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CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Basics Of Financial Mathematics
Quiz 9
$$A$$ borrowed Rs. $$2,500$$ from $$B$$ at $$12$$% per annum compound interest. After $$2$$ years, $$A$$ gave Rs. $$2,936$$ and a watch to $$B$$ to clear the account. Find the cost of the watch.
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0%
Rs. $$100$$
0%
Rs. $$200$$
0%
Rs. $$300$$
0%
Rs. $$400$$
Explanation
$$\Rightarrow$$ $$P=$$Rs. $$2500,\,R=12\%$$ and $$T=2\,$$years
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$
$$\Rightarrow$$ $$A=2500\times (1+\dfrac{12}{100})^2$$
$$\Rightarrow$$ $$A=2500\times \dfrac{28}{25}\times \dfrac{28}{25}$$
$$\Rightarrow$$ $$A=$$Rs. $$3136.$$
$$\Rightarrow$$ Cost of Watch $$= $$Rs. $$(3136-2936) =$$ Rs. $$200$$
Which of the following statement is/are NOT correct?
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0%
Provision for bad debts appears as a liability on the balance sheet
0%
The provision for bad debts is owed to the proprietor
0%
Bad debts could be less than the provision for bad debts
0%
Bad debts could exceed the provision for bad debts
Calculate the compound interest for the second year on Rs. $$16,000/-$$ invested for $$3$$ years at $$10$$% per annnum.
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0%
Rs. $$1760$$
0%
Rs. $$1560$$
0%
Rs. $$1660$$
0%
Rs. $$1860$$
Explanation
$$\Rightarrow$$ $$P=Rs.16000,\,R=10\%$$
$$\Rightarrow$$ C.I. for second year = $$P\times \dfrac{R}{100}\times (1+\dfrac{R}{100})$$
$$\Rightarrow$$ = $$16000\times \dfrac {10}{100}\times (1+\dfrac{10}{100})$$
$$\Rightarrow$$ = $$1600\times \dfrac{11}{10}$$
$$\Rightarrow$$ C.I. for second year = $$Rs.1760.$$
Calculate the principal when time $$= 10$$ years, interest = Rs. $$3000$$; rate $$= 5\%$$ p.a.
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0%
Rs. $$5000$$
0%
Rs. $$6000$$
0%
Rs. $$7000$$
0%
Rs. $$8000$$
Explanation
simple interest formula is given by:
$$S.I.=\dfrac{P\times R\times T}{100}$$
where $$S.I = $$ simple interest
$$P=$$ principle
$$R=$$ rate of interest
$$T=$$ number of years
Given, $$T=10$$ years, $$S.I=$$ Rs. $$3000$$, $$R=5\%$$
by substituting the values in the formula we get:
$$\Rightarrow$$ $$3000=\dfrac{P\times 5\times 10}{100}$$
$$\Rightarrow$$ $$P=\dfrac{3000\times 100}{5\times 10}=$$ Rs. $$6000$$
Therefore, the principle is Rs. $$6000$$.
A sum of money at simple interest amounts to Rs. $$600$$ in $$2$$ years and to Rs. $$800$$ in $$4$$ years. The sum is:
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0%
$$100$$
0%
$$200$$
0%
$$300$$
0%
$$400$$
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%
$$5$$
0%
$$6$$
0%
$$7$$
0%
$$8$$
Explanation
$$\Rightarrow$$ $$P=Rs.5000,\,T=5\,years $$ and $$S.I.=Rs.1500$$
$$\Rightarrow$$ $$S.I.=\dfrac{P\times R\times T}{100}$$
$$\Rightarrow$$ $$1500=\dfrac{5000\times R\times 5}{100}$$
$$\Rightarrow$$ $$R=\dfrac{1500}{250}$$
$$\therefore$$ $$R=6\%$$
Joshita borrowed Rs. $$3,000$$ for $$3$$ years and had to pay Rs.$$ 1,000$$ simple interest at the end of that time. What rate of interest did she pay?
