- The average of n terms equals $$\frac{x_1+x_2+. . .+x_n}{n}$$
- The weighted average of n terms equals $$\frac{w_1*x_1+ w_2*x_2+. . .+ w_n*x_n}{ w_1+ w_2+ . . . w_n}$$
Averages
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Question 1
3 times the first of four consecutive even numbers is 6 more than twice the last number. What is the average of the four even integers?
Solution
As the numbers are consecutive even numbers, we can say they are x, x+2, x+4, x+6. So,
3*x = 6 + 2*(x+6). So, x=18.
Average of 18, 20, 22 and 24 is 21
Question 2
X, Y and Z are three consecutive odd natural numbers in ascending order. If five times X is three more than four times Z, find the average of X, Y and Z
Solution
Let the three odd numbers be 2n-1, 2n+1 and 2n+3
It is given that 5 times X is 3 more than 4 times Z.
So, 5*(2n-1) = 3+4*(2n+3)
Or, 10n - 5 = 3+8n+12
Or, 2n = 20 and n=10
So, the three numbers are 19, 21 and 23 and their average is 21.
Question 3
All the natural numbers from 1 to 'N' were written on a board. When one of the numbers was erased, the average of the remaining numbers was 25.375. Which number was erased?
Solution
Let the number of natural numbers be N and the number deleted be P.
new average = $$\frac{\left(New\ Sum\right)}{N-1}$$ because 1 number is erased, so total number left will be N-1.
Now new sum = $$\frac{N\left(N+1\right)}{2}-P$$
The new average = $$\frac{\left(N^2+N-2P\right)}{2\left(N-1\right)}$$ = $$\frac{\left(N^2-N+2N-2P\right)}{2\left(N-1\right)}$$ = $$\frac{N\left(N-1\right)+2N-2P}{2\left(N-1\right)}$$ = $$\frac{N\left(N-1\right)+2N-2+2-2P}{2\left(N-1\right)}$$ = $$\frac{N}{2}+1-\frac{P-1}{N-1}$$
So, new average = $$\frac{N}{2} + 1 - \frac{(P-1)}{(N-1)}$$. Notice that, $$\frac{(P-1)}{(N-1)}$$ lies between 0 and 1.
$$\Rightarrow$$ $$\frac{N}{2} + 1 - \frac{(P-1)}{(N-1)}$$ = 25.375
$$\Rightarrow$$ $$\frac{N}{2}$$ = 24.375 + $$\frac{(P-1)}{(N-1)}$$
We know that 0 $$\leq$$ $$\frac{(P-1)}{(N-1)}$$ $$\leq$$ 1
$$\Rightarrow$$ 0+24.375 $$\leq$$ 24.375 + $$\frac{(P-1)}{(N-1)}$$ $$\leq$$ 1+24.375 (Adding 24.375 in the inequality)
$$\Rightarrow$$ 24.375 $$\leq$$ $$\frac{N}{2}$$ $$\leq$$ 25.375
But we know that N is an integer therefore $$\frac{N}{2}$$ can take only two values in the given interval.
Where $$\frac{N}{2}$$ = 24.5 or 25
Case 1: when $$\frac{N}{2}$$ = 25 or $$N$$ = 50 then $$\frac{(P-1)}{(N-1)}$$ will be equal to (25 - 24.375)
$$\Rightarrow$$ $$\frac{(P-1)}{(N-1)}$$ = 0.625
$$\Rightarrow$$ $$P = 31.625$$ Which is not possible.
Case 2: when $$\frac{N}{2}$$ = 24.5 or $$N$$ = 49 then $$\frac{(P-1)}{(N-1)}$$ will be equal to (24.5 - 24.375)
$$\Rightarrow$$ $$\frac{(P-1)}{(N-1)}$$ = 0.125
$$\Rightarrow$$ $$P = 7$$
This is a correct solution hence we can say that P=7 is the missing number. Hence option B is the correct answer.
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