Sum of 50 member = (50 x 38 )
= 1900
Sum of 48 member = [1900- (45 + 55 ) ]
= 1800
? Required Average = 1800/48
= 225/6
= 37.5
Average will increase by 8 times.
So average of new members = (21 x 8 ) = 168
Average of first n natural numbers = ( n + 1 ) / 2
= ( 79 + 1 ) / 2
= 40
Age of friend left = Previous average of the age - Number of persons present x Increase in average age
= 15 - ( 10-1 ) x ( 16-15 )
= 15- 9 x 1
= 15 - 9 = 6yr
sum of observations
= 308 + 142 + 160 245 + 25 = 880
Average = Sum of all observations / Number of observations
=880 / 5
=176
Age E = ( A + B + C + D + E ) - [ ( A + B ) + ( C + D ) ]
= ( 40 x 5 ) - [ ( 35 x 2 ) + ( 42 x 2 )]
=200 - ( 70 + 84)
= 200 - 154
= 46
Let the average for last 4 matches be y.
Sum of score for 10 matches = sum of score for first 6 matches + sum of score of last 4 matches.
So 53 x 6 + 4y = 10 x 43.9
? 4x= 439-318
?4x= 121
? x= 30.25
Average height of the whole class = sum of height of 40 girls / 40
= (30 x 160 + 10 x 156) / 40
= 159 cms.
Since a, b, c, d, e are 5 consecutive numbers so
b = a + 2.
c = a + 4.
d = a + 6.
e = a + 8.
Average = (a + b + c + d + e)/ 5
= [a + (a + 2 ) + (a +4 ) + (a+6 )+ (a+8)] / 5
= (5a+20) / 5
= a + 4
Let second number = x
So, first number = 2x and third number= 4x
? ( x + 2x + 4x ) / 3 = 56
? 7x = 168
? x = 24
So the numbers are 48,24, 96,
Average = ( 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 ) / 9
= 3(1+2+3+4+5+6+7+8+9)/9
= ( 45 / 3 )
= 15
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