Test: Elementary Mathematics - 9 - CDS MCQ

Formula used:
Time = Diatance/ Speed
If speed is constant,
Distance ∝ Time
tan θ = Perpendicular / Base
(a² - b²) = (a - b)(a + b)
Calculation:

Let the Height of the tower is 1 unit.

In an Δ ABC

tan 45° = 1/BC

⇒ BC = 1 unit

In an Δ ABD

tan 30° = 1/BD = 1/√3

⇒ BD = √3 unit

Since, CD = BD - BC

⇒ CD = √3 - 1 unit

We know that speed is constant. Therefore

Distance ∝ Time

⇒ √3 - 1 unit = 6 min

⇒ 1 unit = 6/(√3 - 1) = t min

⇒ t =

⇒ t = [∵ (a² - b²) = (a - b)(a + b)]

⇒ t = 3(√3 + 1) = 3 (1.732 + 1)

⇒ t = 3 × 2.732 = 8.19

∴ 8 < t < 8.3

Concept:

Relative speed:

If two trains are traveling with speeds of magnitude u & v respectively, then the relative speed will be

u + v, if both trains are traveling in the opposite direction.

u - v, if both trains are traveling in the same direction.

Formula used:

Speed = distance/ times

Calculation:

Let the speed of train 1 (of length 200 m) is 'u' m/s & speed of train 2

(of length 150 m) be 'v' m/s.

Given that, the man is sitting in the second train & a train of length 200 meters across a man (here, 150 m is irrelevant data).

By using the above concept,

When both trains are traveling in the same direction,

u - v = 200/20

u - v = 10 -----(1)

When both trains are traveling in the opposite direction,

u + v = 200/10

u + v = 20 -----(2)

Adding both equations, we will get

2u = 30 or u =15 m/s

Since, 1 m/s = 3.6 km/hr

u = 15 × 3.6 = 54 km/hr

∴ The correct answer is 54 km/hr.

Alternate Method
Formula used:
Pythagoras theorem:
Hypotenuse² = Base² + Perpendicular²
Area of square = (side)²
Area of equilateral triangle = (√3/4) (side)²
Calculation:

Case:1

By using the Pythagoras' theorem
⇒ a² + a² = (2R)²
⇒ 2a² = 4R²
⇒ a = R√2
Area of square = a²
⇒ x = 2R²
⇒ R = √(x/2) -----(1)

Case:2

In an Δ BOD
Since a center angle will always be double of the angle on the edge of a circumference. Therefore, ∠BOC = 120°
ΔBOC is isosceles triangle. Therefore, ∠BOD = ∠COD = 60°
BD = R sin 60° = R (√3/2)
Therefore, side of triangle = 2 × R(√3/2) = R√3
⇒ y = (√3/4) (side)²
⇒ y = (√3/4) (R√3)²
⇒ y = (3√3/4)R²
From equation (1)
⇒ y = (3√3/4) × (x/2)
⇒ (3√3)x = 8y
∴ The correct answer is 27x² = 64y².

Concept:
HCF: It is the product of the smallest power of the common factors involved in
the prime factorization.
HCF of (a^m - 1) & (aⁿ - 1) is a^{HCF (m, n)} -1
Calculation:
We have 2³⁵ - 1 & 2⁹¹ - 1
By using the above concept
HCF = [2^{HCF (35, 91)} - 1]
Let's find the HCF of (35, 91) first
35 = 5 × 7
91 = 13 × 7
HCF of (35, 91) = 7
Now, the required value
HCF = (2⁷ - 1) = 128 - 1 = 127.

Formula used:

Perimeter = Base of cylinder = 2π × radius

Volume of cylinder = π(radius)² × h

Calculation:

When the square sheet is rolled along one of its sides

The circumference of the base of the cylinder = Length of the side of the square sheet

Therefore, the circumference of the base of the cylinder = 44 cm

⇒ 2πr = 44

⇒ 2 × (22/7) × r = 44

⇒ r = 7 cm

The height of the cylinder is equal to the length of the side of the square sheet, which is 44 cm.

Volume = π × radius² × height

⇒ V = (22/7) × 7² × 44

⇒ V = 22 × 7 × 44

⇒ V = 6776 cm³

Therefore, the volume of the cylinder is 6776 cm³.