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0%
$$11.11$$
0%
$$22.22$$
0%
$$33.33$$
0%
$$44.44$$
Explanation
$$\Rightarrow$$ $$P=Rs.3000,\,T=3\,years$$ and $$S.I.=Rs.1000$$.
$$\Rightarrow$$ $$S.I.=\dfrac{P\times R\times T}{100}$$
$$\Rightarrow$$ $$1000=\dfrac{3000\times R\times 3}{100}$$
$$\Rightarrow$$ $$R=\dfrac{1000}{90}$$
$$\therefore$$ $$R=11.11\%$$
Ajay had purchased a second hand scooter for Rs. $$18000$$ and spent Rs $$1800$$ for repairs. After $$1$$ year he wanted to sell the scooter. At what price should he sell it to gain $$\cfrac{100}{9}$$ % , if $$\cfrac {100}{11}$$ % is to be deducted at the end of every year on account of depreciation?
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0%
$$10,000$$
0%
$$20,000$$
0%
$$30,000$$
0%
$$40,000$$
Explanation
$$C.P$$ of the scooter=Rs.$$18000$$
Gain=$$\dfrac{100}{9}\%$$
$$SP=\dfrac{(100+\dfrac{100}{9})}{100}\times 18000$$
$$=\dfrac{100}{9}\times 180=Rs.20000$$
The annual production in cement industry is subject to business cycles. The production increases for two consecutive years consistently by
18
18
?
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0%
$$24$$ % increase
0%
$$25$$ % increase
0%
$$42$$ % increase
0%
$$45$$ % increase
Explanation
After every $$2$$ years $$18$$% increase/year third year $$12$$% decrease
Base year is $$2008$$
$$2008+2=2010$$
In $$2010,$$ effect$$=18+18=36$$
$$2010+1=2011$$
In $$2011$$, effect$$=36-12=24$$
$$2012$$ effect$$=24+18=42$$
$$\therefore$$ In $$2012$$, total effect will be $$42$$% increase.
A sum of $$Rs.\ 15,000$$ is invested for $$3$$ years at $$10 \%$$ per annum compound interest. Calculate the interest for the second year.
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0%
$$Rs.\ 1,680$$
0%
$$Rs.\ 1,650$$
0%
$$Rs.\ 1,710$$
0%
$$Rs.\ 1,640$$
Principal $$= 2500$$, rate $$= 6$$%, time$$= 4$$ years. Calculate the interest.
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0%
Rs.$$500$$
0%
Rs.$$600$$
0%
Rs.$$300$$
0%
Rs.$$400$$
Explanation
Simple interest $$=\dfrac{P \times R \times T}{100}$$
$$=\dfrac{2500 \times 6 \times 4}{100} =600$$
Therefor the interest is $$Rs \ 600$$
Calculate the compound interest for the third year on Rs. $$15,000$$ invested for $$5$$ years at $$10\%$$ per annum.
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0%
Rs. $$1815$$
0%
Rs. $$1825$$
0%
Rs. $$1845$$
0%
Rs. $$1875$$
Explanation
Here $$P=$$ Rs. $$15,000,\,R=10\%$$
C.I. for 2 years $$=$$ $$P\times \left (1+\dfrac{R}{100}\right)^2 - P$$ ......(1)
C.I. for 3 years $$=$$ $$P\times \left (1+\dfrac{R}{100}\right)^3-P$$ ......(2)
By subtracting (1) from (2):
C.I. for third year $$=$$ C.I. for three years $$-$$ C.I. for two years $$= P\times \dfrac{R}{100}\times \left (1+\dfrac{R}{100}\right)^2$$
C.I. for third year $$=$$
$$15000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)^2$$
C.I. for third year $$=$$ $$1500\times \dfrac{121}{100}$$
C.I. for third year $$=$$ Rs. $$1815$$
Find the compound interest approximately rate applied to a principal over $$5$$ years if the total interest paid equals the borrowed principal.
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0%
$$14$$%
0%
$$24$$%
0%
$$34$$%
0%
$$7$$%
Explanation
We know,
$$CI=P(1+\cfrac{r}{100})^n-P$$
$$CI=P$$ ....... given,
solving the above equations
$$\left(1+\cfrac{r}{100}\right)^5=2$$
$$r=(2^{\tfrac 1 5}-1)\times 100$$
$$\implies r=14.86\%\approx 14\%$$
Principal + Amount - $$2 \times $$ principle =
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0%
Principle
0%
Interest
0%
Amount
0%
Rate
Explanation
Principal + Amount - $$2 \times $$ principle
$$= \,P+A-2P$$
$$=P+(P+I)-2P$$
$$=P+P+I-2P$$
$$=I$$
Find the compound interest on Rs. $$70,000$$ for $$2$$ years, compounded annually at $$10\%$$ per annum.
