The sum of two numbers is 15 . If sum of their reciprocals is 3 10 , find the two numbers.

Let two numbers are x and y x + y = 15 . . . ( i ) and 1 x + 1 y = 3 10 x + y x y = 3 10 ⇒ 15 x y = 3 10 ⇒ x y = 50 ⇒ y = 50 x . . . ( ii ) Putting value of eq (ii) in (i) x + 50 x = 15 x 2 - 15 x + 50 = 0 x 2 - 10 x - 5 x + 50 = 0 x - 10 x - 5 = 0 x = 10 o r x = 5 ⇒ x = 10 a n d y = 5 hence numbers are 10 , 5

The sum of two numbers is 15 . If sum of their reciprocals is 3 10 , find the two numbers.

Let two numbers are x and y x + y = 15 . . . ( i ) and 1 x + 1 y = 3 10 x + y x y = 3 10 ⇒ 15 x y = 3 10 ⇒ x y = 50 ⇒ y = 50 x . . . ( ii ) Putting value of eq (ii) in (i) x + 50 x = 15 x 2 - 15 x + 50 = 0 x 2 - 10 x - 5 x + 50 = 0 x - 10 x - 5 = 0 x = 10 o r x = 5 ⇒ x = 10 a n d y = 5 hence numbers are 10 , 5

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 .

Find a quadratic polynomial whose zeros are reciprocals of the zeros of the polynomials f ( x ) = a x 2 + b x + c , a ≠ 0 , c ≠ 0 .

Let p and q be the zeroes of the quadratic polynomial a x 2 + b x + c . Then, we have, p + q = - b a and p q = c a Let P and Q be the zeroes of the required quadratic polynomial. Then, given in question that P = 1 p and Q = 1 q Then, P + Q = 1 p + 1 q = p + q p q = - b a c a = - b c And P Q = 1 p × 1 q = 1 c a = a c Thus, the required polynomial is x 2 − ( P + Q ) x + P Q = x 2 + b c x + c a

For what values of k , the roots of the equation x 2 + 4 x + k = 0 are real?

We have, x 2 + 4 x + k = 0 Now we know that, For a quadratic equation a x 2 + b x + c = 0 to have real roots b 2 - 4 a c ≥ 0 That is, b 2 ≥ 4 a c In the given quadratic equation a = 1 , b = 4 and c = k Then, 4 2 ≥ 4 1 k 4 ≥ k Therefore, the values of k ≤ 4 .

A motorboat whose speed is 18 km / h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Let the speed of the stream be x km / h Then, the speed of the boat upstream = 18 - x km / h And the speed of the boat downstream = 18 + x km / h Now the time taken to go upstream = 24 18 - x hours [since Time = Distance Speed ] Similarly, the time taken to go downstream = 24 18 + x hours According to the question, 24 18 - x - 24 18 + x = 1 24 ( 18 + x ) – 24 ( 18 – x ) = ( 18 – x ) ( 18 + x ) 24 × 18 + 24 x - 24 × 18 + 24 x = 324 - x 2 x 2 + 48 x – 324 = 0 Now using the quadratic formula, we get x = - 48 ± 48 2 - 4 1 - 324 2 1 x = - 48 ± 48 2 + 1296 2 x = - 48 ± 3600 2 x = - 48 ± 60 2 Then, x = - 48 + 60 2 and x = - 48 - 60 2 x = 12 2 and x = - 108 2 x = 6 and x = - 54 Since x is the speed of the stream, it cannot be negative. So x ≠ - 54 Therefore, x = 6 gives the speed of the stream as 6 km / h .

Two water taps together can fill a tank in 1 7 8 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Let the smaller diameter tap fills the tank alone in x hours then the larger diameter tap fills the tank alone in ( x – 2 ) hours. In 1 hour, the smaller diameter tap can fill 1 x part of the tank. In 1 hour, the larger diameter tap can fill 1 x - 2 part of the tank. Two water taps together can fill a tank in 15 8 hours. But in 1 hour the taps fill 8 15 part of the tank. Acc. to question: 1 x + 1 ( x − 2 ) = 8 15 ⇒ x − 2 + x x ( x − 2 ) = 8 15 ⇒ 2 ( x − 1 ) x 2 − 2 x = 8 15 ⇒ 4 x 2 – 8 x = 15 x – 15 ⇒ 4 x 2 – 23 x + 15 = 0 ⇒ 4 x 2 – 20 x – 3 x + 15 = 0 ⇒ 4 x ( x – 5 ) – 3 ( x – 5 ) = 0 ⇒ ( 4 x - 3 ) ( x – 5 ) = 0 ⇒ x = 5 or x = 3 4 If x = 3 4 then x – 2 = - 5 4 which is not possible since time cannot be negative. Therefore, x = 5 and x – 2 = 3 . Hence, larger diameter tap can fill the tank in 5 hours and smaller diameter tap can fill the tank in 3 hours.