Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares.

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  Q) Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares. Solution) Let the sides of the two squares be x m and y m. Thus, their perimeters will be 4x and 4y and areas will be x² and y². Hence, according to question – =4x-4y=24 ⇒x-y=6 ⇒x=y+6 And x²+y² = 468 Substituting value of x- (y+6)²+y²=468 ⇒36+y²+12y+y²=468 ⇒2y²+12y-432=0 ⇒y²+6y-216=0 ⇒y²+18y-12y-216=0 ⇒y(y+18)-12(y+18)=0 ⇒(y+18)(y-12)=0 ⇒y=-18 or 12 Since, side cannot be negative. Therefore, the sides of the square are 12 m and (12+6)m i.e 18 m.
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A train travels 360 kmkm at a uniform speed. If the speed had been 5km/h more, it would have taken 1hour less for the same journey. Find the speed of the train.

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 Q) A train travels 360 kmkm at a uniform speed. If the speed had been 5km/h more, it would have taken 1hour less for the same journey. Find the speed of the train.

Solution) Let the speed of the train be x km / h .
 Time taken to cover 360 km/h be 360/x

According to question ( x + 5 ) (360/X --1 ) = 360

=  360 - x + 1600/x -5 = 360 
 = x² + 5x - 1800 = 0 
= x² + 45x - 40x - 1800 = 0 
= x ( x + 45 ) -40 ( x + 45 ) = 0 
= ( x + 45 ) ( x - 40 ) = 0 
⇒x = 40 , -45 X 

Since , the speed cannot be negative .
 Therefore , the speed of the train is 40 km / h .




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