IPMAT Mock Test - 9 (New Pattern) - Commerce MCQ

⇒ 3 (tan²A + cot A – 3)² = 3 (tan²(315°) + cot (315°) – 3)²

⇒ 3(tan²(315°) + cot (315°) – 3)² = 3(tan²(360° - 45°) + cot (360° - 45°) – 3)²

We know that, tan (360° - θ) = - tan θ and cot (360° - θ) = - cot θ then,

⇒ 3(tan²(315°) + cot (315°) – 3)² = 3(tan² 45° - cot 45° - 3)²

As we know that, tan 45° = 1 = cot 45°,

⇒ 3(tan²⁡ A + cot⁡A − 3)² = 3 (9) = 27

Hence, the correct option is (D).

We know that,

tan (90⁰ - A) = cot A

= 0

Hence, the correct option is (D).

A matrix having m rows and n columns is called a matrix of order m×n or simply m×n matrix.

Given,

As we can see that the given matrix A has 3 rows and 2 columns.

∴ The order of the given matrix A is 3 × 2.

Hence, the correct option is (B).

= 2405

Hence, the correct option is (A).

= 800 + 845 + 710 + 900 + 1100 + 950 = 5305

Therefore, the required average

= 5305/6 = 884.17 ≈ 884

Hence, the correct option is (A).

The total number of students who qualified in the exam from schools C and D together in the year 2015 = 900 + 400 = 1300

Therefore, required difference = 1300 - 1170 = 130

Hence, the correct option is (D).

Set theory:

A ∪ B means set of all the values in the set A and B.

A ∩ B is the set of common elements of A and B.

Binomial theorm:

(a+b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁a^{n − 1}b¹ + ⁿC₂a^{n − 2}b² +…+ ⁿC_{n − 2}a²b^{n − 2} + ⁿC_n−1a¹bⁿ⁻¹ + ⁿC_na⁰bⁿ

Calculation:

Given X = {4ⁿ − 3n−1, n ∈ N} and Y = {9n−9, n ∈ N}

x = 4ⁿ − 3n−1 = (3 + 1)n − 3n−1

(∵ ⁿC_n−1 = n and ⁿC_n = 1)

⇒ X = 3²(ⁿC₀3ⁿ⁻²+ⁿC₁3ⁿ⁻³+ⁿC₂3ⁿ⁻⁴+…+ⁿC_n−2)

⇒ x = 9(ⁿC₀3ⁿ⁻²+ⁿC₁3ⁿ⁻³+ⁿC₂3ⁿ⁻⁴+…+ⁿC_n−2)

For n ≥ 2, X is some multiple of 9.

And Xn = 1 = 0

∴ X = {0, some multiples of 9 but not all} .....(i)

Y = 9n − 9 = 9(n − 1)

∴ Y = { All multiple of 9 starting from 0} .......(ii)

From (i) and (ii) we can say,

∴ X ∪ Y = Y

Hence, the correct option is (A).

Here, a = 4, r = 2

Sum of n numbers = S_n = 2044

As we know that, Sum of n terms of GP

= S_n = a(rⁿ − 1)/r − 1

Where r > 1

∴S_n = 4(2ⁿ − 1)/2 − 1

2044 = 4×(2ⁿ − 1)

⇒ 511 = (2ⁿ − 1)

⇒ 2ⁿ = 512

⇒ 2ⁿ = 29

∴ n = 9

Hence, the correct option is (D).

Average salary of all employees = Rs. 1500

Total salary increased by Rs. (3000 after 5 new employees

Decrease in average after new employees induction = Rs. 100

Formula used:

Average of n numbers = sum of total numbers/n

Let the total number of employees before new employees induction = x

⇒ Total salary of the employees = 1500x

⇒ After new induction, total salary = 1500x + 3000

Then, according to the question:

1400 = ((1500x + 3000)/x) + 5

⇒ 1400x + 7000 = 1500x + 3000

⇒ x = 40

⇒ x + 5 = 45

∴ Now the number of employees = 45

Hence, the correct option is (D).

So,

(x + y + z/3) = 15

x + y + z = 45

Given that x, y = 12, 15

⇒ 12 + 15 + z = 45

⇒ z = 18

Therefore, age of third sister is 18 years.

Hence, the correct option is (C).

Loss % = ((CP − SP)/CP) × 100

Let the Cost price of article be x

Loss =4%

Selling price of article = x − (4/100) × x = 0.96x

0.96x = 265 ⇒ x = Rs. 276

Now the article is sold at gain

Gain = 12%

Selling price of article = 276 + (12/100) × 276 = Rs. 308

Hence, the correct option is (B).

Quotient is 16 and the divisor is 25 times the quotient.

⇒ Divisor = 25 × 16 = 400

Also, divisor is 5 times the remainder

⇒ Remainder = divisor/ 5

= 400/5 = 80

We know that,

Dividend = quotient × divisor + remainder

⇒ Dividend = 16 × 400 + 80

= 6480

Hence, the correct option is (B).

Then,Unit digit be 2x - 1

Original number:

= 10x + 2x - 1

= 12x - 1

New number:

= 10(2x - 1) + x

= 20x - 10 + x

= 21x - 10

According to the question,

(21x - 10) - (12x - 1)

= (12x - 1) - 20

9x - 9 = 12x - 21

3x = 12

x = 4

Original number:

= 12x - 1

= 12 × 4 - 1

= 47

Hence, the correct option is (D).

Now 4 + 3 + 2 + P + 1 should be divisible by 9.

