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CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 7 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Basics Of Financial Mathematics
Quiz 7
Naman purchased an old bike for Rs. $$20000$$. If the cost of his bike is depreciated at a rate of $$5\%$$ per annum, then find the cost of the bike after $$2$$ years?
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0%
Rs. $$18050$$
0%
Rs. $$15000$$
0%
Rs. $$1900$$
0%
Rs. $$18000$$
Explanation
Initial price of bike is Rs. $$20000$$, time $$=2$$ years, rate od depreciation $$=5\%$$
Cost of bike after $$2$$ years $$=$$ initial price of bike $$\times\left (1-\dfrac {\text{rate of depreciation} }{100}\right)^{\text{time}}$$
Therefore, cost of bike after $$2$$ years $$=20000\left (1-\dfrac {5}{100}\right)^2$$
$$=20000\left (\dfrac {19}{20}\right)^2$$
$$=18050$$
Thus, cost of bike after $$2$$ years is Rs. $$18050$$.
Identify the two statements by which the principal sum can be calculated?
P : The sum amounts to Rs. $$690$$ in $$3$$ yrs at Simple Interest.
Q : The sum amounts to Rs. $$750$$ in $$5$$ yrs at Simple interest.
R : The rate of interest is $$5\%$$ per annum.
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0%
P & Q will give the answer
0%
R & Q will give the answer
0%
P & R will give the answer
0%
Any two will give the answer
Explanation
Any two of three will give the answer because if amount, rate and number of years are given, we can easily find out the principal sum.
Priya invested a certain amount of money and got back an amount of Rs. $$15000$$. If the bank paid an interest of Rs. $$5000$$, what is the amount Priya invested?
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0%
$$5000$$
0%
$$8000$$
0%
$$15000$$
0%
$$10000$$
Explanation
Amount $$=$$ Rs. $$15000$$
Simple interest $$=$$ Rs. $$5000$$
Principal $$=$$ Amount $$-$$ simple interest
$$= 15000 - 5000 $$
Principal $$= 10000$$
Therefore, priya invested Rs. $$10000$$.
Depreciation increase the value of asset.
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0%
True
0%
False
Explanation
Depreciation is the process of reduction in the value of asset.
Depreciation is the permanent and continuous decrease in the book value of a depreciable fixed asset due to use, effluxion of time, obsolescence, expiration of legal rights or any other cause.
The present population of a city is $$18522000$$. If it has been increased at the rate of $$5\%$$ per annum, find the population(approx) after $$3$$ years.
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0%
$$2522200$$
0%
$$16000000$$
0%
$$21441530$$
0%
$$252200$$
Explanation
Given, Principal amount $$18522000$$, rate of interest $$=5\%$$ and period $$=3$$ years
Total Population $$=18522000\left [1+\dfrac{5}{100}\right]^3$$
$$=18522000\left [1+\dfrac{1}{20}\right]^3$$
$$=21441530$$(approx)
Oscar earned Rs. $$2400$$ as simple interest on Rs. `$$4500$$ for $$3$$ months. What is the annual rate of interest?
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0%
$$16.66\%$$
0%
$$17.77\%$$
0%
$$18.88\%$$
0%
$$19.99\%$$
Explanation
Given, principal $$=4500$$, Interest $$=$$ Rs. $$2400$$, period $$=3$$ months
We know the formula,
Rate of interest, $$R =$$ $$\dfrac{I\times 100}{PT}$$
$$\Rightarrow R =$$ $$\dfrac{2400\times 100}{4500\times 3}$$
$$\Rightarrow R = 17.77\%$$
Therefore, the annual rate of interest is $$17.77\%$$.
To start a grocery shop, a woman borrowed Rs. $$1,500$$. If the loan was for four years and the amount of interest was Rs. $$150$$, what simple interest rate was she charged?
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0%
$$1.5\%$$
0%
$$2.5\%$$
0%
$$3.5\%$$
0%
$$4.5\%$$
Explanation
Given, principal $$=$$ Rs. $$1500$$, Interest $$=$$ Rs. $$150$$, period $$=4$$ years
We know the formula,
Rate of interest, $$R =$$ $$\dfrac{I\times 100}{PT}$$
$$\Rightarrow R =$$ $$\dfrac{150\times 100}{1500\times 4}$$
$$\Rightarrow R = 2.5\%$$
Therefore, the simple interest is $$2.5\%$$.
You borrow Rs. $$4,000$$ from a loan shark. If you owe Rs. $$7,200$$ in $$4$$ years, what would be the simple interest rate?
