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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

1. 27.2%
2. 50.25%
3. 51.13%
4. 37.4%

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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.

by
27 95 296

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)

by
340 828 1616

This is wrong :(
No @Sir it is not wrong I think. It needs more accurate value :)
yes, (y)

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

by
1 1 3

### 1 comment

If we have M Things with an average of P and  N  Things with an average of Q , now whatever is the value of M and N , when both groups are merged ,the average will always be somewhere between P and Q .

( I am expecting that we don’t have negative no. of things , like -3 peoples with average height 160 cm .)

In this particular question we have only one option in that range so if we know this fact we don’t need to calculate .

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