In this post we will discuss problems of compound interest. There are mainly two forms of money compounding mechanism in quantitative aptitude syllabus of major competition exas:
1. Simple Interest
2. Compound Interest
Compound interest is some what more technical than simple interest but if you understand the basics, none of their problems is difficult to solve.
In this post, we have compiled important problems of compound interest and all of them are solved step by step for your understanding. Our aim is to make you comfortable solving math related questions without any external help.
Compound Interest Questions
(01) In what time will Rs.1000 becomes Rs.1331 at 10% per annum compounded annually
Given:
Principal (P) = Rs 1000
Amount (A) = Rs. 1331
Rate (r) = 10% per annum
we have to find time (n)
According to question:
\ \quad A\quad =\quad P(1+\frac { r }{ 100 } )^{ n\quad }\\\ \\ =>\quad 1331=1000(1+\frac { 10 }{ 100 } )^{ n }\\\ \\ =>\quad \frac { 1331 }{ 1000 } =(\frac { 100+10 }{ 100 } )^{ n }\\\ \\ =>\quad \frac { 1331 }{ 1000 } ={ (\frac { 110 }{ 100 } ) }^{ n }\\\ \\ =>\quad ({ \frac { 11 }{ 10 } })^{ 3 }=(\frac { 11 }{ 10 } )^{ n }\\\ \\ L.H.S\quad =\quad R.H.S\\\ \\ So\quad time(n)=\quad 3\Hence at 3 years time, Rs. 1000 becomes Rs. 1331
(02) The difference between compound interest and simple interest on a sum of Rs. 1000 at a certain rate of interest for 2 years is Rs. 10. The rate of interest per annum is?
In this type of question we will use following formula for fast calculation
Sol:\quad Principal=\frac { Difference\quad \times \quad { 100 }^{ 2 } }{ { R }^{ 2 } } \\\ \\ Where;\ R=Rate\\\ \\ Principal\quad =\quad ₹1000\\ \\ Time\quad =\quad 2\quad Years\\ \\ Difference=\quad ₹10\\\ \\ Now,put\quad the\quad value\quad in\quad formula,\\\ \\ 1000=\frac { 10{ \quad \times \quad 100 }^{ 2 } }{ { R }^{ 2 } }\\\ \\ \ 1000=\frac { 100000 }{ { R }^{ 2 } }\\\ \\ \ { R }^{ 2 }=\frac { 100000 }{ 1000 } \\\ \\\ { R }^{ 2 }=100\\\ \\ R=\sqrt { 100 } \\\\ \\ R=\quad 10\\\ \\ rate\quad of\quad interest\quad is\quad 10\quad percent.\(03) The difference between simple interest and compound interest on a sum of money at 4% per annum for 2 years is Rs. 8. Find the initial principal
Principal\quad =\quad \frac { Difference\quad \times \quad { 100 }^{ 2 } }{ { R }^{ 2 } } \\\ \\ Where,\\\ \\ Rate(R)=\quad 4\quad percent\\ \\ Principal(P)=?\\ \\ Time=\quad 2\quad years\\ \\ Difference=₹8\\\ \\ Put\quad the\quad value\quad in\quad formula\\\ \\ P=\frac { 8\quad \times \quad { 100 }^{ 2 } }{ { 4 }^{ 2 } } \\\ \\ P=\frac { 8\quad \times \quad 10000 }{ 16 } \\\ \\ P=\frac { 10000 }{ 2 } \\\ \\ P=5000\\\ \\ \ So\quad the\quad Principal\quad is\quad 5000\(04) A sum of money becomes 8 times of itself in 3 years at compound interest. Find the rate of interest per annum
Let\quad P\quad be\quad the\quad sum.\\\ \\ Amount(A)=8P\\ \\ Time(t)=3\quad years\\ \\ Rate\quad Of\quad Interest(R)=?\\\ \\ As\quad we\quad know\quad that,\\ \\ A=P(1+\frac { R }{ 100 } )^{ t }\\\ \\ 8P=P(1+\frac { R }{ 100 } )^{ 3 }\\\ \\ (8)^{ \frac { 1 }{ 3 } }=\frac { P(1+\frac { R }{ 100 } ) }{ P } \\\ \\ (8)^{ \frac { 1 }{ 3 } }=1+\frac { R }{ 100 } \\\ \\ \frac { R }{ 100 } =\sqrt [ 3 ]{ 8 } -1\\\ \\ \frac { R }{ 100 } =2-1\\\ \\ R=1\quad \times \quad 100\\\ \\ R=100\quad percent\Hence at 100% rate of interest, the money becomes 8 times itself in 3 years
(05) The difference between simple and compound interest on a certain sum of money for 2 years at 4% per annum is Rs. 10. Find the sum of money
