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Given:

A real number M is squared to give the value N. Find the minimum value of (M + N).

Formula used:

N = M²

Minimize (M + N) = M + M²

Calculation:

Let f(M) = M + M²

To find the minimum value, take the derivative and set it to zero:

f'(M) = 1 + 2M

⇒ 1 + 2M = 0

⇒ M = -0.5

Substitute M = -0.5 into f(M):

f(M) = M + M²

⇒ f(-0.5) = (-0.5) + (-0.5)²

⇒ f(-0.5) = -0.5 + 0.25

⇒ f(-0.5) = -0.25

∴ The minimum value of (M + N) is -0.25, and the correct answer is option 1.

Given:

We need to find the sum of all 3-digit numbers that give a remainder of 5 when divided by 50.

Formula used:

Sum of arithmetic progression (AP): S_n = (n/2) × (a + l)

Where,

a = First term, l = Last term, n = Number of terms

Calculation:

The 3-digit numbers that give a remainder of 5 when divided by 50 are: 105, 155, 205, ..., 955.

a = 105, l = 955

Common difference (d) = 50

n = (l - a)/d + 1

⇒ n = (955 - 105)/50 + 1

⇒ n = 850/50 + 1

⇒ n = 17 + 1

⇒ n = 18

Sum of all terms:

S_n = (n/2) × (a + l)

⇒ S₁₈ = (18/2) × (105 + 955)

⇒ S₁₈ = 9 × 1060

⇒ S₁₈ = 9540

∴ The correct answer is option (2).

Given:

The average of 64, 69, 72, 75, x lies between 62 and 76 (excluding 62 and 76).

Formula used:

Average = $\dfrac{\text{Sum of all numbers}}{\text{Total numbers}}$

Calculation:

Let the sum of the given numbers = 64 + 69 + 72 + 75 + x = 280 + x

The average = $\dfrac{280 + x}{5}$

Since the average lies between 62 and 76:

⇒ $62 < \dfrac{280 + x}{5} < 76$

⇒ 310 < 280 + x < 380

⇒ 30 < x < 100

The possible integer values of x are from 31 to 99 (both inclusive).

Total number of integers = 99 - 31 + 1 = 69

∴ The correct answer is option (2).

Given:

(x + y + z) = k

(x + z – y) × (x – z + y) ∝ yz

Formula used:

Algebraic manipulation and properties of proportionality.

From (x + y + z) = k, we can derive:

x + z – y = (x + y + z) – 2y = k – 2y

x – z + y = (x + y + z) – 2z = k – 2z

Calculations:

Given that (x + z – y) × (x – z + y) ∝ yz

⇒ (k – 2y) × (k – 2z) = C × yz (where C is the constant of proportionality)

⇒ k² – 2kz – 2ky + 4yz = C × yz

⇒ k² – 2k(z + y) + 4yz = C × yz

⇒ k² – 2k(y + z) = C × yz – 4yz

⇒ k² – 2k(y + z) = (C – 4)yz

Now, let's rearrange to find what (y + z – x) is proportional to.

From (x + y + z) = k, we have y + z = k – x.

Substitute (y + z) in the equation k² – 2k(y + z) = (C – 4)yz:

⇒ k² – 2k(k – x) = (C – 4)yz

⇒ k² – 2k² + 2kx = (C – 4)yz

⇒ –k² + 2kx = (C – 4)yz

⇒ k(2x – k) = (C – 4)yz

We are interested in (y + z – x).

We know (y + z – x) = (k – x) – x = k – 2x

From k(2x – k) = (C – 4)yz, we can write:

⇒ –k(k – 2x) = (C – 4)yz

⇒ k – 2x = $\dfrac{(C – 4)yz}{-k}$

⇒ k – 2x = $\dfrac{(4 – C)yz}{k}$

Since k, C are constants, $\dfrac{(4 – C)}{k}$ is also a constant.

Therefore, (k – 2x) is proportional to yz.

Since (y + z – x) = k – 2x,

∴ (y + z – x) is proportional to yz.

Given:

p = remainder when 7⁸⁴ is divided by 342

q = remainder when 7⁸⁴ is divided by 344

Formula used:

Modular arithmetic and properties of remainders.

If a ≡ b (mod m), then aⁿ ≡ bⁿ (mod m)

aⁿ - bⁿ is divisible by a - b

Calculations:

For p:

We need to find the remainder when 7⁸⁴ is divided by 342.

Notice that 342 = 7³ - 1 (since 7³ = 343).

We know that aⁿ - bⁿ is divisible by a - b.

Consider 7³ ≡ 1 (mod 342) because 7³ - 1 = 343 - 1 = 342, which is divisible by 342.

Now, we have 7⁸⁴ = (7³)²⁸.

⇒ 7⁸⁴ ≡ 1²⁸ (mod 342)

⇒ 7⁸⁴ ≡ 1 (mod 342)

So, p = 1.

For q:

We need to find the remainder when 7⁸⁴ is divided by 344.

Notice that 344 = 7³ + 1 (since 7³ = 343).

We know that aⁿ + bⁿ is divisible by a + b when n is odd.

We also know that aⁿ - bⁿ is divisible by a + b when n is even.

Consider 7³ ≡ -1 (mod 344) because 7³ + 1 = 343 + 1 = 344, which is divisible by 344.

Now, we have 7⁸⁴ = (7³)²⁸.

⇒ 7⁸⁴ ≡ (-1)²⁸ (mod 344)

⇒ 7⁸⁴ ≡ 1 (mod 344)

So, q = 1.

Value of (p – q):

⇒ p – q = 1 – 1

⇒ p – q = 0

∴ (p – q) is equal to 0.

