Mathematics: CUET Mock Test - 6 - Commerce MCQ

Calculation:
Rate of interest for Shyam = 6% p.a.
Rate of interest for Sushil = 10% p.a.
∴ Mean rate of interest = (6 + 10)/2 = 8% p.a.

Calculation:
Let money lent to Shyam = x
⇒ Money lent to Sushil = 200000 - x
∴ Interest to be paid by Shyam = 6x/100
and, Interest to be paid by Sushil =
⇒ Total interest = + = 18000 (given)
⇒ 6x + 10(200000 - x) = 1800000
⇒ 6x + 2000000 - 10x = 1800000
⇒ 200000 = 4x
⇒ x = 50,000
∴ Amount lent to Shyam = x = Rs. 50,000
and, amount lent to Sushil = 200000 - x = 1,50,000
∴ Required ratio = = 1 ∶ 3
∴ Ratio at which Sitaram lent the money at 6% p.a. and 10% p.a. respectively is 1 ∶ 3.

Calculation:
Let money lent to Shyam = x
⇒ Money lent to Sushil = 200000 - x
∴ Interest to be paid by Shyam = 6x/100
and, Interest to be paid by Sushil =
⇒ Total interest = = 18000 (given)
⇒ 6x + 10(200000 - x) = 1800000
⇒ 6x + 2000000 - 10x = 1800000
⇒ 200000 = 4x
⇒ x = 50,000
∴ Amount lent to Shyam = x = Rs. 50,000
and, amount lent to Sushil = 200000 - x = 1,50,000
∴ Shyam borrowed Rs. 50,000.

Calculation:
Let money lent to Shyam = x
⇒ Money lent to Sushil = 200000 - x
∴ Interest to be paid by Shyam = 6x/100
and, Interest to be paid by Sushil =
⇒ Total interest = = 18000 (given)
⇒ 6x + 10(200000 - x) = 1800000
⇒ 6x + 2000000 - 10x = 1800000
⇒ 200000 = 4x
⇒ x = 50,000
∴ Amount lent to Shyam = x = Rs. 50,000
and, amount lent to Sushil = 200000 - x = 1,50,000
∴ Amount of money is lent at 10% p.a. simple interest is Rs. 1,50,000.

Calculation:
Let money lent to Shyam = x
⇒ Money lent to Sushil = 200000 - x
∴ Interest to be paid by Shyam = 6x/100
and, Interest to be paid by Sushil =

⇒ Total interest = = 18000 (given)
⇒ 6x + 10(200000 - x) = 1800000
⇒ 6x + 2000000 - 10x = 1800000
⇒ 200000 = 4x
⇒ x = 50,000
∴ Amount lent to Shyam = x = Rs. 50,000
and, amount lent to Sushil = 200000 - x = 1,50,000
∴ Interest paid by Shyam = = Rs. 3000
and, Interest paid by Sushil = = Rs. 15000
∴ Required ratio = = 1 ∶ 5
∴ Ratio at which Sitaram lent the money at 6% p.a. and 10% p.a. respectively is 1 ∶ 5.

CONCEPT:

• If , and , then .
• If vectors are coplanar then

​CALCULATION:

Given: The vectors are coplanar.

As we know that, if vectors are coplanar then

⇒

⇒ 2(0 - 4p) -(-1)(-15 - 0) + 1(12 - 0) = 0

⇒ -8p - 15 + 12 = 0

⇒ -8p - 3 =0

⇒ -8p = 3

⇒ p = -3/8

Hence, correct option is 3.

Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.
 Probability o fgetting the event E₁ when E₂ is already occured.

Calculation:
Given:
P (E₁) = Probability of stopping at the gas station and ask for tyre checked = 0.12
P (E₂) = Probability of stopping at the gas station and ask for oil checked = 0.29
P (E₁∩ E₂) = Probability of both checked = 0.07
 = Probability of person who has his tyre checked will also have oil checked

Given -

Given differential equation is

Concept -

Order : The order of a differential equation is the highest power of the derivative present in the equation.

Explanation -

The highest derivative in the given differential equation is second order. Hence the order = 2

Hence option (ii) is correct.

Given:
Let E₁ be the event the bike is from the first plant
Let E2 be the event the bike is from the second plant
Let A be the event the bike is standard quality

Formula Used:
Bayes Theorem:- Let E1, E2,... En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or... or En such that P(A) ≠ 0. Then

P(Ei | A) = , i = 1, 2, ... n

Calculation:
P(E₁ | A) = , i = 1, 2, ... n
P(E₁) = 60/100 = 0.6
P(E₂) = 40/100 = 0.4
P( A/E₁) = 80/100 = 0.8
P( A/E₂) = 90/100 = 0.9
As we know that according to bayes' theorem: 

⇒

⇒
The Correct Answer is 4/7

Calculation:

Given y = aemx + be−mx

Differentiating with respect to x:

⇒ ame^mx - mbe^-mx

Differentiating again with respect to x:

⇒ am²e^mx + bm²e^-mx

⇒ m2(aemx + be-mx)

⇒ m2y

⇒ - m2y = 0

The correct answer is option 3.

