CoCubes Averages Quiz-1

Average = (45+50+33+22+15)/5= 165/5
= 33

If fixed cost = x and cost per invitee = y, then
x + 40y = 200*40 or x + 40y = 8000 ----(i)
x + 50y = 180*50 or x + 50y = 9000 ----(ii)
On solving (i) & (ii),
x=4000, y=100

Let the number of units sold for X be x.

The units sold for y is thrice this.

Therefore :

Units for y are : 3x

The total revenue is :

27 × x + 51 × 3x = 180x

Average per unit sold :

Total revenue / total units

Total units = x + 3x = 4x

Average revenue = 180x/4x = 45
= Rs 45 lacs

127+63+81+44+95=410
Average=410/5
total average=82

Let presently the members be x1,x2,…..x8.
Their average age = (x1+x2+x3……x8)/8……(1)
Average age of 8 members years ago
= x1 + x2 + x3……+ x8 - 3*8/8 [here, 3*8 comes for age of 8 members 3 years ago]
= x1 + x2+…. X8 -24/ 8…..(2)
Consider y1 be the young member
= y1+x2+x3…..+x8 – 24/ 8…..(3) [average of 8 people when sachin is replaced by old member]
It is given in the question that when new member is substituted the average age becomes same as it was 3 years ago
From equation (1) and equation (3)
(x1 + x2 +……x8)/8 = (y1 + x2 + x3……+x8)/8
x1 = y1-24
y1-x1 = 24

Let, the sum of Paul’s two Daughter age = a
Then, Paul's age = a² (given, Paul’s age is square of the sum of his daughters age)
After 4years, Paul’s daughter's age = a + 8 (2 daughter’s age)
Paul\'s age = a² + 4
Given, the difference between his age and his daughters ages will be 26 years.
= (a² + 4) - (a +8) = 26
= a² + 4 – a – 8 = 26
= a² - a - 30 = 0
= a² - 6a + 5a - 30 = 0
(on solving the above equation we get this)
= a(a-6) + 5(a-6)
= (a +5) (a-6)
= a = -5, 6
We know that, age can't be in negative form,
So, the sum of two daughter’s age will be, 6
Then, Father's age =a² = 6² = 36

Correct answer: B

Solution:

Average= 40+19+75+83+56+33/6

               =306/6 = 51

Lets say that the students are named a, b, c, d and e, in increasing order of weights. The average of a, b, c and d is 40 kg, whereas the average of b, c, d and e is 45 kg.
The sum of a, b, c and d is 160 kg, and the sum of b, c, d and e is 180 kg.

What is the total weight of all the students?
There are two ways of looking at this.
a) 160 + e
b) 180 + a
Or, e is 20 more than a.

The total weight is 160 + e. So, the highest value of e will correspond to the highest possible average. The highest possible value of e occurs when it is 20 higher than the highest possible value for a, which is 40 (all the first 4 scores are equal to 40).
So, the highest possible average is 160+605 = 44
This will be the case when the weights are 40 kgs, 40kgs, 40 kgs, 40 kgs and 60 kgs.

Conversely, the least possible value for the average occurs when a is the least. This happens when e is the least too (since a is 20 less than e).
The least possible value for e is 45 = 1804
So, the least possible value for a would be 25.
The least possible average = 180+255 = 41

This will be the case when the weights are 25 kgs, 45kgs, 45 kgs, 45 kgs and 45 kgs. So, the difference between maximum possible and minimum possible average = 3 kgs.

Correct Option: C

Solution:

Average of any series = Sum of numbers in given series/ number of items in the series.

Hence 96+24+102+45+63/5

= 330/5 = 66

Hence C is the correct Option.

Correct Option: D

Solution: Let the total weight of 10 students be X

X/10=55

Therefore X = 550

If 2 more students are added then the new average is 58. Lets assume the total weight of 2 new students as Y

58=550+ Y/12

=58*12=550+Y

=696-550 =Y

Hence Y = 146

avg weight= 146/2=73