Test: Elementary Mathematics - 3 - CDS MCQ

Given:
The central angles of the sectors are in the ratio 2 ∶ 3 ∶ 7 ∶ 5 ∶ 1
Radius = 7 cm
Calculation:
Total ratio of all the sectors = 2 + 3 + 7 + 5 + 1 = 18
Angle of smallest sector = 360∘ /180 = 20∘
Perimeter of smallest sector = 2 × π × 7 × (20^∘/360^∘) + 2 × 7
⇒ P = 2 × π × 7 × (20∘/360∘) + 14 = 2.44 + 14 = 16.444cm
9p = 9 × 16.44 = 148 cm
∴ The correct answer is 148 cm.

Given:

t = cos 79°

Formula Used:

sin²θ + cos²θ = 1

a² - b² = (a - b)(a + b)

Calculation:

cosec 79°cosec 79° (1 - cos 79°)

⇒ (1 - t)

⇒

⇒

⇒

⇒

∴ The correct answer is .

Given:

two parallel line segments PQ = 5

RS = 3 cm

Concept Used:

If three lines are parallel as shown below,

then,

Area of Trapezium =

Calculation:

Using the above formula,

​

MN = 15/8 cm

ΔPQS and ΔMNS are similar,

So,

h2 = 3 and h1 = 5

Area of trapezium PQNM =

Area of trapezium RSNM =

Ratio of area of both trapezium = = 275/117

∴ The correct answer is 275/117.

Given:

m ∶ n = 1 ∶ 2

p ∶ q = 3 ∶ 4

Concept Used:

To find the ratio (2m + 4p) ∶ (n + 3q), we need to express m, n, p, q in terms of a common variable.

Calculation:

Let m = 1k and n = 2k (since m ∶ n = 1 ∶ 2)

Let p = 3l and q = 4l (since p ∶ q = 3 ∶ 4)

Calculate 2m + 4p:

⇒ 2m + 4p = 2(1k) + 4(3l)

⇒ 2k + 12l

Calculate n + 3q:

⇒ n + 3q = 2k + 3(4l)

⇒ 2k + 12l

Therefore, the ratio (2m + 4p) ∶ (n + 3q) is:

⇒ (2k + 12l) ∶ (2k + 12l)

⇒ 1 ∶ 1

∴ The correct answer is option 1

Given:

+ + + ...

Calculation:

+ + + ...

⇒ + + + ...

⇒ + + + ...... -

⇒ 1 -

⇒ 1

∴ The correct answer is 1

Given:

The work done by x men in (x + 1) days = The work done by (x + 5) men in (x - 2) days

Calculation:

Let W denote the work done by one man in one day.

The total work done by x men in (x + 1) days is W × X × (x + 1).

The total work done by (x + 5) men in (x - 2) days is: W × (x + 5) × (x - 2).

According to the problem, these two amounts of work are equal:

W × x × (x + 1) = W × (x + 5) × (x - 2)

⇒ x × (x + 1) = (x + 5) × (x - 2)

⇒ x² + x = x² - 2x + 5x - 10

⇒ x = 3x - 10

⇒ 2x = 10

⇒ x = 5

∴ The correct answer is 5.

Concept :

Area of non-overlapping region = 2 [Area of rectangle] – 2 [area of overlapping region]

Calculation:

We have to find first Area of overlapping region

Let the DG = a = GB

In Δ AGB

⇒ a² = (2 - a)² + 1²

⇒ a² = 4 + a² – 4a + 1

⇒ 4a = 5

⇒ a = 5/4 cm.

So,

Area of overlapping region = Area of rectangle – 2 (area of Δ AGB)

⇒ (1 × 2) - 2 × [1/2 × 1 × (2 - 5/4)]

⇒ (2 - 3/4) = 5/4 cm².