The equation is cos^-1 x + cos^-1 y + cos^-1 z = 3π
This means cos^-1 x = π, cos^-1 y = π and cos^-1 z = π
This will be only possible when it is in maxima.
As, cos^-1 x = π so, x = cos^-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

The number of elements for a matrix with the order m×n is equal to mn, where m is the number of rows and n is the number of columns in the matrix.

A square matrix is a matrix in which the number of rows(m) is equal to the number of columns(n). Therefore, the matrix which follows the condition m = n is a square matrix.

We need to evaluate the determinant of the 2x2 matrix:

| 2 5 |
| -1 -1 |

The determinant of a 2x2 matrix
|a b|
|c d|
is given by ad - bc.

Here, a = 2, b = 5, c = -1, and d = -1.

So, determinant = (2 × -1) - (5 × -1) = -2 - (-5) = -2 + 5 = 3.

Final Answer: 3

Determinant of the matrix A= is not possible as it is a rectangular matrix and not a square matrix. Determinants can be calculated only if the matrix is a square matrix.

sin^-1sin(6 π/7)
Now, sin(6 π/7) = sin(π – 6 π/7)
= sin(π/7)
Therefore, sin^-1sin(6 π/7) = sin^-1sin(π/7) = π/7

a₁₁=1+1=2, a₁₂=1+2=3, a₂₁=2+1=3, a₂₂=2+2=4
∴ 

The matrix in which number of rows is smaller than the number of columns is called is called a horizontal matrix. In the given matrix A=  m = 3 and n = 2 i.e.
3<2. Hence, it is a horizontal matrix.

Evaluating along R₁, we get
∆ = 5(√3)-(-4)1 = 5√3+4.

A relation in a set A is said to be symmetric if (a₁, a₂)∈R implies that (a₁, a₂)∈R,for every a₁, a₂∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a₁, a2)∈R and (a₂, a₃)∈R implies that (a₁, a₃)∈ R for every a₁, a₂, a₃∈R.

Let, θ = sin^-1(-x)
So, -π/2 ≤ θ ≤ π/2
⇒ -x = sinθ
⇒ x = -sinθ
⇒ x = sin(-θ)
Also, -π/2 ≤ -θ ≤ π/2
⇒ -θ = sin^-1(x)
⇒ θ = -sin^-1(x)
So, sin^-1(-x) = -sin^-1(x)

The number of rows (m) and the number of columns (n) in the given matrix A=   is 2. Therefore, the order of the matrix is 2×2(m×n).

The given matrix A =  is of the order 3×1. The matrix has only one column (n=1). Hence, it is a column matrix.

Expanding along R₁, we get
∆=-sinθ(sinθ)-(-1)1=-sin²⁡θ+1=cos^2⁡θ.

For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3, 4)∈R and (4, 6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a, a) ∈ R, for every a ∈ A for a relation R in the set A.

We know that sin(x) = sin(2A * π + x) where A can be positive or negative integer.
If A is -1, then sin(6) = sin(-2π + 6);
If A is 1, then sin(6) = sin(2π + 6);

The given determinant is f(r) =  
Now, _{r = 1}Σⁿ (2r) = 2[(n(n + 1))/2] ……….(1)
= n² + n
_{r = 1}Σⁿ(6r² – 1) = 6[((n + 1)(2n + 1))/6] – n ……….(2)
= n(2n² + 2n + n + 1) – n
= 2n³ + 2n² + n² + n – n
= 2n³ + 3n²
= _{r = 1}Σⁿ(4r³ – 2nr) = n³ (n + 1) ……….(3)
From (1), (2) and (3) we get
_{r = 1}Σⁿ f(x) = 0 (If two rows of a matrix A are identical, then the determinant of A is 0. Row 1 and row 3 are identical)

For x in the interval [-1, 1], let
θ = cos⁻¹(−x)

So,
cos(θ) = −x

Now, multiply both sides by −1:
x = −cos(θ)

But we also know:
cos(π − θ) = −cos(θ)

So,
x = cos(π − θ)

This means:
cos⁻¹(x) = π − θ

Rewriting this, we get:
θ = π − cos⁻¹(x)

Since θ was originally defined as cos⁻¹(−x), we conclude:
cos⁻¹(−x) = π − cos⁻¹(x)

Final Answer: cos⁻¹(−x) = π − cos⁻¹(x) for x in [−1, 1]

Solving the given system of equation by Cramer’s rule, we get,
x = D₁/D, y = D₂/D, z = D₃/D where,


Now, performing, C₃ = C₃ – C₂ and C₂ = C₂ – C₁ we get,

As two columns have identical values, so,
D = 0
Similarly,

Now, performing, C₁ = C₁ – C₃

Now, performing, C₃ = C₃ – C₂

As two columns have identical values, so,
D₁ = 0

Now, performing,

Now, performing, C₂ = C₂ – C₃ and C₃ = C₃ – C₁

As two columns have identical values, so,
D₂ = 0


Now, performing, C₂ = C₂ – C₂ and C₃ = C₃ – C₂

As two columns have identical values, so,
D₃ = 0
Since, D = D₁ = D₂ = D₃ = 0, thus, it has infinitely many solutions.

The two matrices  and  are equal matrices. Comparing the two matrices, we get a+b=3, c=2, a-b=1, 2c+d=6
Solving the above equations, we get a=2, b=1, c=2, d=2.