Q. 33

# A motorboat whose

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.

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