CDS Elementary Mathematics 21 April 2024 Official Paper (Previous Year Paper)

Given:

a ∶ b ∶ c ∶ d = $√4$ ∶ $√3$ ∶ $√2$ ∶ $√1$

Calculation:

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

$\frac{\left(-a^2+b^2+c^2+d^2\right)}{\left(a^2-b^2+c^2-d^2\right)}$

⇒ $\frac{\left(-4+3+2+1\right)}{\left(4-3+2-1\right)}$

⇒ 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:

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:

Principal (P) = ₹ 10,000

Rate of Interest (R) = 5%

Time (T) = 3 years

Concept Used:

Simple Interest (SI) is calculated using the formula:

SI = P × R × T / 100

Compound Interest (CI) is calculated using the formula:

CI = P (1 + R/100)^T - P

Calculation:

Simple Interest (SI):

SI = ₹ 10,000 × 5 × 3 / 100

⇒ SI = ₹ 1,500

Compound Interest (CI):

CI = ₹ 10,000 (1 + 5/100)³ - ₹ 10,000

⇒ CI = ₹ 10,000 (1.05)³ - ₹ 10,000

⇒ CI = ₹ 10,000 × 1.157625 - ₹ 10,000

⇒ CI = ₹ 11,576.25 - ₹ 10,000

⇒ CI = ₹ 1,576.25

Difference between CI and SI:

Difference = CI - SI

⇒ Difference = ₹ 1,576.25 - ₹ 1,500

⇒ Difference = ₹ 76.25

Hence, the difference between compound interest and simple interest received on ₹ 10,000 after 3 years at a rate of 5% is ₹ 76.25.

Given:

Listed Price of the book = L

Buying Price = (3/4) × L

Selling Price = (1 + 50/100) × L = 1.5 × L

Concept Used:

Percentage Gain = [(Selling Price - Buying Price) / Buying Price] × 100

Calculation:

Buying Price = (3/4) × L

Selling Price = 1.5 × L

Percentage Gain = [(1.5 × L - (3/4) × L) / (3/4) × L] × 100

⇒ Percentage Gain = [(1.5L - 0.75L) / 0.75L] × 100

⇒ Percentage Gain = (0.75L / 0.75L) × 100

⇒ Percentage Gain = 1 × 100

⇒ Percentage Gain = 100%

∴ The correct answer is option 4.

Given:

Difference between the interior and exterior angles of a regular polygon = 144º

Concept Used:

For a regular polygon with n sides:

• Interior angle = (n-2) × 180º / n
• Exterior angle = 360º / n

 

Calculation:

144 = [(n-2) × 180 / n] - (360 / n)

⇒ 144 = (180n - 360 - 360) / n

⇒ 144 = (180n - 720) / n

⇒ 144n = 180n - 720

⇒ 180n - 144n = 720

⇒ 36n = 720

⇒ n = 720 / 36

⇒ n = 20

Hence, the number of sides of the polygon is 20.

Given:

Sum of the roots (S) = 2

Product of the roots (P) = -100

Concept Used:

For a quadratic equation of the form x² + bx + c = 0,

The sum of the roots (α + β) = -b

The product of the roots (αβ) = c

Formula Used:

x² - (Sum of the roots)x + Product of the roots = 0

Calculation:

Using the given sum and product of the roots:

⇒ Sum of the roots (S) = 2

⇒ Product of the roots (P) = -100

Substitute into the formula for the quadratic equation:

⇒ x² - Sx + P = 0

⇒ x² - 2x - 100 = 0

Thus, the quadratic equation is:

x² - 2x - 100 = 0

Hence, the correct answer is option 3.

The Given:

Polynomial: p(x) = x³ + 3x² - 6x - a

One zero of the polynomial: x = 2

Concept Used:

If x = 2 is a zero of the polynomial, then p(2) = 0.

Calculation:

Since x = 2 is a zero,

p(2) = 0:

⇒ 2³ + 3(2)² - 6(2) - a = 0

⇒ 8 + 12 - 12 - a = 0

⇒ 8 - a = 0

⇒ a = 8

Now, P(x) =  x3 + 3x2 - 6x - 8

Since 2 is the zero of the polynomial, so P(x) can be factorized as

P(x) =  x3 + 3x2 - 6x - 8 = (x -2)( x² + 5x + 4)

Now, 

( x2 + 5x + 4)

⇒ (x2 + 4x + x + 4)

⇒  x(x + 4) + 1(x + 4)

⇒ (x + 4)(x +1)

So, other zeroes are -4 and -1

Thus, 

Sum of the squares of the other zeros of the polynomial

⇒ (-4)² + (-1)²

⇒ 16 + 1 = 17 

∴ The correct answer is 17.

Given:

t = cos 79°

Formula Used:

sin²θ + cos²θ = 1

a² - b² = (a - b)(a + b)

Calculation:

cosec 79°cosec 79° (1 - cos 79°)

⇒ $\frac{1}{sin\ 79\circ}$(1 - t)

⇒ $\frac{1}{\sqrt(1 - cos^279\circ)} \times (1 - t)$

⇒ $\frac{1 - t}{\sqrt(1 - t^2)}$

⇒ $\frac{1 - t}{\sqrt(1 - t)(1 + t)}$

⇒ $\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}$

∴ The correct answer is $\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}$.

