Question
Which of the following can be true? i. R
reached the destination in 5hrs and P reached the destination in 3 hrs ii. R reached the destination is 3 hrs and P reached the destination in 5 hrs iii. If the distance is increased by 90 km and the speed of P is 1/5th of the speed of Q, then R reached the new distance at 3.30 pmSolution
Speed of Q = √625 m/sec = 25 m/sec = 25 x 18/5 = 90 km/hr Time taken to reach point B by Q = 8 hrs (11.30 am – 7:30 pm) Distance between A and B = 90 x 8 = 720 km Speed of P = 0.8 x 90 = 72 km/hr Now, From i: We know that speed of R is more than that of P. Hence time taken by R will be less than that of P. So, statement I is false. From ii: Time taken by Q to reach at point B is 8 hrs but, in the statement, it is written that R will reach the destination in 3 hrs, which is not possible as speed of R is more than that of Q. Hence, statement ii is false. From iii: If distance is increased by 90 km, then new distance = 720 + 90 = 810 km Speed of P = 1/5 x 90 = 18 km/hr Time taken by P = 810/18 = 45 hrs Now distance is increased, then time to reach the destination by P will also increase. And it is also given that the person who is fastest will reach by 7:30 pm. So, R will not cover the new distance by 3:30 pm. Hence statement iii is also false.
- Suppose both the roots of q² + kq + 49 = 0 are real and equal, then determine the value of 'k'.
I. 84x² - 167x - 55 = 0
II. 247y² + 210y + 27 = 0
l). 2p² + 12p + 18 = 0
ll). 3q² + 13q + 12 = 0
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
I. 6x2 - 47x + 77 =0
II. 6y2 - 35y + 49 = 0
I. 12a2 – 55a + 63 = 0
II. 8b2 - 50 b + 77 = 0
...I. 2y2 – 19y + 35 = 0
II. 4x2 – 16x + 15 = 0
I. 5x² - 24 x + 28 = 0
II. 4y² - 8 y - 12= 0
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer
I. x2 – ...
I.8(x+3)+ 8(-x)=72
II. 5(y+5)+ 5(-x)=150