Concept:

If in a right-angle triangle,

The ratio of angle are 30° : 60° : 90°

The ratio of sides length opposite to the angle = 1 : √3 : 2

Formula used:

Area of a circle = πr²

Area of a rectangle = Lenght × Breadth

Calculation:

If in a right-angle triangle,

The ratio of angle are 30° : 60° : 90°

The ratio of sides length opposite to the angle = 1 : √3 : 2

In an ΔADC, side AC is opposite to the ∠D which is 90°

We know that, if in the right-angle triangle

The ratio of angle are 30° : 60° : 90°

⇒ AD : DC : AC = r : r√3 : 2r

Area of ΔAEC = Area of ΔADC - Area of ΔADE

Area of ΔAEC = 1/2 (AD × DC) - 1/2 (AD × DE)

Area of ΔAEC = 1/2 (r × r√3) - 1/2 (r × DE)

But, in the ΔADE,

tan 30° = DE/AD = DE/r

= 1/√3 = AD/r

⇒ AD = r/√3

Area of ΔAEC = 1/2 (r × r√3) - 1/2 (r × r/√3)

Area of ΔAEC = r² - r² =

∴ The required area is π/√3.

Statement I:
The length of the line segment joining the mid-points of two sides of Δ is half of the third side of Δ

From this, we cannot say it is a right-angle triangle
So, the statement I is not sufficient to answer the question,
Statement II:
The angles of Δ are in the ratio 1 : 2 : 3
As we know in a triangle sum of all the angles is 180°
So, The angles of Δ are 30°, 60°, 90°
So, from statement II we can say the triangle is a right-angle triangle.
∴ The required answer is Option 1.

Given:

The perimeter of a sector of a circle of radius = 16.4 cm

The radius of a sector of a circle of radius 5.2 cm.

Formula used:

The perimeter (P) of a sector of a circle with radius (r) and central angle (θ) is

P = θr + 2r

Area of sector = (θ/2π) πr²

Calculation:

Using the given values, we have:

16.4 = θ(5.2) + 2(5.2)

16.4 = 5.2θ + 10.4

θ = 6/5.2 ≈ 1.153 radians (rounded to three decimal places)

Area of sector = (θ/2π) πr²

A = (θ/2π)πr²

A = (1.153/2π) × π(5.2)²

A = (1.153 × 5.2 × 5.2)/2

A = 15.6 cm² (rounded to one decimal place)

Therefore, the area of the sector is approximately 15.6 cm².

Formula used:
1 + cot² θ = cosec² θ
1 + tan² θ = sec² θ
cosec θ = 1/sin θ and sec θ = 1/cos θ
Calculation:
(1 + cot² θ) (1 + cos θ) (1 - cos θ) - (1 + tan² θ) (1 + sin θ) (1 - sin θ)
⇒ cosec² θ (1 - cos² θ) - sec² θ (1 - sin² θ)
⇒ cosec²θ . sin² θ - sec²θ . cos² θ
⇒ 1 - 1 = 0

Alternate Method
Formula used
sin²θ + cos²θ = 1
Calculation:
⇒ 2cos²θ + sinθ - 2 = 0
⇒ 2(1 - sin²θ) + sinθ - 2 = 0
⇒ 2 - 2sin²θ + sinθ - 2 = 0
⇒ -2sin²θ + sinθ = 0
⇒ sinθ(-2sinθ + 1) = 0
⇒ sinθ = 0 or -2sinθ + 1 = 0
⇒ sinθ = 0, sinθ = 1/2
But 0 < θ ≤ π/2, Therefore,
⇒ sin θ = 1/2
∴ The correct answer is θ = π/6.

Concept:

The sum of the length of the two sides of a triangle is greater than the length of the third side.

Explanation:

Let's first check whether these points can form a triangle or not (i.e. collinear point).

Statement:1 AB = 5 cm, BC = 5 cm, CA = 6 cm

5 + 5 > 6

5 + 6 > 5

So, A, B & C can form the tirangle.

Therefore, the circle can be drawn from these points.

Statement:2 AB = 3 cm, BC = 4 cm, CA = 7 cm.

3 + 4 = 7

So A, B & C are collinear points & a circle can not be drawn.

Clearly, we can see, we are able to answer the given question with Statement I but not with Statement II.

