** Simple Interest Shortcut Tricks:**

- Simple Interest = PTR/100, (P=Principle Amount, T=Time, R=Rate).

- The annual payment that will discharge a debt of A Rs/- in t years at the rate of interest r% per annum.

Then Payment = (100A) / [(100t )+[(rt(t-1)]/2].

- Sum = (100I)/(r1t1+r2t2+….)

If a sum of money become x times in t years at simple interest. Then Rate = [[100(x-1)]/t]%.

- Principle = 100A/100+RT.

- Sum= [Interest difference/(r1t1-r2t2)]*100.

- If a sum amounts to A1 in t1 years and A2 in t2 years.

Then Rate of Interest = 100[A1-A2]/ [A1T2-A2T1].

**Simple Interest problems for Practice:**

**Problem 1:** Simple Interest on an amount after 24 months at the rate of 2% per quarter is 960. What is the sum?

**Sol:** R= 2% per quarter.

=(2% * 4)per year. = 8 % p.a.

Simple Interest (S.I.)=1^{st} year+ 2^{nd} year= 8%+ 8%= 16%

Here understand the logic i.e. 16% amount à960

Then (sum means 100%)so for 100% à?

With cross multiplication, Sum= 100*960/16=**6000**.

**Problem 2: **The simple interest of money is Rs.300 after 5 years. In next 5 years, If the principle is tripled what will be the total interest at the end of the 10^{th} year?

**Sol: **1st 5 years àSI = 300.

Next 5 years à New P^{|} = 3P

New SI^{|} = 3 SI

= 3 *300 =900.

Total Interest = 300+900. = 1200.

**Problem 3: **Varun borrows 1500 Rs. From a lender. He pays interest at the rate of 12% p.a. for one loan and rate of 14 % for another loan. How much does he borrow at 12% p.a? If the interest paid at the end of the year is 186 Rs.

**Sol:** Given, SI 1 + SI 2=186. P1 is 12% p.a. P2 is at 14% p.a.

And P1 +P2= 1500 àequation 1

From above information 12%of P1 +14% of P2 =186 àequation 2.

Solve these 2 equations 1&2 we get P2=300.

And P1 = 1200.