**Compound Interest Shortcuts and Tricks**

### CI Formulae:

- Amount = P[1+(r/100)]

- Compound Interest (CI) = P [ (1+r/100)
^{t}-1 ].

- Simple Interest (SI) = (CI * rt) / 100 [(1+r/100)
^{t}-1].

- When the difference between SI and CI on the certain sum for 2 years at r% is d Rs. Sum = (d/r
^{2})*100^{2.}

^{ }When the difference between SI and CI on a certain sum for 3 years at r% is d Rs.

Sum = [d(100)^{3 }/ r^{2}(300+r)].

- If a sum A becomes B in t1 years at CI. Then After t2 years,

Sum = (B)^{t2 / t1 }/(A)(^{t2 / t1)-1}.

- The rate of Interest = [ (Difference of amount after n years and (n+1) years )/ Amount after n years ]*100.

## Shortcut Tricks to Solve Compound Interest Problems

**Solving CI Problems using smart tricks:**

^{ }**Problem 1:** What will be the compound interest on 5000 Rs. For 2 years at the rate of 12% p.a.?** **

**Sol:** From percentage formula Net percentage = (a+b+ab/100)%

= [12+12+(12*12)/100]%.

= [24 +1.44]%

= 25.44%.

We know, 100%(means principle) à 5000.

25.44% à ?

è(25.44 *5000)/100 = 1272.

**Problem 2:** A sum of money invested at compound interest amounts to 800Rs. in 3 years and Rs. 882 in 5 years. What is the rate of interest?**Sol:** [ In the case of Simple interest

[A_{m} –A_{n}] years. ]** **But Here, In the case of Compound Interest[ A_{m} / A_{n }] years = A^{m-n }years.

- P(1+R/100)
^{5}/ P(1+R/100)^{3}= 882/800. - (1+R/100)
^{2 }years =441/400. - 1+R/100 =21/20.
- 100+ R= (21/20)*100
- R=5%.

**Problem 3:** On a particular amount the CI at the end of the year is 40Rs/- and in the 2^{nd} year is 42 Rs/-.How much money was deposited?**Sol: **Rate of interest = [(Final- Initial)/Initial] *100.**= [(**42 **– **40)/40] *100 =(2/40)*100 = 5%.

We know, 5% à 40,

100% à?

P = (100 *40)/5 = 800.

**Problem 4:** When time period T is given infraction at CI i.e. T= 3 ¾ years.

Amount = P[(1+R/100)^{3 }(1+(3/4 R)/100 )].**Problem 5****: **On What is the difference between CI and SI for the sum of 20000 over 2 years period. If compound interest is calculated at 20% p. a. and SI at 23% p.a.**Sol:**_{ }R_{SI} = 23% , Net % for 2 years = 23+23 =46%.** ** R_{CI} = 20% , Net % for 2 years = 20+20+(20*20)/100 = 44%

Difference of SI –CI = 46-44 = 2%.

If 100% à 20000 ( principle amount is always taken as 100%)

Then 2% à ?

i.e. (2 * 20000)/100 = 400.

**Problem 6: **A bank offers 5% CI on half year basis customer deposit 1600 Rs each on 1^{st} July of a year at the end of the year What is the interest?**Sol:** CI = [ [2.5 + 2.5 +(2.5* 2.5)/100]+[2.5] ]*1600