Percentage Part - 3
Rule : If the sides of a triangle, rectangle, square, circle any two dimensional figure are increased by x %, its area is increased by
Rule : If one side is taken x % in excess and the other y % in deficit. The % error in area will be ( x - y - x y / 100 ) in excess or deficit according to the + ve or - ve sign.
Ex. If the one side of a rectangle is take in 10 % in excess while other taken as in 5 % deficit. Find the error percentage in area calculated in from the measurment.
Solution : % error = 10 - 5 - (10 * 5 / 100 ) = 5 - 1/ 2 = 9/ 2 % in excess
Rule : A student scores p % in an examination, fail by x marks, while other student scores q % of marks and gets y marks more than the minimum pass marks. Then the maximum pass marks for the examination are :
Ex. A student scores 30 % and fail by 25 marks , while other candidate who scores 55 % marks, gets 30 marks more than the minimum passed marks. Find the maximum passed marks for the examination.
Solution : here x = 25 , p = 30 % , y = 30 , q = 55 %
Rule : In an examination x % failed in Hindi and y % failed in Science, If z % of students failed in both the subjects, the percentage of students who passed in both the subjects will be : 100 - ( x - y - z )
Proof :
% of students failed in Hindi = ( x - z ) %
% of students failed in Science = ( y - z ) %
% of students failed in both sujects = z %
So % of students passed in both the subject = 100 - [ ( x - z ) + ( y - z ) - z ] = 100 - [ x + y + z ]
Ex. In an examination 35 % students failed in English, 25 % students failed in hindi and 10 % failed in both the subjects then find the % of students who passed in both the subjects.
Solution : 100 - ( 35 + 25 + 10 ) = 50 %
Ex. There are 600 boys in a hostel. Each plays either hockey or football or both. If 75 % play hockey and 45 % play football, How many play both ?
Solution : Let boys playing hockey be denoted by A and boys playing football be denoted by B then their number be n(A) and n(B)
so, 120 students plays both games
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