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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Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

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 Q) Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution) Let the time taken by the smaller pipe to fill the tank be x hr . 

So , time taken by larger pipe be ( x - 10 ) hr .

 Part of the tank filled by smaller pipe in 1 hour is 1/x

 Part of the tank filled by larger pipe in 1 hour is 

1/x-10 

 So , according to the question










 

Case 1- If time taken by smaller pipe be 30/8 i.e 3.75 hours . 

So , Time taken by larger pipe will be negative which is not possible . 

Hence , this case is rejected . 

Case 2- If the time taken by smaller pipe be 25. Then , time taken by larger pipe will be 15 hours . 

Therefore , time taken by smaller pipe be 25 hours and time taken by larger pipe will be 15 hours .



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