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0%
$$10\%$$
0%
$$20\%$$
0%
$$30\%$$
0%
$$40\%$$
Explanation
Given:
Principal $$=$$ Rs. $$ 4,000$$
Interest $$= 7200 - 4000 = 3200$$
We know the formula,
Rate of interest, $$R =$$ $$\dfrac{I\times 100}{PT}$$
$$\Rightarrow R =$$ $$\dfrac{3200\times 100}{4000\times 4}$$
$$\Rightarrow R = 20\%$$
Thus, the simple interest rate is $$20\%$$.
Sharmila got a Rs. $$1300$$ loan for $$5$$ years. She paid Rs. $$100$$ in interest. What was the interest rate?
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0%
$$1.1\%$$
0%
$$1.3\%$$
0%
$$1.5\%$$
0%
$$1.7\%$$
Explanation
Given, principal $$=$$ Rs. $$1300$$, Interest $$=$$ Rs. $$100$$, period $$=5$$ years
We know the formula,
Rate of interest, $$R =$$ $$\dfrac{I\times 100}{PT}$$
$$\Rightarrow R =$$ $$\dfrac{100\times 100}{1300\times 5}$$
$$\Rightarrow R = 1.5\%$$
Therefore, the interest rate is $$1.5\%$$.
The present population of a town is $$14000$$. If it increases at the rate of $$10\%$$ per annum, what will be its population after $$4$$ years?
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0%
$$10497$$
0%
$$20497$$
0%
$$30497$$
0%
$$40497$$
Explanation
Population after $$4$$ years $$=$$ $$\text{Present Population}$$ $$\left (1+\dfrac{r}{100}\right)^n$$
Population after $$4$$ years $$=$$ $$14000 \times \left (1+\dfrac{10}{100}\right)^4$$
$$=14000\times1.4641$$
$$=20497$$
Population after $$4$$ years $$= 20497$$
The present population of a city is $$15000$$. If it increases at the rate of $$4\%$$ per annum. Find its population after $$2$$ years.
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0%
$$16224$$
0%
$$26224$$
0%
$$36224$$
0%
$$46224$$
Explanation
Given, $$P=15000, r=4\%, n=2$$ years
Population after $$2$$ years $$=$$ $$P[(1+\frac{R}{100})^2]$$
$$=15000\left [(1+\dfrac{4}{100}\right)^2]$$
$$=15000\times 1.0816$$
$$=16224$$
Population after $$2$$ years is $$ 16224$$.
The cost of a machine is Rs. $$24500$$. After two year the value of that machine was depreciated by $$6\%$$. Find the value of machine after two year.
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0%
$$11560$$
0%
$$21560$$
0%
$$31560$$
0%
$$41560$$
Explanation
Given: Principal $$= 24500$$, $$r = 6\%, n = 2$$
Reduction $$= 6\%$$ of Rs. $$24500$$ per year
$$= \dfrac{24500\times 6\times 2}{100}=2940$$
Value at the end of $$2$$ years $$= 24500 - 2940 =$$ Rs. $$21560$$
Jenn borrowed Rs. $$5,000$$ for $$5$$ years and had to pay Rs. $$1,500$$ simple interest at the end of that time. What rate of interest did she pay?
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0%
$$4\%$$
0%
$$5\%$$
0%
$$6\%$$
0%
$$7\%$$
Explanation
Given, $$I=1500,P=5000, T=5$$ years
We need to find rate of interest $$R$$.
We know the formula,
Rate of interest, $$R =$$ $$\dfrac{I\times 100}{PT}$$
$$\Rightarrow R =$$ $$\dfrac{1500\times 100}{5000\times 5}$$
$$\Rightarrow R = 6\%$$
A machine was purchased $$3$$ years ago. Its value decreases by $$5\%$$ every year. Its present value is $$Rs.23000$$. For how much money was the machine purchased?
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0%
$$Rs.16828.45$$
0%
$$Rs.26826.07$$
0%
$$Rs.36828.56$$
0%
$$Rs.46828.80$$
Explanation
Given,
Depreciated value $$A=Rs. 23,000$$
Rate of depreciation $$R=5\%$$
Time $$T=3$$ years
So, $$n=3$$
let the machine was purchased in $$Rs. P$$
Depreciated value,$$ A = P\left( 1-\dfrac { R }{ 100 } \right) ^{ n } $$
$$\Rightarrow 23000 = P \times \left( 1-\dfrac { 5 }{ 100 } \right) ^{ 3 } $$
$$ \Rightarrow 23000= P \times \left( 1-\dfrac { 1 }{ 20 } \right) ^{ 3 } $$
$$\Rightarrow 23000 = P \times \left(\dfrac { 19}{ 20 } \right) ^{ 3 } $$
$$\Rightarrow 23000 = P \times \dfrac { 19}{ 20 } \times\dfrac{19}{20}\times\dfrac{19}{20} $$
$$\Rightarrow P=23,000 \times \dfrac { 20}{ 19 } \times\dfrac{20}{19}\times\dfrac{20}{19} $$
$$\Rightarrow P=26,826.07$$
Therefore, the machine was purchased for $$Rs.26,826.07$$.
