A sum of money becomes 79/16 times of itself in 35 years at r% of SI. What is the value of r?

Options

A. `10.5%`
B. `11.25%`
C. `10.8%`
D. `12.5%`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
We need to find the rate \( r \) at which a sum of money becomes \( \frac{79}{16} \) times itself in 35 years using simple interest (SI).

### Step 2: Define the Principal
Let the principal amount be \( P = 16x \). This is a convenient choice since the amount becomes \( \frac{79}{16} \) times the principal.

### Step 3: Calculate the Amount
The amount \( A \) after 35 years is given by:
\[ A = \frac{79}{16} \times P = \frac{79}{16} \times 16x = 79x \]

### Step 4: Calculate Simple Interest (SI)
The simple interest (SI) can be calculated using the formula:
\[ SI = A - P = 79x - 16x = 63x \]

### Step 5: Use the SI Formula
The formula for simple interest is:
\[ SI = \frac{P \times r \times t}{100} \]
Substituting the values we have:
\[ 63x = \frac{16x \times r \times 35}{100} \]

### Step 6: Simplify the Equation
To eliminate \( x \) from both sides (assuming \( x \neq 0 \)):
\[ 63 = \frac{16r \times 35}{100} \]

### Step 7: Cross Multiply to Solve for \( r \)
Cross multiplying gives:
\[ 63 \times 100 = 16r \times 35 \]
\[ 6300 = 560r \]

### Step 8: Solve for \( r \)
Now, divide both sides by 560:
\[ r = \frac{6300}{560} \]
\[ r = 11.25 \]

### Final Answer
The value of \( r \) is \( 11.25\% \).

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