# PE 2 - SICI | Arithmetic - Simple & Compound Interest

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**PE 2 - SICI | Arithmetic - Simple & Compound Interest**

A sum was invested at simple interest. At the end of two years, the total interest was equal to the sum. At the end of four years the total interest was Rs. 6,250. Find the interest on the sum at the end of three years (in Rs.)

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.

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**PE 2 - SICI | Arithmetic - Simple & Compound Interest**

A sum was divided into two equal parts. One part was lent at 20% p.a. simple interest. The other part was lent at 20% p.a. compound interest, interest being compounded annually. The difference in the interests fetched by the parts in the second year is Rs. 500. Find the difference in the interests fetched by the parts in the fourth year (in Rs.).

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.

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**PE 2 - SICI | Arithmetic - Simple & Compound Interest**

In an investment scheme the investment triples in 5 years when invested at a certain rate of compound interest. How long will it take for accumulated interest to become 80 times the original investment.

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.

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**PE 2 - SICI | Arithmetic - Simple & Compound Interest**

Sonam took a loan of Rs 1,00,000 at 10% rate for 2 years compounded annually. He repays a sum of Rs 40,000 at the end of first year. What is the amount that he must pay at the end of the second year?

- (a)
Rs. 77,000

- (b)
Rs. 82,000

- (c)
Rs. 61,000

- (d)
Rs. 73,000

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).

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**PE 2 - SICI | Arithmetic - Simple & Compound Interest**

Achal invested a certain amount for three years compounded annually, where he received 5% interest for the first year, 6% for the second year and 7% for the third year. Had he invested the same amount at 6% compounded annually for all the three years he would have earned Rs. 3286 more. Find the amount he invested.

- (a)
Rs. 2,50,000

- (b)
Rs. 3,00,000

- (c)
Rs. 3,10,000

- (d)
None of these

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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