Test: Elementary Mathematics - 2 - CDS MCQ

Given:

The first scheme is a series of discounts of 10% and 20%

The second scheme is a series of discounts of 15% and 15%

Formula Used:

Effective discount = x + y - xy/100

where x and y are

Calculations:

Using the above formula,

Scheme (1)

Effective Discount = 10 + 20 - (10 × 20)/100

= 30 - 2

= 28%

Scheme (2)

Effective Discount = 15 + 15 - (15 × 15)/100

Effective Discount = 30 - 2.25 = 27.75%

Difference = 28 - 27.75 = 0.25%

The answer is first scheme, 0.25%.

As we know, when a line intersects two or more parallel lines are corresponding intersecting angles are equal,

⇒ ∠APQ = ∠ARS = ∠ATU = (∠APQ + ∠ARS + ∠ATU) / 3

= 300°/3 = 100°

Similarly,

⇒ ∠DUT = ∠DSR = ∠DQP = 210°/3 = 70°

Now, if the two lines AB and CD intersect at point O, then in ΔTOU,

⇒ ∠OTU = 180°– 100° = 80°

⇒ ∠OUT = ∠DUT = 70°

But,

⇒ ∠OUT + ∠OTU + ∠TOU = 180°

⇒ 70° + 80° + ∠TOU = 180°

⇒ ∠TOU = 180°– 150° = 30°

∴ Angle of intersection of two lines = 30°

Given:

PQR an isosceles right angle triangle with the center O touches the side QR at W, PR at X and PQ at Y. If PR = 3√2 cm

Concept Used:

In-circle of an isosceles right angle triangle touches the hypotenuse at its midpoint.

Calculation:

PR = 3√2 cm

PQ = QR = 3 cm

In-circle of an isosceles right angle triangle touches the hypotenuse at its midpoint.

So, PX = XR = 1.5√2 cm

WR = XR = 1.5√2 cm

QW = QR − WR = (3 − 1.5√2) cm

So the required ratio is:

1.5√2 ∶ 3 − 1.5√2 ∶ 1.5√2

1 ∶ (√2 − 1) ∶ 1

∴ The ratio PX ∶ QW ∶ PY is 1 ∶ (√2 − 1) ∶ 1.

Given:

Statement I: The radius of the circle is one of the roots of the quadratic equation x² - 5x + 4 = 0 for x > 0 in cm.

Statement II: The radius of the circle is the value of cosec² θ - cot² θ in cm.

Formula used:

Area of the circle = πr²

Where, r is the radius of the circle.

Calculation:

Using statement I:

Radius of the circle is one of the roots of the quadratic equation,

⇒ x2 - 5x + 4 = 0

⇒ x2 - 4x - x + 4 = 0

⇒ (x - 4)(x -1) = 0

⇒ x = 4 and 1

Area of the circle is = π(1)² = π cm²

Area of the circle is = π(4)2 = 16π cm2

So, the statement I is not sufficient to answer the question.

Using statement II:

Radius of the circle is the value of cosec2 θ - cot2 θ in cm

As, we know cosec2 θ - cot2 θ = 1

So, the radius of the circle is 1 cm.

Area of the circle is = π(1)2 = π cm2

So, the statement II is sufficient to answer the question.

∴ The the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

Given:

N = (24³ + 25³ + 26³ + 27³)

Concept:

(a³ + b³) = (a + b)(a² - ab + b²) and

(a + b) always divides a³ + b³

Calculation:

N = (24³ + 25³ + 26³ + 27³)

⇒ N = (24³ + 25³ + 26³ + 27³)

By using the above formula

⇒ N = (24 + 27)(24² - 24 × 27 + 27²) + (25 + 26)(25² - 25 × 26 + 26²)

⇒ N = 51(odd number) + 51(odd number)

As the sum of two odd numbers is even.

⇒ N = 51 × (even number)

Now, required remainder

R = 51 × even number/(51 × 2)

R = even number / 2

Hence, the remainder will be zero.

