NDA Mock Test: Mathematics - 8 - NDA MCQ

Given:

ABC is a right-angled triangle with ∠ABC = 90°.

The radius of circle having radius O is 2 cm.

MA ∶ MC = 2 ∶ 3.

Concept:

1. In a right-angled triangle, Hypotenuse² = Base² + Perpendicular²

2. Tangents drawn from an external point to a circle are equal.

Calculation:

In an ΔABC, using the Pythagoras theorem

⇒ (5x)² = (2x + 2)² + (3x + 2)²

⇒ 25x² = 4x² + 4 + 8x + 9x² + 4 + 12x

⇒ 25x² = 13x² + 20x + 8

⇒ 12x² - 20x - 8 = 0

⇒ 3x² - 5x - 2 = 0

⇒ 3x² - 6x + x - 2 = 0

⇒ 3x(x - 2) + 1(x - 2) = 0

⇒ (3x + 1)(x - 2) = 0

x = 2 & x = -1/3 (not possible)

Now,

AB + BC = 2x + 2 + 3x + 2

⇒ AB + BC = 5x + 4

⇒ AB + BC = 5 × 2 + 4

⇒ AB + BC = 14

∴ AB + BC equal to 14.

Given:

ABC is a right-angled triangle with ∠ABC = 90°.

The radius of circle having radius O is 2 cm.

MA ∶ MC = 2 ∶ 3.

Concept:

1. In a right-angled triangle, Hypotenuse2 = Base2 + Perpendicular2

2. Tangents drawn from an external point to a circle are equal.

3.

Calculation:

Let the radius of circle of center O₁ is r₁

In an ΔABC, using the Pythagoras theorem

⇒ (5x)² = (2x + 2)² + (3x + 2)²

⇒ 25x² = 4x² + 4 + 8x + 9x² + 4 + 12x

⇒ 25x² = 13x² + 20x + 8

⇒ 12x² - 20x - 8 = 0

⇒ 3x² - 5x - 2 = 0

⇒ 3x² - 6x + x - 2 = 0

⇒ 3x(x - 2) + 1(x - 2) = 0

⇒ (3x + 1)(x - 2) = 0

x = 2 & x = -1/3 (not possible)

Now,

2x = 4

Using the Pythagoras theorem

AO² = 2² + 4²

AO = 2√5

Let, ∠ OAM = θ

sin θ = 2/2√5 = 1/√5

We know that

r₁ = 3 - √5

Given:

Two identical semicircles and one circle inscribed in a rectangle of length 10 cm

Calculation:

DC is the diameter of semi circle,

DO = OC = EO = FO all radius of semicircle,

∠OEF = ∠ AOQ = 45°

Using pythagorous in ΔEOF,

EF = 5√2

AB = 10 (given)

Height of the trapezium AEFB = EF/2 = 5/√2

Area of trapezium = Height × (Sum of parallel sides)/2

Area of trapezium =

⇒ 120 × 4 = 30

∴ The correct answer is 30 square cm.

Given:
ABCD be the diameter of a circle of radius 6 cm
The lengths AB, BC and CD are equal
Calculation:

Perimeter of semi circle without Diameter = πR
Required perimeter = sum of perimeter of all three semi circle.
Required Perimeter = 6π + 2π + 4π = 12π cm
∴ The correct answer is 12π.

Calculation:

Statement I:

⇒ r! means the product of 1 × 2 × 3 × ...× r.

⇒ We can see that in this sequence r will occur at least once.

⇒ So, we can say that "The product of r consecutive number is divisible by r!"

Hence, Statement I is true.

Statement II:

This will not always be true.

⇒ Let us take r = 6

⇒ Here, 6! = 1 × 2 × 3 × 4 × 5 × 6

⇒ For this to be divisible by 7! we need 7 at least once in the above sequence which is not satisfied.

⇒ So, we can say that "The product of r consecutive number is divisible by (r + 1)!" is not always true.

Hence, Statement II is not true.

∴ The correct answer is Option 1).

