There are three quantities A, B and C. B is `16(2)/(3)%` less than A and C is `14(2)/(7)%` more than B. By wht per cent is A more than C?

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem step by step, we need to find out by what percentage A is more than C, given the relationships between A, B, and C.

### Step 1: Define the relationships
We know:
- B is \(16 \frac{2}{3}\%\) less than A.
- C is \(14 \frac{2}{7}\%\) more than B.

### Step 2: Convert percentages to fractions
First, we convert the mixed percentages to improper fractions:
- \(16 \frac{2}{3}\% = \frac{50}{3}\%\)
- \(14 \frac{2}{7}\% = \frac{104}{7}\%\)

### Step 3: Express B in terms of A
Since B is \(16 \frac{2}{3}\%\) less than A, we can express B as:
\[ B = A - \left(\frac{50}{3}\% \text{ of } A\right) \]
This can be simplified as:
\[ B = A - \frac{50}{300}A = A - \frac{1}{6}A = \frac{5}{6}A \]

### Step 4: Express C in terms of B
Next, since C is \(14 \frac{2}{7}\%\) more than B, we express C as:
\[ C = B + \left(\frac{104}{7}\% \text{ of } B\right) \]
This can be simplified as:
\[ C = B + \frac{104}{700}B = B + \frac{52}{350}B = \frac{8}{7}B \]

### Step 5: Substitute B in terms of A into C
Now, substitute \(B = \frac{5}{6}A\) into the equation for C:
\[ C = \frac{8}{7} \left(\frac{5}{6}A\right) = \frac{40}{42}A = \frac{20}{21}A \]

### Step 6: Find the percentage by which A is more than C
Now we need to find out by what percentage A is more than C:
\[ \text{Percentage} = \left(\frac{A - C}{C}\right) \times 100 \]
Substituting \(C = \frac{20}{21}A\):
\[ \text{Percentage} = \left(\frac{A - \frac{20}{21}A}{\frac{20}{21}A}\right) \times 100 \]
This simplifies to:
\[ \text{Percentage} = \left(\frac{\frac{1}{21}A}{\frac{20}{21}A}\right) \times 100 = \left(\frac{1}{20}\right) \times 100 = 5\% \]

### Final Answer
A is 5% more than C.

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