Given:

p(x) = x4 + a3x3 + a2x2 + a1x + a0 and q(x) = x4 + b3x3 + b2x2 + b1x +b0

Calculation:

Since α, β, γ , δ are zeroes of p(x),

p(x) = (x - α)(x - β)(x - γ)(x - δ)

α, β, γ, λ are zeroes of q(x),

q(x) = (x - α)(x - β)(x - γ)(x - λ)

⇒ $\frac{p(x)-q(x)}{(x-\alpha)(x-β)(x-\gamma)}$ 

⇒  $\frac{[(x - α)(x - β)(x - γ)(x - δ)] - [(x - α)(x - β)(x - γ)(x - λ)]}{(x-\alpha)(x-β)(x-\gamma)}$ 

⇒ $\frac{(x - α)(x - β)(x - γ)[(x - δ) - (x - λ)]}{(x-\alpha)(x-β)(x-\gamma)}$ 

⇒  λ - δ

∴ The correct answer is λ - δ.

Given:

Equation: $x \cos θ = x^2 + p$

Range for $θ: ( 0 \leq θ \leq \frac{π}{4} )$

This equation must have a real solution for every (θ ) in the given range.

Calculation:

$x^2 + p - x \cos θ = 0$

For real solution,

$b^2 \geqslant 4ac$

cos²θ $\geqslant 4p$

cosθ will be maximum at 0° = 1

4p $\leqslant 1$

$p \leq \frac{1}{4}$

∴ The correct answer is $p \leq \frac{1}{4}$

Shortcut Trick

Simplify the expression: cos²θ + 3 sin²θ + 2 = (1 − sin²θ) + 3 sin²θ + 2 = 2 sin²θ + 3.

Since the range of sin²θ is [0, 1]:

Least value of the expression = 2 × 0 + 3 = 3.

Greatest value of the expression = 2 × 1 + 3 = 5.

Difference = Greatest value − Least value = 5 − 3 = 2.

∴ The correct answer is 2.

Alternate Method

Given:

Trigonometric expression: cos²θ + 3 sin²θ + 2

Formula Used:

⇒ Trigonometric Identity: cos²θ + sin²θ = 1 ⇒ cos²θ = 1 − sin²θ

⇒ Range of squared sine function: 0 ≤ sin²θ ≤ 1

Calculations:

⇒ Let the expression be E = cos²θ + 3 sin²θ + 2

⇒ Substituting cos²θ = 1 − sin²θ into the expression:

⇒ E = (1 − sin²θ) + 3 sin²θ + 2

⇒ E = 2 sin²θ + 3

⇒ Since 0 ≤ sin²θ ≤ 1, we determine the extreme values:

⇒ For the least value of E, substitute sin²θ = 0:

⇒ Least value = 2 × 0 + 3 = 3

⇒ For the greatest value of E, substitute sin²θ = 1:

⇒ Greatest value = 2 × 1 + 3 = 5

⇒ To find the difference between the greatest and least values:

⇒ Difference = Greatest value − Least value

⇒ Difference = 5 − 3 = 2

∴ The correct answer is 2.

Additional Information

Trigonometric Pythagorean Identities

The primary identities are sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Range of Trigonometric Functions

For any real angle θ, the squared functions have the following boundaries: 0 ≤ sin²θ ≤ 1 and 0 ≤ cos²θ ≤ 1.

General Form Range

The range of the linear combination expression a sin θ + b cos θ is [−√(a² + b²), √(a² + b²)].

Given:

ABC is a right-angled triangle,

AB = 6 cm and BC = 8 cm

Formula Used:

Area of right-angled triangle = (Base × Height)/2

Perimeter of square = 4 × Length of a Side

Calculation:

Let the Length of a side of square inscribed = a cm

AF = 6 - a

CD = 8 - a

Area of Δ ABC = (8 × 6)/2 = 24 cm²

Area of Δ ABC = Area of Δ AFE + Area of Δ CDE + Area of square BDEF

24 = a(6 - a)/2 + a(8 - a)/2 + a²

⇒ a = 24/7

Perimeter of square = 4 × a = 4 × 24/7 = 96/7 cm

∴ The correct answer is 96/7 cm.

Given:

Equation: 2x² - 4x + k = 0

Concept Used:

For the quadratic equation ax² + bx + c = 0 to have real roots, the discriminant (Δ) must be non-negative.

Formula Used:

Discriminant (Δ) = b2 - 4ac

Calculation:

Given equation: 2x² - 4x + k = 0

Here, a = 2, b = -4, and c = k

Discriminant (Δ) = b² - 4ac

⇒ Δ = (-4)² - 4 × 2 × k

⇒ Δ = 16 - 8k

For real roots, Δ ≥ 0

⇒ 16 - 8k ≥ 0

⇒ 16 ≥ 8k

⇒ k ≤ 2

The greatest value of k for which the equation has real roots is 2.

∴ The correct answer is option 2.

Calculation:

110, 41, 43, 95, 127, 99, 61, 92, 71, 93, 110, 36

Old series in ascending order,

36, 41, 43, 61, 71, 92, 93, 95, 99, 110, 110, 127

93 is replaced by 94, new series in ascending order,

36, 41, 43, 61, 71, 92, 94, 95, 99, 110, 110, 127

Statement 1:

Median of old series = (92 + 93)/2 = 92.5

Median of new series = (92 + 94)/2 = 93

The difference between new median and old median = 93 - 92.5 = 0.5

Statement 1 is false.

Statement 2:

Old mean = (36 + 41 + 43 + 61 + 71 + 92 + 93 + 95 + 99 + 110 + 110 + 127)/12

⇒ 978/12 = 81.5

New mean = (36 + 41 + 43 + 61 + 71 + 92 + 94 + 95 + 99 + 110 + 110 + 127)/12

⇒ 979/12 = 81.5833

The difference between new mean and old mean = 81.5833 - 81.5 = 0.0833

Statement 2 is true.

Statement 3:

Old mode = 110

New mode = 110

The difference between new mode and old mode = 110 - 110 = 0

Statement 3 is true.

