Test Level 2: Averages - 2 - CAT MCQ

Here, total sum of x + y should be equal to 8 and for that, the pairs of x and y are (1, 7), (2, 6), (3, 5), (4, 4), (7, 1), (6, 2) and (5, 3).

Hence, there are total 7 pairs.

Direct formula:

Average speed = 3abc/(ab+ bc+ca) 
(where a, b, c are the different speeds for different km)

Average age of class of n boys = X
Total age of n boys = nX
After 3 students join the class, total age of (n + 3) boys
= nX + X - 1 + X + 1 + X + 2
= nX + 3X + 2
= (n + 3)X + 2

Sum of ages of 5 members = 5 × 22 = 110 years
The youngest member is of 10 years.
Now, excluding the youngest member, sum of ages of the rest 4 members = 110 - 10 = 100 years
10 years ago, sum of ages = 100 - (4 × 10) = 60
Thus, average of 4 members = 60/4 = 15 years

Average weight of 35 students = 47.5 kg
Total weight of 35 students = 35 × 47.5 = 1662.5 kg
Average weight of 36 people = 47.5 + 0.5 = 48 kg
Total weight of 36 people = 36 × 48 = 1728
Therefore, weight of the teacher = 1728 - 1662.5 = 65.5 kg

A + B + C = 48 × 3 = 144 ............(i)
A + B + C + D = 44 × 4 = 176 ............(ii)
(ii) - (i) gives, D = 176 - 144 = 32
Marks of E = 32 + 3 = 35
B + C + D + E = 43 × 4 = 172
B + C + D = 172 - 35 = 137 .........(iii)
(ii) - (iii) gives,
A = 176 - 137
A = 39

Total persons = 630

Male : Female = 5 : 4

Females = 630 - 350 = 280

Bachelor males = 350 - 100 = 250

Unmarried females = 280 - 70 = 210

If half of the first number is removed, the average decreases by 1, that means, the total decreases by 3.We assume that the number is z.

Therefore, if we take 2z instead of z, the total will increase by 6 and the average increases by 2.
So, it will be x + 2.

Let x be the fixed cost and y be the variable cost.
17,500 = x + 25y ……..(i)
30,000 = x + 50y ……..(ii)
Solving the equations (i) and (ii), we get
x = 5,000, y = 500
Now, if the average expense of 100 boarders be `A`, then
100 × A = 5,000 + 500 × 100
∴ A = 550

It is given that average of the four numbers is 27.
Let A, B, C and D be the four numbers.

A + B + C + D = 108 … (1)
Also, A is increased by K; A + K.
B is increased by K + 7; B + K + 7.
C is increased by 2K + 3; C + 2K + 3.
D is decreased by 2; D – 2.

⇒ A + B + C + D + 4K = 128 – 8 = 120
⇒ 108 + 4K = 120 {From equation (1)}
⇒ 4K = 12
K = 3

The original number is ab.
Let the sum of the rest nine terms be x.
Average before the digits were interchanged = [(x/10) + (10a + b)/10]
Average after the digits were interchanged = [(x/10) + (10b + a)/10]
Given;
[(x/10) + (10b + a)/10] - [(x/10) + (10a + b)/10] = 1.8
(b - a) - (b - a)/10 = 1.8
9(b - a)/10 = 1.8
(b - a) = 2

Let there be n students in the class.
Total marks = 60n
After deducting computational errors,
we have:

⇒ 60n - 3000 = 50n
⇒ 10n = 3000
⇒ n = 300
So, there were 300 students in the class.

Let the three numbers be x, y and z.
Now, according to the question,
y = 2z and x = 2y = 4z
Also, average of the reciprocal = 7/36

After solving this, we get:
z = 3
Then,
x = 12
y = 6
Sum of all three numbers = 21

Fixed expense = Rs. 2,000
Variable cost per book = Rs. 2.25
Then, variable cost for 40,000 books = Rs. 2.25  40,000 = Rs. 90,000
Total cost of 40,000 books = Rs. 90,000 + Rs. 2,000 = Rs. 92,000

Average cost per book = 92,000/40,000 = Rs. 2.30

OR

The average cost per book = 2.25 + 2000/40000= 2.30

In both the cases, the number of items is the same.
Since the average is the middle number, the middle number in the first case is just one more than that in the second case.
So, x - y = 1

2n + 1 is always odd, what ever may be the value of n.
So, P is a set of consecutive numbers starting and ending with odd numbers.
Out of total (2n + 1) terms, we have (n + 1) number of odd terms and n number of even terms.
Let us take n = 2002
Total odd numbers = 2003 and their average = B = Sum of 1st 2003 odd numbers, divided by 2003

Total even numbers = 2002

So, difference between them = 0
Or,
A is the average of 2, 4, 6,..........., 2n = [n(n + 1)]/n = n + 1
B is the average of 1, 3, 5,........., 2n + 1 = (n + 1)2/(n + 1) = n + 1
A - B = 0

Let total bonus = x and number of workers = n
Each worker gets x/n
Condition (1):

Condition (2):

Solving (2), we get
x = 25n - 15
From (1), we get

25n² - 15n = 25n² - 15n + 75n - 45 - 4n² - 12n
4n² - 63n + 45 = 0
n = 15 (as n cannot be in fractions)
Total workers now = n + 3 = 18

Given average age of class of n boys is Y years.
Total age of n boys = n × Y = nY years
Three new boys that join the class are of ages (Y - 1), (Y - 2) and (Y + 3) years.
New total age of the boys will be

[nY + (Y - 1) + (Y - 2) + (Y + 3)] years = nY + 3Y = (n + 3)Y

The students will be in Class X after 2 years.
So, average age of the class after 2 years = (Y + 2) years

Alternate method:
Sum of the ages of the three new boys = 3Y
⇒ The average age of the class does not change.
⇒ Average age will be Y years.
So, when they are promoted to Class X (i.e. gap of 2 years), the average age will be (Y + 2) years.

His bonus in January = 3 × 10 + 7 = Rs. 37
His bonus in February = 3 × 20 + 7 = Rs. 67
His bonus per month from March to December = 3 × 30 + 7 = Rs. 97
Total bonus = 37 + 67 + 10 × 97 = Rs. 1074
Total basic salary = 50 × 1074 = Rs. 53,700

Assume a, b and c are the numbers. a > b > c

a + b + c = 3x 

a + c = 3x – b

2y = 3 x – b 

2(x – 2) = 3x – b

2x – 4 = 3x – b  

x – b = - 4

So, the difference between x and b is 4.