

\({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + \frac{1}{{{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}}}} \)

\(\Rightarrow {\rm{y}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^3} + 1\)

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative

Hence, the degree of the differential equation is 3.

Mistake PointsNote that, there is a term (dx/dy) which needs to convert into the dy/dx form before calculating the degree or order. 

Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

Mean = Sum of all the terms/Total number of terms

Median = {(n + 1)/2}^th term when n is odd 

Median = 1/2[(n/2)^th term + {(n/2) + 1}^th] term when n is even

Range = Maximum value – Minimum value 

Calculation:

Arranging the given data in ascending order 

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so 

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd 

⇒ {(15 + 1)/2}^th term 

⇒ (8)^th term

⇒ 6 

Now, Range = Maximum value – Minimum value 

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3 

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Formula used:

The mean of grouped data is given by,

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

Where, \(u_i \ = \ \frac{X_i\ -\ a}{h}\)

Xi = mean of ith class

fi = frequency corresponding to ith class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Then,

We know that, mean of grouped data is given by

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

= \(\frac{1535}{43}\)

= 35.7

Hence, the mean of the grouped data is 35.7

Concept:

• Rational numbers are those numbers that show the ratio of numbers or the number which we get after dividing it with any two integers.
• Irrational numbers are those numbers that we can not represent in the form of simple fractions a/b, and b is not equal to zero.
• When we add any two rational numbers then their sum will always remain rational.
• But if we add an irrational number with a rational number then the sum will always be an irrational number.

Explanation:

Case:1 Take two irrational numbers π and 1 - π

⇒ Sum =  π +1 - π = 1

Which is a rational number.

Case:2 Take two irrational numbers π and √2 

⇒ Sum =  π + √2

Which is an irrational number.

Hence, a sum of two irrational numbers may be a rational or an irrational number.

Given:

tan0° tan1° tan2° tan3° tan4° …… tan89°

Formula:

tan 0° = 0

Calculation:

tan0° × tan1° × tan2° × ……. × tan89°

⇒ 0 × tan1° × tan2° × ……. × tan89°

⇒ 0

Concept:

a² - b² = (a - b) (a + b)

sec x = 1/cos x and cosec x = 1/sin x

a³ + b³ = (a + b) (a² + b² - ab)

Calculation:

 \(\frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)}}{{{\rm{cosA}}\left( {{\rm{secA}} - {\rm{cosecA}}} \right)\left( {{\rm{si}}{{\rm{n}}^3}{\rm{A}} + {\rm{co}}{{\rm{s}}^3}{\rm{A}}} \right)}}\) 

⇒ \( \frac{{\left( {{\rm{1}} - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{1}{{{\rm{cosA}}}} - \frac{1}{{{\rm{sinA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} + {\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{sinAcosA}}} \right)}}\)

⇒ \( \frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{{{\rm{sinA}} - {\rm{cosA}}}}{{{\rm{sinA}}.{\rm{cosA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {1 - {\rm{sinAcosA}}} \right)}}\)

⇒ \(\frac{sinA - cosA}{cosA[\frac{sinA - cosA}{sinA.cosA}]}\)

⇒ \(\frac{(sinA - cosA)\times sinA.cosA}{cosA[sinA - cosA]}\)

⇒ \(\frac{ sinA.cosA}{cosA}\)

⇒ sin A

∴ The correct answer is option (1).

Concept:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z =  = x – iy

Calculation:

Let z = (1 + i) ³

Using (a + b) ³ = a³ + b³ + 3a²b + 3ab²

⇒ z = 1³ + i³ + 3 × 1² × i + 3 × 1 × i²

= 1 – i + 3i – 3

= -2 + 2i

So, conjugate of (1 + i) ³ is -2 – 2i

NOTE:

The conjugate of a complex number is the other complex number having the same real part and opposite sign of the imaginary part.

Concept:

cosec² x – cot² x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec² x – cot² x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

\(\frac1q \times p=1\)

Concept:

sin² x + cos² x = 1

Calculation:

Given: sin θ + cos θ = 7/5 

By, squaring both sides of the above equation we get,

⇒ (sin θ + cos θ)² = 49/25

⇒ sin² θ + cos² θ + 2sin θ.cos θ = 49/25

As we know that, sin2 x + cos2 x = 1

⇒ 1 + 2sin θcos θ = 49/25

⇒ 2sin θcos θ = 24/25