Given:

f(x) = 3x⁴ - 9x³ + x² + 15x + k

Solution:

k + 10 = 0

k = -10

Hence, the correct option is 1.

Given:

Ratio of monthly incomes of Ravi and Shiv = 1 : 2

Ratio of monthly expenditures = 1 : 3

Savings per month each = Rs 4,000

Concept Used:

Income – Savings = Expenditure

Calculation:

⇒ Let the monthly incomes of Ravi and Shiv be x and 2x.

⇒ According to the question,

⇒

⇒ 3(x – 4000) = 1(2x – 4000)

⇒ 3x – 12000 = 2x – 4000

⇒ 3x – 2x = 12000 – 4000

⇒ x = 8000

⇒ Income of Shiv = 2x = 2 × 8000 = Rs 16,000

Therefore, the monthly income of Shiv is Rs 16,000.

Given:

K + (1/K) = -3

Concept Used:

If X + 1/X = a, then

X³ + 1/X³ = a³ - 3a

X² + 1/X² = a² - 2

Calculation:

K + (1/K) = -3

-36 + 7 = -29

⇒ 123⁴⁵× 544⁵⁶ + 786⁹⁸

⇒ 3¹ × 4⁴ + 6²

⇒ 3 × 6 + 6 = 14

∴ unit digit is 4

As we know,

⇒ 125! = 125 × 124 × 123 × ……. × 1

⇒ (6 × 3 + 5 × 1)³⁴⁰

⇒ (8 + 5)³⁴⁰

⇒ 3³⁴⁰

⇒ 3⁴

Given:

AM is perpendicular to BC and AM is the bisector of ∠A.

∠A = 60°.

Concept used:

The sum of interior angles of a triangle = 180°

Exterior angle = 180° - interior angle

Calculation:

⇒ ∠BAM = (60º/2) = 30°

⇒ ∠BAM + ∠AMB + ∠B = 180°

⇒ 30° + 90° + ∠B = 180°

⇒ ∠B = 180° - 120°

⇒ ∠B = 60°

Exterior ∠B = 180° - 60° = 120°

∴ The measure of exterior ∠B is 120°.

Given:

AM is the bisector of ∠A.

∠A = 60°.

Concept used:

A bisector divides an angle in two equal halves.

Calculation:

Since, AM is the bisector of ∠A

⇒ ∠MAC = 60º/2

⇒ 30°

∴ The measure of ∠MAC is 30°.

Given:

AM is perpendicular to BC and AM is the bisector of ∠A.

∠A = 60°.

Concept used:

The sum of interior angles of a triangle = 180°

Calculation:

⇒ ∠BAM = 60º/2 = 30°

⇒ ∠BAM + ∠AMB + ∠B = 180°

⇒ 30° + 90° + ∠B = 180°

⇒ ∠B = 180° - 120°

⇒ ∠B = 60°

∴ The measure of ∠B is 60°.

Given :

Two circles touch each other externally.

Concept used:

In a triangle PAC, ∠CAP = ∠APC = α

Similarly CB = CP and

∠CPB = ∠PBC = β

Calculation:

let ∠CAP = α and ∠CBP = β.

CA = CP [lengths of the tangents from an external point C].

Now in the triangle APB,

∠PAB +∠PBA +∠APB = 180° [sum of the interior angles in a triangle]

α + β + (α+β) = 180°

2α + 2β = 180°

α + β = 90°

Therefore, ∠APB is

α + β = 90°

∴ The correct option is 1.

Concept:

Secant Property

If chord AB and chord CD of a circle intersect at a point P, then

PA × PB = PC × PD

Calculation:

Let the AB = x

PA × PB = PC × PD

⇒ (8 + x) × 8 = 18 × 10

⇒ (8 + x) = 180/8

⇒ (8 + x) = 22.5

⇒ x = 22.5 - 8

= 14.5

∴ The answer is 14.5.

Given:

Diameter of First Pizza = 8 cm; Radius = 4 cm

Price = Rs 240.