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0%
$$14,400$$
0%
$$14,700$$
0%
$$14,300$$
0%
$$14,100$$
Explanation
Here,$$P=$$ Rs $$70000$$ , $$R=10\%\ p.a$$
interest for the first year $$=\cfrac{P\times R\times T}{100}$$
$$=\cfrac{70000\times 10\times 1}{100}$$
$$=$$ Rs $$7000$$
The amount after the first year $$=$$ Rs $$77000$$
Principal for the second year $$=$$ Rs $$77000$$
Interest for the second year $$=$$ Rs $$\cfrac{77000\times 10\times 1}{100}$$
$$=$$ Rs $$7700$$
The final amount $$=$$ Rs $$77000$$ + Rs $$7700$$
$$=$$ Rs $$84700$$
Compound interest $$= $$ Rs $$84700 -70000$$
$$=$$ Rs $$14700$$
In what time will Rs. $$1000$$ amount to Rs. $$1210$$ at $$10$$% p.a. in CI?
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0%
$$2$$
0%
$$1$$
0%
$$1.5$$
0%
$$2.1$$
Explanation
$$\Rightarrow$$ $$P=Rs.1000,\,A=Rs.1210$$ and $$R=10\%$$
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$
$$\Rightarrow$$ $$1210=1000\times (1+\dfrac{10}{100})^T$$
$$\Rightarrow$$ $$\dfrac{121}{100}=(\dfrac{11}{10})^T$$
$$\Rightarrow$$ $$(\dfrac{11}{10})^2=(\dfrac{11}{10})^T$$
$$\therefore$$ $$T=2\,years$$
Calculate the compound interest for the Second year on Rs. $$15,000$$ invested for $$5$$ years at $$10$$% per annum.
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0%
$$1500$$
0%
$$1650$$
0%
$$1800$$
0%
$$2000$$
Explanation
$$\Rightarrow$$ $$P=Rs.15,000,\,R=10\%$$
$$\Rightarrow$$ C.I. for second year = $$15000\times \dfrac{10}{100}\times (1+\dfrac{10}{100})$$
$$\Rightarrow$$ C.I. for second year = $$1500\times \dfrac{11}{10}$$
$$\therefore$$ C.I. for second year = $$Rs.1650.$$
What amount will sum up to Rs. 12100 at 10$$\%$$ per annum, in CI in $$2$$ years?
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0%
$$Rs.\ 10,000$$
0%
$$Rs.\ 11,000$$
0%
$$Rs.\ 9,000$$
0%
$$Rs.\ 8,000$$
Explanation
Given that, $$A=Rs.\ 12100,\,R=10\%$$ and $$T=2\,years$$
To find out: Principal amount, $$P$$.
For compound interest, we know that, $$A=P\left(1+\dfrac{R}{100}\right)^T$$
$$\Rightarrow$$ $$12100=P\times \left(1+\dfrac{10}{100}\right)^2$$
$$\Rightarrow$$ $$12100=P\times \left(\dfrac{11}{10}\right)^2$$
$$\therefore$$ $$P=12100\times \dfrac{100}{121}$$
$$\therefore$$ $$P=Rs.\ 10,000.$$
Hence, $$Rs.\ 10,000$$ will sum up to $$Rs. \ 12,100$$ at $$10\%$$ per annum, in $$2$$ years.
If $$1000$$ dollars is deposited in a bank account for $$4$$ years at $$8$$% per annum.
Calculate interest using compound interest formula.
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0%
$$340.5$$ dollars
0%
$$350.5$$ dollars
0%
$$360.5$$ dollars
0%
$$370.5$$ dollars
Explanation
$$CI=P[(1+\cfrac{r}{100})^n-1]$$
$$\implies CI=1000[(1+\cfrac{8}{100})^4-1]=360.48\approx 360.5$$
$$\implies CI=360.5$$dollars.
Calculate the compound interest for the third year on Rs. $$30,000$$ invested for $$5$$ years at $$10$$% per annum.