So, 10 + P should be divisible by 9.

Also P is a single-digit number.

The number after 10 which is divisible by 9 is 18.

So P = 18 − 10 = 8

Hence, the correct option is (B).

= 16/81

Hence, the correct option is (C).

Average speed

= 25 km/h

Hence, the correct option is (A).

Using number line rule:

1 ≤ |x| < />

⇒ x ∈ (-2, -1) ∪ (1, 2)

Hence, the correct option is (C).

So, Sum invested by Shikha in another scheme = (1500 - x)

Thus,

= 100

= 100

On solving,

x = Rs. 1100

Another amount = 1500 − 1100

= Rs. 400

Thus, the required ratio = 11 : 4

Hence, the correct option is (B).

The number of ways to select r things out of n things is given by, ⁿC_r

Calculation:

Here, we have 7 boys and 4 girls out of which we have to select an equal number of boys and girls to form a team of 6 (i.e., 3 boys and 3 girls)

∴ Required number of ways ⁷C₃ x ⁴C₃

= 35 x 4

= 140

Hence, the correct option is (D).

⇒ y = x + 1 and z = y + 1

⇒ z = x + 2

Consider, log (1 + xz)

= log [1 + x(x + 2)]

= log [1 + x² + 2x]

= log (1 + x)²

= 2 log (1 + x)

= 2 log y

So, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y.

Hence, the correct option is (D).

Diameter = 8

Radius = 4

Also diameters equations:

y + 2x − 5 = 0 ...(i)

2y + 3x − 8 = 0 ..(ii)

Subtracting 2× (i) from (ii)

−x + 2 = 0

x = 2

Putting back in (i):

y + 2 × 2 − 5 = 0

y = 1

So center is (2,1) and radius is 4.

The equation of circle:

(x − 2)² + (y − 1)² = 42

x² + 4 − 4x + y² + 1 − 2y = 16

x² + y² − 4x − 2y − 11 = 0

Hence, the correct option is (B).

Total time = 10 hours

Half journey at the rate of speed = 21 km/hr

and

Half journey at the rate of speed = 24 km/hr

We know that,

Time = Distance/Speed

Let distance = x

Hence, the correct option is (B).

Dividend = Divisor × Quotient + Remainder

Let the quotient be =b

n = 5 × b + 2

n² = (5b + 2)²

⇒ n² = 25b² + 4 + 2 × 5b × 2

⇒ n² = 25b² + 4 + 20b

⇒ n² = 5(5b² + 4b) + 4

∴ When n2 is divided by 5, the remainder will be 4.

Hence, the correct option is (C).

log₆√⁡2 + log₆⁡√3

= log_6⁡(√2 × √3) (∵ log⁡m + log⁡n = log⁡mn)

= log_6⁡(√6)

= log₆⁡6^1/2

=(½)log₆⁡6 (∵ log⁡mⁿ = nlog⁡m)

As we know,

log_m⁡m = (log_a⁡m/log_a⁡m) = 1

∴ log₆⁡√2 + log₆⁡√3 = ( ½)log₆⁡6

= (½) ×1 = 1/2

Hence, the correct option is (A).

Number of students who qualified in the exam from school D in the year 2017 = 600

Required ratio = 710 : 600

= 71 : 60

Hence, the correct option is (B).

The number of students who qualified in the exam from school A in the year 2012 = 450

Increase = 625 - 450 = 175

Therefore, required percentage

= (175/450) × 100 = (350/9) = 38.88% ≈ 39%

Hence, the correct option is (A).

A = [2, 3]

Then,

A × A = [2, 3] × [2, 3]

= [(2, 2), (2, 3), (3, 2), (3, 3)]

Therefore,

A × A × A = [2, 3] × [2, 3] × [2, 3]

= [(2, 2), (2, 3), (3, 2), (3, 3)] × [2, 3]

= [(2, 2, 2), (2, 2, 3), (2, 3, 2), (2, 3, 3), (3, 2, 2), (3, 2, 3), (3, 3, 2), (3, 3, 3)]

So, n(A × A × A) = 8

Therefore, cardinality = 8

Hence, the correct option is (C).

Equation of ellipse is x² + 2y² = 4

Here, we have to find the equation of the tangent to the given ellipse at the point where ordinate is 1 such that the point lies in the first quadrant.

Let the point be P = (c, 1)

Now this point P lies on the ellipse. So, x = c and y = 1 will satisfy the equation x² + 2y² = 4

c/2 + 2 = 4

c = ± √2

So, the point P is (√2, 1) because the point P lies in the first quadrant.

The given equation of ellipse we can be re-written as: (x²/4) + (y²/2) = 1

As we know that, the equation of the tangent to the horizontal ellipse x²/a² + y²/b² = 1 where 0 < b="" />< a="" at="" the="" point="" />₁, y₁) is given by:

So, the required equation of tangent is √2x + 2y - 4 = 0

Hence, the correct option is (D).

Now, we can get transpose of AB by switching its rows with its columns.

Hence, the correct option is (B).

Total surface area of cube = 108 cm²

Formula used:

TSA of cube = 6a²

Volume of cube = a³

Let, the side of the cube be a.

∵ TSA of cube = 6a²

⇒ 108 cm² = 6a²

⇒ a² = 108/6

⇒ a² = 18 cm²

⇒ a = √18 = 3√2 cm

∴ Volume of cube = a³

= (3√2)³

= 54√2 cm³

Hence, the correct option is (B).