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The arithmetic mean of attendance of $49$ students of class $\text{A}$ is $40%$ and that of $53$ students of class $\text{B}$ is $35%.$ Then the percentage of arithmetic mean of attendance of class $\text{A}$ and $\text{B}$ is

  1. $27.2\%$
  2. $50.25\%$
  3. $51.13\%$
  4. $37.4\%$
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3 Answers

Best answer
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13 votes

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.

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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)

–1 votes
–1 votes

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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