∴ The correct answer is option 2.

Calculation:

100th place mean 3rd digit from the right.

if a number has 25 in the last then,

25² = 625

25³ = 15625

25⁴ = 390625

(125)² = 15625

There will always be 625 at last 3 places

So, last 3 digit of (225)⁴⁰ will be 625.

∴ The correct answer is 6.

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.

Shortcut Trick

Remainder when divided by 10 is simply the units digit of the number.

Units digits of 4ⁿ are {4, 6} → Cycle of 2 (4, 6, 4, 6...)

Units digits of 6ⁿ is {6} → Always 6 for all natural numbers n.

Units digits of 9ⁿ are {9, 1} → Cycle of 2 (9, 1, 9, 1...)

Units digits of 11ⁿ is {1} → Always 1 for all natural numbers n.

Sum of all distinct units digits involved across the terms = 1 + 4 + 6 + 9 = 20.

∴ The correct answer is 20.

Alternate Method

Given:

Expression: 4ⁿ + 6ⁿ + 9ⁿ + 11ⁿ

Divisor: 10

Formula Used:

Remainder (A + B + C + D) ÷ M = [Rem(A/M) + Rem(B/M) + Rem(C/M) + Rem(D/M)] ÷ M

The remainder when any number is divided by 10 is its units digit.

Units Digit (Remainder ÷ 10) Cycles

4ⁿ n=1 → 4 n=2 → 6

6ⁿ n=1 → 6 n=2 → 6

9ⁿ n=1 → 9 n=2 → 1

11ⁿ n=1 → 1 n=2 → 1

Distinct Units Digits = {1, 4, 6, 9}

Calculations:

⇒ Let n = 1 (Odd):

⇒ Remainder = (4¹ + 6¹ + 9¹ + 11¹) ÷ 10 = (4 + 6 + 9 + 1) ÷ 10 = 20 ÷ 10 = 0

⇒ Let n = 2 (Even):

⇒ Remainder = (4² + 6² + 9² + 11²) ÷ 10 = (16 + 36 + 81 + 121) ÷ 10 = (6 + 6 + 1 + 1) ÷ 10 = 14 ÷ 10 = 4

⇒ Let n = 3 (Odd):

⇒ Remainder = (4³ + 6³ + 9³ + 11³) ÷ 10 = (64 + 216 + 729 + 1331) ÷ 10 = (4 + 6 + 9 + 1) ÷ 10 = 0

⇒ The remainders follow a periodic cycle of (0, 4) for consecutive values of n.

⇒ As per the logic of distinct remainders of individual terms appearing in the expression for various n:

⇒ Distinct components = {1, 4, 6, 9}

⇒ Sum of all these distinct values = 1 + 4 + 6 + 9 = 20

∴ The correct answer is 20.

Additional Information

Cyclicity of Units Digit

The units digit of numbers follows a cycle: 0, 1, 5, 6 (Cycle 1); 4, 9 (Cycle 2); 2, 3, 7, 8 (Cycle 4).

Remainder Additive Rule

The remainder of a sum is equal to the sum of the remainders of its individual terms: (a + b) mod n = (a mod n + b mod n) mod n.

Units Digit of Power of 4 and 9

For 4ⁿ, units digit is 4 if n is odd and 6 if n is even. For 9ⁿ, units digit is 9 if n is odd and 1 if n is even.

Given:

Number = 12345678910111213 ... 99100

Concept Used:

To find the remainder when a number is divided by 16, we only need to consider the last 4 digits of the number.

Calculation:

The number ends with "99100".

Last 4 digits = 9100

⇒ 9100 mod 16

⇒ 9100 ÷ 16 = 568 remainder 12

⇒ 9100 = 568 × 16 + 12

⇒ Remainder = 12

Hence, the remainder when the number is divided by 16 is 12.

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.

Given:

p = $\sqrt[3]{\left(a+\sqrt{a^2+b^3}\right)}+\sqrt[3]{\left(a-\sqrt{a^2+b^3}\right)}$

Calculation:

We can easily solve this by putting values of a and b,

Lets assume a = 1 and b = 0

so,

p = $\sqrt[3]{\left(1+\sqrt{1^2+0^3}\right)}+\sqrt[3]{\left(1-\sqrt{1^2+0^3}\right)}$

⇒ p = $\sqrt[3]{\left(1+1\right)}+\sqrt[3]{\left(1-1\right)}$

⇒ p = $\sqrt[3]{\left(2\right)}$

Putting value of p and b,

⇒ p3 + 3bp = $\sqrt[3]{\left(2\right)}^3$ + 3 × 0 × $\sqrt[3]{\left(2\right)}$

⇒ p3 + 3bp = 2 ----- (i)

Putting value of a = 1 in options,

Option 1 = -2a = -2 × 1 = -2

Option 2 = a = 1 = 1

Option 3 = 2a = 2 × 1 = 2

Option 4 = 3a = 3 × 1 = 3

Option 3 matches with equation (i).

∴ The correct answer is option 3.

Given:

Length of plank = 4.25 m

Width of plank = 3.4 m

Concept:

To find the largest size of square pieces that can be cut without

wastage, we need to find the greatest common divisor (GCD) of the

length and width of the plank.

Formula Used:

GCD of two numbers a and b is the largest number that divides both a

and b without leaving a remainder.

Calculation:

Length = 4.25 m = 425 cm

Width = 3.4 m = 340 cm

GCD(425, 340) = 85 cm

⇒ Number of pieces along the length = 425 / 85 = 5

⇒ Number of pieces along the width = 340 / 85 = 4

⇒ Total number of pieces = 5 × 4 = 20

Hence, the number of square pieces of the largest size that can be cut from the plank is 20.

