### Type 2: How to Solve Problems On Simple Interest (SI) When Rates are different for different years

**Question 1. Nisha borrowed some money at the rate of 5% p.a. for the first two years. She again borrowed at the rate of 10% p.a. for the next three years. Later at the rate of 15% p.a. for the rest of the years. Total interest paid by her was Rs. 15000 at the end of 10 years. Calculate the amount of money she borrowed?**

**Options:**

**A. Rs. 15005**

**B. Rs. 20000**

**C. Rs. 15000**

**D. Rs. 10000**

**Solution: **According to the question,

r_{1} = 5% , T_{1} = 2 years

r_{2} = 10% , T_{2} = 3 years

r_{3} = 15% , T_{3} = 5 years

(since, beyond 5 years rate is 14%)

Simple interest = 15000

Therefore, P = \frac{1500 * 100}{5*2 +10*3 +15*4 }

=\frac{1500000}{10 + 30 + 60 }

= \frac{1500000}{100}

= Rs. 15000

**Correct option: C**

**Question 2. Rahul invests some amount of money in three different schemes for 5 years, 10 years and 15 years at 10%, 12% and 15% Simple Interest respectively. At the completion of each scheme, he gets the same interest. Find out the ratio of his investment?**

**Options:**

**A. 3: 9: 15**

**B. 10: 24: 45**

**C. 10: 24: 40**

**D. 9: 24: 45**

**Solution: ** If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are 10 %, 12 %,15% respectively and time periods are 5 years , 10 years , 15 years respectively.

Let the three amounts be Rs. x, Rs. y and Rs. z,

Then , According to question

\frac{x × 10 × 5}{100} = \frac{y× 12 × 10}{100} = \frac{z × 15 × 15}{100} [/latex</span>.</p><div>50 x = 120 y = 225 z = k(say)</div><div>10 x = 24 y = 45 z = k</div><div>x= <span style="font-size: 14.157px">\frac{k}{10}

Amount = Principal + Rate of Interest.

Principal is 100% of the amount.

Therefore, the amount that Ram has to pay back to the bank after three years Rs 12100.

IN Type 2 Question 2 Answer should be 20:15:8

please check