The median of the following data is 525 . Find the values of x and y , if the total frequency is 100 . Class Interval Frequency 0 - 100 2 100 - 200 5 200 - 300 x 300 - 400 12 400 - 500 17 500 - 600 20 600 - 700 y 700 - 800 9 800 - 900 7 900 - 1000 4

Cumulative frequency for the given data; Class interval Frequency f Cumulative frequency CF 0 - 100 2 2 100 - 200 5 7 200 - 300 x x + 7 300 - 400 12 x + 19 400 - 500 17 x + 36 500 - 600 20 x + 56 600 - 700 y x + y + 56 700 - 800 9 x + y + 65 800 - 900 7 x + y + 72 900 - 1000 4 x + y + 76 total Σ f = 76 + x + y = 100 Given that the median is 525 . So median class is 500 - 600 Now median = l + N 2 - C F f × h where l = 500 , h = 100 , N = 100 , C F = 36 + x , f = 20 so, 525 = 500 + 100 2 - ( 36 + x ) 20 × 100 ⇒ 525 - 500 = ( 50 - ( 36 + x ) ) × 5 ⇒ 25 = ( 14 - x ) × 5 ⇒ x = 9 Also, Σ f = 76 + x + y = 100 ⇒ 9 + y = 100 - 76 ⇒ y = 15 Hence, x = 9 and y = 15

In a flight of 600 km , an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km / hr and time of flight increased by 30 minutes. Find the original duration of flight.

Let the original duration is x hours & original speed is y km / h . Then, x × y = 600 .....(1) The aircraft was slowed due to bad weather. Its average speed is reduced by 200 km / h and time of flight increased by 30 minutes or 1 2 h . Then, x + 1 2 ( y - 200 ) = 600 .....(2) From equation (1) and (2), we have x + 1 2 ( y - 200 ) = x y ⇒ x y - 200 x + y 2 - 100 = x y ⇒ 200 x - y 2 + 100 = 0 From equation (1), put y = 600 x then, we have ⇒ 200 x - 600 2 x + 100 = 0 ⇒ 200 x - 300 x + 100 = 0 ⇒ 2 x - 3 x + 1 = 0 ⇒ 2 x 2 - 3 + x = 0 ⇒ 2 x 2 + x - 3 = 0 From middle term factorization, we have ⇒ 2 x 2 - 2 x + 3 x - 3 = 0 ⇒ ( x - 1 ) ( 2 x + 3 ) = 0 ⇒ x = 1 , - 3 2 Here, we consider x = 1 hour . Hence, the original duration of the flight is 1 hour .