Question

There are 4 consecutive odd numbers (p1, P2, P3 and p4) and three consecutive even numbers (m^1 m^2
and m^3). The average of the odd numbers is 6 less than the average of the even numbers. If the sum of
the three even numbers is 16 less the sum of the four odd numbers, what is the average of (pl, p2, p3 and
p4).​

Answer 1

Answer:

The average of p1, p2, p3 and p4 is 34

Step-by-step explanation:

Let the odd numbers be denoted as: n, n+2, n+4 and n+6

Let the even numbers be denoted as: m, m+2, m+4

The sum of the odd numbers is (4n + 12) .......................(A)

The average of the odd numbers is (4n+12)/4 = (n+3).............(B)

The sum of the even numbers is (3m + 6).........................(C)

The average of the even numbers  is (3m + 6)/3 = (m+2)................(D)

From the information provided, we have:

(B) is 6 less than (D). This means:

(n+3) + 6 = m + 2

n - m = -7...............(Eqn 1)

Also, it is given that (C) is 16 less than (A). This means:

3m + 6 + 16 = 4n + 12

4n - 3m = 10...............(Eqn 2)

We now have two equations in two unknowns that need to be solved simultaneously.

Multiply Eqn 1 by 4, we get:

4n - 4m = -28...........(Eqn 3)

Subtracting (Eqn 3) from (Eqn 2), we get:

m = 38

Substituting for m in (Eqn 1), we get:

n = 31

Thus, the odd numbers are: 31, 33, 35 and 37

and the even numbers are: 38, 40 and 42

The required average of the odd numbers is (31+33+35+37)/4 = 34

Verify:

Sum of odd numbers = 136

average of odd numbers = 34

sum of even numbers = 120

average of even numbers = 40

Using information given in the problem, we can see:

34 is 6 less than 40, and

120 is 16 less than 136

Note: You may denoted the odd numbers as p1, p2, p3 and p4 and assume that p1 = n; p2=n+2; p3= n+4 and p4 = n+6. Similarly, for m1, m2 and m3. What is most important is that consecutive odd or even numbers have a common difference of "2".