Given y is inversely proportional to √x, and x = 36 when y = 36. What is the value of x when y = 54?
16
This problem involves the concept of inverse proportionality. When one quantity is inversely proportional to another (or its root), their product (or the product of one quantity and the root of the other) is a constant.
The question states that y is inversely proportional to \(\sqrt{x}\). Mathematically, this relationship can be written as:
\(y \propto \frac{1}{\sqrt{x}}\)
To turn this proportionality into an equation, we introduce a constant of proportionality, let's call it \(k\). So, the equation is:
\(y = \frac{k}{\sqrt{x}}\)
We are given that when \(x = 36\), \(y = 36\). We can use these values to find the value of \(k\).
Substitute \(x = 36\) and \(y = 36\) into the equation:
\(36 = \frac{k}{\sqrt{36}}\)
Calculate the square root of 36:
\(\sqrt{36} = 6\)
So the equation becomes:
\(36 = \frac{k}{6}\)
To find \(k\), multiply both sides by 6:
\(k = 36 \times 6\)
\(k = 216\)
Now we have the specific relationship between \(y\) and \(x\):
\(y = \frac{216}{\sqrt{x}}\)
We need to find the value of \(x\) when \(y = 54\). Use the relationship equation we just found:
\(y = \frac{216}{\sqrt{x}}\)
Substitute \(y = 54\) into the equation:
\(54 = \frac{216}{\sqrt{x}}\)
To solve for \(\sqrt{x}\), we can rearrange the equation:
\(\sqrt{x} = \frac{216}{54}\)
Perform the division:
\(\sqrt{x} = 4\)
To find \(x\), we need to square both sides of the equation:
\((\sqrt{x})^2 = 4^2\)
\(x = 16\)
So, the value of \(x\) when \(y = 54\) is 16.
Here is a summary of the steps taken:
| Step | Calculation | Result |
|---|---|---|
| Find k | \(36 = \frac{k}{\sqrt{36}} \implies k = 36 \times 6\) | \(k = 216\) |
| Find x when y=54 | \(54 = \frac{216}{\sqrt{x}} \implies \sqrt{x} = \frac{216}{54} = 4\) | \(x = 4^2 = 16\) |
| Concept | Description | Mathematical Form |
|---|---|---|
| Direct Proportion | As one quantity increases, the other increases proportionally. | \(y \propto x\) or \(y = kx\) |
| Inverse Proportion | As one quantity increases, the other decreases proportionally. | \(y \propto \frac{1}{x}\) or \(y = \frac{k}{x}\) |
| Inverse Proportion to Root | As one quantity increases, the other decreases proportionally to the root. | \(y \propto \frac{1}{\sqrt{x}}\) or \(y = \frac{k}{\sqrt{x}}\) |
| Constant of Proportionality (k) | A constant value relating two quantities in a proportion. | \(k = \frac{y}{x}\) (Direct) or \(k = yx\) (Inverse) or \(k = y\sqrt{x}\) (Inverse to √x) |
Proportionality describes how two quantities change in relation to each other. There are two main types:
In this specific problem, the inverse relationship is not with \(x\) directly, but with its square root, \(\sqrt{x}\). This is why the relationship is written as \(y = \frac{k}{\sqrt{x}}\).
Problems involving proportionality usually require you to first use a given pair of values to find the constant of proportionality (\(k\)), and then use that constant with a new value of one variable to find the corresponding value of the other variable.
When x is added to each of the numbers 11, 18, 27, 42, the numbers so obtained are in proportion. What is the mean proportional between (11x + 3) and (9x - 2)?
The ratio of the third proportion of 5 and 12 with the fourth proportion of 5, 8 and 9 is:
A, B and C start a business. A invests for 3 months, B for 4 months, and C for 6 months. C invests Rs. 2400. If A's share of profit is $\frac{2}{3}$ of B's share, and C's share of profit is $\frac{1}{2}$ of the total profit, how much money did A and B invest?
Divide Rs. 156 in the ratio 1 : 2 : 4 : 5. The rupees in the respective ratios are given by:
A. 13, 26, 53 & 64
B. 13, 26, 51 & 66
C. 13, 26, 52 & 65
D. 13, 25, 53 & 65Divide Rs. 368 in the ratio 1:5:8:9. The rupees in the respective rations are give by.
A. 16, 80, 127 & 145
B. 16, 80, 129 & 143
C. 16, 80, 128 & 144
D. 16, 80, 128 & 143