Diameter of Second Pizza = 12 cm; Radius = 6 cm

Price = Rs 360.

Concept:

The price is proportional to the size (i.e., the area), we should find the cost per square cm.

Calculation:

Area of the first pizza = π × (4)² = 16π cm².

Price per cm² for the first pizza = 240/16π ≈ Rs 4.77/cm²

Area of the second pizza = π × (6)² = 36π cm².

The expected price for the second pizza = 4.77 × 36π ≈ Rs 540.

The discount on the second pizza is Rs 540 - Rs 360 = Rs 180.

∴ The correct answer is Rs 180.

Given:

The ratio of the length, width, and height of a cuboid = 6:3:2.

Total surface area of the cuboid = 1800 cm².

Formula Used:

Surface Area of a Cuboid = 2(lw + lh + wh)

Volume of a cuboid = l × w × h

Calculation:

Let the length = 6x, width = 3x, and height = 2x.

⇒ Surface area = 2[(6x × 3x) + (6x × 2x) + (3x × 2x)] = 1800

⇒ 2[18x² + 12x² + 6x²] = 1800

⇒ 2[36x²] = 1800

⇒ 72x² = 1800

⇒ x² = 1800 / 72

⇒ x² = 25

⇒ x = 5

⇒ Volume = l × w × h = (6x) × (3x) × (2x)

⇒ Volume = 6 × 3 × 2 × x³

⇒ Volume = 36 × 5³

⇒ Volume = 36 × 125

⇒ 4500 cm³

∴ The volume of the cuboid is 4500 cm³.

Given:

Radius of big ladoo, R = 810 cm

Radius of small ladoo, r = 90 cm

Formula used:

The volume of a sphere = (4/3)πR³

The surface area of a sphere = 4πR²

Calculation:

Volume of big ladoo = Volume of all small ladoos together

⇒ (4/3)πR³ = n × (4/3)πr³

⇒ n = (R/r)³ (Where n is the number of small ladoos)

Total surface area of all small ladoos = n × 4πr² = (R/r)³ × 4πr²

The ratio of total surface area of all small ladoos to the surface area of big ladoo

[(R/r)³ × 4πr²] : 4πR² = R/r = 810 : 90 = 9 : 1

∴ The correct answer is 9 : 1

Given:

The total surface of a solid right circular cylinder = 372π cm²

Height of cylinder (h) = 25 cm.

Volume of cylinder = (1/5) × (volume of sphere)

Formula Used:
Total surface of a cylinder A = 2πr × (r + h) square units

The volume of a cylinder V = πr² h cubic units

Total surface area of sphere A = 4πR²

The volume of sphere V = (4/3)πR³

Where h is height of cylinder and r is the radius.

Calculation:

Let radius of cylinder = r & radius of sphere = R

According to question-

⇒ 2πr (r+ h) = 372π

⇒ r (r + 25) = 186

⇒ r² + 25 r = 186

⇒ r² + 25 r - 186 = 0

⇒ r² + 31 r - 6 r - 186 = 0

⇒ (r + 31) × (r - 6) = 0

⇒ r = - 31, 6

But radius is always positive. So, r = 6 cm.

Radius of cylinder r = 6 cm.

According to question-

⇒ π r²h = (1/5) × (4/3)π R³

⇒ π(6)² × 25 = (1/5) × (4/3)π R³

⇒ 36 × 25 × 5 × 3 = 4 × R³

⇒ 9 × 125 × 3 = R³

⇒ R³ = 3³ × 5³

Cube root on both sides, R = 15 cm

Surface area of sphere (A) = 4π (15)²

A = 4π × 225 = 900π cm²

∴ The correct answer is 900π cm².

Given:

PA = x² + x - 8

PB = x² - x

OP = 13 cm

Concept used:

The two tangents drawn from a common external point to a circle are of equal length.

Each tangent to a circle is perpendicular to the radius at the point of tangency.

Circumference of the circle = 2πr

Area of the equilateral triangle = (√3/4)a²

Perimeter of the equilateral triangle = 3a

Where,

r is the radius of the circle and a is the side of the equilateral triangle.