∴ The correct answer will be option A.

Alternate MethodConcepts or formulas:

A circle can be drawn through three points if they form a triangle, and the sum of any two sides must be greater than the third side (Triangle Inequality Theorem).
Calculation steps:

Statement I:
⇒ 5 cm + 5 cm > 6 cm (True)
⇒ 5 cm + 6 cm > 5 cm (True)
⇒ 6 cm + 5 cm > 5 cm (True)

All inequalities hold; hence, a circle can be drawn.
Statement II:
⇒ 3 cm + 4 cm > 7 cm (False)

The inequality does not hold; hence, a circle cannot be drawn.
Conclusion:

From Statement I, a circle can be drawn. Statement II fails to satisfy the triangle inequality.
Therefore, the correct answer is: 1) Choose this option if the Question can be answered by one of the Statements alone but not by the other.

Calculation:

Statement:1

⇒ m > n

Statement:2 m > 2n

Let n = 1

If m > 2 × 1 then m > 1

∴ Option 2 is correct.

Formula used:

tan θ = Perpendicular / Base

Calculation:

From the figure,

AB = DE & BD = AE

∠EAD = ∠ADB = 30° & ∠CAE = 60°

In an Δ ABD

tan 30° = h/BD = 1/√3

⇒ BD = h√3 = AE

In an ΔAEC

tan 60° = CE/AE = CE/h√3

⇒ √3 = CE/h√3

⇒ CE = 3h

Therefore, the height of the tower

CD = CE + ED = 3h + h = 4h

∴ The height of the tower is 4h.

Concept:
For the number "a", which can be written as,

The total number of positive factors is given as,
T(a) = (a₁ + 1)(a₂ + 1)
Here, p₁ and p₂ are the prime numbers.
Sum of factors =
Calculation:
360 = 2³ × 3² × 5¹
Here, 2, 3 & 5 are the only prime number.
Total number of factor = (3 + 1) × (2 + 1) × (1 + 1) = 24

Sum of factors =

Sum of factors = (2⁴ - 1) × [(3³ - 1)/2] × [(5² - 1)/4]
Sum of factors = (15) × (13) × 6
Sum of factors = 1170
∴ The number of factors of 360 is 24 and the sum of all factors is 1170.

Formula used:
The area of each surface of the cube = (side)² [each surface of a cube is a square]
Calculation:
Let the initial surface area of the cube = 600
then the initial area of each surface = 100
Initial length of side = √100 = 10 unit
Final Surface area of the cube = 600 × 125/100 = 750
Then the final area of each surface = 125
(Final length)² = 125
(Final length) = √125
Therefore, according to the question
P = [(√125 - 10)/10] × 100
P = 10(5√5 - 10)
P = 10(5 × 2.236 - 10) = 11.8
∴ 10 < P < 12

Alternate Method
Given:

Increased length = Original length + (2/3) of original length

Concept used:

New area = Original area

% Change = [(Final value - Initial value)/Initial value ] × 100

Calculation:

Let the length and width of the rectangle be L and W respectively.

Original area = Length × Width = L × W

New length = Original length + (2/3) of original length

New length = L + (2/3)L = (5/3)L

New area = New length × New width

New area = (5/3)L × New width

According to the question, New area = Original area

(5/3)L × New width = L × W
New width = (3/5)W

Now, the percent decrease in width

⇒ [(W - 3W/5) / W] × 100

Percent decrease in width = (2/5) × 100 = 40%

∴ The required value is 40%.

Formula used:

(x + )² = (x² + ) + 2

(x + ) =

Calculation:

Given that

tan⁸ θ + cot⁸ θ = m

Since, cot θ = 1/tan θ. Therefore,

tan⁸θ + (1/tan²θ) = m

Let tan θ = n then

n⁸ + 1/n⁸ = m ---(1)

We know that , (x + ) =

⇒ (n⁴ + 1/n⁴) =

⇒ (n⁴ + 1/n⁴) =

Again applying the same identity

⇒ (n² + 1/n²) =

Again applying the same identity

⇒ (n + 1/n) =

But, n = tan θ

⇒ tan θ + 1/tan θ =

∴ tan θ + cot θ =

Calculation:

The set X is given by

{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98}.

We want to find the maximum size of a subset S of X such that no two elements sum to 100.

