i) Let the fixed monthly hostel charges be Rs. x and the cost of food per day taken from the mess be y.As per the given condition,x+20y=1000        ...(1)x+26y=1180         ...(2)Subtracting eq(1) from eq (2), we get⇒6y=180⇒y=30Substitute y=30 in eq (1), we get⇒x+20(30)=1000⇒x+600=1000⇒x=400So, the fixed monthly hostel charge is Rs. 600 and the cost of food per day is Rs. 30.ii) Let the numerator be x and the denominator be y, so the fraction will be yx​.yx−1​=31​   ⇒3x−y=3⇒3x−3=y       ...(1)    y+8x​=41​       ⇒4x=y+8       ...(2)Substituting for y in (2) we get,4x=3x−3+8x=5putting x=5 in eq(1) we get y=12∴fraction=yx​=125​iii) Let x be the number of correct answers and y be the number of wrong answers.So, 3x−y=40   ...(1)4x−2y=50       ...(2)Multiplying (1) by 2, we get6x−2y=80       ...(3)4x−2y=50      ...(4)Subtracting eq(4) from eq(3), we get⇒2x=30⇒x=15Putting x=15 in eq (1), we get ⇒3(15)−y=40⇒y=5So, Correct answers =15 and wrong answers =5.Total number of questions in the test =x+y=20.iv) Let the speeds of the first and second cars be x km/hr and y km/hr respectively.Distance between places A and B =100km.Moving in the same direction, relative speed will be =x−y5×(x−y)=100⇒x−y=20            ...(1)  Moving in opposite directions, relative speed will be =x+y⇒1×(x+y)=100⇒x+y=100         ...(2)Adding (1) and (2), we get ⇒2x=120⇒x=60Substitute in eq (1), we get⇒y=40The speeds of the first and second cars are 60 km/hr and 40 km/hr respectively.v) Let the length of rectangle be l and the breadth be b. So, the area will be A=l×b.Case 1: l′=l−5b′=b+3A′=A−9=lb−9⇒(l−5)(b+3)=lb−9⇒lb+3l−5b−15=lb−9⇒3l−5b=6    ....(1)    Case 2:l′′=l+3b′′=b+2A′′=A+67=lb+67(l+3)(b+2)=lb+67⇒lb+2l+3b+6=lb+67⇒2l+3b=61  ...(2)Multiplying (1) by 3 and (2) by 5 and adding both equations, we get9l−15b=18        ...(3)10l+15b=305   ...(4)Adding eq (3) and eq (4), we get19l=323⇒l=17Putting l=17 in eq(1), we get ⇒3(17)−5b=6⇒45=5b⇒b=9The length and breadth of the rectangle are 17 and 9 units respectively.