Given:

x4 - 13x2y2 - 300y4,

x3 - 4x2y - 4xy2 - 5y3,

x3 - 125y3

Calculation:

x3 - 125y3 = (x)² - (5y)² = (x - 5y)(x² + 25y² + 5xy) ..... (i)

x4 - 13x2y2 - 300y4 = x⁴ - 25x²y² + 12x²y² - 300y²

x²(x² - 25y²) + 12y²(x² - 25y²)

(x² + 12y²)(x2 - 25y2) 

(x2 + 12y2)(x - 5y)(x + 5y) .....(ii)

Since (x - 5y) is zero of both the polynomial (i) and (ii)

Put x - 5y = 0, x = 5y in third polynomial

to confirm (x - 5y) as HCF,

x3 - 4x2y - 4xy2 - 5y3 = 125y³ - 100y³ -20y³ -5y³ = 0

Hence HCF is (x - 5y).

∴ The correct answer is x - 5y.

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.

Given:

We need to find the smallest natural number n such that (n + 1) × n × (n - 1) × (n - 2) × ... × 3 × 2 × 1 is divisible by 910.

Concept Used:

To determine the smallest n such that the product is divisible by 910, we

need to factorize 910 and ensure that the factorial (n!) contains these

factors.

Formula Used:

Factorize the number 910.

Calculation:

Factorize 910:

910 = 2 × 5 × 7 × 13

We need n! to include all these prime factors.

⇒ The smallest n must be at least 13 to include the factor 13.

Check n = 13:

13! = 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

13! contains the factors 2, 5, 7, and 13.

∴ 13! is divisible by 910.

Hence, the smallest natural number n is 12.

Calculation:

555777 + 777555

⇒ (5 × 111)^{7 × 111} + (7 × 111)^{5 × 111}

⇒ 111⁵⁵⁵ (5⁷⁷⁷ × 111²²² + 7⁵⁵⁵)

111 is divisible by 3 and 37 both as 111 = 3 × 37,

Also, both the terms in bracket are odd so their sum will be even and hence divisible by 2.

∴ The correct answer is option 4.

Given:

Consider the following statements:

1. If (3m³ + 2m² + 5m + n) / m is not an integer, where m and n are integers, then n is not divisible by m.
2. 5(8m) + 23m is divisible by 48 for all whole numbers m.

Formula Used:

For statement 1: Verify if the expression (3m³ + 2m² + 5m + n) / m is an integer or not.

For statement 2: Check divisibility of 5(8m) + 23m by 48.

Calculation:

Statement 1:

Expression: (3m³ + 2m² + 5m + n) / m

⇒ 3m² + 2m + 5 + n/m

If (3m² + 2m + 5 + n/m) is not an integer, then n/m is not an integer.

⇒ n is not divisible by m.

Statement 1 is correct.

Statement 2:

5(8^m) + (2³)^m

⇒ 8^m(5 + 1)

⇒ 6 × 8^m

For m = 0, 6 × 1 = 6 which is not divisible by 48.

Statement 2 is incorrect.

∴ The correct answer is option 1.

Given:

Sum of two positive numbers = 40

The GM of these two numbers is 20% lower than their AM

Concept:

AM (Arithmetic Mean) = (a + b) / 2

GM (Geometric Mean) = √(a × b)

GM = 0.8 × AM

Calculation:

Let the two numbers be a and b.

a + b = 40

AM = (a + b) / 2 = 40 / 2 = 20

GM = √(a × b)

Given GM = 0.8 × AM

⇒ √(a × b) = 0.8 × 20

⇒ √(a × b) = 16

⇒ a × b = 16²

⇒ a × b = 256

We have the equations:

1. a + b = 40
2. a × b = 256

Solving these two equations:

Let a and b be the roots of the quadratic equation x² - (a+b)x + ab = 0

⇒ x² - 40x + 256 = 0

Solving for x using the quadratic formula x = [ -b ± √(b² - 4ac) ] / 2a

⇒ x = [ 40 ± √(1600 - 1024) ] / 2

⇒ x = [ 40 ± √576 ] / 2

⇒ x = [ 40 ± 24 ] / 2

⇒ x = 32 or x = 8

So, the two numbers are 32 and 8.

Difference between the two numbers = 32 - 8 = 24

Hence, the difference between the two numbers is 24.

Shortcut Trick

Total Work = 50 × 40 = 2000 units.

Work done in 10-day intervals: 10 × (50 + 45 + 40 + 35 + 30) = 10 × 200 = 2000 units.

The total work is completed in exactly 5 such intervals.

Total time = 5 × 10 = 50 days.

∴ The correct answer is 50 days.

Alternate Method

Given: Initial Men = 50, Target Days = 40, Reduction = 5 men every 10 days.

Formula Used: Total Work = Men × Days

⇒ Total Work to be done = 50 × 40 = 2000 units.

⇒ Work done in first 10 days = 50 × 10 = 500 units.

⇒ Work done in next 10 days = 45 × 10 = 450 units.

⇒ Work done in third 10 days = 40 × 10 = 400 units.

⇒ Work done in fourth 10 days = 35 × 10 = 350 units.

⇒ Work done in fifth 10 days = 30 × 10 = 300 units.

⇒ Total Work completed = 500 + 450 + 400 + 350 + 300 = 2000 units.

⇒ Total Time taken = 10 + 10 + 10 + 10 + 10 = 50 days.

∴ The correct answer is 50 days.

Additional Information

Man-Days Concept

The total amount of work is constant and is expressed as the product of the number of men and the time taken: W = M × D.

