^x⁄_y = 4 ∴ y = ^x⁄_y ∴ 2x - y = 2x - ^x⁄₄ = \(\frac{7x}{4}\)

∴ Required % = \(\frac{2x-y}{x}=\frac{\frac{7x}{4}}{x}\times 100\)

= ⁷⁄₄ × 100 = 175%

\(\frac{(AB)\times (BE)}{2}=\frac{(BE)^{2}}{2}=7\) ∴ BE = \(\sqrt{14}\)

∴ EC = 3\(\sqrt{14}\) ∴ BC = 4\(\sqrt{14}\)

Area \(□\) ABCD = \(\sqrt{14}\) x 4\(\sqrt{14}\) = 56

Ans: (D)

Alternatively since EC is 3 times of BE, therefore area of rectangle will be 8 times of triangle ABE.

∴ Area of rectangle will be 56.

Paul and Linda are coming towards each other.

∴ Adding their speeds, we get 57 + 63 = 120 floors per minute i.e. 2 floors per second.

They are separated by 40 (51 - 11) floors, so they require 20 seconds to reach each other.

∴ In 20 seconds, i.e. 1/3^rd minute, Paul crosses \(\frac{57}{3}\) = 19 floors and Linda \(\frac{63}{3}\) = 21 floors.

∴ 11 + 19 or 51- 21 = 30^th floor.

∴ They meet on the 30^th floor.

To maximize the value of expression we must have max. value for ^x⁄_y and for z²

Put z = -3 x = 2 Y= ¹⁄₂ ∴ Max. value = 36

We can keep one or two empty desks between students.

(1) One empty desk → x in position 1,

y in position 3 and vice versa.

x in position 2 and y in position 4 and vice versa.

∴ 2 + 2 = 4 ways.

(2) Two empty desks → x in position 1,

Y in position 4 and vice versa. i.e. 2 ways

∴ Total ways = 2 + 4 = 6

Volume of tank = \(\frac{22}{7}\) x 7 x 7 x 10.5

= 154 x 10.5 cubic meter.

Half of this volume = 77 x 10.5 cubic meter.

Rate of inflow is 10.5 cubic meter per hour.

∴ It requires 77 hours to completely fill the tank.

Use following formulae to find out angle between two hands.

Theoretical time difference x 6° +time x\(\iota _{2}^{0}\)

(O x 6°) + (30 x ¹⁄₂) = 15°

Note: When minute hand is before the hour hand, we have to add. But if minute hand is after the hour hand, then we have to subtract.

Let the original length and breadth be ℓ and b.

∴ Area = ℓb ∴ Increased area = 1.21 (ℓb)

= 1.1ℓ x 1.1b ∴ New length = 10% more.

New width = 10% more

Old perimeter = 2(ℓ + b)

New perimeter = 2.2(ℓ + b) = 10% increase.

Average price of 25 products = $1200

∴ Total price of 25 products = $1200 x 25 = $30,000

To maximize the price of the 25th products, we need to minimize the price of the other 24 products.

∴ The minimum price of the 24 products, according to the two conditions is (420 x 10) + (1000 x 14) = $18,200.

∴ Maximum price of 251 product = $30,000 - $18,200 = $11,800

\(x-y=\frac{x^{2}-y^{2}}{y^{2}-x^{2}}\, ∴ x-y=-1\left ( \frac{y^{2}-x^{2}}{y^{2}-x^{2}} \right )\)

Since y² - x² ≠ 0, it can be cancelled.

∴ x - y = -1 ∴.y - x = +1