Mathematics Questions and Answers

Perimeter of a triangle = sum of all sides

⇒ \(P = y + x + x = 2\)

⇒ \(y + 2x = 2\)

⇒ \(y= 2 - 2x\)-----(i)

Using Pythagoras theorem

\(y^2 = x^2 + x^2\)

⇒ \(y^2 = 2x^2\)

⇒ \(y = \sqrt2x^2\)

⇒ \(y = x\sqrt2\)-----(ii)

Equate \(y\)

⇒ \(2 - 2x = x\sqrt2\)

Square both sides

⇒ \((2 -2x) ^2 = (x\sqrt2)^2\)

⇒ \(4 - 8x + 4x^2 = 2x^2\)

⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\)

⇒ \(2x^2 - 8x + 4 = 0\)

⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\)

⇒ \(x = \frac{8\pm\sqrt32}{4}\)

⇒ \(x = \frac{8\pm4\sqrt2}{4}\)

⇒ \(x = 2\pm\sqrt2\)

⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\)

∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter)

∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\)

∴ The length of the longer side = -2 + 2\(\sqrt2\)m

Speed = \(\frac{Distance}{Time}\)

⇒ Time = \(\frac{Distance}{Time}\)

Let D = distance between the two airports

∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\)

Since total time of flight= 3hours,

⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3

⇒ \(\frac{15D + 12D}{1800}\) = 3

⇒ \(\frac{27D}{1800}\) = 3

⇒ \(\frac{3D}{200} = \frac{3}{1}\)

⇒ 3D = 200 x 3

∴ D =\(\frac{ 200\times3}{3}\)= 200km

∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary)

⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180

⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180

⇒ 3\(x\) + 2\(y\) + 155 = 180

⇒ 3\(x\) + 2\(y\) = 180 - 155

⇒ 3\(x\) + 2\(y\) = 25 ----- (i)

∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary)

⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180

⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180

⇒ -2\(x\) + \(y\) + 255 = 180

⇒ -2\(x\) + y = 180 - 255

⇒ -2\(x\) + \(y\) = -75 ------- (ii)

⇒ \(y\) = -75 + 2\(x\) -------- (iii)

Substitute (-75 + 2\(x\)) for \(y\) in equation (i)

⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25

⇒ 3\(x\) - 150 + 4\(x\) = 25

⇒ 7\(x\) = 25 + 150

⇒ 7\(x\) = 175

⇒ \(x = \frac{175}{7} = 25\)

From equation (iii)

⇒ \(y\) = -75 + 2(25) = -75 + 50

⇒ \(y\) = -25

∴ \(x\) + \(y\) = 25 + (-25) = 0

Area of trapezium = \(\frac{1}{2}(a + b) h\)

⇒ \(\frac{1}{2} (a + b)\times 16 = 200\)

⇒ 8(a + b) = 200

⇒ a + b = \(\frac{200}{8}\) = 25 -----(i)

⇒ a : b = 2 : 3

⇒ \(\frac{a}{b} = \frac{2}{3}\)

⇒ 3a = 2b

⇒ a = \(\frac{2b}{3}\) -------(ii)

Substitute \(\frac{2b}{3}\) for a in equation (i)

⇒ \(\frac{2b}{3}\) + b = 25

 \(\frac{5b}{3}\) = 25

⇒ b = 25 ÷ \(\frac{5}{3} = 25\times\frac{3}{5} = 15cm\)

From equation (ii)

⇒ a = \(\frac{2 \times 15}{3} = 2\times5 = 10cm\)

∴ Lengths of each parallel sides are 10cm and 15cm

Total number of students that scored at most 50 marks = 100 + 80 + 60 + 40 + 80 = 360