Welcome to Coaching Genie's series-11 of Quantitative Aptitude. The Best free IBPS PO Clerk coaching , study materials and practice tests.
This is a series on SIMPLE INTEREST
When a person(borrower) borrows (lends) money from other person (money lender) or from a bank for a certain time and after that specified time he returns the money back to money lender or bank with extra amount, then that extra amount is called interest.
If a certain amount and at a certain rate interest are same for all the given years then it is called Simple Interest. It is changed on principal only.
Important Formulas for Simple Interest are as follows:
\[\blacksquare S.I = A - P\]
\[\blacksquare S.I=\frac{P\times R\times T}{100}\]
\[\blacksquare P =\frac{ S.I\times 100}{T\times R}\]
\[\blacksquare R =\frac{ S.I\times 100}{P\times T}\]
\[\blacksquare T =\frac{ S.I\times 100}{P\times R}\]
Where A ➙ Amount
P ➙ Principal
T ➙ Time
R ➙ Rate
S.I ➙ Simple Interest
Principal - Initial amount or the amount borrowed is called the principal.
Amount - Sum of principal and extra amount which a borrower has to return is called amount.
◉ Relation of Amount and Principal
Amount ✖ 100 = Principal(100 + RT)
✪ Type-1
101. Find Amount on ₹8500 at 7% p.a for 6 years?
a)₹14120
b)₹13250
c)₹12070
d)₹11160
e)₹10880
102. At what rate % of S.I ₹3600 will become ₹5976 in 6 years?
a)10%
b)11%
c)12%
d)13%
e)14%
✪ Type-2:
\[\blacksquare Difference-of-S.I =\frac{P\times T\times Difference-of-Rate}{100}\]
103. The difference of simple interest on ₹6000 from two banks in 3 years is ₹540. Find difference of their rate?
a)7%
b)6%
c)5%
d)4%
e)3%
✪ Type-3:
Trick : If S.I on a sum of money is x/y of the principal, and the number of years is equal to the rate percent per annum. Then
\[\blacksquare Time=Rate=\sqrt{\frac{x}{y}\times 100}\]
104.The S.I on a sum of money is 16/25 of the principal, and the number of years is equal to the percent per annum.Find rate ?
a)5%
b)6%
c)7%
d)8%
e)9%
✪ Type-4:
Trick : If a certain sum of amount becomes x times in t years then Rate in % is
\[\blacksquare Rate=[\frac{(x-1)\times100}{t}]\%\]
Trick : If a certain sum of money becomes x times at the rate of r% then
\[\blacksquare Time=[\frac{(x-1)\times100}{r}]\space years\]
105.A sum of money becomes 4 times in 12 years at a certain rate of S.I. Find rate percent per annum?
a)15%
b)18%
c)22%
d)25%
e)27%
106.In how many years a certain sum of money becomes 3 times at the rate of 20% per annum at S.I?
a)4
b)6
c)8
d)10
e)12
✪ Type-5
Trick : If a certain sum becomes x times at the rate of r1% and becomes y times at the rate of r2% then,
\[\blacksquare \space r_{2}=[\frac{(y-1)}{(x-1)}]\times r_{1}\space\%\]
Trick : If a sum becomes x times in t1 years and y times in t2 years at a S.I then,
\[\blacksquare \space t_{2}=[\frac{(y-1)}{(x-1)}]\times t_{1}\space years\]
107.A sum becomes 4 times in 10 years. In how many years will it amount to 10 times at the same rate of S.I?
a)18 years
b)22 years
c)26 years
d)28 years
e)30 years
✪ Type-6
If rate of interest are r1%, r2%, r3%, r4% for t1, t2, t3, t4 years respectively then
\[\blacksquare S.I=P[\frac{t_{1}r_{1}+t_{2}r_{2}+t_{3}r_{3}+t_{4}r_{4}+....}{100}]\]
\[\blacksquare Principal=[\frac{S.I\times100}{t_{1}r_{1}+t_{2}r_{2}+t_{3}r_{3}+t_{4}r_{4}+....}]\]
\[\blacksquare Amount=P[1+\frac{t_{1}r_{1}+t_{2}r_{2}+t_{3}r_{3}+t_{4}r_{4}+....}{100}]\]
108.Find the S.I for a principal of ₹5000 at a rate for interest 7% for the first 2 years then 8% for the next 3 years then 9% for the next 4 years and 10% for the next 5 years?
a)5600
b)5800
c)6000
d)6200
e)6400
✪ Type-7
Trick : If at a S.I a sum becomes A1 at the rate of r1% and becomes A2 at the rate of r2% then
\[\blacksquare P=[\frac{A_{2}r_{1}-A_{1}r_{2}}{r_{1}-r_{2}}]\]
\[\blacksquare Time=[\frac{A_{2}-A_{1}}{A_{2}r_{1}-A_{1}r_{2}}]\times100\]
109.If at a S.I a sum becomes ₹6500 at the rate of 8% and becomes ₹5000 at the rate of 5% then what is the principle invested?
a)2500
b)3500
c)4500
d)5500
e)6500
✪ Type-8
Trick : If a sum becomes A1 in t1 years and A2 in t2 years at S.I then
\[\blacksquare P=\{A_{1}-\frac{t_{1}}{(t_{2}-t_{1})}\times(A_{2}-A_{1})\}\]
110.Find the sum which amount to ₹1250 in 4 years and becomes ₹2000 in 8 years at rate of simple interest.
a)250
b)500
c)750
d)1000
e)1250
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