Test: Elementary Mathematics - 6 - CDS MCQ

Given:

= and x + y = 2a³

Formula used:

a³ + b³ = (a + b) (a² + b² - ab)

(a + b)² = a² + b² + 2ab

Calculation:

We have

=

⇒ =

Put the value of (a2 + b2 - ab) = (a3 + b3)/(a + b)

⇒ =

Similarly,

=

⇒ =

Now,

Here, x = a3 - b³ and y = a3 + b3

⇒ x - y = (a3 - b³) - (a3 + b³) = -2b³

Hence the value of x - y is -2b³.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan β = H/x and tan γ = h/x

⇒ tan β = H/x

⇒ x = H/tan β -----(1)

⇒ tan γ = h/x

⇒ x = h/tan γ -----(2)

From equation (1) & (2), We know that

⇒ H/tan β = h/tan γ

⇒ H tan γ = h tan β

⇒ H tan γ - h tan β = 0

∴ The correct option is A.

Given:

Circumference of a circle is 16.8 cm more than the diameter

Calculation:

Let the diameter of the circle be D

Circumference = 2πr, r = D/2

D + 16.8 = 2π(D/2)

⇒ D + 16.8 = 22/7 × D

⇒ (22/7 -1)D = 16.8

⇒ 15/7 × D = 16.8

⇒ D = (16.8 × 7)/15 = 7.84 cm

∴ The correct answer is 7.84 cm

Given:

Circle C₁ subtend angle = 60°

Circle C₂ subtend angle = 75°

Length of arc L₁ = Length of arc L₂

Formula used:

Length of Arc L = r × θ

Where r = radius and θ = angle subtended

1° = π/180° radians

Calculation:

Length of arc of C₁ = r₁ × 60°

Converting into radians we get

⇒ L₁ = r1 × 60° × π/180° = r₁ × π/3

⇒ L₂ = r₂ × 75° × π/180 = r₂ × 5π/12

Here L₁ = L₂

⇒ r1 × π/3 = = r2 × 5π/12

⇒ r₁/r₂ = (5π/12)/(π/3) = 5/4

Hence, the ratio of radius r₁ : r₂ = 5 : 4

Given:

Statement 1 The cost of 7 pens, 6pencils, and 5 notebooks is 200

Statement 2 The cost of 3 pens, 8 pencils, and 7 notebooks is 210

Calculation:

Let, the cost of 1 pen, 1 pencil, and 1 notebook be Rs a, Rs b, and Rs c respectively

According to the question,

⇒ 7a + 6b + 5c = 200 ....(i)

⇒ 3a + 8b + 7c = 210 ....(ii)

If we add both equations

⇒ 10a + 14b + 12c = 410 ...(iii)

If we multiply 1.5 in the equation (iii)

⇒ 15a + 21b + 18c = 605

⇒ 15 pen + 21 pencil + 18 notebook = 615

∴ The correct answer is option C

Given:

Sinθ = (a + b)/2√ab

Concept:

-1 ≤ sinθ ≤ +1

Calculation:

-1 ≤ (a + b)/2√ab ≤ +1

We solve one of them

(a + b)/2√ab ≤ +1

⇒ a + b ≤ 2√ab

⇒ a + b - 2√ab ≤ 0

⇒ (√a - √b)² ≤ 0

⇒ √a = √b = a = b

∴ The correct answer is a = b

Shortcut Trick We can solve this question by option to satisfy equation

put a = b

then Sinθ = (a + a)/ 2√(a × a)

Sinθ = 2a/2a = 1 (sinθ ≤ 1)

Given:

tan 2θ tan 4θ - 1 = 0, 0 < θ < π/2

Concept Used:

If tan A + tan B = 1

Then A + B = 90⁰

Calculation:

According to the question

⇒ tan 2θ tan 4θ - 1 = 0

⇒ tan 2θ tan 4θ = 1

⇒ 2θ + 4θ = 90⁰

⇒ θ = 15⁰ =  π/12

∴ The value of θ is  π/12 which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Alternate Method We have to check for which value of θ,  0 < θ < π/2

The equation tan 2θ tan 4θ - 1 = 0 satisfies

Option (A): θ =  π/12 = 15⁰

⇒ tan 2θ tan 4θ - 1 = tan 2(15⁰) tan 4(15⁰) - 1

⇒ tan 2θ tan 4θ - 1 = tan 30⁰ tan 60⁰ - 1

⇒ tan 2θ tan 4θ - 1 = (1/√3× √3) - 1 = 0

∴ The value of θ is π/12 which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Given:

An isosceles triangle has its base length 2a & its height is h.

