simple interest
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
Simple Interest
Interest: (I) Interest is money paid to the lender by the borrower for using his money for
a specified period of time. Various terms and their general representation are as follows:
Principal(P): The original sum borrowed
Time (t): Time for which money is borrowed.
Rate of Interest(r): Rate at which interest is calculated on the original sum.
Amount (A): Sum of Principal and Interest. (P+I).
Simple Interest (SI): When interest is calculated every year (or every time period) on
the original Principal, such interest is called Simple Interest.
SI =P r t
100
A=P+I
A=P+
Type I:
Exp 1: A Sum of Rs. 1600 gives a SI of Rs. 252 in 2 years and 3 months. The rate of interest is:
Solution:
P=1600
SI=252
t= 2 years 3 months=2+3
12 =
9
4
r=?
SI =P r t
100
252 =1600 r 9
100 4
=7%
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
Type II: Times
Part A: Simple Interest becomes x times of Principal
Take
P=1
SI=x
Exp2: In what time will the SI be2
5 of the principal at 8 ppa.
Solution:
SI =P r t
100
2
5=
1 8 t
100
t=5 years
Part B: A sum of money becomes x times of itself.
Take
P=1
SI=x-1
Exp3: A sum of money becomes 7
6 times of itself in 3 years at a certain rate of SI .The rate ppa
(percent per annum) is:
Solution:
P=1
SI=7
6 -1 =
1
6
SI =P r t
100
1
6=
1 r 3
100
r=55
9 %
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
Part C: A sum of money becomes x times of Amount.
A=P+I
1=P+x1
P=1-x
Take
SI=x
P=1-x
Exp 4: At what rate ppa will the SI on a sum of money be 2
5 of the amount in 10 years.
Solution:
SI=2
5
P=1-2
5 =
3
5
SI =P r t
100
2
5=
3
5
r 100
100
r=62
3%
Part D: x times in t1 years
y times in t2 years
1
1=
t1
t2
Exp 5: A certain sum of money becomes 3 times of itself in 20 years at SI. In How many years
does it become double of itself at SI?
Solution:
1
1=
t1
t2
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
3 1
2 1=
20
t2
t2 = 10 years
Type III
Part A: A sum of money P amounts to A1 in t years, it will amount to A2 if it was put at r %
higher.
A2= A1+ Prt
100
Exp 6: A sum of 800 amounts to 920 in 3 years at SI. If the rate is increased by 3ppa .What will
be the sum amount to in the same period.
A2= 920+ 80033
100
=992
Part B: A sum was put at SI at a certain rate for t years .Had it been put at r ppa higher, it would
have fetched D Rs. more find the sum:
P=D100
Exp 7: A sum of money was lent at SI at certain rate for 3 years .Had it been lent at 2.5 % higher
rate, it would have fetched 540 more. The money lent was:
P=D100
P=540100
32.5
=7200
Type IV: The SI on certain sum of money at r1 ppa for t1 years is Rs. D more than the interest
on the same sum for t2 years at r2 ppa.
P =D100
(r1t1r2t2)
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
Type V: Annual Payment
Annual Payment= P100
100+ (1)
2
Exp 8: What annual payment will discharge a debt of Rs. 848 in 4 years at 4ppa.
Annual Payment= P100
100+ (1)
2
Annual Payment= 848100
4100+4(41)4
2
=200
Type VI: If certain money amounts to A1 in t1 years, A2 in t2 years. Find Principal and rate of
interest:
Solution:
A1= P+ t1 x(i)
A2= P+ t2 x..(ii)
(ii)-(i)
X=A2A1
t2t1 (iii)
( iii) in (i)
P =A1t2A2t1
t2t1
From
SI =P r t
100
R =100(A2A1)
(A1t2A2t1)
Exp 9: A certain sum of money amounts to 756 in 2 years and 873 in 3.5 yeas at a certain rate
find P and SI:
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Black Holes of Aptitude-Simple Interest
Mayank [email protected],9990123477
Solution:
756=P+2x..(i)
873=P+3.5x(ii)
X=78(iii)
(iii)in (i) or (ii)
P=600
from
SI =P r t
100
R=13%
Type VII Partly at x % partly at y %,avg. is r %
Ratio of principal=()
()
Exp10: A sum of Rs. 10000 lent partly at 8 % and remaining at 10ppa.If the yearly interest on
avg. is 9.2%, the two parts are:
Solution: Ratio of principal=()
()
Ratio of principal=(109.2)
(9.28)
=2:3
Parts will be 4000 and 6000