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Simple interest on a certain sum at a ce...

Simple interest on a certain sum at a certain rate is `9/16` of the sum. If the number representing rate percent and time in years be equal, then the time is-

A

`5 1/2`years

B

`6 1/2`years

C

`6 1/4`years

D

`7 1/2`years

Text Solution

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The correct Answer is:
To solve the problem step by step, we need to use the formula for Simple Interest (SI), which is given by: \[ SI = \frac{P \times R \times T}{100} \] Where: - \( SI \) = Simple Interest - \( P \) = Principal amount (the sum) - \( R \) = Rate of interest per annum - \( T \) = Time in years ### Step 1: Understand the given information We know that the Simple Interest is \( \frac{9}{16} \) of the sum (Principal amount, \( P \)). Therefore, we can write: \[ SI = \frac{9}{16} P \] ### Step 2: Set up the equation using the SI formula From the formula of Simple Interest, we can substitute \( SI \): \[ \frac{9}{16} P = \frac{P \times R \times T}{100} \] ### Step 3: Cancel \( P \) from both sides Assuming \( P \neq 0 \), we can divide both sides by \( P \): \[ \frac{9}{16} = \frac{R \times T}{100} \] ### Step 4: Use the condition that rate and time are equal According to the problem, the rate \( R \) and time \( T \) are equal. Let’s denote them both as \( T \): \[ R = T \] Substituting \( R \) with \( T \) in the equation gives: \[ \frac{9}{16} = \frac{T \times T}{100} \] ### Step 5: Simplify the equation This simplifies to: \[ \frac{9}{16} = \frac{T^2}{100} \] ### Step 6: Cross-multiply to solve for \( T^2 \) Cross-multiplying gives: \[ 9 \times 100 = 16 \times T^2 \] \[ 900 = 16 T^2 \] ### Step 7: Solve for \( T^2 \) Now, divide both sides by 16: \[ T^2 = \frac{900}{16} \] ### Step 8: Simplify the fraction Simplifying \( \frac{900}{16} \): \[ T^2 = 56.25 \] ### Step 9: Take the square root to find \( T \) Taking the square root of both sides: \[ T = \sqrt{56.25} \] Calculating the square root: \[ T = 7.5 \] ### Conclusion Thus, the time \( T \) is \( 7.5 \) years, which can also be expressed as \( 7 \frac{1}{2} \) years.
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