A is 20% more than B, B is 10% less than C and C is 15% more than D. Which of the following is true ?

Options

A. A. A is 8% more than C
B. B. D is 24.2% less than A
C. C. D is 3.5 less than B
D. D. C is 11.5% more than B

Answer

To solve the problem step by step, we will define the variables based on the relationships given in the question.

### Step 1: Define Variables
Let:
- \( B = x \) (Let B be some value \( x \))

### Step 2: Find A in terms of B
According to the question, A is 20% more than B.
- \( A = B + 0.2B = 1.2B \)
- Substituting \( B = x \):
\[ A = 1.2x \]

### Step 3: Find C in terms of B
B is 10% less than C.
- This means \( B = C - 0.1C = 0.9C \)
- Rearranging gives:
\[ C = \frac{B}{0.9} = \frac{x}{0.9} = \frac{10x}{9} \]

### Step 4: Find D in terms of C
C is 15% more than D.
- This means \( C = D + 0.15D = 1.15D \)
- Rearranging gives:
\[ D = \frac{C}{1.15} = \frac{\frac{10x}{9}}{1.15} = \frac{10x}{9 \times 1.15} = \frac{10x}{10.35} = \frac{1000x}{1035} = \frac{200x}{207} \]

### Step 5: Summary of Relationships
Now we have:
- \( A = 1.2x \)
- \( B = x \)
- \( C = \frac{10x}{9} \)
- \( D = \frac{200x}{207} \)

### Step 6: Compare A and C
To check if A is 8% more than C:
- Calculate 8% of C:
\[ 8\% \text{ of } C = 0.08 \times C = 0.08 \times \frac{10x}{9} = \frac{0.8x}{9} \]
- Now calculate \( C + 8\% \text{ of } C \):
\[ C + 8\% \text{ of } C = \frac{10x}{9} + \frac{0.8x}{9} = \frac{10x + 0.8x}{9} = \frac{10.8x}{9} \]

### Step 7: Check if A equals C + 8% of C
Now, we need to check if:
\[ A = C + 8\% \text{ of } C \]
Substituting the values:
\[ 1.2x = \frac{10.8x}{9} \]
Cross-multiplying gives:
\[ 1.2x \times 9 = 10.8x \]
\[ 10.8x = 10.8x \]
This is true, confirming that A is indeed 8% more than C.

### Conclusion
Thus, the statement that A is 8% more than C is true.