Calculate the compound interest on a sum of Rs. $$20000$$ at the end of $$3$$ years at the rate of $$10\%$$ p.a. compounded annually.
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0%
$$6620$$
0%
$$5620$$
0%
$$3410$$
0%
$$2386$$
Explanation
Given, $$P = 20000, r = 10\%, n = 3$$ years
$$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
$$A = 20000\left [\left (1+\dfrac{10}{100}\right)^3\right]$$
$$A = 20000\times\dfrac{11}{10}\times\dfrac{11}{10}\times\dfrac{11}{10}$$
$$A = \text{Rs.} 26620$$
Compound interest $$=$$ Amount $$-$$ Principal
$$C.I. = 26620 - 20000$$
$$C.I. = \text{Rs.} 6620$$
A machine depreciates at the rate of $$10\%$$ of its value at the beginning of a year. If the present value of a machine is $$\text{Rs. }4000$$, find its value after $$3$$ years.
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0%
$$\text{Rs. }1916$$
0%
$$\text{Rs. }2916$$
0%
$$\text{Rs. }3916$$
0%
$$\text{Rs. }4916$$
Explanation
Given:-
$$P = \text{Rs. }4000$$
$$r = 10\%$$
$$n = 3$$ years
Now, as the value of the machine depriciates every year by $$10\%,$$
$$\begin{aligned}{}A& = P{\left( {1 - \frac{r}{{100}}} \right)^n}\\ &= 4000{\left( {1 - \frac{{10}}{{100}}} \right)^3}\\& = 4000{\left( {0.9} \right)^3}\\& = \text{Rs. }2916\end{aligned}$$
Hence, the value of the machine after $$3$$ years is $$\text{Rs. }2916.$$
A company bought a car that had a value of Rs. $$10000$$. each year the value of the car depreciates by $$15\%$$. What is the value of the car at the end of $$2$$ years.
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0%
$$7225$$
0%
$$6225$$
0%
$$5225$$
0%
$$4225$$
Explanation
Given, $$P = 10000, r = 15\%$$ depreciation, $$n = 2$$ years
We know $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 10000\left [\left (1-\dfrac{15}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$7225$$
Therefore, the value of car after $$2$$ years is Rs. $$7225$$.
The annual growth of capital of a company is $$12\%$$. At present, the capital is Rs. $$35000$$. What will be the capital after $$4$$ years?
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0%
$$55500$$
0%
$$55073$$
0%
$$45500$$
0%
$$56430$$
Explanation
Given: $$P = 35000, r = 12\%, n = 4$$ years
We know $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
Putting all the given values, we get
$$A = 35000\left [\left (1+\dfrac{12}{100}\right)^4\right]$$
$$A =$$ Rs. $$55073$$
The price of a new car is Rs. $$2500$$. The price depreciates by $$12\%$$ each year (p.a). Find its value at the end of $$3$$ years.
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0%
$$1703.68$$
0%
$$2703.68$$
0%
$$3703.68$$
0%
$$4703.68$$
Explanation
Given, $$P = 2500, r = 12\%, n = 3$$ years
We know, $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 2500\left [\left (1-\dfrac{12}{100}\right)^3\right]$$
$$\Rightarrow A =$$ Rs. $$1703.68$$
Thus, the value of a car at the end of $$3$$ years is Rs. $$1703.68$$.
The population of a town increases every year at rate of $$10\%$$. Find the population of the town three years hence if it is $$13000$$ now.
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0%
$$17303$$
0%
$$27303$$
0%
$$37303$$
0%
$$47303$$
Explanation
Given, $$P = 13000, r = 10\%, n = 3$$ years
We know, $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 13000\left [\left (1+\dfrac{10}{100}\right)^3\right]$$
$$\Rightarrow A= 17303$$
Hence, the population of the town $$3$$ years is $$17303$$.
Grace bought a new van for Rs. $$15000$$. Each year, the value of her car depreciated by $$10\%$$. Calculate the value of car before $$2$$ years.