Calculation:

As we know, PA = PB

⇒ x² + x - 8 = x² - x

⇒ 2x = 8

⇒ x = 4

​⇒ PA = 4² + 4 - 8 = 12

​⇒ PA = 12 cm

Using Pythagoras theorem,

⇒ PA² + OA² = OP²

⇒ 12² + OA² = 13²

⇒ OA² = 13² - 12²

⇒ OA = √(13² - 12²)

⇒ OA = 5 cm

Radius of the new circle = (OA + 16)

⇒ (5 + 16) = 21 cm

According to the question,

Perimeter of equilateral triangle = circumference of the circle

⇒ 3a = 2 × (22/7) × 21

⇒ a = 44 cm

Area of the equilateral triangle is = (√3/4)× 44²

⇒ 484√3 cm²

∴ The area of the equilateral triangle is 484√3 cm².

Given -

Length of direct common tangent = 8 cm

Formula used -

If two circle having radius r and R respectively, touches each - other

length of direct common tangent = 2 × √(rR)

If a semi - circle having radius r,

area = (1/2) × πr²

Solution -

Let the radius of circles be R and r respectively and radius of bigger semi - circle be (r + R).

⇒ 2 × √(r × R) = 16

⇒ √(r × R) = 8

⇒ r × R = 64

⇒ Area of shaded - region = (1/2) × π × (R + r)² - (1/2) × π × R² - (1/2) × π × r²

⇒ Area of shaded region = (1/2) × π × {R² + r² + 2Rr - R² - r²)

⇒ Area of shaded region = R × r × π = 64π

∴ Area of shaded region = 64π cm²

Given

sin α + cos α = ,

Formula used

sin² α + cos² α = 1

Calculation

sin α + cos α =

Squaring both sides,

sin2 α + cos2 α + 2sin α.cos α = (2/√3)²

1 + 2sin α.cos α = 4/3

2sin α.cos α = 4/3 - 1 = 1/3

sin α.cos α = 1/6

tan α + cot α = (sin α/cos α) + (cos α/sin α)

= (sin2 α + cos2 α )/(sin α.cos α)

= 1/(sin α.cos α) = 6

The answer is 6.

Formula used:

Tan (90° - θ) = cot θ

Cos (90° - θ) = sin θ

Calculation:

cos2 36° + cos 54° ⋅ sin 36° +

⇒ cos2 36° + cos 54° ⋅ sin 36° + (tan 26°/cot 64°)

⇒ cos2 36° + cos (90° - 36) ⋅ sin 36° + tan 26°/cot (90° - 26)

⇒ cos2 36° + sin 36° ⋅ sin 36° + tan 26°/tan 26°

⇒ cos2 36° + sin² 36° + 1

⇒ 1 + 1 = 2

∴ The correct answer is 2.

Given:

Ravi spends 70% of his income. His income increases by 25% and his expenditure also increases by 10%

Concept used:

Income - Expenses = Saving

Calculation:

Let the income be 100p

According to the question:

Ravi spends 70% of his income

Expenses = 70p

⇒ 100p - 70p = 30p

Savings = 30p

Income increases by 25%

⇒ 100p + 25p = 125p

Expenditure increases by 10%

⇒ 70p + 70p × 10/100 = 77p

Savings = 125p - 77p = 48

Saving Increase% = (48 - 30)/30 × 100

= 60%

The answer is 60%

Given:

The rate of Interest is 10% and in 't' years received interest is 5/8^th of the principal amount

Concept Used:

Simple Interest (SI) = (P × R × t) / 100

Where, P = Principal, R = rate of interest and, t = time

Calculation:

Let, the principal = 8a and the interest received = 5a

Now, if the rate of interest is 3/4^th of 10% then R = 10% × 3/4 = 7.5%

According to the formula,

Required time (t) = (5a × 100) / (8a × 7.5) = 25/3 = 8(1/3) years

∴ The correct answer is option 3.