Alternate Method

Concept:

The remainder when a number is divided by 102 is the same as the remainder when the sum of the individual remainder of its parts divided by 102 is taken.

Calculation:

First, calculate the value of N:

N = (24³ + 25³ + 26³ + 27³) = 13824 + 15625 + 17576 + 19683 = 66708

⇒ (66708) ÷ 102 = 654

So the given expression is completely divided by 102.

Therefore, N divided by 102 leaves a remainder 0.

(A) If N = (2378)² + 2378 + 2379, then the value of √N is 2379.

⇒ N = 2378² + 2378 + 2378 + 1

⇒ N = 2378² + (2 × 2378) + 1

⇒ N = 2378² + (2 × 2378) + 1²

⇒ N = (2378 + 1)²

⇒ √N = 2379

∴ (A) is correct statement.

(B) The unit digit of 349 × 468 × 637 × 698 × 436 is 6.

⇒ 9 × 8 × 7 × 8 × 6

⇒ (72) × (56) × 6

⇒ (2 × 6) × 6

⇒ 72

⇒ 2

∴ The unit digit of the given expression is 2.

So, (B) is the wrong statement.

(C) The total number of positive factors of 3024 is 40.

⇒ 3024 = 2⁴ × 3³ × 7¹

Let there be a composite number N and its prime factors be a, b, c, d, … etc. and p, q, r, s, … etc. be the indices (or powers) of the a, b, c, d, … etc respectively i.e., if N can be expressed as –

N = a^p × b^q × c^r × d^s…..

⇒ The number of positive number of factors = (p + 1) (q + 1) (r + 1) (s + 1)

∴ The total number of positive factors of 3024 is (4 + 1) (3 + 1) (1 + 1) = 40

∴ (C) is the correct statement.

Identity used:

cosec²θ – cot²θ = 1

Calculation:

⇒ cosec θ + cot θ = m ---- (1)

⇒ (cosec θ - cot θ)(cosec θ + cot θ) = 1

⇒ (cosec θ - cot θ) =1/m

⇒ cosec θ - cot θ = 1/m ---- (2)

By adding (1) and (2), we'll get

2cosecθ = m + 1/m

cosecθ = (m² + 1)/2m ...(3)

By subtracting (1) and (2), we'll get

2cotθ = m - 1/m ...(4)

cotθ = (m2 - 1)/2m

⇒

Putting eq. (3) annd (4) in above expression:

⇒ 2mcotθ/2mcosecθ

⇒ cotθ/cosecθ

⇒ cosθ

The value is cosθ.

Given:

Two statements are given.

Calculations:

Considering statement 1:

Let y = √(6 + y)

Squaring both sides,

y² = 6 + y

⇒ y² - y - 6 = 0

⇒ y² - 3y + 2y - 6 = 0

⇒ y(y - 3) + 2(y - 3) = 0

⇒ (y+ 2)(y - 3) = 0

y = -2, y = 3

Value of y cannot be negative, so, y = 3.

Hence, statement - 1 is true.

Considering statement 2:

Let, y =

y = √(x + y)

⇒ y² = x + y

⇒ y² - y - x = 0

On comparing with y² + by + c = 0

We have a = 1, b = - 1, c = - x

y =

=

= ≠ x

Hence, Statement - 2 is false.

∴ The answer is statement 1 is true and statement 2 is false.

Given:

Mean of eight observations. a, a - 7, a - 4, a - 9, a - 8, a - 5, a - 12, a - 1, is 34.

Concept used:

Mean = Sum of observation / Number of observations

Calculation:

Sum of observation = a + a - 7 + a - 4 + a - 9 + a - 8 + a - 5 + a - 12 + a - 1 = 8a - 46

Number of observations = 8

⇒ 34 = (8a-46)/8

⇒ 272 = 8a - 46

⇒ 8a = 272 + 46

⇒ a = 318/8 = 159/4

∴ The value of a is 159/4.

