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Formulas,
shortcuts, tricks and solved examples:

If P =
Principal, A = Amount, R = Rate percent per year, T = T years, S.I = Simple
interest, C.I = Compound interest, Then,

**1. For Simple interest:**

(A). S.I =
(P × R × T)/100

(B) A = P +
S.I

**2. For Compound interest:**
(A) When
interest is compounded yearly,

(B) When
interest is compounded half-yearly,

(C) When
interest is compounded quarterly

(D) When
time is in fraction of a year, say 2⅕

(E) If rate
of interest in 1st year, 2nd year …………..… nth year are R_{1}%, R_{2}%
…………. R_{n}% respectively, then,

**3. Equivalent or Successive interest:**

Single
equivalent interest rate or Successive interest rate of 20% and 10% is

Single
equivalent interest rate or Successive interest rate of 10%, 20% and 30% would
be

Since,
equivalent interest 10% and 20% is 28%

So,
equivalent interest of 28% and 30% will be,

__Example: 01__
**A sum of Rs. 15,500 is lent out into two
parts, one at 8% and another one at 6%. If the total annual income is Rs. 1060,
the money lent at 8% is:**

__Solution:__

Let the
money lent at 8% be Rs. x, then

[(x × 8 ×
1)/100] + [15500 - x) × 6 × 1/100] = 1060

or, 2x +
93000 = 106000

or, x = 6500

Therefore, the
money lent at 8% is Rs. 6500

__Example: 02__

**A sum of money at compound interest amounts
to Rs. 10,580 in 2 years and to Rs. 12,176 in 3 years. The rate of interest per
annum is:**

__Solution:__

Interest on
Rs. 10580 for 1 year = Rs. (12176 - 10580) = Rs. 1587

∴ Rate = [(100 × 1587)/10580]% = 15%

Hence, the
rate of interest per annum is 15%.

__Example: 03__

**A sum of money becomes Rs. 13,380 after
3 years and Rs. 20,070 after 6 years on compound interest. The sum is:**

__Solution:__

Let, the sum
be x, then

x[1 + (R/100)]^{3} =
13380 and, x[1 + (R/100)]^{6} = 20070

On dividing,
we get, [1 + (R/100)]^{3} = (20070/13380) = 3/2

∴ x (3/2) = 13380

or, x = 13380 × (3/2)
= 8920

Hence, the
sum is Rs. 8920.
**SOLVE EXAMPLE **
__Question No. 01__

**A sum
fetched a total simple interest of Rs. 4016.25 at the rate of 9% p.a. in 5
years. What is the sum?**

(A) Rs.
4462.50

(B) Rs.
8032.50

(C) Rs. 8900

(D) Rs. 8925

Answer:
Option D

__Explanation:__

Principal =
Rs. (100 × 4016.25)/(9 × 5)

= Rs. (401625/45) = Rs. 8925.

__Question No. 02__

**Calculate
the amount on Rs. 4480 at 8% per annum for 3 years.**

(A) Rs.
5555.20

(B) Rs. 5545.20

(C) Rs. 5000

(D) Rs. 6555

Answer:
Option A

__Explanation____:__

S.I. = (P × N
× R)/100

= Rs. (4480 × 3 × 8)/100

= Rs. 1075.20

∴ Amount
= Rs. (4480 + 1075.20)

= Rs. 5555.20

__Question No. 03__

**A certain
sum of money at simple interest amounts to Rs. 1260 in 2 years and to Rs. 1350
in 5 years. The rate percent per annum is?**

(A) 35 %

(B) 25 %

(C) 50 %

(D) 45 %

Answer:
Option B

__Explanation____:__

S.I. for 3
years = Rs. (1350 - 1260) = Rs. 90

∴ S.I.
for 2 years = Rs. (90/3) × 2 = Rs. 60

Principal =
Rs. (1260 - 60) = Rs. 1200

Rate, R = (100
× 60)/(1200 × 2) % = 25 %

__Question No. 04__

**Mr. Roy
invested an amount of Rs. 13,900 divided in two different schemes A and B at
the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total
amount of simple interest earned in 2 years be Rs. 3508, what was the amount
invested in Scheme B?**

(A) Rs. 6400

(B) Rs. 6500

(C) Rs. 7200

(D) Rs. 7500

Answer:
Option A

__Explanation:__

Let the sum
invested in Scheme A be Rs. *x* and that in Scheme B be Rs.
(13900 - *x*).

Then, [*x* ×
14 × 2)/100] + [{(13900 - *x*) × 11 × 2}/100] = 3508

=> 28*x* -
22*x* = 350800 - (13900 × 22)

=> 6*x* =
45000

=> x =
7500.

So, sum
invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

__Question No. 05__

**How much
time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5%
per annum of simple interest?**

(A) 3.5
years

(B) 4 years

(C) 4.5
years

(D) 5 years

Answer:
Option B

__Explanation:__

Time = [(100
× 81)/( 450 × 4.5)] years = 4 years.

__Question No. 06__

**A sum of Rs.
725 is lent in the beginning of a year at a certain rate of interest. After 8
months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At
the end of the year, Rs. 33.50 is earned as interest from both the loans. What
was the original rate of interest?**

(A) 3.6%

(B) 4.5%

(C) 5%

(D) None of
these

Answer:
Option D

__Explanation:__

Let the
original rate be R%. Then, new rate = (2R)%.

Here,
original rate is for 1 year(s); the new rate is for only 4 months i.e. for ^{1}/_{3}
year.

∴
[(725 × R × 1)/100] + [362.50 × 2R × 1)/(100 × 3)] = 33.50

=> (2175
+ 725) R = 33.50 × 100 × 3

=> (2175
+ 725) R = 10050

=> (2900)
R = 10050

=> R = (10050/2900)
= 3.46

∴ Original
rate = 3.46%

__Question No. 07__

**A certain
amount earns simple interest of Rs. 1750 after 7 years. Had the interest been
2% more, how much more interest would it have earned?**

(A) Rs. 35

(B) Rs. 245

(C) Rs. 350

(D) Cannot be
determined

Answer:
Option D

__Explanation:__

We need to
know the S.I., principal and time to find the rate.

Since the
principal is not given, so data is inadequate.

__Question No. 08__

**Ravi took
a loan of Rs. 1200 with simple interest for as many years as the rate of
interest. If she paid Rs. 432 as interest at the end of the loan period, what
was the rate of interest?**

(A) 3.6

(B) 6

(C) 18

(D) None of
these

Answer:
Option B

__Explanation:__

Let rate =
R% and time = R years.

Then, (1200 ×
R × R)/100 = 432

=> 12R^{2} =
432

=> R^{2} =
36

=> R = 6.

__Question No. 09__

**S.I. on Rs.
1500 at 7% per annum for a certain time is Rs. 210. Find the time;**

(A) 3 years

(B) 5 years

(C) 2 years

(D) 1½ years

Answer:
Option C

__Explanation____:__

Time, N = (210
× 100)/(1500 × 7) = 2 years

__Question No. 10__

**A sum of
money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4
years. The sum is:**

(A) Rs. 650

(B) Rs. 690

(C) Rs. 698

(D) Rs. 700

Answer:
Option C

__Explanation:__

S.I. for 1
year = Rs. (854 - 815) = Rs. 39.

S.I. for 3
years = Rs. (39 × 3) = Rs. 117.

∴ Principal
= Rs. (815 - 117) = Rs. 698