Average Questions

Answer: Option C

sum of ages of girls = 35
sum of ages of rest of 4 girls=20
So, Average = 20/4 = 5

Answer: Option D

Given Avg of age of 5 family = 32
Total Member age / 5 = 32
Total age of 5 = 160

After grandma died, total age will be
Total Member age / 4 = 18
Total age of 4 Member = 72

Age of grandma = 160-72 = 88 Years.

Answer: Option A

Let the average age of the whole team by x years.

11x - (26 + 29) = 9(x -1)

11x - 9x = 46

2x = 46

x = 23.

So, average age of the team is 23 years.

Answer: Option B

total age of students = 22 x 21 = 462
if teacher is included no of persons will be 23 avg will be 22 yrs
total age (if teacher included) = 23 x 22 = 506
Age of teacher = 506 - 462 = 44

Answer: Option C

Actually u wrote the que. wrong. 225 Rs was not given.
s=8000
s=8(2*275+15d)
by solving
d=30
so 1st team prize=275+(15*30)=725

Answer: Option A

5 subjects each subject 100 marks total = 500

Mark = average * no. of paper
=> 52*5 = 260
=> 260+15 = 275
Mark= (275/500)*100 = 55%

Answer: Option A

Total weight is M+D+I=74 given that M-D-I=46
solving above 2 eq's we get Mother weight = 60
Now reaming weight=14 which is sum of daughter and Infant weight. Given that Infant weight is 60% less than Daughter.
i.e If daughter weight is 100 then infant weight is 40. So weight ratio Of D and I is 100 : 40 i.e ... 5 : 2

So (5+2)=7 parts equal to 14. then 5 parts equal to 10.
2 parts equal to 4. daughter weight is =10 and infant weight is = 4

Answer: Option C

Total Initial Salary : 17*45000 = 765000
Average Salary After removal of 3 Teachers = 45000-2500 = 42500
Total Final Salary : 14*42500 = 595000
Sum of Salaries of 3 teachers who left : 765000 - 595000 = 170000

Answer: Option C

1+3+5+7+9+11+13.........+199 i.e 100 terms and
sum of n odd natural numbers is n^2 i.e 100^2 =10,000

-(2+4+6+8+10+12...........+200) i.e 100 terms and
sum of n even natural numbers is n(n+1) i.e -(100(100+1))= -10,100
average of first 200 terms = (10,000+(-10,100))/200
=-100/200
=-1/2 Ans.

Answer: Option D

Transaction I

Thomas have bought X items for Rs.9000.
Then the cost of each item = Rs.9000/X ...(1)

Transaction II

If each item was cheaper by Rs.30, he could have bought 50 more items.
Therefore, with Rs.30 discount, the amount of each item = Rs.9000 / X + 50 ...(2)

We know that the cost of each item in transaction II will be lesser than that of transaction I by Rs. 30.

i.e (1) - (2) = 30

Since he bought 50 more items when cost of each item is less by Rs.30 then we have
9000/X - 9000/X + 50 = 30
1/X - 1/X + 50 = 30/9000
50 / X(X + 50) = 1/300
X(X + 50) = 15000
X^2 + 50X = 15000
X^2 + 50X - 15000 = 0
(X+150)(X-100) = 0
X = -150 or X = 100
X cannot be a negative value. Therefore, X = 100.
Hence the answer is 100.