When x is substracted from each of 55, 50, 23 and 22, the numbers so obtained in this order, are in proportion. What is the fourth proportional of 3, 7 and x?
35
The question involves two main concepts: numbers in proportion and finding the fourth proportional. First, we are told that when a specific value, let's call it 'x', is subtracted from four given numbers (55, 50, 23, and 22), the resulting numbers are in proportion in that specific order. This means the ratio of the first resulting number to the second is equal to the ratio of the third resulting number to the fourth.
The second part asks us to find the fourth proportional of three given numbers: 3, 7, and the value of x that we found in the first part.
Let the four numbers after subtracting x be:
Since these numbers are in proportion, we can write the relationship as:
$$\frac{55 - x}{50 - x} = \frac{23 - x}{22 - x}$$
To solve for x, we can cross-multiply:
$$(55 - x)(22 - x) = (23 - x)(50 - x)$$
Now, let's expand both sides of the equation:
Left side: \(55 \times 22 - 55x - 22x + x^2 = 1210 - 77x + x^2\)
Right side: \(23 \times 50 - 23x - 50x + x^2 = 1150 - 73x + x^2\)
So, the equation becomes:
$$1210 - 77x + x^2 = 1150 - 73x + x^2$$
We can subtract \(x^2\) from both sides:
$$1210 - 77x = 1150 - 73x$$
Now, let's isolate the terms with x on one side and the constant terms on the other side:
$$1210 - 1150 = 77x - 73x$$
$$60 = 4x$$
Finally, solve for x:
$$x = \frac{60}{4}$$
$$x = 15$$
So, the value of x is 15.
The second part of the question asks for the fourth proportional of 3, 7, and x. We found that \(x = 15\). Let the fourth proportional be 'd'.
For four numbers a, b, c, and d to be in proportion, we have the relationship:
$$\frac{a}{b} = \frac{c}{d}$$
In our case, a = 3, b = 7, and c = x = 15. We need to find d.
So, we have:
$$\frac{3}{7} = \frac{15}{d}$$
To solve for d, we can cross-multiply:
$$3 \times d = 7 \times 15$$
$$3d = 105$$
Now, divide by 3:
$$d = \frac{105}{3}$$
$$d = 35$$
The fourth proportional of 3, 7, and 15 is 35.
| Step | Description | Calculation |
|---|---|---|
| 1 | Set up the proportion equation | \(\frac{55 - x}{50 - x} = \frac{23 - x}{22 - x}\) |
| 2 | Cross-multiply | \((55 - x)(22 - x) = (23 - x)(50 - x)\) |
| 3 | Expand both sides | \(1210 - 77x + x^2 = 1150 - 73x + x^2\) |
| 4 | Solve for x | \(60 = 4x \Rightarrow x = 15\) |
| 5 | Set up equation for fourth proportional | \(\frac{3}{7} = \frac{15}{d}\) |
| 6 | Solve for d (fourth proportional) | \(3d = 105 \Rightarrow d = 35\) |
The value of x is 15, and the fourth proportional of 3, 7, and x (which is 15) is 35.
| Concept | Definition/Formula | Example |
|---|---|---|
| Ratio | Comparison of two quantities (\(a:b\) or \(\frac{a}{b}\)) | 3:7 or \(\frac{3}{7}\) |
| Proportion | Equality of two ratios (\(a:b :: c:d\) or \(\frac{a}{b} = \frac{c}{d}\)) | 3:7 :: 15:35 because \(\frac{3}{7} = \frac{15}{35}\) |
| Terms of Proportion | In \(a:b :: c:d\), a and d are extremes; b and c are means | In 3:7 :: 15:35, 3 and 35 are extremes; 7 and 15 are means |
| Product of Extremes and Means | In a proportion, product of extremes equals product of means (ad = bc) | \(3 \times 35 = 105\), \(7 \times 15 = 105\). So \(105 = 105\). |
| Fourth Proportional | In \(a:b :: c:d\), d is the fourth proportional of a, b, and c | Fourth proportional of 3, 7, 15 is d such that \(\frac{3}{7} = \frac{15}{d}\) |
Solving equations like \((55 - x)(22 - x) = (23 - x)(50 - x)\) is a fundamental skill in algebra. It involves expanding binomials, combining like terms, and isolating the variable.
In this problem, we encountered a quadratic term (\(x^2\)) which conveniently cancelled out, leaving a linear equation that was straightforward to solve.
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