find three consecutive largest positive integers such that the sum of one third of first,one fourth of second and one fifth of third is atmost 20.
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let no. 1= x
no. 2= x+1
no. 3= x+1+1=x+2
A/Q :-
x/3+ (x+1)/4+ (x+2)/5= 20
20x + 15x+15+ 12x + 24/60=20
47x+ 39= 20×60
47x + 39= 1200
47x = 1161
x=24.7
x = 25 approx.
therefore no. 1= 25
no. 2=25+1=26
no. 3=25+2=27
hope it helps u
no. 2= x+1
no. 3= x+1+1=x+2
A/Q :-
x/3+ (x+1)/4+ (x+2)/5= 20
20x + 15x+15+ 12x + 24/60=20
47x+ 39= 20×60
47x + 39= 1200
47x = 1161
x=24.7
x = 25 approx.
therefore no. 1= 25
no. 2=25+1=26
no. 3=25+2=27
hope it helps u
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