Formula used:

Area of Triangle = 1/2 × base × height

Area of Square = side²

Calculation:

According to the figure,

Area of isosceles triangle = 1/2 × b × h

⇒ 1/2 × 2a × h ( base = 2a) .....(1)

Again, according to the figure,

By Pythagoras theorem

⇒ AC = √DC² + AD²

⇒ AC = √a² + h²

Then, Area of square drawn on AB & AC

Side² = (√a² + h²)² = (a² + h²)

Now, both side will be equal then,

Area = 2(a² + h²) .....(2)

Again, according to the figure,

Area of square drawn on BC which is (2a)²

Area of square = (2a)² = (4a²) .....(3)

By adding above equation (1), (2) & (3) we get,

Area of the figure,

(1/2 × 2a × h) + 2(a² + h²) + (4a²)

⇒ ah + 2a² + 2h² + 4a²

⇒(2a² +4a²) + 2h² + ah

⇒ 6a² + 2h² + ah

∴ The area of the figure is 6a² + 2h² + ah

Given:

tan x = sin θ + cos θ/sin θ - cos θ ,π/4 < θ < π/2

Formula Used:

1. tan θ = P/B

2. sin θ = P/H

3. sin²θ + cos²θ = 1

Calculation:

By Pythagoras Theorem

⇒ H =

⇒ H =

⇒ H = = √2

According to the figure

⇒  √2 sin x = √2 × sin θ + cos θ /√2 

∴ The value of  √2 sin x is equal to sinθ + cosθ.

Given:

L : H = 3 : 2

Sum of two numbers = 45

Concept used:

Numbers are multiples of HCF.

Product of numbers = LCM × HCF

HCF (Highest Common Factor) is the largest number that can divide the given set of two or more numbers.

Calculation:

Let, the common factor of the ratio of L and H be m.

Now, LCM = 3m and HCF = 2m

According to the concept of HCF,

The numbers considered are 2ma and 2mb.

According to the question,

⇒ 2ma × 2mb = 3m × 2m

⇒ ab = 3/2 ---(i)

According to the question,

⇒ 2ma + 2mb = 45

⇒ 2m(a + b) = 45 ---(ii)

Now, using equations (i) and (ii), we can't find the value of all three variables a, b, and m.

∴ The answer cannot be determined due to insufficient data.

Given:

The arch of a bridge is in the form of an arc of a circle. Span of the bridge is 40m & height in the middle is 8m.

Formula used:

Pythagoras theorem:

(Hypotenuse)² = (Base)² + (Perpendicular)²

Calculation:

According to the question,

In Δ OAP

According to the formula,

⇒ R² = 20² + (R - 8)²

⇒ R² = 400 + R² + 64 - 16R

⇒ 400 + 64 = 16R

⇒ 464 = 16R

⇒ R = 29

∴ The radius of the curvature of the bridge is 29m.

Concept:

Componendo Dividendo Rule:

If (a + b ) : (a – b) = (c + d) : (c – d), than

a : b = c : d

Calculation:

According to the question,

⇒ =

⇒ =

By componendo dividendo method,

⇒ =

⇒ =

⇒ =

∴ The equation is equal to .

Given:

Average weekly wages earned by a worker = Rs. 3,520

Formula used:

Average Mean =

x(i) = (L₁ + L₂)/2

L₁ = upper limit & L₂ = lower limit

Where f(i) = median of class & x(i) = Mean of class

Calculation:

3520 =

123200 + 3520k = 126500 + 3300k

220k = 3300

k = 15

∴ Correct answer is 15.

