Answer :

Given :- Speed of boat =18km/hr

Distance =24 km

Let x be the speed of stream.

Let t_{1} and t_{2} 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 = t_{1}

Therefore,

t_{1} =

For downstream,

Speed = (18 + x) km/hr

Distance =24 km

Time = t_{2}

Therefore, t_{2} =

Now according to the question-

t_{1} = t_{2} + 1

⇒

⇒ 48x = (18 − x)(18 + x)

⇒ 48x = 324 + 18x − 18x − x^{2}

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + 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 = x^{2} +(x + 2)^{2}

= x^{2} + x^{2} + 4x +4

= 2x^{2} + 4x + 4

Given that their sum of squares = 290

⇒ 2x^{2} + 4x + 4 = 290

⇒ 2x^{2} +4x = 290 – 4 = 286

⇒ 2x^{2} + 4x – 286 = 0

⇒ 2(x^{2} + 2x – 143) = 0

⇒ x^{2} + 2x – 143 = 0

⇒ x^{2} + 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.

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