Given:

N is the smallest 5-digit number which leaves a remainder of 1 when divided by 2, 2², 2³, 2⁴, ..., 2ⁿ

Formula used:

N ≡ 1 (mod 2^k) for all k = 1 to n

⇒ N - 1 is divisible by LCM of all 2^k for k = 1 to n

LCM(2, 4, 8, ..., 2ⁿ) = 2ⁿ

Calculations:

N - 1 = multiple of 2ⁿ

N = 2ⁿ × k + 1

We want the smallest N ≥ 10000

Try values of n:

n = 13 ⇒ 2¹³ = 8192

⇒ N = 8192 × 2 + 1 = 16385

n = 14 ⇒ 2¹⁴ = 16384

⇒ N = 16384 × 1 + 1 = 16385

n = 14 follows the condition.

∴ The value of n is 14.

Given:

The expression is:

$\frac{1}{532900} + \frac{p^{2}}{266450} + \frac{p^{4}}{532900}$

Calculation:

$\frac{1}{532900} + \frac{p^{2}}{266450} + \frac{p^{4}}{532900}$

⇒ $\frac{1}{532900} + \frac{p^2}{266450} + \frac{p^4}{532900} = \frac{1}{532900} + \frac{2p^2}{532900} + \frac{p^4}{532900}$

⇒ $\frac{1 + 2p^2 + p^4}{532900}$

⇒ $\frac{(p^2 + 1)^2}{(730)^2}$

Since both numerator and denominator both are perfect square and the result is an integer, minimum possible value of expression is 1

So. p² + 1 = 730

p2 = 730 - 1 = 729

p = 27

The minimum value of p for which the expression is an integer is 27.

Given:

Equation: x + a + b = $(\dfrac{abx}{ab + ax + bx})$

Roots of the equation are α and β.

Formula used:

For a quadratic equation of the form Ax2 + Bx + C = 0, with roots α and β:

Sum of roots: α + β = $-\dfrac{B}{A}$

Product of roots: αβ = $\dfrac{C}{A}$

Calculations:

First, rearrange the given equation into the standard quadratic form Ax2 + Bx + C = 0.

x + a + b = $(\dfrac{abx}{ab + ax + bx})$

Multiply both sides by (ab + ax + bx):

⇒ (x + a + b)(ab + ax + bx) = abx

⇒ x(ab) + x(ax) + x(bx) + a(ab) + a(ax) + a(bx) + b(ab) + b(ax) + b(bx) = abx

⇒ abx + ax² + bx² + a²b + a²x + abx + ab² + abx + b²x = abx

⇒ ax² + bx² + abx + a²x + abx + abx + b²x − abx + a²b + ab² = 0

⇒ (a + b)x² + (a²x + b²x + 2abx) + (a²b + ab²) = 0

⇒ (a + b)x² + (a² + b² + 2ab)x + ab(a + b) = 0

Recognize that (a² + b² + 2ab) = (a + b)²:

⇒ (a + b)x² + (a + b)²x + ab(a + b) = 0

⇒ x² + (a + b)x + ab = 0

Sum and product of the roots (α and β):

Sum of roots: α + β = $(-\dfrac{B}{A}) = -\dfrac{(a+b)}{1}$ = −(a + b)

Product of roots: αβ = $\dfrac{C}{A} = \dfrac{ab}{1}$ = ab

Finally, calculate (αβ + α + β):

⇒ (αβ + α + β) = ab + (−(a + b))

⇒ (αβ + α + β) = ab − a − b

∴ The correct answer is option 2.

Given:

Let the 2-digit number be N = 10×a + b, where:

a = digit in tens place

b = digit in units place

P = a × b

Formula used:

Condition 1: P + a² = 84

Condition 2: P + b² = 60

Calculations:

From (1): P = 84 − a²

From (2): P = 60 − b²

⇒ 84 − a² = 60 − b²

⇒ 24 = a² − b²

⇒ (a − b)(a + b) = 24

Try factor pairs of 24: (1,24), (2,12), (3,8), (4,6)

Try a − b = 2 and a + b = 12

⇒ Add: 2a = 14 ⇒ a = 7, b = 5

Check P = a × b = 7 × 5 = 35

P + a² = 35 + 49 = 84 

P + b² = 35 + 25 = 60

N = 10×7 + 5 = 75

P = 35

P + N = 35 + 75 = 110

∴ The correct answer is: 110

Given:

The initial mixture contains kerosene and turpentine oil in the ratio 3:2.

Total mixture = 100 L

Kerosene = (3/5) × 100 = 60 L

Turpentine oil = (2/5) × 100 = 40 L

Final kerosene percentage = 20%

Formula used:

Final kerosene in the mixture = Total mixture × (Final kerosene percentage / 100)

Additional kerosene to be added = Final kerosene - Initial kerosene

Calculation:

Let the additional kerosene to be added = x L.

Total mixture after adding kerosene = 100 + x L

Kerosene in the new mixture = 60 + x L

For 20% kerosene in the final mixture:

⇒ (60 + x) / (100 + x) = 20 / 100

⇒ (60 + x) = (20 / 100) × (100 + x)

⇒ 60 + x = 20 × (100 + x) / 100

⇒ 60 + x = (2000 + 20x) / 100

⇒ 100(60 + x) = 2000 + 20x

⇒ 6000 + 100x = 2000 + 20x

⇒ 6000 - 2000 = 20x - 100x

⇒ 4000 = -80x

⇒ x = -4000 / 80

⇒ x = -50

∴ It is not possible to achieve 20% kerosene in the final mixture. The correct answer is option (4).