The pairs in X that sum to 100 are

(2, 98), (6, 94), (10, 90), (14, 86), (18, 82), (22, 78), (26, 74), (30, 70), (34, 66), (38, 62), (42, 58), (46, 54), (50, 50).

Therefore,

The maximum possible number of elements in S = 13

∴ The maximum possible number of elements in S be 13.

Formula used:

Mode = L +

L = lower limit,

D₁ = ​​The difference between the frequency of the modal class and the

frequency of the class preceding the modal class,

D₂ = The difference between the frequency of the modal class and the

frequency of the class following the modal class.

i = The class width,

Where l = lower limit of median class,

n = number of observations,

i = class size, f = frequency of median class,

cf = cumulative frequency of class preceding the median class.

Calculation:

Since class 60 - 90 has the highest frequency. Therefore,

Model class = 60 - 90

L = 60, D₁ = 7 - 5 = 2

D₂ = 7 - 4 = 3, i = 90 - 60 = 30

Mode = 60 + = 72

Let's calculate the median

Here, n = 20. (n/2) = 20/2 = 10

So, the 10th observation will be in class 60 - 90.

Median = 60 + (30/7) = 450/7

Median = P = 450/7

Mode = Q = 72

According to the question, 7(Q - P) = 9R

⇒ 7 (72 - 450/7) = 9R

⇒ 7 (504 - 450)/7 = 9R

⇒ 54 = 9R

∴ R = 6

Calculation:
Let, x number of persons taking milk only.

According to the question
60 + 15 + (x - 5) + (50 - x) + x = 150
120 + x = 150
x = 150 - 120 = 30
∴ The required value is 30.

Formula used:

Area of circle = πr²

Calculation:

To achieve the maximum coverage, we will arrange the circles such that the

larger circle (with a radius of 4 cm) is inscribed within the rectangle, and the

two smaller circles (each with a radius of 2 cm) are positioned at opposite

corners of the rectangle.

As shown in the figure, radius of three circle will be 4 cm, 2 cm, 2 cm.

Therefore, total area

⇒ 16π cm² + 8π cm²

⇒ 24π cm²

∴ Required area is 24π cm².

Concept:

Even Number: Any number that can be exactly divided by 2 is called an even number. Even numbers always end up with the last digit as 0, 2, 4, 6, or 8.

Calculation:

Let's assume the odd number = 7

Even power of 7 = 7² = 49

49 = 8 × 6 + 1

Similarly, we can also check for another value.

∴ The correct answer is 1.

Concept:

For the quadratic equation ax² + bx + c = 0

Sum of root (α + β) = -b/a

Produt of root (αβ) = c/a

The quadratic equation can also be written as

k[x² - x (α + β) + α β] = 0

Where k is any integer.

Calculation:

We have

ax² + bx + c = 0 ----(1)

One of the roots is 2 (let α)

Statement: 1 Ratio of c to a is 1.

Product of root

αβ = c/a = 1

2 × β = 1

⇒ β = 1/2

The quadratic equation will be

k[x² - x (2 + 1/2) + 2 × (1/2)] = 0

k[x² - 5x/2 + 1] = 0

k[2x² - 5x + 2] = 0

On comparing it to equation (1)

a = 2k, b = -5k, c = 2k

So there is no unique value of a, b & c.

Statement:2 Ratio of b to a is (-5/2)

2 + β = 5/2

β = 1/2

So again there is no unique value of a, b & c.

∴ Option 4 will be correct.

Calculation:
Statement I: (x + y)4 = 256
(x + y)⁴ = 16⁴
(x + y) = ± 4
So, no unique values of x, and y are possible from this statement alone.
Statement II: (x + y)³ < 16
No unique values of x, and y are possible from this statement alone.
Let's combine both statements
Let x = 4
(4)³ < 16
64 < 16 which is wrong
If x = -4
-64 < 16 which is correct.
So, (x + y) = - 4
∴ Optipon 3 is correct.

Given:

AB = x

∠APO = θ

Concept used:

Diagonal of square = √2a

cotθ = base/perpendicular

Calculation:

According to the above concept,

Diagonal of square = √2a

⇒ AC = x√2

⇒ OA = =

Now, according to the above figure,

In ΔOAP,

⇒ cotθ = =

⇒ cotθ = 2√2

∴ The required value of cotθ is 2√2

Given:

The average mark in Mathematics of Section A comprising 30 students is 65

The average mark in Section B comprising 35 students is 70.