Given:

A, B, and C together can do a piece of work in 36 days

A and B together can do five times as much work as C alone

B and C together can do as much work as A alone

Calculation:

Ratios of Efficiency

(A + B) : C = 5 : 1

⇒ A + B = 5a, C = a --(1)

⇒ A + B + C = 6a

(B + C) : A = 1 : 1

⇒ (B + C) = a, A = a

⇒ A + B + C = 2a

Efficiency should be equal so

A + B + C = 2a × 3 = 6a now

(B + C) = 3a, A = 3a --(2)

By solving (1) and (2)

A = 3a , C = a and B = 3a - a = 2a

Efficiency of (A + C) = n(B)

(3a + a) = n × 2a

n = 4a/2a = 2

∴ The correct answer is n = 2

Shortcut Trick

OD = OB = r

X² + 40² = 30² + (X + 10)²

Clearly X = 30

then OD² = 30²+40²

OD = 50

Diameter = 2r = 100

Alternate Method

Given:

AB = 60 cm and CD = 80 cm

The distance between AB and CD is 10 cm

Formula Used:

In right angles triangle

Hypotenuse2 = Base² + Perpendicular²

Calculation:

Let OP be x cm

⇒ OQ = (x - 10) cm

OD = OB = Radius -----(1)

In ΔOQD, by Pythagoras Theorem

OD² = OQ² + QD² -----(2)

In ΔOPB, by Pythagoras Theorem

OB² = OP² + PB² -----(3)

From (1), (2) and (3)

OQ² + QD² = OP² + PB2

⇒ (x - 10)2 + 402 = x2 + 302

⇒ x² + 100 - 20x + 1600 = x² + 900

⇒ 20x = 800

⇒ x = 40 cm

In ΔOPB, by Pythagoras Theorem

OB² = 40² + 30²

⇒ OB2 = 1600 + 900 = 2500

⇒ OB = 50 cm

Diameter = 2 × Radius = 2 × OB = 100 cm

∴ The diameter of the circle is 100 cm.

Given:

(sin² θ - 4 sin θ + 3) (4 - cos² θ + 4 sin θ) = 0, 0 < θ <  π/2

Calculation:

According to the question

(sin² θ - 4 sin θ + 3) (4 - cos² θ + 4 sin θ) = 0

⇒ (sin² θ - 4 sin θ + 3) = 0 OR (4 - cos² θ + 4 sin θ) = 0

Solving the first equation:

(sin² θ - 4 sin θ + 3) = 0

⇒ sin² θ - sin θ - 3 sin θ + 3 = 0

⇒ sin θ (sin θ - 1) - 3 (sin θ - 1) = 0

⇒ (sin θ - 1) (sin θ - 3) = 0

⇒ sin θ = 1 OR sin θ = 3

⇒ θ = 90⁰ =  π/2 (∵ -1 < sin θ < 1, neglecting sin θ = 3)

Solving the another equation:

(4 - cos² θ + 4 sin θ) = 0

⇒ 4 - (1 - sin² θ) + 4 sin θ = 0

⇒ sin² θ + 4 sin θ + 3 = 0

⇒ sin² θ + sin θ + 3 sin θ + 3 = 0

⇒ sin θ (sin θ + 1) + 3 (sin θ + 1) = 0

⇒ (sin θ + 1) (sin θ + 3) = 0

⇒ sin θ = -1 OR sin θ = -3

⇒ θ = 270⁰ =  3 π/2 (∵ -1 < sin θ < 1, neglecting sin θ = -3)

Again, according to the question 0 < θ <  π/2

∴ The given equation does not satisfy for any value of θ, 0 < θ <  π/2.

Formula used:

a3 + b3 + c3 - 3abc = (a + b +c)(a2 + b2 + c2 - ab - bc - ac)

Calculation:

a2 - bc = α ] × a

a3 - abc = aα ------(1)

b2 - ac = β ] × b

b3 - abc = bβ ------(2)

c2 - ab = γ ] × c

c3 - abc = cγ ------(3)

Adding equation (1),(2) and (3)

a3 + b3 + c3 - 3abc = aα + bβ + cγ

a2 - bc + b2 - ac + c2 - ab = α + β + γ

⇒ =

∴ = 1

Formula Used:

1. (a - b)² = a² + b² - 2ab

2. a² - b² = (a + b)(a - b)

Calculation:

a2 - b2 - c2 + 2bc + a + b - c

⇒ a2 - (b2 + c2 - 2bc) + a + b - c

⇒ a2 - (b - c)² + a + b - c

⇒ (a + b - c) (a - b + c) + (a + b - c)

⇒ (a + b - c) [(a - b + c) + 1]

∴ Factors of the given equation are (a + b - c) and (a - b + c + 1).