Now, we have to find Area of non-overlapping region

Area of non-overlapping region = 2 [Area of rectangle] – 2 [area of overlapping region]

⇒ 2 × 1 × 2 - 2 × 5/4

⇒ 4 - 5/2 = 3/2 cm²

∴ Area of non-overlapping region is 3/2 cm²

Given:

3 sin θ + 5 cos θ = 5

Concept Used:

We can use the trigonometric identities and algebraic manipulation to solve the given equation.

Calculation:

Let 3 sin θ + 5 cos θ = R

Given R = 5

We need to find the value of 5 sin θ - 3 cos θ

Square both sides of the given equation:

⇒ (3 sin θ + 5 cos θ)² = 5²

⇒ 9 sin² θ + 30 sin θ cos θ + 25 cos² θ = 25

Using the identity sin² θ + cos² θ = 1:

⇒ 9 (1 - cos² θ) + 30 sin θ cos θ + 25 cos² θ = 25

⇒ 9 - 9 cos² θ + 30 sin θ cos θ + 25 cos² θ = 25

⇒ 9 + 16 cos² θ + 30 sin θ cos θ = 25

⇒ 16 cos² θ + 30 sin θ cos θ = 16

Now, consider the expression we need to find:

Let 5 sin θ - 3 cos θ = K

Square both sides of this equation:

⇒ (5 sin θ - 3 cos θ)² = K²

⇒ 25 sin² θ - 30 sin θ cos θ + 9 cos² θ = K²

Using the identity sin² θ + cos² θ = 1:

⇒ 25 (1 - cos² θ) - 30 sin θ cos θ + 9 cos² θ = K²

⇒ 25 - 25 cos² θ - 30 sin θ cos θ + 9 cos² θ = K²

⇒ 25 - 16 cos² θ - 30 sin θ cos θ = K²

Since we know from earlier:

16 cos² θ + 30 sin θ cos θ = 16

Substitute this into the equation:

⇒ 25 - 16 = K²

⇒ 9 = K²

⇒ K = ±3

Since the options provided are negative, the correct answer is:

⇒ K = -3

∴ The value of 5 sin θ - 3 cos θ is -3.

Given:

Speed of train A = 60km/h

Speed of train B = 90km/h

Distance between P and Q = 800km

Train A started at 7 pm and Train B started at 4 am the next day.

Calculation:

Train A traveled for 9 hours before train B started,

Distance covered by train A = Time × Speed = 9 × 60 = 540km

Remaining Distance = 800 - 540 = 260km

Since the trains are moving toward each other,

The relative speed = 60 + 90 = 150 km/h

Time required for both the trains to meet after 4 am = Remaining distance/Relative Speed

⇒ Time = 260/150 = 1.733 hours

Distance traveled by train B in 1.733 hours = Speed × time

⇒ 90 × 1.733 = 156km

∴ The correct answer is 156km.

Given:

HCF of 768 and x⁶y² is 32xy for natural numbers x ≥ 2, y ≥ 2.

Concept Used:

1. Highest Common Factor (HCF)

2. Prime factorization

Calculation:

Prime factorization of 768:

768 = 2⁸ × 3

Prime factorization of 32xy:

32xy = 2⁵ × x × y

Prime factorization of x⁶y²:

x⁶y² = x⁶ × y²

For HCF to be 32xy, the minimum power of each prime factor in both numbers should be considered:

2⁵ × x × y = 32xy

Here, x and y must be such that:

2⁵ × x × y is a factor of both 768 and x⁶y².

Since 32 = 2⁵,

Therefore, x and y must be:

x = 2

y = 3

Sum of x and y:

⇒ x + y = 2 + 3 = 5

∴ The value of (x + y) is 5.

Total amount = ₹ 9400

After deductions, the ratio of shares is 3 ∶ 4 ∶ 5

Deductions: ₹ 93 from P's share, ₹ 24 from Q's share, ₹ 55 from R's share

Concept Used:

Use the concept of ratios and linear equations to solve for individual shares.

Calculation:

Let the shares of P, Q, and R be p, q, and r respectively.

Given, (p - 93) ∶ (q - 24) ∶ (r - 55) = 3 ∶ 4 ∶ 5

Assume k is the constant of proportionality.