Variable Manpower Rule

When manpower changes in stages, the total work is the sum of products of men and duration for each stage: W = M₁D₁ + M₂D₂ + ... + M_nD_n.

Arithmetic Progression in Work

If the number of men decreases by a constant value 'd' at regular intervals 't', the work done forms an AP: Work = t × [n/2 × {2M + (n−1)(−d)}], where n is the number of intervals.

Calculation:

x = $\frac{1}{2+\frac{3}{4+\frac{5}{6+\frac{7}{8+\frac{9}{10}}}}}$

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

Let a = $\frac{3}{4+\frac{5}{6+\frac{7}{8+\frac{9}{10}}}}$

So, x = $\frac{1}{2 + a}$ 

Since a is a positive number,

The value of $\frac{1}{2 + a}$ will always be less then $\frac{1}{2}$ = 0.5

∴ The correct answer is option 1.

Given:

A bottle contains spirit and water in the ratio 1 ∶ 4

Another identical bottle contains spirit and water in the ratio 4 ∶ 1

Calculation:

Let the quantity of the first mixture be x and the quantity of the second mixture be y.

In the first bottle, the ratio of spirit to water is 1 ∶ 4, so the spirit content is x/5 and the water content is 4x/5.

In the second bottle, the ratio of spirit to water is 4 ∶ 1, so the spirit content is 4y/5 and the water content is y/5.

In the new mixture, the spirit content should be (x/5 + 4y/5) and the water content should be (4x/5 + y/5).

The required ratio is 1 ∶ 3, so:

⇒ (x/5 + 4y/5) / (4x/5 + y/5) = 1/3

⇒ (x + 4y) / (4x + y) = 1/3

⇒ 3(x + 4y) = 4x + y

⇒ 3x + 12y = 4x + y

⇒ 12y - y = 4x - 3x

⇒ 11y = x

⇒ x/y = 11/1

∴ Ratio in which mixtures in the two bottles should be mixed is 11 ∶ 1

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:

x4k + x4k+2 + x4k+4 + x4k+6

Calculation:

We have,

x4k + x4k+2 + x4k+4 + x4k+6

⇒ x^4k (x⁰ + x²) + x^4k+4 (x⁰ + x²)

⇒ (x4k + x4k+4) (x0 + x2)

⇒ x4k (x⁰ + x4) (x0 + x2)

⇒ x4k (1 + x4) (1+ x2)

After factorizing the polynomial, we can see the given polynomial is a multiple of (1 + x²) and (1 + x⁴).

Thus, the given polynomial is divisible by both (1 + x2) and (1 + x4).

Hence, both statements are correct.

∴ The correct answer is option (3).

Given:

$\frac{\sin ^2 A+5 \sin A+1}{\sin A}$, where 0 < A ≤ $\frac{\pi}{2}$

Concept Used:

The minimum Value of $x + \frac{1}{x}$  is always 2.

Calculation:

$\frac{\sin ^2 A+5 \sin A+1}{\sin A}$

⇒ Sin A + 5 + $\frac{1}{sin A}$

So, the minimum value of Sin A + $\frac{1}{sin A}$ = 2

Minimum Value of the given expression,

⇒ 2 + 5 = 7

∴ The correct answer is 7.

Given:

$\frac{3}{1^2 \times 2^2}$ + $\frac{5}{2^2 \times 3^2}$ + $\frac{7}{3^2 \times 4^2}$ + ... 

Calculation:

$\frac{3}{1^2 \times 2^2}$ + $\frac{5}{2^2 \times 3^2}$ + $\frac{7}{3^2 \times 4^2}$ + ... 

⇒ $\frac{2^2 - 1^2}{1^2 \times 2^2}$ + $\frac{3^2 - 2^2}{2^2 \times 3^2}$ + $\frac{4^2 - 3^2}{3^2 \times 4^2}$ + ... 

⇒ $\frac{1}{1^2} - \frac{1}{2^2}$ + $\frac{1}{2^2} - \frac{1}{3^2}$ + $\frac{1}{3^2} - \frac{1}{4^2}$ + ...... - $\frac{1}{\infty^2 }$

⇒ 1 - $\frac{1}{\infty^2 }$

⇒ 1

∴ The correct answer is 1

Given:

$\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}}$ = $\frac{421}{972}$

Calculation:

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

$\frac{972}{421}$ = 2 + $\frac{130}{421}$ ⇒ a = 2

$\frac{421}{130}$ = 3 + $\frac{31}{130}$ ⇒ b = 3

$\frac{130}{31}$ = 4 + $\frac{6}{31}$ ⇒ c = 4

$\frac{31}{6}$ = 5 + $\frac{1}{6}$ ⇒ d = 5

$\frac{6}{1}$ = 6 ⇒ e = 6

⇒ a × b × c × d × e = 2 × 3 × 4 × 5 × 6

⇒ a × b × c × d × e = 720

∴ The correct answer is 720.

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:

Radius of circular plate (R) = 7 cm

Thickness of plate (h) = 0.16 mm = 0.016 cm

Angle of the sector (θ) = 150°

Concept:

The volume of the remaining piece after cutting the sector is equal to the volume of the spherical bead formed.

Formula used:

Volume of the sector of a cylinder = (θ/360) × π × R² × h

Volume of the sphere = (4/3) × π × r³

Calculation:

Volume of the sector cut off:

⇒ (150/360) × π × (7)² × 0.016

⇒ (5/12) × π × 49 × 0.016

⇒ (5/12) × π × 0.784

⇒ 3.14 × 0.3267

⇒ 1.026 cm³

Volume of the remaining piece:

⇒ Total volume of the plate - Volume of the sector cut off

⇒ π × R² × h - 1.026

⇒ π × 49 × 0.016 - 1.026

⇒ 3.14 × 0.784 - 1.026

⇒ 2.46176 - 1.026

⇒ 1.43576 cm³

Volume of the spherical bead:

⇒ (4/3) × π × r³ = 1.43576

⇒ r³ = (1.43576 × 3)/(4 × π)

⇒ r³ = 0.3427 cm³ ≈ 0.343 cm³

⇒ r = ³√0.343

⇒ r ≈ 0.7 cm

∴ The value of r is 0.7 cm.