Given:

PQR an isosceles right angle triangle with the center O touches the side QR at W, PR at X and PQ at Y. If PR = 3√2 cm

Calculation:

∠XOY = 180° − ∠YPX

= 180° − 45° = 135°

Radius of in-circle = OY = QW = cm

Area of PYOX = area of ΔPYO + area of PXO = 2(area of ΔPYO)

=

=

= cm²

∴ The area of quadrilateral PYOX is cm².

Given:

Radius of cone = (a + 15) cm

Height of the cone = 20 cm

Volume of the cone = 9240 cm3

Radius of hemisphere = (a + 15) cm

Formula used:

Volume of the right circular cone =

Volume of the hemisphere =

Where, r is the radius of the cone and h is the height of the cone

Calculation:

Volume of the right circular cone,

⇒ = 9240

⇒ = 441

⇒ = 212

⇒ (a + 15) = 21

⇒ a = 6

Radius of hemisphere = (a + 15)

⇒ (6 + 15) = 21 cm

⇒ Volume of the hemisphere =

⇒ 19404 cm³

∴ The volume of the hemisphere on top of cone 19404 cm³.

Given:

If AB ∶ AC = 1 ∶ 3 and DC = 6 cm

Concept Used:

using Internal Bisector theorem.

Calculation:

Direction: In the diagram given below ED is a tangent to the circle and line AE intersects the circle at point C. B is a point on AC such that DB is angle bisector of ∠ADC. If ∠ADB = 30° and ∠EDC ∶ ∠ECD = 2 ∶ 5, then answer the following questions.

According to the internal angle bisector theorem:

⇒

AD = = 3 cm

∴ The AD is 3 cm.

From the venn diagram:

a + c + d + 40 = 0.5 × 500 = 250

⇒ a + c + d = 210 ----(i)

a + b + d + 50 = 0.64 × 500 = 320

⇒ a + b + d = 270 ----(ii)

b + c + d + 70 = 0.74 × 500 = 370

⇒ b + c + d = 300 ----(iii)

a + b + c + d + 40 + 50 + 70 = 500

⇒ a + b + c + d = 340 ----(iv)

Solving all 4 equations:

⇒ a = 40, b = 130, c = 70 & d = 100

∴ Number of students participated in both 2^nd and 3^rd years but not in 1^st year = b = 130

Given:

Sum of roots of the equation x² - bx + c - 5/2 = 0 is b

Product of roots of the equation x2 - bx + c - 5/2 = 0 is c - 5/2

Concept Used:

For the quadratic equation ax² + bx + c = 0

Sum of roots of the equation (α + β) = -b/a

Product of roots of the equation (α × β) = c/a

(a + b)² = (a - b)² + 4ab

Calculation:

For the quadratic equation x2 - bx + c - 5/2 = 0

Sum of roots of the equation (α + β) = -b/a = b

Product of roots of the equation (α × β) = c/a = c - 5/2

Statement I- if the roots of the equation differ by 5

⇒ α - β = 5

⇒ (α + β)² = (α - β)² + 4αβ

⇒ b² = 5² + 4(c - 5/2)

⇒ b² = 25 + 4c - 10

⇒ b² = 15 + 4c

⇒ b² - 4c = 15

∴ Statement I is incorrect.

Statement II- If the roots of the equation differ by 10

⇒ α - β = 10

⇒ (α + β)2 = (α - β)2 + 4αβ

⇒ b² = 10² + 4(c -5/2)

⇒ b² = 100 + 4c - 10

⇒ b2 - 4c = 90

∴ Statement II is correct.

∴ The question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

Given:

x + y + z = 1(11/12)

z/x = 5/2

z/x = y + 7/4

Calculation:

x + y + z = 1(11/12) = 23/12

z/x = 5/2

Again,

5/2 = y + 7/4

So, y = 5/2 - 7/4

⇒ y = (10 - 7)/4

⇒ y = 3/4

So, x + z = 23/12 - 3/4

⇒ x + z = (23 - 9)/12

⇒ x + z = 7/6

So, x = 7/6 × 2/7 [∵ z : x = 5 : 2]

⇒ x = 1/3

And z = 7/6 × 5/7 [∵ z : x = 5 : 2]

⇒ z = 5/6

Now, To find (z – x) : (y – x)

⇒ (5/6 - 1/3) : (3/4 - 1/3)

⇒ 3/6 : 5/12 = 6 : 5

∴ Required answer is 6 : 5

GIVEN:

CONCEPT:

Mean of grouped data.

