Explanation

(a) \(\frac{1}{x} + \frac{1}{y} = 5 ..... (1)\)

\(\frac{1}{y} - \frac{1}{x} = 1 ........ (2)\)

Substitute p and q for \(\frac{1}{x}\) and \(\frac{1}{y}\) respectively.

\(p + q = 5 ......... (1a)\)

\(q - p = 1 ....... (2a)\)

From (2a), q = 1 + p.

Putting into (1a), we have:

\(p + (1 + p) = 5\)

\(2p + 1 = 5 \implies 2p = 5 - 1 = 4\)

\(p = 2\)

\(q = 1 + p = 1 + 2 = 3\)

\(p = \frac{1}{x} = 2\)

\(\implies x = \frac{1}{2}\)

\(q = \frac{1}{y} = 3\)

\(\implies y = \frac{1}{3}\)

Hence, \(x = \frac{1}{2} ; y = \frac{1}{3}\).

(b) Total distance travelled = 48 km

Total time taken = 45 minutes = \(\frac{45}{60} = 0.75 hr\)

Let the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.

Let the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.

Using \(speed = \frac{distance}{time}\) in each case

On good surface : \(72 = \frac{x}{t}\)

\(\implies x = 72t ... (1)\)

On bad surface : \(48 = \frac{48 - x}{0.75 - t}\)

\(48(0.75 - t) = 48 - x\)

\(\implies 36 - 48t = 48 - x ...... (2)\)

Substitute 72t = x in (2) :

\(36 - 48t = 48 - 72t \implies 72t - 48t = 48 - 36\)

\(24t = 12 \implies t = \frac{12}{24} = 0.5 hours\)

From (1), x = 72t

\(x = 72 \times 0.5 = 36 km\)

Hence, the man travels a distance of 36 km on good road surface.

Correct Answer: C. x² + x - 6 = 0

Explanation

The roots are -3 and 1

x = -3 and x = 1

y = (x + 3)(x - 1)

= x² - x + 3x - 3

y = x² + x - 6 = 0

Correct Answer: (c) 19/32

Explanation

a. a = (10t - 4t\(^2\))m/s\(^2\), t = t seconds, u = 0

from first equation of motion,

v = u + at

v = 0 + (10t - 4t\(^2\))t

v = (10t\(^2\) - 4t\(^3\))m/s\(^2\)

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4

Δa4 = 10(t4-t3) - 4(t4-t3)\(^2\)

Δa4 = 10(4-1) - 4(4-3)\(^2\)

Δa4 = 10(1) - 4(12)

Δa4 = (10-4)m/s\(^2\)

Δa4 = 6ms-2

b. R = m(g+a)

When the lift moves with a constant velocity, a = 0

R = mg

R = 120 × 10

R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s\(^2\)

R = m(g+a)

R = 120(10+3)

R = 120 × 13

R = 1560N

Correct Answer: A. a 'b = b 'a

Explanation

A binary operation '*' defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A.

If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A.

If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

Correct Answer: C. \((\frac{5}{8}, -\frac{3}{8})\)

Explanation

Equation : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding : \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Given, \(4x^{2} + 4y^{2} - 5x + 3y - 2 = 0\)

Divide through by 4 to make the coefficient of \(x^{2}\) and \(y^{2}\) to be 1.

\(x^{2} + y^{2} - \frac{5}{4}x + \frac{3}{4}y - \frac{1}{2} = 0\)

Comparing, \(2a = \frac{5}{4} \implies a = \frac{5}{8}\)

\(2b = -\frac{3}{4} \implies b = -\frac{3}{8}\)

\((a, b) = (\frac{5}{8}, -\frac{3}{8})\)

Correct Answer: A. 18

Explanation

\(\sqrt{12} ( \sqrt{48} - \sqrt{3}\))

\(\sqrt{4 \times 3} (6 \times 3 - \sqrt{3}) = 2 \sqrt{3}(4 \sqrt{3} - \sqrt{3})\)

= 2\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\)

Correct Answer: A. (10 units, 053°)

Explanation

|→PQ| = √[(2-(-4))\(^2\) + (3-(-5))\(^2\)]

|→PQ| = √\(6^2 + 8^2\)

|→PQ| = √100

|→PQ| = 10units

tanθ = \(\frac{3--5}{2--4}\)

tanθ = \(\frac{4}{3}\)

θ = \(tan^{-1}\frac{4}{3}\)

= 53º

Correct Answer: B. 28

Correct Answer: B. \( \frac{3\pi x^2}{8} \)

Explanation

A =^?₃₆₀ x πr² =¹³⁵₃₆₀ x π x x² = (3πx²)/8