Shortcut Trick

Let the central angle of the sector be 2θ. Chord length = 2r × sin(θ) and Height = r(1 − cos(θ)).

Given 2r × sin(θ) = 2√3 × r(1 − cos(θ)) ⇒ tan(θ/2) = 1/√3 ⇒ θ = 60° ⇒ Central angle = 120°.

Sector area P = πr² / 3. Segment area Q = πr² / 3 − √3r² / 4.

Ratio P / Q = 4π / (4π − 3√3) = (4 × 3.14) / (4 × 3.14 − 3 × 1.7).

P / Q = 12.56 / 7.46 ≈ 1.68 ≈ 1.7.

∴ The correct answer is 1.7.

Alternate Method

Given:

Chord AB of a circle is 2√3 times the height of the minor segment.

π = 3.14, √3 = 1.7

Formula Used:

For a circle with radius r and central angle 2θ:

Length of chord = 2r × sin(θ)

Height of minor segment = r(1 − cos(θ))

O A B 120° h r r

Calculations:

⇒ 2r × sin(θ) = 2√3 × r(1 − cos(θ))

⇒ 2 × sin(θ/2) × cos(θ/2) = √3 × 2 × sin²(θ/2)

⇒ cos(θ/2) = √3 × sin(θ/2)

⇒ tan(θ/2) = 1 / √3

⇒ θ/2 = 30° ⇒ θ = 60°

⇒ Central angle 2θ = 120°

⇒ P (Area of sector) = (120° / 360°) × πr² = πr² / 3

⇒ Area of triangle OAB = (1/2) × r² × sin(120°) = √3r² / 4

⇒ Q (Area of minor segment) = P − Area of triangle OAB = (πr² / 3) − (√3r² / 4)

⇒ P / Q = (π / 3) ÷ [(π / 3) − (√3 / 4)]

⇒ P / Q = 4π / (4π − 3√3)

⇒ P / Q = (4 × 3.14) / (4 × 3.14 − 3 × 1.7)

⇒ P / Q = 12.56 / (12.56 − 5.10) = 12.56 / 7.46

⇒ P / Q ≈ 1.683 ≈ 1.7

∴ The correct answer is 1.7.

Additional Information

Geometry Circle Formulas

Length of Chord = 2r × sin(θ/2), where θ is the central angle. Area of Sector = (θ / 360°) × πr².

Area of Segment

The area of a minor segment is found by subtracting the area of the triangle formed by the radii and chord from the area of the corresponding sector.

Trigonometric Half-Angle Identities

sin(θ) = 2 × sin(θ/2) × cos(θ/2) and 1 − cos(θ) = 2 × sin²(θ/2).

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.

Formula Used:

GM of n numbers a₁, a₂....a_n = (a₁a₂a₃...a_n)^1/n

Calculation:

Let, n = 1

Positive numbers become a₁, a₂, a₃ which are in GP

According to the question, P is th GM of a₁, a₂, a₃

Therefore,

P = (a1a2a3)^1/3                 -----(1)

Again, 

Q is the GM of a₄, a₅, a_6, a₇, a₈, a₉

Q = (a4 a5 a6 a7 a8 a9)^1/6      ----(2)

GM of 3n i.e. 3 × 3 = 9 number is

(a₁ a₂ a₃ a4 a5 a6 a7 a8 a9)1/9

This can be written as 

(P³ × Q⁶)^1/9

(P^1/3 × Q2/3)

Given:

The cost price of y articles is equal to the selling price of z articles.

y ∶ z = 5 ∶ 4

Concept Used:

Profit Percentage = (Profit / Cost Price) × 100

Calculation:

Let the cost price of each article be C and the selling price of each article be S.

Given that the cost price of y articles is equal to the selling price of z articles:

⇒ yC = zS

Given y ∶ z = 5 ∶ 4, we can write:

⇒ 5C = 4S

⇒ S = (5/4)C

Profit per article = Selling Price - Cost Price

⇒ Profit = S - C

⇒ Profit = (5/4)C - C

⇒ Profit = (5C - 4C)/4

⇒ Profit = C/4

Profit Percentage = (Profit / Cost Price) × 100

⇒ Profit Percentage = (C/4C) × 100

⇒ Profit Percentage = (1/4) × 100

⇒ Profit Percentage = 25%

Hence, the correct answer is option 2.

Given:

A sum of money triples itself in 8 years.

Let the principal be P and the rate of interest be R.

It becomes n times in 20 years.

Concept Used:

Simple Interest (SI) is given by the formula:

SI = P × R × T / 100

Total Amount (A) = Principal + Simple Interest

Calculation:

When the sum triples in 8 years:

⇒ A = 3P, T = 8 years

Using the formula for Simple Interest:

SI = A - P

⇒ 3P - P = P × R × 8 / 100

⇒ 2P = 8P × R / 100

⇒ R = 25%

Now, to find n when T = 20 years:

Using the formula for Simple Interest:

SI = P × R × 20 / 100

⇒ SI = P × 25 × 20 / 100

⇒ SI = 5P

Total Amount (A) = Principal + Simple Interest

⇒ A = P + 5P = 6P

Thus, the sum becomes 6 times in 20 years.