⇒ p - 93 = 3k

⇒ q - 24 = 4k

⇒ r - 55 = 5k

Total amount: p + q + r = 9400

Substitute the values of p, q, and r:

⇒ (3k + 93) + (4k + 24) + (5k + 55) = 9400

⇒ 3k + 4k + 5k + 93 + 24 + 55 = 9400

⇒ 12k + 172 = 9400

⇒ 12k = 9400 - 172

⇒ 12k = 9228

⇒ k = 9228 / 12

⇒ k = 769

Find the share of P:

⇒ p = 3k + 93

⇒ p = 3 × 769 + 93

⇒ p = 2307 + 93

⇒ p = 2400

∴ The share of P is ₹ 2307.

We use the concepts of areas of circles and areas of triangles to solve the problem.

CA and BD are diameters of the circle with center O.

∴ OC = OA = OD = OB = Radius of the circle R = 7 cm

BD = 2R = 14 cm

Radius of shaded circular region, r = OC/2 = 7/2 cm.

Area of the shaded smaller circular region = πr²

= π (7/2 cm)²

= 22/7 × 7/2 × 7/2 cm²

= 77/2 cm²

= 38.5 cm²

Area of the shaded segment of larger circular region = Area of semicircle DAB - Area of ΔDBA

= 1/2 π(OD)² - 1/2 × DB × OA

= 1/2 πR² - 1/2 × 2R × R

= 1/2 × 22/7 × 7² - 1/2 × 14 × 7

= 77 - 49

= 28 cm²

Area of the shaded region = Area of the shaded smaller circular region + Area of the shaded segment of larger circular region.

= 38.5 cm² + 28 cm²

= 66.5 cm²

Calculation:

Let the radius of the bigger circle = R

The radius of a smaller circle = r

AB = 14 cm.

AM = 14/2 = 7 cm

In triangle OAM

OA² = OM² + AM²

R² = r² + 7²

R² - r² = 49

Area of between two triangle = π(R² - r²)

⇒ (22/7) × 49 = 154 square cm

∴ The correct answer is 154 square cm

Given:
AB = 15 cm, BC 20 cm and AC = 25 cm
BP is perpendicular on AC.
Calculation:
Since, 15, 20, and 25 Pythagorean triplet the triangle is a right-angle triangle.

We know that,

AP = AB²/AC

⇒ AP = 15²/25 = 9cm

PC = BC²/AC

⇒ PC = 20²/25 = 16cm

BP = (AB × BC)/AC

⇒ BP = (15 × 20)/25 = 12cm

Area of ΔAPB = (AP × BP)/2

⇒ 9 × 12/2 = 54cm²

Area of ΔBPC = (BP × PC)/2

⇒ 12 × 16/2 = 96cm²

Difference between the area of both triangles,

⇒ 96 - 54 = 42cm²

∴ The correct answer is 42cm².

Concept Used:
The sum of the length of the two sides of a triangle is greater than the length of the third side
and the difference between the two sides of a triangle is less than the length of the third side.
Calculation:

In ΔABD,
⇒ AD + AB > x
⇒ 10 + 6 > x
⇒ 16 > x
Also,
⇒ AD - AB < x
⇒ 10 - 6 < x
⇒ 4 < x
⇒ 4 < x < 16 ..... (i)
In ΔDBC,
⇒ BC + DC > x
⇒ 18 + 6 > x
⇒ 24 > x
Also,
⇒ BC - DC > x
⇒ 18 - 6 > x
⇒ 12 > x
⇒ 12 < x < 24 ...... (ii)
Combining both (i) and (ii),
12 < x < 16
∴ The correct answer is 12 < x < 16.

Given:

a ∶ b ∶ c ∶ d = ∶ ∶ ∶

Calculation:

Directly putting the values of a = √4, b = √3, c = √2, d = √1 in the question,

⇒

⇒ 2/2 = 1

∴ The correct answer is 1.

Given:

Speeds of four cars are 2u, 3u, 4u, and xu.

