The arithmetic mean of attendance of 49 students of class A is 40% and that of 53 students of class B is 35%. Then the percentage of arithmetic mean of attendance of class A and B is
AM of attendance of 49 students of class A = 40%
AM of attendance of 53 students of class B = 35%
AM of attendance of class A and B= $\frac{49*40+35*53}{( 49+53)}$= $\frac{3815}{ 102}$ = 37.4019%
Hence,Option(D)37.4% is the correct choice.
40% of 49 students are 49⨉40/100 =19.6=20
35% of 53 students are 53⨉35/100 =19
Among 102 students present 39 students
" 1 " " 39/102 "
" 100 " " 39/102⨉100 =38
Ans D)
AM OF A=40%
AM OF B=35%
by doing mean of these two class also we can get the answer
=(a+b)/2
=(40%+35%)/2
= 37.5%
option d is correct
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