Given:

One of the roots is

Statement-1 The product of the roots is

Statement-2 The sum of roots of the quadratic equation is -1.

Calculation:

According to the first statement

Let, the other root = a

⇒ a × =

⇒ a = ×

This means the first statement is sufficient to answer.

According to the second statement

Let, the other root = a

⇒ + a = -1

⇒ a = -1 -()

This means the second statement is sufficient to answer.

∴ The correct answer is option B

Given:

A pendulum swings through an angle = 9°

Length of arc = 14.3 cm

Formula used:

Length of acr of sector of a angle = (θ/360°) × 2πr

Calculation:

Angle of AOB = 9°

14.3 = (9°/360°) × 2 × (22/7) × r

r = 14.3 x 70/11

r = 91 cm

∴ The length of the pendulum is 91 cm

Given:

Weight relations:

Weight of B = Weight of C

A = B - 10, D = E + 4, E = F + 4 and F = G + 4

Age : Weight of D = 9 : 20

Age : weight of A = 2 : 5

Formula used:

Average = Sum of observations/Number of observations

Calculations:

Age : weight of D = 9 : 20

So, Weight will be a multiple of 20 and weight can not exceed 40

Let the weight of D = 40 kg

So, the weight of A = 5 × 3 = 15 kg

Weight of B = A + 10 = 25 kg

Weight of B = Weight of C = 25 kg

Now, E = D - 4

⇒ 40 - 4 = 36

Similarly, F = 32 and D = 28 years

Now weight of G - weight of C = 28 - 25 = 3

Hence, the difference in weight of G and C is 3 kg.

Given:

A solid sphere of radius 3 cm is melted to form a hollow cylinder of height 4cm & external diameter 10cm.

Formula used:

Volume of sphere = 4/3π r³

Volume of hollow cylinder = π(R² - r²)h

Thickness of cylinder = External radius - Internal radius

Calculation:

According to the question,

External diameter = 10cm

External radius = 10/2= 5cm

Volume of sphere = Volume of hollow cylinder

⇒ 4/3π (3³) = π(5²-r²)4

⇒ 9 = 25 - r²

⇒ r² = 16

⇒ r = 4

Thickness of cylinder = 5 - 4 = 1 cm

∴ The thickness of the cylinder is 1 cm.

Shortcut Trick

Clearly we can see, at x = 6,

LHS = RHS

Therefore,

x2 - 3x + 2 = 6² - 3 × 6 + 2

x2 - 3x + 2 = 36 - 18 + 2

∴ x2 - 3x + 2 = 20

Alternate MethodGiven:

Calculate

Concept Used:

Recurrence and simplification of nested fractions followed by algebraic simplification.

Calculation:

Let's solve the recurrence:

⇒ Let .

⇒ Then

⇒ Substitute back in, where

⇒

⇒

⇒

⇒

⇒

⇒ or (Since , )

Now, calculate :

⇒ Substitute into ,

⇒

⇒

⇒

∴ .

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

∠ CAB = 2θ

Calculation:

Let, the radius of the base = BP

⇒ tanθ = BP/

⇒ 2R² × tanθ /(R - r) = BP

⇒ BP = 2R2 × tanθ /(R - r)

∴ The answer is option B.

Given:

AB is a straight road leading to the foot P of a tower of height h.

Q is at distance x from P and R is at a distance y from Q.

The angle of elevation of the top of the tower at Q is twice of that at R.

Concept Used:

Use the formula: tan 2θ = 2tan θ/1 - tan²θ

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan θ = h/(x + y) and tan 2θ = h/x

⇒ tan 2θ = h/x

⇒ 2θ = 2tan θ/1 - tan²θ = h/x

⇒ 2 tan θ /1 - tan² θ (h/x)

⇒ 2h/(x + y) = [1 - (h²/(x + y)²)] (h/x)

⇒ 2h/(x + y) = [(x + y)² - h²]/(x + y)² (h/x)

⇒ (2x) (x + y) = x² + y² + 2xy - h²

⇒ 2x² + 2xy = x² + y² + 2xy - h²

⇒ x² = y² - h²

⇒ h² = y² - x²

To find the height we need y greater than x.