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0%
$$3630$$
0%
$$3530$$
0%
$$3330$$
0%
$$3230$$
Explanation
$$\Rightarrow$$ $$P=Rs.30000,\,R=10\%$$
$$\Rightarrow$$ C.P. for third year = $$P\times \dfrac{R}{100}(1+\dfrac{R}{100})^2$$
$$\Rightarrow$$ C.P. for third year = $$30000\times \dfrac{10}{100}\times (1+\dfrac{10}{100})^2$$
$$\Rightarrow$$ C.P. for third year =$$3000\times \dfrac{121}{100}$$
$$\therefore$$
C.P. for third year = $$Rs.3630.$$
What amount will sum up to Rs. $$6,655$$ at $$10$$% p.a. in C.I. in $$3$$ years?
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0%
$$4000$$
0%
$$6000$$
0%
$$5000$$
0%
$$7000$$
Explanation
$$\Rightarrow$$ Here, $$A=Rs.6655\,R=10\%$$ and $$T=3years$$
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$
$$\Rightarrow$$ $$6655=P\times (1+\dfrac{10}{100})^3$$
$$\Rightarrow$$ $$6655=P\times (\dfrac{11}{10})^3$$
$$\therefore$$ $$P=\dfrac{6655\times 1000}{1331}=Rs.5000$$
In what time will a sum of Rs. $$1600$$ at $$5$$% p.a. $$C.I$$ amounts to Rs. $$1764$$?
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0%
$$1$$
0%
$$1.5$$
0%
$$2$$
0%
$$3$$
Explanation
$$A=P(1+\dfrac{R}{100})^T$$
where, $$P= $$ principle
$$A=$$ amount
$$R=$$ rate of interest
$$T=$$ number of years
Given, $$P=Rs.1600\ , R=5\%$$ and $$A=Rs.1764$$
by substituting the values we get ,
$$\Rightarrow$$ $$1764=1600\times (1+\dfrac{5}{100})^T$$
$$\Rightarrow$$ $$\dfrac{1764}{1600}=(\dfrac{21}{20})^T$$
$$\Rightarrow$$ $$\dfrac{441}{400}=(\dfrac{21}{20})^T$$
$$\Rightarrow$$ $$(\dfrac{21}{20})^2=(\dfrac{21}{20})^T$$
$$\therefore$$ $$T=2\,years.$$
What sum lent out at CI will amount to Rs. $$1936$$ in $$2$$ years at $$10$$% p.a. interest?
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0%
$$1500$$
0%
$$1600$$
0%
$$1700$$
0%
$$1800$$
Explanation
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$
$$\Rightarrow$$ $$1936=P\times (1+\dfrac{10}{100})^2$$
$$\Rightarrow$$ $$1936=P\times (\dfrac{121}{100})$$
$$\Rightarrow$$ $$P=\dfrac{193600}{121}=Rs.1600$$
$$\therefore$$ Sum lent out money is $$Rs.1600.$$
The amount on Rs. $$20,500$$ at $$7%$$ per annum compunded annually for $$2$$ years, is:
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0%
$$22470$$
0%
$$23470$$
0%
$$24470$$
0%
$$25470$$
Explanation
$$\Rightarrow$$ $$P=Rs.20,500,\,R=7\%$$ and $$T=2\,years$$
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$
$$\Rightarrow$$ $$A=20500\times (1+\dfrac{7}{100})^2$$
$$\Rightarrow$$ $$A=20500\times (\dfrac{107}{100})^2$$
$$\Rightarrow$$ $$A=20500\times (1.07)^2$$
$$\Rightarrow$$ $$A=20500\times 1.1449$$
$$\therefore$$ $$A=Rs.23470.$$
A sum of Rs $$15,000$$ is invested for $$3$$ years at $$13$$ % per annum compound interest. Calculate the approx interest for the second year.
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0%
$$2100$$
0%
$$2200$$
0%
$$2300$$
0%
$$2400$$
Explanation
Interst for the first year
$$=Rs \cfrac{15000\times 13\times 1}{100}$$
$$=Rs1950$$
Amount after the first year
$$=Rs15000+Rs1950$$
$$Rs16950$$
Interest for the second year
$$=Rs\cfrac{16950\times 13\times 1}{100}$$
$$=Rs2203.5$$
$$=Rs2200$$(approx)
A certain sum is to be invested for a certain time at Compound interest at $$12$$% p.a., such that there is an increase of $$40.4$$% in the sum. Find the time.