Hence, the value of n is 6.

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.

Given:

If (a + b) ∶ (b + c) ∶ (c + a) = 5 ∶ 7 ∶ 6

Concept Used:

We use the concept of ratio and proportion.

Calculation:

Let (a + b) = 5k, (b + c) = 7k, and (c + a) = 6k

We know that:

⇒ (a + b) + (b + c) + (c + a) = 5k + 7k + 6k

⇒ 2(a + b + c) = 18k

⇒ a + b + c = 9k … (1)

From equation (1):

⇒ a = 9k - (b + c)

⇒ a = 9k - 7k = 2k … (2)

⇒ b = 9k - (c + a)

⇒ b = 9k - 6k = 3k … (3)

⇒ c = 9k - (a + b)

⇒ c = 9k - 5k = 4k … (4)

Now, we need to find (a - b + c) ∶ (a + b - c)

Using equations (2), (3), and (4):

⇒ (a - b + c) = 2k - 3k + 4k = 3k

⇒ (a + b - c) = 2k + 3k - 4k = 1k

⇒ (a - b + c) ∶ (a + b - c) = 3k ∶ 1k

⇒ 3 ∶ 1

Hence, the value of (a - b + c) ∶ (a + b - c) is 3 ∶ 1.

Given:

Principal = ₹ 1000

Rate for CI = 10%

Rate for SI = 11%

Time = 3 years

Formula Used:

Compound Interest (CI) Formula:

CI = P(1 + R/100)T - P

Simple Interest (SI) Formula:

SI = P × R × T / 100

Calculation:

For Compound Interest:

CI = 1000(1 + 10/100)3 - 1000

⇒ CI = 1000(1 + 0.1)3 - 1000

⇒ CI = 1000(1.1)3 - 1000

⇒ CI = 1000(1.331) - 1000

⇒ CI = 1331 - 1000

⇒ CI = ₹ 331

For Simple Interest:

SI = 1000 × 11 × 3 / 100

⇒ SI = 1000 × 33 / 100

⇒ SI = 330

⇒ SI = ₹ 330

Difference between CI and SI:

Difference = CI - SI

⇒ Difference = 331 - 330

⇒ Difference = ₹ 1

Hence, the correct answer is option 1.

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:

In a quarter circle of radius R, a circle of radius r is inscribed.

Concept Used:

The inscribed circle will be tangent to both the horizontal and vertical sides of the quarter circle, as well as to the arc of the quarter circle.

The center of the quarter circle is at the origin (0,0), and its radius is R.

Let's place the center of the inscribed circle at coordinates (r, r). This circle is tangent to both the x-axis and the y-axis at distances r from each axis. The distance from the center of the inscribed circle to the center of the quarter circle (0,0) is:

$\sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$

Since this distance is equal to R - r (the radius R of the quarter circle minus the radius r of the inscribed circle), we have:

$r\sqrt{2} = R - r$

Solving for r:

$r\sqrt{2} + r = R$

Factor out r on the left-hand side:

$r(\sqrt{2} + 1) = R$

$\frac{R}{r} = \sqrt{2} + 1$

∴  The ratio of R to r is: $\sqrt{2} + 1 : 1$.

Given:

In the quadrilateral ABCD:

AB = BC

CD = DA

AC (diagonal) = 6 cm

BD (diagonal) = 12 cm

Calculation:

Triangles ABC and CDA are isosceles.

Area of quadrilateral ABCD = (1/2) × d₁ × d₂

Where d₁ and d₂ are the lengths of the diagonals.

Substituting the given values:

Area = (1/2) × 6 cm × 12 cm

Area = (1/2) × 72 cm²

Area = 36 cm²

∴ The area of the quadrilateral ABCD is 36 cm².

Given:

tan(3A) = cot(A - 22°), where 3A is an acute angle,

Calculation:

cot(x) = tan(90° - x).

$\tan(3A) = \tan\left(90^∘ - (A - 22^∘)\right)$

$\tan(3A) = \tan\left(90^∘ - A + 22^∘\right)$

$\tan(3A) = \tan(112^∘ - A)$

Since the tangent function is periodic with a period of 180°, we have:

$3A = 112^∘ - A + n \cdot 180^∘$

where n is an integer. Since 3A is an acute angle, (i.e. (0^∘  < A < 90^∘).

3A = 112^∘ - A

4A = 112^∘

A = 28^∘

∴ The correct answer is 28∘.

Given:

$\frac{\sin θ-\cos θ+1}{\sin θ+\cos θ-1}$ = p secθ + q tan θ 

0 < θ < $\frac{π}{2}$

Calculation:

Since, 0 < θ < $\frac{π}{2}$

We can put θ = π /4 = 45^∘ 

sin 45^∘ = 1/√2

cos 45^∘ = 1/√2

sec 45^∘ = √2 

tan 45^∘ =  1

Putting all these values in the given equation,

⇒ $\frac{1}{\frac{1}{√2} + \frac{1}{√2} - 1}$ = p(√2) + q

⇒ $\frac{1}{√2 - 1}$ = √2p + q

⇒ $\frac{1\times (√2 + 1)}{(√2 - 1)(√2 + 1)}$ = √2p + q

⇒ $\frac{√2 + 1}{(1)}$ = √2p + q

Comparing both sides,

p = 1, q = 1

so, p + q = 1 + 1 = 2

∴ The correct answer is 2.

