Answer: Option C
Answer: (c)Fixed distance = x km and certain speed = y kmph (let).Case I,$x/{y +10} = x/y$ - 1$x/{y +10} + 1 = x/y$ --- (i)Case II,$x/{y + 20} = x/y - 1 - 3/4$= $x/y - {4 + 3}/4$$x/{y + 20} + 7/4 = x/y$ --- (ii)From equations (i) and (ii),$x/{y +10} + 1 = x/{y + 20} + 7/4$$x/{y +10} - x/{y + 20} = 7/4$ - 1$x({y + 20 - y - 10}/{(y + 10)(y + 20)})$= ${7 - 4}/4 = 3/4$${x × 10}/{(y + 10)(y + 20)} = 3/4$3 (y + 10) (y + 20) = 40 x${3(y + 10)(y + 20)}/40$ = x --(iii)From equation (i),${3(y + 10)(y + 20)}/{40(y + 10)}$ + 1= ${3(y + 10)(y + 20)}/{40y}$3 (y +20) + 40 = ${3(y + 10)(y + 20)}/y$$3y^2 + 60y + 40y$= $3(y^2 + 30y + 200)$$3y^2 + 100y = 3y^2 + 90y + 600$10y = 600 ⇒ y = 60Again from equation (i),$x/{y +10} + 1 = x/y$$x/{60 + 10} + 1 = x/60$$x/70 + 1 = x/60$${x + 70}/70 = x/60$6x + 420 = 7x7x - 6x = 420$x$ = 420 km.