The average of three numbers p, q and r is k. p is as much more than the average as q is less than the average. What is the value of r?
K
Understanding the Average Problem
The problem asks us to find the value of one number, r, given the average of three numbers p, q, and r, and a specific relationship between p, q, and the average.
We are given that the average of p, q, and r is k. The formula for the average of three numbers is:
$$\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of items}}$$
In this case, the average is k, and the numbers are p, q, and r. So, we can write the equation:
$$\frac{p + q + r}{3} = k$$
Multiplying both sides by 3, we get the sum of the three numbers:
$$p + q + r = 3k$$
This is our first key relationship.
Analysing the Relationship between p, q, and the Average k
The problem states: "p is as much more than the average as q is less than the average."
Let the amount by which p is more than the average (k) be 'x'.
- p is more than k by x means: $$p = k + x$$
The problem says q is less than the average (k) by the *same* amount 'x'.
- q is less than k by x means: $$q = k - x$$
So, we have expressions for p and q in terms of k and a variable 'x'.
Solving for the Value of r
Now, we use the sum equation we derived earlier: $p + q + r = 3k$.
Substitute the expressions for p and q into this equation:
$$(k + x) + (k - x) + r = 3k$$
Let's simplify the left side of the equation:
$$k + x + k - x + r = 3k$$
The '+x' and '-x' terms cancel each other out:
$$k + k + r = 3k$$
$$2k + r = 3k$$
To find the value of r, we need to isolate r on one side of the equation. Subtract 2k from both sides:
$$r = 3k - 2k$$
$$r = k$$
Thus, the value of r is equal to k.
Summary of the Solution
Given that the average of p, q, and r is k, we have $p+q+r = 3k$. The condition that 'p is as much more than the average as q is less than the average' implies that the deviations of p and q from the average (k) are equal in magnitude but opposite in sign. If $p = k+x$, then $q=k-x$. Substituting these into the sum equation leads to $(k+x) + (k-x) + r = 3k$, which simplifies to $2k + r = 3k$. Solving for r gives $r = k$.
| Given Information | Mathematical Representation |
|---|---|
| Average of p, q, r is k | $$(p+q+r)/3 = k \implies p+q+r = 3k$$ |
| p is 'x' more than average k | $$p = k + x$$ |
| q is 'x' less than average k | $$q = k - x$$ |
| Substitute p and q into sum equation | $$(k+x) + (k-x) + r = 3k$$ |
| Simplify and solve for r | $$2k + r = 3k \implies r = k$$ |
The final value of r is k.
Revision Table: Key Concepts
| Concept | Definition/Formula | Relevance to Problem |
|---|---|---|
| Average (Mean) | Sum of values divided by the number of values. | The initial relationship given: $(p+q+r)/3 = k$. |
| Algebraic Equation | A mathematical statement with an equals sign. | Used throughout to represent relationships and solve for the unknown (r). |
| Substitution | Replacing a variable with an expression it is equal to. | Substituting expressions for p and q into the sum equation. |
| Solving Equations | Finding the value(s) of the variable(s) that make the equation true. | The final step to isolate and find the value of r. |
Additional Information: Properties of Average
The average, or arithmetic mean, is a fundamental concept in statistics and mathematics. It represents a central value of a set of numbers.
- Sum of Deviations: A key property of the arithmetic mean is that the sum of the deviations of each number from the mean is always zero. In this problem, the deviations are $(p-k)$, $(q-k)$, and $(r-k)$. We found that $(p-k) = x$ and $(q-k) = -x$. If the sum of deviations is zero, then $(p-k) + (q-k) + (r-k) = 0$. Substituting the values: $x + (-x) + (r-k) = 0$, which simplifies to $r-k = 0$, meaning $r=k$. This confirms our result using a property of the mean.
- Impact of Values: The average is influenced by the value of every number in the set. If any number changes, the average generally changes.
- Applications: Averages are used widely in daily life, from calculating test scores to understanding economic data.
This problem demonstrates how algebraic manipulation and understanding the definition of average can be used to solve quantitative aptitude questions.
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