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0%
$$12150$$
0%
$$22150$$
0%
$$32150$$
0%
$$42150$$
Explanation
Given, $$P = 15000, r = 10\% $$ depreciated, $$n = 2$$ years
We know, $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 15000\left [\left (1-\dfrac{10}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$12150$$
The value of car before $$2$$ years is Rs. $$12150$$.
The population of an invasive species of moth doubles every $$5$$ years. If the initial population is $$300$$, what will be the population after $$15$$ years?
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0%
$$900$$
0%
$$1200$$
0%
$$2000$$
0%
$$2400$$
Explanation
Population of an invasive species of moth doubles every $$5$$ years.
Let the initial population be $$x$$
So after $$5$$ years,
population
will be $$2\times x=2x$$
After $$10$$ years,
population
will be $$2\times 2x=4x$$
After $$15$$ years,
population
will be $$2\times 4x=8x.$$
Here, initial population is $$x=300$$.
After $$15$$ years,
population
will be $$8x=8\times 300=2400$$
Hence, the population after $$15$$ years is $$2400$$
If a $ $$12,000$$ car loses $$10$$% of its value every year, what is the worth after $$3$$ years?
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0%
8748
0%
8750
0%
8752
0%
8744
Explanation
After $$1^{st}$$ year the value of car is $$12000\left (1- \dfrac{10}{100}\right) = 10800$$
The value of car after $$2^{nd}$$ year is $$10800\left (1-\dfrac{10}{100}\right) = 9720$$
The value of car after $$3^{rd}$$ year is $$9720\left (1-\dfrac{10}{100}\right) = 8748$$
Which of the following is true about annuity?
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0%
It is sequence of equal instalments.
0%
It is sequence of unequal instalments.
0%
It is paid at unequal interval of time.
0%
None of these
Explanation
$$\Rightarrow$$ The true statement about annuity is $$It\,\,is\,\,sequence\,\,of\,\,equal\,\,instalments.$$
$$\Rightarrow$$ Series of payments at fixed intervals, guaranteed for a fixed number of years or the lifetime of one or more individuals.
$$\Rightarrow$$ Annuities are insurance products that provide long-term income through a stream of future payments.
$$\Rightarrow$$ While investment annuities save money for retirement and beneficiaries, structured settlement annuities stem from personal-injury legal cases, wrongful-death claims or lottery payouts. When unexpected circumstances arise and require immediate funds, you can sell these payments for a lump sum of cash.
The compound interest on a certain sum of money for 2 years at 10$$\%$$ per annum is RsFind the simple interest on the same sum at the same rate and for the same time.
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0%
$$Rs. \ 350$$
0%
$$Rs. \ 375$$
0%
$$Rs. \ 380$$
0%
$$Rs. \ 400$$
Explanation
Given that, the compound interest on certain sum of money for $$2$$ years at $$10\%$$ per annum is $$Rs.\ 420$$
To find out: The simple interest on the same amount at the same rate and for the same time.
Let the principal amount be $$Rs.\ x$$
We know that, $$CI=P\left[\left(1+\cfrac{R}{100}\right)^T-1\right]$$
Here, $$CI=Rs.\ 420, \ R=10\%\ and\ T=2$$
$$\therefore \ 420=x\left[\left(1+\cfrac{10}{100}\right)^2-1\right]$$
$$\Rightarrow \ 420=x\left[\left(\cfrac{11}{10}\right)^2-1\right]$$
$$\Rightarrow \ 420=x\left[\left(\cfrac{121}{100}\right)-1\right]$$
$$\Rightarrow \ 420=x\left(\cfrac{121-100}{100}\right)$$
$$\Rightarrow \ x=\dfrac{420\times 100}{21}$$
$$\therefore\ x=Rs.\ 2000$$
Now, we also know that,
$$SI=\dfrac{P\times R\times T}{100}$$
Here, $$P=Rs.\ 2000,\ R=10\%\ and\ T=2$$
$$\therefore \ SI=\dfrac{2000\times 10\times 2}{100}$$
$$\Rightarrow SI=\dfrac{40000}{100}$$
$$\therefore \ SI=Rs.\ 400$$
Hence, the simple interest on the same amount, at the same rate of interest and for the same time is $$Rs.\ 400$$.
Annuity where payments are made at the end of each payment period, i.e. 1st
payment is made at the end of the 1
st
payment interval, and so on, is known as
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0%
Perpetual annuity
0%
Contingent annuity
0%
Ordinary annuity
0%
Immediate annuity
Explanation
Answer is Ordinary or Immediate Annuity.