FORMULA USED:

Mean = ΣF_iX_i/ΣF_i

CALCULATION:

20 + 25 + X + Y + 10 = 100

⇒ X + Y = 45 ----(1)

Mean = ΣF_iX_i/ΣF_i

21.2 = (24X + 32Y + 960)/100

⇒ 24X + 32Y + 960 = 2120

⇒ 24X + 32Y = 1160

⇒ 3X + 4Y = 145 ----(2)

From (1) and (2):

⇒ X = 35 and Y = 10

∴ The value of X is 35

Given:

tan²A + 2tanA - 63 = 0

Formula used:

Pythagoras Theorem,

h =

Where, h = Hypotenuse, p = Perpendicular, and b = Base

Calculation:

tan²A + 2tanA - 63 = 0

Let, tan A = x

⇒ x² + 2x - 63 = 0

⇒ x² + 9x - 7x - 63 = 0

⇒ x (x + 9) - 7 (x + 9) = 0

⇒ (x + 9) (x - 7) = 0

⇒ x + 9 = 0 ⇒ x = - 9 ["-" will be neglected because 0 < A < π/2]

⇒ x - 7 = 0 ⇒ x = 7

So, tan A = 7/1 = P/b

Using Pythagoras Theorem,

h =

⇒ h = = √50

So, 2sinA + 5cosA

= [2 × ] + [5 × ]

= + = =

∴ The value of (2sinA + 5cosA) is .

Given :

AC and BD are the chords of the circle

AB and CD are the diameters

∠OAC = 50°

Calculations :

OA and OC are the radius of circle

They subtends equals angles at A and C

⇒ ∠OAC = ∠OCA = 50°

⇒ Area of triangle AOC = ∠OAC + ∠OCA + ∠AOC = 180°

⇒ 50° + 50° + ∠AOC = 180°

⇒ ∠AOC = 180° - 100°

⇒ ∠AOC = 80°

⇒ ∠AOC = ∠DOB (vertically opposite angles)

∴ The value of ∠DOB is 80° .

Given:

P = and Q =

The expression is

Formula used:

(a + b)² = a² + b² + 2 ab

(a - b)² + (a + b)² = 2 (a² + b²)

Calculation:

The expression is given

⇒

⇒

⇒

⇒

⇒

⇒

⇒ ---------- (1)

We will calculate the value of

⇒ =

⇒ =

⇒ =

⇒ = 6

Put the value of in equation (1)

⇒ = ( 6 + 2)²

⇒ = 8²

⇒ = 64

∴ The value of the expression is 64.

Formula used:

p(x) = ax³ + bx² + cx + d

If α, β, γ are the roots of the above cubic polynomial, then

• α + β + γ = -b/a
• α β γ = -d/a

Calculation:

On comparing given equation with p(x) = ax³ + bx² + cx + d

we have, a = 4, b = 3, c = 8, d = -7

⇒ α + β + γ = -b/a = -3/4

⇒ α β γ = -d/a = -(-7)/4 = 7/4

∴ The correct answer is -3/4, 7/4.

Given:

5 men and 12 women can complete a work in 2 days

3 men and 4 women can complete it in 5 days.

Calculation:

According to question,

(5 men + 12 women) × 2 = (3 men + 4 women) × 5

⇒ 10 men + 24 women = 15 men + 20 women

⇒ 5 men = 4 women

⇒ men/women = 4/5

So, men : women = 4 : 5 i.e ratio of efficiency

Now, total work = (5 × 4 + 12 × 5) × 2

⇒ (20 + 60) × 2

⇒ 160 units

So, to complete 160 units in 8 days total men need = 160/(8 × 4) [Each man has an efficiency of 4 so we will multiply 4 with the time]

⇒ 5

∴ Required number of men is 5.