The\quad difference\quad between\\ \\ the\quad simple\quad interest\\ \\and\quad compound\quad interest\\ \\on\quad a\quad certain\quad sum\quad of\quad money'P'\\ \\ for\quad 2\quad years\quad at\quad r\quad percent\quad per\quad annum\\\ \\is\quad given\quad by:\\ \\ =>(CI-SI)_{ 2\quad years }=\frac { P{ r }^{ 2 } }{ { 100 }^{ 2 } } \\\ \\ Using\quad the\quad above\quad formula\\ \\for\quad a\quad sum\quad of\quad money\quad 'P'\quad invested\\ \\for\quad 2\quad years\quad at\quad 4\quad percent \quad p.a,\\\ \\ given\quad that\quad the\quad difference\quad between\\ \\ CI\& SI\quad is\quad ₹10\\\ \\ \therefore 10=\frac { P{ (4) }^{ 2 } }{ { 100 }^{ 2 } } \\\ \\ =>\quad 10\quad =\quad \frac { P(4\quad \times \quad 4) }{ 100\quad \times \quad 100 } \\\ \\ =>10=\frac { 4P }{ 100\quad \times \quad 25 } \\\ \\ =>\frac { 10 }{ 4 } =\frac { P }{ 100\quad \times \quad 25 } \\\ \\ =>2.5=\frac { P }{ 100\quad \times \quad 25 } \\\ \\ =>2.5\quad \times \quad 25\quad \times \quad 100=P\\\ \\ =>P=6250\\\ \\ Hence\quad the\quad sum\quad is\quad ₹6250\(06) A sum of money placed at a compound interest doubles itself in a 15 years. In how much years, it would amount to eight times of itself at the same rate of interest
we have to find number of years in which compound interest amount to eight times
we know that A=P(1+\frac { R }{ 100 } )^{ n }
where
A = Amount
P = Principal
R= rate of interest
n = time in years
Let
A = 2x
P = x
n =15
Now putting the values in above formula
2x=x(1+\frac { R }{ 100 } )^{ 15 }
Cancelling x on both sides
2=(1\frac { R }{ 100 } )^{ 15 }\\\ \\ { 2 }^{ \frac { 1 }{ 15 } }=\quad 1+\frac { R }{ 100 } \quad …………..(a)
Now we have to find number of years in which the sum will become 8 times
Let A = 8x and P=x
8x=x(1+\frac { R }{ 100 } )^{ n }\\\ \\ 8=(1+\frac { R }{ 100 } )^{ n }……………..(b)\
Put the value of equation (a) in equation (b)
8=({ 2 }^{ \frac { 1 }{ 15 } })^{ n }\\\ \\ { 2 }^{ 3 }={ 2 }^{ \frac { n }{ 15 } }\Now comparing both the sides we get
n/15 = 3
n = 45
In 45 years the sum will become 8 times at the same rate of interest
(07) If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of Rs. 10,000 for 3 years will be?
Given that
Rate for first year (R1) = 4%
Rate for second year (R2)= 5%
Rate for third year (R3) = 6%
Principal = Rs. 10,000
A= 10,000 * 1.04 * 1.05 * 1.06
A= 11575.20
We know that
Compound Interest = Amount – Principal
C.I = 11575.20 – 10,000 => 1575.2 Rs
Hence the compound interest is 1575.2 Rs
(08) A person deposited a sum of Rs. 6000 in a bank at 5% per annum simple interest. Another person deposited Rs. 5000 at 8 percent per annum compound interest. After 2 years the difference of their interest will be?
First Person
Principal = Rs. 6000
Rate (R1) = 5% per annum
Time (T) = 2 years
Simple Interest Formula = (P * R * T)/100
S.I = (6000 * 5 * 2)/100
S.I = Rs .600
Second Person
Principal = Rs. 5000
Rate (R2) = 8 percent
Time (n) = 2 years
C.I = Rs 832
Therefore, difference in interest==> 832 – 600 = 232 rs
Hence rs 232 is the right answer
(09) The compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually is?
Principal (P) = Rs. 8000
Interest for first year = 15% of 8000
==> 8000 * 15/100 ==> 1200 Rs.
So, new principal after 1st year = 8000 + 1200 => 9200 Rs
Interest for second year = 15% of 9200
==> 9200 * 15/100 ==> 1380 Rs
So new principal after second year = 9200 + 1380 = 10580 Rs.
Interest for 4 months (1/3 years) ==> 10580 * 1/3 * 15/100 => Rs. 529
So, interest after 2 years 4 months => 1200+1380+529=> 3109 rs.
(10) At what rate per annum will Rs 32000 yield a compound interest of rs. 5044 in 9 months interest being compounded quarterly?
Here
Principal (P) = Rs. 32,000
Compound Interest (C.I) = 5044 Rs
Time (n)= 9 months ==> 9/12 year ==>4/3 year
Let Rate of interest be R% per annum
If the interest is compounded quarterly then Rate of interest ==> R/4 % compounded quarterly
Hence the required rate of interest is 20% per annum