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
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.
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'.
The problem says q is less than the average (k) by the *same* amount 'x'.
So, we have expressions for p and q in terms of k and a variable 'x'.
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.
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.
| 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. |
The average, or arithmetic mean, is a fundamental concept in statistics and mathematics. It represents a central value of a set of numbers.
This problem demonstrates how algebraic manipulation and understanding the definition of average can be used to solve quantitative aptitude questions.
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When a child reaches adolescence, there is apt to be a conflict between the parents and the child, since
the latter considers himself to be by now quite capable of managing his own affairs, while the former
are filled with parental solicitude, which is often a disguise for love of power. Parents consider, usually,
that the various moral problems which arise in adolescence are peculiarly their province. The options
they express, however, are so dogmatic that the young seldom confide in them, and usually go their
own way in secret.