Time taken by them to cover the same distance is xt, 4t, 3t, and 2t respectively.

Where x, u, t are real numbers.

Concept Used:

Distance = Speed × Time

Calculation:

Since the distance covered by all cars is the same:

Distance₁ = Distance₂ = Distance₃ = Distance₄

Let d be the distance, then:

d = Speed₁ × Time₁

⇒ d = 2u × xt

⇒ d = 2uxt

d = Speed₂ × Time₂

⇒ d = 3u × 4t

⇒ d = 12ut

d = Speed₃ × Time₃

⇒ d = 4u × 3t

⇒ d = 12ut

d = Speed₄ × Time₄

⇒ d = xu × 2t

⇒ d = 2xut

Equating the distances:

2uxt = 12ut

⇒ 2ux = 12u

⇒ x = 12 / 2

⇒ x = 6

Hence, the correct answer is option 2.

Given:
If the people coming from a particular State belonging to S are 15% and 24,000 in number
Calculation:
According to the question,
Total People × 20% × 15% = 24000
⇒ Total People = 24000 × 10000/300
⇒ 800000
The total number of migrating people belonging to the age group B 15%
⇒ 800000 × 15/100
⇒ 1.2 lac
∴ The total number of migrating people belonging to the age group B is 1.2 Lac

Given:

Edge of the cube = 14 cm

Radius of each circle = 7 cm

π = 22/7

Concept:

To find the unpainted surface area of the cube, we need to calculate the total surface area of the cube and then subtract the total area of the painted circles.

Formula Used:

Total surface area of the cube = 6a²

Area of one circle = πr²

Total area of painted circles = 6 × Area of one circle

Unpainted surface area = Total surface area of the cube - Total area of painted circles

Calculation:

⇒ Total surface area of the cube = 6 × (14 cm)²

⇒ Total surface area of the cube = 6 × 196 cm²

⇒ Total surface area of the cube = 1176 cm²

⇒ Area of one circle = π × (7 cm)²

⇒ Area of one circle = (22/7) × 49 cm²

⇒ Area of one circle = 154 cm²

⇒ Total area of painted circles = 6 × 154 cm²

⇒ Total area of painted circles = 924 cm²

⇒ Unpainted surface area = 1176 cm² - 924 cm²

⇒ Unpainted surface area = 252 cm²

Hence, the total area of the unpainted surface is 252 cm².

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

Area of Shaded area = Area of semi circle - area of 2 small semi circles
⇒ π × 36/2 - [2π + 8π] = 8π
Unshaded Area = area of semi circle + area of other small semi circles
⇒ 18π + 2π + 8π = 28π
Required Ratio = 8π/28π = 2/7
∴ The correct answer is 2 : 7.

Calculation:

Unit place of 1^a = 1

Possible unit place of 2^b = 2, 4, 6, 8

Possible unit place of 3^d = 3, 9, 7, 1

Possible unit place of 4^d = 4, 6

Sum of all the above unit places will always be an even number because

sum of two odd and two even numbers will always be even.

Possible even unit digits are = 0, 2, 4, 6, 8

We have to find the combination to get all these numbers for conformation,

1 + 2 + 3 + 4 = 10 (0 at unit place)

1 + 4 + 3 + 4 = 12 (2 at unit place)

1 + 2 + 7 + 4 = 14 (4 at unit place)

1 + 2 + 9 + 4 = 16 (6 at unit place)

1 + 2 + 1 + 4 = 8 (8 at unit place)

All the five unit digit can be found using these combinations.

∴ The correct answer is 5.

Given:

Speed of train A = 60km/h

Speed of train B = 90km/h

Distance between P and Q = 800km

Train A started at 7 pm and Train B started at 4 am the next day.

