Averages Quiz -1

Average age group of 15 persons = 25 \times 12 + 5 = 305 months.

Age of people who left the group in months = 40\times 12 months = 480 months

Average age of the remaining persons in the group = \frac{ 15 \times 305 - 2 \times 480}{13} = 278.07 months

= \frac{278.07}{12} = 23.17 years.

In One hour the total number of sheets typed will be \frac{60}{10} +\frac{60}{20} + \frac{60}{30} = 11

Hence the average number of sheets per hour per typist = \frac{11}{3}.

Let the original expenditure of the mess be A

= 30 \times A + 1600 = 50 \times (A-8)

= 20 \times A = 2000

=  A = 100

= Total expenditure original = 100 \times 30 =3000

Let the sum of the ages of 6 members be x.

=> \frac {x}{6} = 28

Therefore , x = 168

Between 2001 & 2004, he got married, total members = 7

In 2004, there was an addition of of child, thus total members = 8

In 2006, avg age of family = 32,

In 2001, sum of ages of 6 members in the family = x

In 2006, their sum would be = x + 6 \times 5 = x + 30

Let present age of wife be w

Therefore according to the question

\frac {x + 30 + w + 2}{8} = 32

=> 168 + 30 + w + 2 = 256

=> w = 256 - 200 = 56

=> w = 56 years

Required average = \frac{68 \times 56 + 56 \times 50 + 42 \times 68}{56 + 42 + 68}

= \frac{3808 + 2800 + 2856}{166}

= \frac{9464}{166}

= 57.01

25 <= \frac{21 + 23 + x + 24 + 27 }{5}<=28

= 125 <= 95+x <=140

= 30 <= x <= 45

Number of Possible values of x = 16.

Total number of balls in all 10 boxes= 10 \times 25 = 250

Minimum possible number of balls in any box = 8+9+10+11+12+13+14+15+16 = 108

Maximum possible number of balls in any box = 250 - 108 = 142

Total possible groups = (w,x) , (w,y) , (w,z) , (x,y), (x,z) ,(y,z)

According to the questions = w+ x+w+y+w+z+x+y+x=z+y+z =1440.

Hence (w+x+y+z)= 480.

Required average = \frac{480}{4} = 120