Given:

Total surface area : Curve surface area = 7 : 4

Volume of cylinder = 4851 m³

Breadth of rectangle is half the height of cylinder

Length of rectangle is 17 m more than breadth

Formula used:

For cylinder

• CSA = 2πr × h (r = radius of cylinder)
• TSA = CSA + 2 × Base
• TSA = (2πr × h) + (2πr²) = 2πr × (h + r)
• Volume (V) = πr² × h

For rectangle,

• Perimeter of rectangle (P) = 2 × (L + B)
• Diagonal of rectangle (D) = √(L² + B²)
• Diagonal of square (d) = √2a (a = side of square)

Where, TSA = Total surface area, CSA = curved surface area (CSA),

Height of cylinder (h), Breadth of rectangle (B), Length of rectangle (L)

Calculation:

⇒ TSA / CSA = 7/4

⇒ (2πr × (h + r)) / (2πr × h) = 7/4

⇒ (h + r) / h = 7/4

⇒ 4h + 4r = 7h

⇒ 4r = 3h

⇒ h = 4r/3

Now,

⇒ V = πr² × (4r/3)

⇒ 4851 m³ = (22/7) × r² × (4r/3)

⇒ (441 × 21) / 8 = r³

⇒ r = 21/2 m

⇒ h = 14 m

Now,

⇒ B = h/2 = 7 m

⇒ L = B + 17 m = 24 m

⇒ P = 2 × (24 + 7) m = 62 m

∴ The correct answer is 62 m.

Volume of the pyramid = 400 cm³

⇒ 1/3 × Area of the base × height = 400

⇒ 1/3 × 10² × height = 400

⇒ Height = 12 cm

When the pyramid will be opened, it will look like as follows:

In the figure, AC is the slant height of the pyramid.

So, AC² = OA² + OC²

⇒ AC² = 12² + 5²

⇒ AC = 13 cm

∴ Height of the maximum possible rod placed in the pyramid will be the slant height of the pyramid.

So, volume of rod = πr²h = 22/7 × (0.7)² × 13 = 20.02 cm³

Volume of all the 4 rods = 4 × 20.02 = 80.08 cm³ = 80 cm³ (Approx)

∴ Required volume = 400 – 80 = 320 cm³

Given:

43x × 47y = (2021)2, x ≠ 0, y ≠ 0

To find

Calculation:

We have 43x × 47y = (2021)2

On factorization of 2021 we get

2021 = 43 × 47

So, we can write the equation (i) as

43x × 47y = (43 × 47)2

⇒ 43x × 47y = 432 × 472

On comparing we get

x = 2 and y = 2

Now we have to find the value of

⇒ =

∴ The required value is 5.

Given:

Ω is directly proportional to cube of x.

Ω is inversely proportional to fourth power of y.

Concept used:

In such questions we assume a proportionality constant k and find its value to determine the answer.

Calculation:

Let the proportionality constant be k then we have:

Ω = kx³ / y⁴

When x = 4 and y = 2 we have Ω = 68 thus

68 = k(4)³ / 2⁴

68 = 64k / 16

68 = 4k

k = 68 / 4

k = 17.

Thus we have the relation:

Ω = 17x³ / y⁴

When x = 9 and y = 3 we have

Ω = 17(9)³ / 3⁴

Ω = 17(729) / 81

Ω = 17(9)

Ω = 153.

∴ The value of Ω is 153.

Given:

A tangent AB touches a circle of diameter FG at point F and tangent DC touches the circle at point G.

