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

OR

Find two consecutive odd positive integers, sum of whose squares is 290.

Given :- Speed of boat =18km/hr

Distance =24 km

Let x be the speed of stream.

Let t₁ and t₂ be the time for upstream and downstream.

As we know that,

speed= distance / time

⇒ time = distance / speed

For upstream,

Speed = (18 − x) km/hr

Distance = 24 km

Time = t₁

Therefore,

t₁ =

For downstream,

Speed = (18 + x) km/hr

Distance =24 km

Time = t₂

Therefore, t₂ =

Now according to the question-

t₁ = t₂ + 1

⇒

⇒ 48x = (18 − x)(18 + x)

⇒ 48x = 324 + 18x − 18x − x²

⇒ x² + 48x – 324 = 0

⇒ x² + 54x − 6x – 324 = 0

⇒ x(x + 54) − 6(x + 54) = 0

⇒ (x + 54)(x − 6) = 0

⇒ x = − 54 or x = 6

Since speed cannot be negative.

⇒ x = − 54 will be rejected

∴ x = 6

Thus, the speed of stream is 6km/hr.

OR

Let one of the odd positive integer be x

then the other odd positive integer is x + 2

their sum of squares = x² +(x + 2)²

= x² + x² + 4x +4

= 2x² + 4x + 4

Given that their sum of squares = 290

⇒ 2x² + 4x + 4 = 290

⇒ 2x² +4x = 290 – 4 = 286

⇒ 2x² + 4x – 286 = 0

⇒ 2(x² + 2x – 143) = 0

⇒ x² + 2x – 143 = 0

⇒ x² + 13x – 11x – 143 = 0

⇒ x(x + 13) – 11(x + 13) = 0

⇒ (x – 11)(x + 13) = 0

⇒ (x – 11) = 0 , (x + 13) = 0

Therefore , x = 11 or – 13

According to question, x is a positive odd integer.

Hence, We take positive value of x

So, x = 11 and (x + 2) = 11 + 2 = 13.

Therefore , the odd positive integers are 11 and 13.