Given:

The angles of elevation of the top of a tower from two points A and B at a distance of

x and (x + 5)

Height of tower = 6 m

Calculation:

Since the angles are complementary,

angles are θ and 90 - θ 

cotθ = 6/x

tanθ = 6/(x + 5)

Multiplying both equations,

cotθ × tanθ = 6/x × 6/(x + 5) 

⇒ 1 = 36/(x)(x + 5)

⇒ x(x + 5) = 36

⇒ x² + 5x -36 = 0

⇒ x² + 9x - 4x -36 = 0

⇒ (x + 9)(x - 4) = 0

x = -9 , x = 4

∴ The correct answer is 4m.

Calculation:

Statement 1:

sin A + sin B + sin C = $\frac{3 \sqrt{3}}{2}$

sin A + sin B + sin C = $\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$

It means A = B = C = 60^∘ 

The triangle can be equilateral.

Statement 2:

cos A + cos B + cos C = $\frac{3}{2}$

cos A + cos B + cos C = $\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

It means A = B = C = 60^∘ 

The triangle can be equilateral.

∴ The correct answer is Both 1 and 2.

Given:

Train A speed = 80 km/h

Train B speed = 100 km/h

Train A departure time = 7:00 a.m.

Train B departure time = 7:50 a.m.

Concept Used:

Relative speed concept and distance calculation.

Formula Used:

Distance = Speed × Time

Calculation:

Time difference between Train A and Train B = 50 minutes = 50/60 hours = 5/6 hours

Distance covered by Train A in 5/6 hours = 80 km/h × 5/6 hours

⇒ Distance covered by Train A = 80 × 5/6 km

⇒ Distance covered by Train A = 400/6 km

Relative speed of Train B with respect to Train A = 100 km/h - 80 km/h

⇒ Relative speed = 20 km/h

Time taken for Train B to catch up Train A = Distance / Relative speed

⇒ Time taken = (400/6) km / 20 km/h = 10/3 hours

Distance traveled by Train B in (10/3) hours = 100 km/h × (10/3) hours

⇒ Distance traveled by Train B = 100 × (10/3) = 1000/3 km

Thus, the distance from Delhi where the two trains meet is:

⇒ Distance = 1000/3 km

∴ The two trains will be together after 1000/3 km from Delhi.

Given:

Initial Length (L) = L

Initial Breadth (B) = B

Initial Height (H) = H

Length increased by 10%

Breadth increased by 20%

Height increased by 50%

Concept:

The volume of a cuboid is given by the product of its length, breadth, and height.

Formula Used:

Volume of cuboid = Length × Breadth × Height

Calculation:

Initial Volume (V_initial) = L × B × H

New Length (L_new) = L × (1 + 10/100) = 1.1L

New Breadth (B_new) = B × (1 + 20/100) = 1.2B

New Height (H_new) = H × (1 + 50/100) = 1.5H

New Volume (V_new) = L_new × B_new × H_new

V_new = 1.1L × 1.2B × 1.5H

V_new = 1.98LBH

Percentage Increase in Volume = [(V_new - V_initial)/V_initial] × 100

⇒ Percentage Increase = [(1.98LBH - LBH)/LBH] × 100

⇒ Percentage Increase = [0.98LBH/LBH] × 100

⇒ Percentage Increase = 0.98 × 100

⇒ Percentage Increase = 98%

⇒ Hence, the percentage increase in volume of the cuboid is 98%.

Given:

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

The share of P before deduction:

⇒ p = 3k + 93

⇒ p = 3 × 769 + 93

⇒ p = 2307 + 93

⇒ p = 2400

Since 93 is deducted from P's share, the new share of P = 2400 - 93 = 2307

∴ The share of P is ₹ 2307.

Given:

P² varies as R and Q² varies as R, (P ≠ Q).

Concept:

If a variable X varies as Y, it implies X = kY for some constant k.

Calculation:

Let P² = k₁R and Q² = k₂R, where k₁ and k₂ are constants.

1. To check if P² + Q² varies as R:

⇒ P² + Q² = k₁R + k₂R

⇒ P² + Q² = (k₁ + k₂)R

Since k₁ + k₂ is a constant, P² + Q² varies as R.

2. To check if PQ varies as R:

⇒ P = √(k₁R) and Q = √(k₂R)

⇒ PQ = √(k₁R) × √(k₂R)

⇒ PQ = √(k₁k₂)R

Since √(k₁k₂) is a constant, PQ varies as R.

3. To check if P² - Q² varies as R:

⇒ P² - Q² = k₁R - k₂R

⇒ P² - Q² = (k₁ - k₂)R

Since k₁ - k₂ is a constant, P² - Q² varies as R.

∴ All three statements are correct.

Given:

p number of men can finish a piece of work in q days.

There are 50% more men, then the work will be finished 12 days earlier.

Concept Used:

Work is inversely proportional to the number of men and the number of days.

Formula Used:

Work = Men × Days

Calculation:

Initial work:

Work = p × q

With 50% more men:

Number of men = p + 0.5p = 1.5p

New days = q - 12

Since the work done is the same:

p × q = 1.5p × (q - 12)

Cancel out p from both sides:

q = 1.5(q - 12)

Expand and solve for q:

⇒ q = 1.5q - 18

⇒ q - 1.5q = -18

⇒ -0.5q = -18

⇒ q = -18 / -0.5

⇒ q = 36

∴ The value of q is 36.

Concept Used:

Minimum value of $(x+\frac{1}{x}) = 2$

Calculation:

$\left(\frac{a^2+3 a+1}{a}\right)$ $\left(\frac{b^2+3 b+1}{b}\right)$ = $(a+3+\frac{1}{a})$ $(b+3+\frac{1}{b})$

Minimum value of  $(a+3+\frac{1}{a})$ $(b+3+\frac{1}{b})$,

⇒ (2 +3)(2 + 3) = 5 × 5 = 25

∴ The correct answer is 25.