MCQ Questions for 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.

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?

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.

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.

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:

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?

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?

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?

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

1818?

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

$2008$

$2008+2=2010$

$2010,$ effect

$=18+18=36$

$2010+1=2011$

$2011$ , 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.

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.

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.

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.

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 =

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.

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?

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.

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?

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.

$1000$ dollars is deposited in a bank account for

$4$ years at

$8$ % per annum.

Calculate interest using compound interest formula.

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.

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?

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.

$1764$ ?

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?

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:

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.

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.

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 ________.

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.

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.

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.

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

$=Rs.\ 2809.856-Rs.\ 2000$

$=Rs.\ 809.856$

$\approx Rs.\ 810$

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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