Here, x < y

∴ The correct option is B.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

cot θ = B/P, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

cot α = (x + d)/H and cot β = x/H

⇒ cot α = (x + d)/H

⇒ x = H cot α - d -----(1)

⇒ cot β = x/H

⇒ x = H cot β -----(2)

From equation (1) & (2), We know that

H cot β = H cot α - d

⇒ d = H cot α - H cot β

⇒ d = H (cot α - cot β)

∴ The correct option is C.

Shortcut Trick

Given that,

2s = a + b + c

's' is not present in any option.

Therefore, Put s = 0.

⇒ 02 + (0 - a)(0 - b) + (0 - b)(0 - c) + (0 - c)(0 - a)

⇒ (-a)(-b) + (-b)(-c) + (-c)(-a)

⇒ ab + bc + ca

∴ The correct option is B which is ab + bc + ca.

Given:

Radius of Sphere = 5 cm

Radius of Cylindrical Vessel = 10 cm

Concept Used:

1) V_sphere = 4/3πr³​

V = Volume, r = radius of sphere

2) The base area of the cylinder, A_cylinder = πr²cylinder

3) The rise in the water level, h = V_sphere/Base area of Cylinder

Calculation:

Vsphere =  4/3 × π × (5)³ = 500/3π cm³

⇒ Acylinder = π × (10)² = 100π

Now, we can calculate the rise in water level:

⇒ h = = (500π/3) × (1/100π) = 5/3

∴ The level of water rise is 5/3 cm.

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

Concept used:

The Angle-Angle (AA) Similarity Criterion for triangle similarity states that if two angles of one triangle are respectively equal to two angles of another, the triangles are similar. It proves that proves that the third angle of both triangles will also be equal by leveraging the angle sum property of a triangle.

The Side-Side-Side (SSS) Criterion for triangle similarity states that if the sides of one triangle are proportional to the sides of another (i.e., they maintain the same ratio), their corresponding angles will be equal, and the triangles will be similar.

Calculation:

Let, the height of the cone, AP = h; and AD = a

In ∆ ADE and ∆ AFG,

(i) ∠ ADE = ∠ AFG = 90º

(ii) ∠ DAE = ∠ FAG

So, according to the AA Similarity concept,

∆ ADE ≅ ∆ AFG

Now, according to the SSS similarity criterion,

⇒ DE/FG = AD/AF

⇒ r/R = a/(a + r + R)

⇒ R/r = (a + r +R)/a

⇒ R/r = a/a + (r + R)/a

⇒ R/r = 1 + (r + R)/a

⇒ (R - r)/r = (r + R)/a

⇒ a = (r + R) × r/(R - r)

⇒ a = r(r + R)/(R - r)

Now, h = AD + DO + OF + FP

⇒ (a + r + R + R) = (a + r + 2R)

⇒ =

∴ The height of the cone is .

Given:

Age relations: D = 3 × A

B - C = E - G = D - E

(D + G)/2 = 16, (A + E)/2 = 11 and (B + C)/2 = 11

Age : Weight of D = 9 : 20

Age : weight of A = 2 : 5

Formula used:

Average = Sum of observations/Number of observations

Calculations:

Age : weight of D = 9 : 20

So, Weight will be a multiple of 20 and weight can not exceed 40

Let the weight of D = 40 kg

So, age of D = 9 × 2 = 18 years

Also age of A = D/3 = 18/3 = 6 years

So, the weight of A = 5 × 3 = 15 kg

Age of E - Age of G = 16 - 14 = 2 years

Also from the given conditions,

B - C = 2 -----(1)

B + C = 22 -----(2)

From equation(1) and equation(2)

⇒ 2C = 20

⇒ C = 20/2 = 10

So, Age of B = 22 - 10 = 12

Total age of B, C, D, G, and E = 12 + 10 + 18 + 14 + 16 = 70

Average age = 70/5 = 14

Hence, the required average age is 14 years.