Calculation:

X = {x : x = 4(n - 1), n ∈ N}

n = 1

⇒ x = 4(1 -1)

⇒ x = 0

n = 2

⇒ x = 4(2 - 1)

⇒ x = 4

n = 3

⇒ x = 4(3 - 1)

⇒ x = 4(2) = 8

Y = {y : y = 3n - 2n - 1, n ∈ N}

y = 3n - 2n - 1

n = 1

⇒ y = 31 - 2(1) - 1

⇒ y = 0

n = 2

⇒ y = 32 - 2(2) - 1

⇒ y = 9 - 4 -1

⇒ y = 4

n = 3

⇒ y = 33 - 2(3) - 1

⇒ y = 27 - 6 - 1

⇒ y = 20

So, X = {0, 4, 8,...} and Y = {0, 4, 20,...}

∴ X ∪ Y = X = {0, 4, 8, 20,...} ϵ W

W is a set of the whole number.

The correct option is 1 i.e. W

Given:

Akash is thrice as good a workman as Arvind.

Akash can finish a job in 30 days less than Arvind.

Formula Used:

If A can do a work in X days and B can do the same work in Y days, then together they will finish the work in days.

Calculation:

Let the number of days Arvind takes to finish the job be x days.

Therefore, Akash will take (x/3) days to finish the same job.

Given that Akash takes 30 days less than Arvind to finish the job.

⇒ x - (x/3) = 30

⇒ 2x/3 = 30

⇒ x = 45 days

So, Akash takes 45/3 = 15 days to finish the job.

Together, they will finish the job in days.

Therefore, together they can finish the job in 11(1/4) days.

Concept use:

To determine the combined time taken by A and B to type 110 pages, we need to calculate their individual typing rates and then combine these rates to find the total time required.

Calculation:
A's typing rate: A takes 6 hours to type 32 pages.
Typing rate of A = 32 pages / 6 hours = 5.33 pages per hour.
B's typing rate: B takes 5 hours to type 40 pages.
Typing rate of B = 40 pages / 5 hours = 8 pages per hour.

Now, Combined typing rate:
Combined typing rate = Typing rate of A + Typing rate of B
Combined typing rate = 5.33 pages/hour + 8 pages/hour = 13.33 pages/hour

Total time to type 110 pages:
Time = Total pages / Combined typing rate

Time = 110 pages / 13.33 pages/hour ≈ 8.25 hours
8.25 hours is equivalent to 8 hours and 15 minutes.

Hence, the correct answer is 8 hours and 15 minutes (Option 3).

Given:

A can complete a work in 3 hours

B can complete a work in 6 hours

C can complete a work in 9 hours

D can complete a work in 12 hours

Only one person can work at a time in each hour and nobody can work for two consecutive hours.

It is not necessary to engage all.

Concept Used:

To minimize the total time, we need to maximize the work done in each hour by choosing the fastest available worker who hasn't worked in the previous hour.

Calculation:

Work done by A in 1 hour = 1/3

Work done by B in 1 hour = 1/6

Work done by C in 1 hour = 1/9

Work done by D in 1 hour = 1/12

Let's assign work in the following manner:

Hour 1: A (1/3 of work)

Hour 2: B (1/6 of work)

Hour 3: A (1/3 of work)

Hour 4: B (1/6 of work)

Total work done in 4 hours = 1/3 + 1/6 + 1/3 + 1/6

⇒ Total work done = 2/6 + 1/6 + 2/6 + 1/6

⇒ Total work done = 6/6

⇒ Total work done = 1 (which means the entire work is completed)

∴ The minimum number of hours to complete the work is 4 hours.

Solution:

Number of employees in company P = 4560

Number of female employees in company P = 2210

Number of male employees in company P = 4560 - 2210 = 2350

Number of employees in company R = 3052

Number of female employees in company R = 1280

Number of male employees in company R = 3052 - 1280 = 1772

Difference between the number of male employees in company P and that in company R = 2350 - 1772 = 578

∴ Option 2 is the correct answer.