Calculation:

Train A traveled for 9 hours before train B started,

Distance covered by train A = Time × Speed = 9 × 60 = 540km

Remaining Distance = 800 - 540 = 260km

Since the trains are moving toward each other,

The relative speed = 60 + 90 = 150 km/h

Time required for both the trains to meet after 4 am = Remaining distance/Relative Speed

⇒ Time = 260/150 = 1.733 hours = 1 hours 44 min

Time at which both trains meet = 4 am + 1 hours 44 min = 5 : 44 am

∴ The correct answer is 5 : 44 am.

Calculation:

Statement - I : x =

We can not find real values of x and y from this statement.

Statement - II : y =

y =

y³ = x

y⁴ = xy

Power 4 is always a positive number.

Hence statement 2 alone is sufficient to answer the question.

∴ The correct answer is option 1.

Total People = 800000
According to the question,
The maximum of differences between the number of people coming from different groups P, Q, R and S is
40% - 10%
⇒ 30%
⇒ 30% of 800000
⇒ 2.4 lac
∴ The maximum of differences between the number of people coming from different groups P, Q, R and S is 2.4 lac

Calculation:

x =

We know that the value of will decrease if we increase the denominator,

Let a =

So, x =

Since a is a positive number,

The value of will always be less then = 0.5

∴ The correct answer is option 1.

Given:

50 men can complete a work in 40 days.

They begin the work together but a batch of 5 men left after each period of 10 days.

Concept Used:

Work is measured in man-days. The total work remains constant regardless of the number of men or days.

Formula Used:

Total Work = Number of Men × Number of Days

Calculation:

Total work = 50 men × 40 days = 2000 man-days

First 10 days:

⇒ 50 men × 10 days = 500 man-days

Remaining work = 2000 man-days - 500 man-days = 1500 man-days

Next 10 days:

⇒ 45 men × 10 days = 450 man-days

Remaining work = 1500 man-days - 450 man-days = 1050 man-days

Next 10 days:

⇒ 40 men × 10 days = 400 man-days

Remaining work = 1050 man-days - 400 man-days = 650 man-days

Next 10 days:

⇒ 35 men × 10 days = 350 man-days

Remaining work = 650 man-days - 350 man-days = 300 man-days

Next 10 days:

⇒ 30 men × 10 days = 300 man-days

Remaining work = 300 man-days - 300 man-days = 0 man-days

⇒ Total time = 10 days + 10 days + 10 days + 10 days + 10 days = 50 days

∴ The time to complete the work is 50 days.

Hence, the correct answer is option 2.

Given:

Calculation:
Statement - I : AB + DC = 10 cm
If AB + DC = 10 cm than AD + BC = 10 cm
hence we can calculate the perimeter by using statement 1
Statement - II : AD + BC = 10 cm
If AD + BC = 10 cm than AB + DC = 10 cm
hence we can calculate the perimeter by using statement 2.
∴ The correct answer is option 2.

Given:

Calculation:
For this type of question, we solve the reciprocal of the value and the integers will be values of the variables,

⇒ a × b × c × d × e = 2 × 3 × 4 × 5 × 6
⇒ a × b × c × d × e = 720
∴ The correct answer is 720.

Given:

The chord AB of a circle with center at O is 2√3 times the height of the minor segment.

P is the area of the sector OAB and Q is the area of the minor segment of the circle.

Calculation:

Let the radius of circle = r

AC = BC = √3

OC = r - 1

Using Pythagoras theorem on ΔOCA,

r² = (√3)² + (r -1)²

r² = 3 + r² + 1 - 2r

r = 2

OC = 2 -1 = 1

Ratio of sides = 2 : 2 : 2√3 = 1 : 1 : √3

If ratio of sides of a triangle is 1 : 1 : √3 than the angle opposite to side √3 will be 120°.

P = Area of sector OAB

Q = Area of minor segment under AB

P = π × 2² × (120/360)

Q = [π × 2² × (120/360)] - Area of triangle

Q = [π × 2² × (120/360)] - (2 × 2 × sin 120°)/2 = [π × 2² × (120/360)] - (2 × 2 × (√3/2))/2

Q = (π × 4)/3 - √3

P/Q = 1.7

∴ The correct answer is 1.7.

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.