∠FOB = 60º

Formula used:

A pair of tangents drawn from common point P to the circle at A and B,

then, PA = PB (The lengths of tangents drawn from an external point to a circle are equal)

Calculation:

According to the question, the required figure is:

In ΔFBO and ΔBOE,

Since the lengths of tangents drawn from an external point to a circle are equal,

Therefore,

FB = BE

OB (Common line-segment)

OF = OE = radius

So, ΔFBO ∼ ΔBOE

∴ ∠FOB = ∠EOB

In ΔEOC and ΔGOC,

Since the lengths of tangents drawn from an external point to a circle are equal,

Therefore,

EC = GC

OC (Common line-segment)

OE = OG = radius

So, ΔEOC ∼ ΔGOC

∴ ∠EOC = ∠GOC

The straight-line angle is 180∘.

⇒ ∠EOC + ∠GOC + ∠FOB + ∠EOB = 180º

⇒ 2∠GOC + 2∠FOB = 180º

⇒ ∠GOC + ∠FOB = 90º

⇒ ∠GOC + 60^º = 90º

⇒ ∠GOC = 30º

∴ The required angle ∠ GOC is 30º.

Formula used:

(a + b)2 = a2 + 2ab + b2

Calculation:

Let

y = 2 – 2 cos x – cos2 x

⇒ y = 3 - (cos2 x + 2 cos x + 1)

We know that,

(a + b)2 = a2 + 2ab + b2

⇒ y = 3 - (cos x + 1)2 ----(1)

For maxima in the interval x ∈ [-π/2, π/2]

For y to be maximum,

(cos x + 1)² to be minimum & hence, cos x = 0

(y)max = 3 - (0 + 1)2 = 2

Similarly, for minimum,

(cos x + 1)² to be maximum & hence, cos x = 1

⇒ (y)min = 3 - (1 + 1)2 = -1

Hence, the required sum is

ymax + ymin = 2 - 1 = 1

∴ Sum of the greatest to the smallest value is 1.

Given:

pq + qr + rp = 0 ----(1)

Calculation:

Now, We have to find the value of

[p² / (p² - qr)] + [q² / (q² - rp)] + [r² / (r² - pq)]

From equation (i), we get

- qr = rp + pq ----(ii)

- rp = pq + qr ----(iii)

- pq = qr + rp ----(iv)

Now, From equation (ii), (iii) and (iv), we get

⇒ [p² / (p² + pq + rp)] + [q² / (q² + pq + qr)] + [r² / (r² + qr + rp)]

⇒ [p² / p (p + q + r)] + [q² / q (q + p + r)] + [r² / r (r + q +p)]

⇒ [p / (p + q + r)] + [q / (p + q + r)] + [r / (p + q + r)]

⇒ (p + q + r) / (p + q + r)

⇒ 1

∴ The required value of the expression is 1.

Given:

cos (x - y) = √3/2

sin (x + y) = 1/2

Formula:

cos30° = √3/2

sin30° = 1/2

Calculation:

cos(x - y) = √3/2

⇒ cos(x - y) = cos30°

(x - y) = 30° ---- (1)

sin(x + y) = 1/2

⇒ sin(x + y) = sin30°

(x + y) = 30 ---- (2)

Add equation (1) and equation (2), we get

2x = 60°

∴ x = 30°

GIVEN:

Three statements.

CONCEPT:

Mensuration

FORMULA USED:

Base area of cylinder = πR²

Curved surface area of cylinder = 2πRh

Volume of cylinder = πR²h

Volume of sphere 4πR³ / 3

Lateral height of the cone = √(R² + H²)

Curved surface area of cone = πRl

CALCULATION:

A:

Base area of cylinder = πR² = 38.5

R² = 12.25

⇒ R = 3.5 cm

Curved surface area of cylinder = 2πRh = 132

h = 6 cm

Volume of cylinder = πR²h

= 38.5 × 6

= 231 cm³

B:

Volume of large spherical balls = N × Volume of each small spherical ball

4πR³ / 3 = N × 4πr³ / 3

⇒ N × 3³ = 15³

⇒ N = 125

C:

Lateral height of the cone = √(7² + 24²) = 25 cm

Curved surface area of cone = πRl

= (22 / 7) × 7 × 25

= 550 cm²