An ordinary annuity or immediate annuity is where payments are made at the end of each payment period, i.e. 1
st
payment is made at the end of the 1
st
payment interval, and so on. Examples are repayment of car loans, house mortgage etc.
A contingent annuity is one where the term depends upon some event whose occurrence is not fixed. An example is periodic payments of life insurance premiums which stop when the person dies.
A perpetual annuity is an annuity whose term does not end, i.e. it extends till infinity. Thus there is no last payment; they go on forever. An example is freehold property, where you can earn rent in perpetuity.
The C.I. on a certain sum for $$2 $$ years is Rs $$410$$ and S.I. is Rs $$400.$$ What is the rate of interest per annum?
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0%
$$10\%$$
0%
$$8\%$$
0%
$$5\%$$
0%
$$4\%$$
Explanation
$$\Rightarrow$$ It is given that simple interest in 2 years = Rs.400
$$\Rightarrow$$ Simple interest for 1 year = Rs.200
$$\Rightarrow$$ Hence, Compound interest for 1 year = Rs.200
$$\Rightarrow$$ Given compound interest for 2 years = Rs. 410
$$\therefore$$ Compound interest for 2 years = Compound interest 1st year + Compound interest 2nd year
$$\Rightarrow$$ Compound interest for 2nd year = Rs.410 - Rs.200 = Rs.210
$$\Rightarrow$$ That is interest on Rs.200 for 1 year = Rs.210-Rs.200=Rs.10
$$\Rightarrow$$ We know $$I=\dfrac{P\times T\times R}{100}$$
$$\Rightarrow$$ $$10=\dfrac{200\times 1\times R}{100}$$
$$\Rightarrow$$ $$2R=10$$
$$\therefore$$ $$R=5\%$$
The price of commodity $$X$$ increases by $$40$$ paise every year, while the price of commodity $$Y$$ increases by $$15$$ paise every year. If in $$2001$$, the price of commodity $$X$$ was Rs. $$4.20$$ and that of $$Y$$ was Rs. $$6.30$$, in which year commodity $$X$$ will cost $$40$$ paise more than the commodity $$Y$$?
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0%
2010
0%
2011
0%
2012
0%
2013
Explanation
Suppose commodity $$X$$ will cost $$40$$ paise more than $$Y$$ after $$z$$ years.
Then, $$\left( 4.20+0.40z \right) -\left( 6.30+0.15z \right) =0.40$$
$$\Rightarrow 0.25z=0.40+2.10$$
$$\Rightarrow z=\dfrac { 2.50 }{ 0.25 } =\dfrac { 250 }{ 25 } =10$$.
$$\therefore$$ $$X$$ will cost $$40$$ paise more than $$Y$$ $$10$$ years after $$2001 $$ which is $$2011$$.
Three types of annuities are
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0%
Annuity certain
0%
Annuity contingent
0%
Annuity perpetual
0%
All of the above
Explanation
$$\Rightarrow$$ Three types of annuities are :
$$(1)$$ $$Annuity\,\, certain$$ - Annuity that, as a minimum, guarantees a fixed number of payments. It continues over the life of the annuitant, even if he or she lives beyond the number of payments specified in the annuity contract. In case the annuitant dies before exhausting the payments, a named beneficiary continues to receive the remaining number. Also called life annuity certain or life annuity certain and continuous.
$$(2)$$ $$Annuity\,\, contingent$$ - An annuity arrangement in which the beneficiary does not begin receiving payments until a specified event occurs. A contingent annuity may be set up to begin sending payments to a beneficiary upon the death of another individual who wishes to ensure financial stability for the beneficiary, or upon retirement or disablement of the beneficiary.
$$(3)$$ $$Annuity\,\, perpetual$$ - Annuity derived from an asset (such as an income generating security) where the life span of the annuitant (security holder or his or her beneficiary) is of no consequence.
The difference between simple interest and compound on $$Rs. 1200$$ for one year at $$10$$% per annum reckoned half-yearly is:
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0%
$$Rs. 2.50$$
0%
$$Rs. 3$$
0%
$$Rs. 3.75$$
0%
$$Rs. 4$$
0%
None of these
Explanation
$$S.I. = Rs. \left (\dfrac {1200\times 10\times 1}{100}\right ) = Rs. 120$$.
$$C.I. = Rs. \left [1200\times \left (1 + \dfrac {5}{100}\right )^{2} - 1200\right ] = Rs. 123$$.
$$\therefore Difference = Rs. (123 - 120) = Rs. 3$$.
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