Question
The time taken by ‘A’ to build a wall alone is 20%
more than the time taken by him to build it along with ‘B’. ‘B’ will take how much percent more time to build the same wall than the time taken by him to build it along with ‘A’?Solution
Let the work ‘A’ and ‘B’, can do in 1 day be ‘a’ units and ‘b’ units, respectively. Let they takes ‘p’ days to build the wall, together Therefore, Total work = (a + b) × p units …….. (1) According to the question, Time taken by ‘A’ to build the wall alone = 1.20p days Therefore, total work = 1.20p × a units….. (2) From equation (1) and (2), we get 5b = a Therefore, total work to be done = (a + b) × p = 6bp units Time taken by 'B' to build the wall = 6bp/b = 6p days Therefore, required percentage = (6p – p)/p} × 100 = 500% more time
If x - √(3y) =19 Then find the slope in angle?
tan 1˚ × tan 2˚× …………………….tan 88˚ × tan 89˚ = ?
If 2ycosθ = x sinθ and 2xsecθ – ycosecθ = 3, then x 2 + 4y 2 = ?
If cos 2x = 1/2, then the value of x:
A man is standing at a point 40 meters away from a tower. The angle of elevation of the top of the tower from the point is 45°. After walking 20 meters...
If X = 15°, find the value of:
sin²(4X) + cos²(3X) - tan²(2X)if a sin 45 ˚ = b cosec 60 ˚ then find the value of a⁴/b⁴
...
