Two equal sums of money were invested, one at 4% and other at 4½ %, at the end of 7 years, the simple interest received from the later exceeded that received from the former by Rs. 31.50. Each sum was

Questions : Two equal sums of money were invested, one at 4% and other at 4½ %, at  the end of 7 years, the simple interest received from the later exceeded that received from the former by Rs. 31.50. Each sum was

•  900
•  1200
•  1500 
•  2000 

Answer:  

 Let each sum invested be Rs. x.

Given conditions:

• First sum is invested at 4% simple interest.
• Second sum is invested at 4.5% (or 9/2%) simple interest.
• The time period for both investments is 7 years.
• The difference in simple interest between the two investments is Rs. 31.50.

Simple Interest Formula:

$$SI = \frac{P \times R \times T}{100}$$

where:
$P$ = Principal amount,
$R$ = Rate of interest per annum,
$T$ = Time in years.

Interest for the first sum:

$$SI_1 = \frac{x \times 4 \times 7}{100} = \frac{28x}{100} = 0.28x$$

Interest for the second sum:

$$SI_2 = \frac{x \times 4.5 \times 7}{100} = \frac{31.5x}{100} = 0.315x$$

Given that the difference in interest is Rs. 31.50:

$$0.315x - 0.28x = 31.50$$ $$0.035x = 31.50$$ $$x = \frac{31.50}{0.035} = 900$$

Answer:

Each sum invested was Rs. 900.

Additional Techniques 

Yes! You can use a shortcut approach to solve this problem faster.

Shortcut Formula for Difference in Simple Interest

$$\text{Difference} = \frac{P \times (\text{Rate Difference}) \times \text{Time}}{100}$$

Given Data:

• Rate difference = $4.5\% - 4\% = 0.5\% = \frac{1}{2}\%$
• Time = 7 years
• Difference in SI = Rs. 31.50

Applying the Shortcut Formula:

$$31.50 = \frac{P \times \frac{1}{2} \times 7}{100}$$ $$31.50 = \frac{7P}{200}$$ $$P = \frac{31.50 \times 200}{7} = \frac{6300}{7} = 900$$

Final Answer:

Each sum invested was Rs. 900.

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