The entry of 47 marks was wrongly made as 74

Formula used:

Average = Sum of all observations / Number of observation

Calculation:

By using the Average formula,

Total marks of students in Section A = 30 × 65 = 1950

Total marks of students in section B = 35 × 70 = 2450

Total marks of students in section A & B = 1950 + 2450 = 4400

But, the entry of 47 marks was wrongly made as 74

Actual marks = 4400 + 47 - 74 = 4373

Again, using the above formula

Average = 4373 / (30 + 35) = 67.28

∴ The average mark for both sections is 67.28.

Calculation:

XYXYXY = 100000X + 10000Y + 1000X + 100Y + 10X + Y

XYXYXY = (100000X + 1000X + 10X) + (10000Y + 100Y + Y)

XYXYXY = 101010X + 10101Y

XYXYXY = 10101 × (10X) + 10101(Y)

XYXYXY = 10101.[10X + Y]

Prime factors of 10101 are 3 x 7 x 13 x 37

XYXYXY = 3 x 7 x 13 x 37 (10X + Y)

∴ XYXYXY is divisible by 3, 7, 13 & 37.

Formula used:

Volume of sphere = (4/3) π (radius)³

Surface Area of sphere = 4π(radius)²

Calculation:

Let's assume the radius of the larger solid iron ball is R, and the radius of

each of the smaller solid balls is r.

By using the above formula

V_larger = (4/3)πR³

V_smaller = (4/3)πr³

Since the entire volume of iron is used to make the smaller balls, we have:

V_larger = 64V_smaller

(4/3)πR³ = 64 × (4/3)πr³

R³ = 64r³

R = 4r

Now, let's calculate the surface area of the larger ball and the sum of the surface areas of the smaller balls.

Ratio = (4πR²) / (64 (4πr²)

Ratio = (4π(4r)²) / (256πr²)

Ratio = (64πr²) / (256πr²)

Ratio = 64/256 = 1/4

∴ The required ratio is 1/4.

Calculation:
Statement:1 n³ - n is divisible by 6.
n³ - n = n(n² - 1) = n(n - 1)(n + 1)
This product is the product of three consecutive integers which is always divisible by 6
Statement:2 n⁵ - n is divisible by 5
n⁵ - n = n(n⁴ - 1)
We know that (a² - b²) = (a - b)(a + b)
n⁵ - n = n (n² - 1)(n² + 1)
n⁵ - n = n (n² - 1)[(n² - 4) + 5)]
n⁵ - n = n (n² - 1)(n² - 4) + 5 n (n² - 1)
n⁵ - n = n(n - 1)(n + 1)(n - 2)(n + 2) + 5 n (n - 1)(n + 1)
n⁵ - n = [(n - 2)(n - 1) n (n + 1)(n + 2)] + 5 n (n - 1)(n + 1)
Since the first part is the product of five consecutive numbers which will always be divisible by 5. Therefore,
(n² - n) is divisible by 5.
Statement:3 n5 - 5n3 + 4n is divisible by 120.
n⁵ - 5n³ + 4n
n(n⁴ - 5n² + 4)
n(n² - 4)(n² - 1)
n(n - 2)(n + 2)(n - 1)(n + 1)
(n - 2)(n - 1) n (n + 1)(n + 2)
Which is always divisible by 120.
∴ All three statements are correct.

Concept:
1. For the quadratic equation ax² + bx + c = 0
Sum of root (α + β) = -b/a
Product of root = c/a
2. a² + b² = (a + b)² - 2ab
3. a⁴ + b⁴ = (a² + b²)² - 2(ab)²
Calculation:
x² - 7x + 1 = 0
As α & β be roots of the quadratic equation
α + β = -(-7)/1
α + β = 7
αβ = 1
By using the above identity
α² + β² = (α + β)² - 2αβ = 7² - 2
α² + β² = 47
Now we can use the identity:
α⁴ + β⁴ = (α² + β²)² - 2α²β²
Substituting in the value of α² + β² and αβ = 1, we get:
α⁴ + β⁴ = (47)² - 2 = 2207
∴ The value of α⁴ + β⁴ is 2207.