PE 2 - SICI | Arithmetic - Simple & Compound Interest

Answer: 3125

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Explanation :

Let the sum invested be A and rate of interest r% p.a.

At the end of  two years, the total interest was equal to the sum

∴ $\frac{\mathrm{A}\times \mathrm{r}\times 2}{100}$ = A

⇒ r = 50% p.a.

Also, at the end of four years the total interest was Rs. 6,250

⇒ $\frac{\mathrm{A}\times \mathrm{r}\times 4}{100}$ = 6250

⇒ A = $\frac{6250\times 100}{50\times 4}$ = 3125.

Alternately,
Interest for 2 years = Amount invested

Given, Interest for 4 years = Rs. 6250

⇒ 2 × (Amount invested) = 6250

⇒ Amount invested = 3125

Hence, 3125.

Answer: 1820

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Explanation :

Let the total sum be 2P.

Interest accumulated when P was lent at 20% SI for 2 years = (P × 20 × 2)/100 = 2P/5 = 0.4P

Interest accumulated when P was lent at 20% CI for 2 years = 1.22P – P = 1.44P – P = 0.44P

∴ 0.44P – 0.4P = 500

⇒ P = 500/0.04 = 12,500

Interest accumulated when P was lent at 20% SI for 4th  year = (P × 20 × 1)/100 = P/5 = 0.2P

Interest accumulated when P was lent at 20% CI for 4th year = P × 0.2 × 1.2 × 1.2 × 1.2 = 0.3456P 

Difference in total interest for 4th year = 0.3456P – 0.2P = 0.1456P = 0.1456 × 12500 = 1820.

Hence, 1820.

Answer: 20

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Explanation :

When invested at a certain rate of compound interest, the amount triples in 5 years.

For  accumulated interest to become 80 times the invested amount, total amount due must become 81 times the investment.

81 = 3 × 3 × 3 × 3

To triple the amount once, it takes 5 years.
∴ To triple  the amount 4 times it  will take 5 + 5 + 5 + 5 = 20 years.

Hence, 20.

Answer: Option A

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Explanation :

The amount payable after the first year is 1,00,000 × 1.1 = Rs 1,10,000

As he repays Rs 40,000 after first year, the amount left is 1,10,000 – 40,000 = 70,000

So, the second year’s interest will only be on the remaining Rs. 70,000.

Amount payable at the end of the second year = 70,000 × 1.1 = Rs. 77,000.

Hence, option (a).

Answer: Option C

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Explanation :

Let the amount Achal invested be Rs. P.

Amount due at the end of 3 years = $\mathrm{P}\times \left(1+\frac{5}{100}\right)\times \left(1+\frac{6}{100}\right)\times \left(1+\frac{7}{100}\right)$

Amount due at the end of 3 years had he invested at 6% = $\mathrm{P}\times {\left(1+\frac{6}{100}\right)}^3$

The difference = $\mathrm{P}\times {\left(1+\frac{6}{100}\right)}^3$ - $\mathrm{P}\times \left(1+\frac{5}{100}\right)\times \left(1+\frac{6}{100}\right)\times \left(1+\frac{7}{100}\right)$

= $\mathrm{P}\times \left(1+\frac{6}{100}\right)\left[{\left(1+\frac{6}{100}\right)}^2-\left(1+\frac{5}{100}\right)\left(1+\frac{7}{100}\right)\right]$

= $\mathrm{P}\times \left(\frac{106}{100}\right)\left[{\left(\frac{106}{100}\right)}^2-\left(\frac{105}{100}\right)\left(\frac{107}{100}\right)\right]$

= $\mathrm{P}\times \left(\frac{106}{100}\right)\left[\frac{{106}^2-(106-1)(106+1)}{{100}^2}\right]$

= $\mathrm{P}\times \left(\frac{106}{100}\right)\left[\frac{{106}^2-(106-1)(106+1)}{{100}^2}\right]$

= $\mathrm{P}\times \left(\frac{106}{100}\right)\left[\frac{{106}^2-{106}^2-1}{{100}^2}\right]$

= $\mathrm{P}\times \frac{106}{{100}^2}$

Given, $\mathrm{P}\times \frac{106}{{100}^2}$ = 3286

⇒ P = 3,10,000

Hence, option (c).

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