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0%
$$2.5$$ years
0%
$$3$$ years
0%
$$3.5$$ years
0%
$$2.75$$ years
Explanation
$$A=140.4\%\times P=1.404P$$
$$A=P(1+\cfrac{r}{100})^n$$
$$\implies 1.404P=P(1+\cfrac{12}{100})^n\\ \implies n\approx3$$
Time$$=3$$years.
The compound interest on Rs. $$100000$$ at $$20\%$$ per annum for $$2$$ years $$3$$ months, compound annually is ________.
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0%
Rs. $$151200$$
0%
Rs. $$100000$$
0%
Rs. $$51200$$
0%
Rs. $$251200$$
Explanation
Given, $$P=100000$$, $$r=0.2$$,
$$n=2+\dfrac{1}{4}$$ (converting $$2$$ years $$3$$ months to years)
$$=n=2.25$$
Interest is compounded anually,
For first year $$=100000\times 0.2\times 1=20000$$
For second year $$P_{new }= 120000$$
Interest $$=120000\times 0 .2\times 1=24000$$
For last $$0.02544$$ year, $$p_{new}=144000$$
Interest $$ = 144000\times 0.2\times 0.25=7200$$
New amount $$=144000+7200 = 151200$$
$$\text{Interest} = 151200+100000=51200$$
A sum of Rs.$$15,000$$ is invested for $$3$$ years at $$13$$ % per annum compound interest. Calculate the compound interest.
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0%
$$6500.435$$
0%
$$6689.245$$
0%
$$6643.455$$
0%
$$6276.585$$
Explanation
We have,
$$P=Rs.\ 15000$$
$$T=3\ years$$
$$R=13\%$$
$$C.I=?$$
We know that
$$A=P\left(1+\dfrac{R}{100}\right)^T$$
$$A=15000\left(1+\dfrac{13}{100}\right)^3$$
$$A=15000\left(\dfrac{113}{100}\right)^3$$
$$A=15000\left(\dfrac{113\times 113\times 113}{100\times 100\times 100}\right)$$
$$A=15\left(\dfrac{113\times 113\times 113}{1000}\right)$$
$$A=Rs.\ 21,643.455$$
So, the compound interest
$$=21.643.455-15000=Rs.\ 6643.455$$
Raman borrowed Rs.$$1,20,000$$ for $$4$$ years at $$8$$ % per year compound interest. Calculate the final amount at the end of four years.
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0%
$$1,63,250$$
0%
$$1,53,250$$
0%
$$1,63,700$$
0%
$$1,66,250$$
Explanation
We have,
$$P=Rs.\ 120, 000$$
$$T=4\ years$$
$$R=8\%$$
$$A=?$$
We know that
$$A=P\left(1+\dfrac{R}{100}\right)^T$$
So,
$$A=120000\left(1+\dfrac{8}{100}\right)^4$$
$$A=120000\left(\dfrac{108}{100}\right)^4$$
$$A=120000\left(\dfrac{108\times 108\times 108\times 108}{100\times 100\times 100\times 100}\right)$$
$$A=12\left(\dfrac{11664\times 11664}{10000}\right)$$
$$A=Rs.\ 163258.675\approx Rs.\ 163250$$
Hence, this is the answer.
Find the compound interest on Rs. $$2,000$$ for $$3$$ years, compounded annually at $$12\%$$ per annum.
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0%
$$764$$
0%
$$810$$
0%
$$820$$
0%
$$850$$
Explanation
Given: $$P=Rs\ 2000$$
$$T=3$$ years
$$R=12\%$$ p.a.
So, Interst for the first year $$=\cfrac{2000\times 12\times 1}{100}$$
$$=Rs\ 240$$
Amount after the first year $$=P+$$ Interest for the first year
$$=Rs.\ 2000+Rs.\ 240=Rs.\ 2240$$
Interst for the second year $$=\cfrac{2240\times 12\times 1}{100}$$
$$=Rs.\ 268.8$$
Hence, the amount after the second year $$=Rs.\ 2240+Rs.\ 268.8=Rs.\ 2508.8$$
Interst for the third year
$$=\cfrac{2508.8\times 12\times 1}{100}$$
$$=Rs.\ 301.056$$
Amount after the third year
$$=Rs.\ 2508.8+Rs.\ 301.056=Rs.\ 2809.856$$
Compound interest
$$=$$ Final amount $$-$$ original amount
$$=Rs.\ 2809.856-Rs.\ 2000$$
$$=Rs.\ 809.856$$
$$\approx Rs.\ 810$$
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