If 5A = 4B, 7B = 3C, and 2C = 7D, then A:D is:
6:5
This problem asks us to find the ratio of A to D, given a series of equations relating A, B, C, and D. We are given the following relationships:
5A = 4B7B = 3C2C = 7DTo find the ratio A:D, we can first express each given equation as a ratio and then combine them. Let's convert each equation into a ratio form:
From the first equation, 5A = 4B, we can find the ratio A:B:
$$ \frac{A}{B} = \frac{4}{5} $$
So, the ratio A:B is 4:5.
From the second equation, 7B = 3C, we can find the ratio B:C:
$$ \frac{B}{C} = \frac{3}{7} $$
So, the ratio B:C is 3:7.
From the third equation, 2C = 7D, we can find the ratio C:D:
$$ \frac{C}{D} = \frac{7}{2} $$
So, the ratio C:D is 7:2.
Now that we have the individual ratios A:B, B:C, and C:D, we can chain them together to find the ratio A:D. We can do this by multiplying the fractional forms of the ratios:
$$ \frac{A}{D} = \left(\frac{A}{B}\right) \times \left(\frac{B}{C}\right) \times \left(\frac{C}{D}\right) $$
Substitute the values of the ratios we found:
$$ \frac{A}{D} = \left(\frac{4}{5}\right) \times \left(\frac{3}{7}\right) \times \left(\frac{7}{2}\right) $$
Now, we perform the multiplication and simplify the resulting fraction:
$$ \frac{A}{D} = \frac{4 \times 3 \times 7}{5 \times 7 \times 2} $$
We can cancel out common factors in the numerator and denominator. The number 7 appears in both the numerator and denominator. The number 4 in the numerator can be written as $2 \times 2$, allowing us to cancel out one factor of 2 with the 2 in the denominator.
$$ \frac{A}{D} = \frac{4 \times 3 \times \cancel{7}}{5 \times \cancel{7} \times 2} = \frac{4 \times 3}{5 \times 2} $$
Further simplifying:
$$ \frac{A}{D} = \frac{(2 \times 2) \times 3}{5 \times 2} = \frac{2 \times \cancel{2} \times 3}{5 \times \cancel{2}} = \frac{2 \times 3}{5} $$
$$ \frac{A}{D} = \frac{6}{5} $$
The ratio A:D is therefore 6:5.
This can be summarized in a table showing the steps:
| Given Equation | Equivalent Ratio | Fractional Form |
|---|---|---|
5A = 4B |
A:B = 4:5 | $A/B = 4/5$ |
7B = 3C |
B:C = 3:7 | $B/C = 3/7$ |
2C = 7D |
C:D = 7:2 | $C/D = 7/2$ |
| Combined Ratio | A:D = ? | $(A/B) \times (B/C) \times (C/D) = (4/5) \times (3/7) \times (7/2)$ |
| Simplified Ratio | A:D = 6:5 | $A/D = 6/5$ |
The final ratio of A:D is 6:5.
| Step | Action | Formula/Calculation |
|---|---|---|
| 1 | Convert 5A = 4B to ratio |
$A/B = 4/5$ |
| 2 | Convert 7B = 3C to ratio |
$B/C = 3/7$ |
| 3 | Convert 2C = 7D to ratio |
$C/D = 7/2$ |
| 4 | Chain ratios for A:D | $A/D = (A/B) \times (B/C) \times (C/D)$ |
| 5 | Substitute values | $A/D = (4/5) \times (3/7) \times (7/2)$ |
| 6 | Simplify | $A/D = (4 \times 3 \times 7) / (5 \times 7 \times 2) = 84 / 70 = 6/5$ |
A ratio is a comparison of two quantities. It can be written as A:B or as a fraction A/B. Proportions are equations stating that two ratios are equal.
In this problem, we used the property of chaining ratios. If we have ratios A:B, B:C, C:D, and so on, we can find the ratio A:D by multiplying the corresponding fractions:
$$ \frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D} $$
This works because the intermediate terms (B and C in this case) cancel out:
$$ \frac{A}{\cancel{B}} \times \frac{\cancel{B}}{\cancel{C}} \times \frac{\cancel{C}}{D} = \frac{A}{D} $$
This method is very useful for solving problems involving multiple interconnected ratios.
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