Averages Quiz-1

Change in total weight of 10 students = difference in weight of the student who joined and the student
who left = weight of first student who left = 30 + (10×2) = 50
weight of the student who join last = 50 – 10 = 40

Thus change in average weight = (40 – 30)/10 = 1

Average of 5 terms = 20
Hence, the total sum of 5 terms = 100
Let terms be A,B,C,D,E
So, the sum will be A+B+C+D+E =100
Given, 9A = B+C+D+E
So, 10A = 100,
A = 10

Average age group of 15 persons=305 months.
age of people who left the group in months=40*12 months
=480 months
Average age of the remaining persons in the group=(15*305-2*480)/13=278.07 months or 23.17 years.

In One hour the total number of sheets typed will be 60/10 +60/20 +60/30=11

Hence the number of sheets /hour per typist is 11/3.

Let the initial expenditure be A.
According to ques
30 * A + 1600=50 *(A-8)

= 20 * A= 2000

= A=100

= Total expenditure original =100*30 =3000

If present age of Binod\'s wife is x years.

Then according to the question:

(33*6+x+2)/8=32

X+2+198 = 256

X + 2 = 58

X=56

X=56 Years

Binod wife's age in 2001 was 51 years. Therefore Binod's age in 2001 was greater than 51 and minimum possible prime number above 51 is 53.

Therefor minimum possible present age of Binod=53+5=58 years.

25 *5<=21+23+x+24+27<=28*5

= 125<=95+x<=140

30<=x<=45

Number of Possible values of x= 16.

Total number of balls in all 10 boxes= 10 * 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= 480/4=120