MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 14

Find the compound interest to the nearest rupee on Rs.

$7500$ for

$2$ years

$4$ months at

$12\%$ per annum reckoned annually.

0%
Rs. $2240$
0%
Rs. $2250$
0%
Rs. $2284$
0%
Rs. $2276$

Explanation

$P=7500$ ,

$r=12$ ,

$n=2\tfrac{4}{12}=\cfrac{7}{3}$

$CI=P[(1+\cfrac{r}{100})^n-1]$

$CI=Rs.7500[(1+\cfrac{12}{100})^{7/3}-1]$

$=2270.19$

Value to the nearest integer

$=Rs.2276$ .

A manufacturer reckons that the value of a machine which costs him Rs.

$15025,$ will depreciate each year by

$20$ %. Find the estimated value at the end of

$5$ years.

0%
$12020$
0%
$9616$
0%
$6154.24$
0%
$4923.39$

Explanation

Value after 5 years $=15025\times(1-(\dfrac{20}{100})^5)$

$=15025\times\>0.32768=Rs.4923.39$

The time in which

$Rs.1800$ amount to

$Rs.2178$ at

$10 \%$ per annum, compounded annually is

0%
$3$ years
0%
$2$ years
0%
$4$ years
0%
$1 \dfrac{1}{2} years$

Explanation

Interest is compounded annually,

So amount,

$A=P\left ( 1+\dfrac{r}{100} \right )^{t}$

Where,

$P$ is Principal (or sum)

$t$ is time (in years)

$A$ is the amount, person will get after time

$t$

$r$ is rate of interest per annum

Here,

$A=2178$

$P=1800$

$t=?$

$r=10\%$

Now apply the formula,

$\Rightarrow$

$A=P\left ( 1+\dfrac{r}{100} \right )^{t}$

$\Rightarrow$

$2178=1800\left ( 1+\dfrac{10}{100} \right )^{t}$

$\Rightarrow$

$\dfrac{2178}{1800}=\left ( \dfrac{110}{100} \right )^{t}$

$\Rightarrow$

$\dfrac{121}{100}=\left ( \dfrac{11}{10} \right )^{t}$

$\Rightarrow$

$\left ( \dfrac{11}{10} \right )^{2}=\left ( \dfrac{11}{10} \right )^{t}$

$\Rightarrow t=2$ years

The price of a scooter is

$Rs.14000$ . If it depreciates by

$2\%$ of its value at the beginning of each year. Find the sale value after

$4$ years.

0%
$Rs\,\,1213.154$
0%
$Rs\,\,12000.154$
0%
$Rs\,\,12913.154$
0%
None of these

Explanation

Given that

Original value of the scooter $=Rs.14000$

Rate of depreciation $=2$ %

The value of the scooter after $4$ years

$=14000\times {{\left( 1-\dfrac{2}{100} \right)}^{4}}$

$=14000\times {{\left( 0.98 \right)}^{4}}$

$=14000\times 0.92236816$

$=12913.154$

Thus, the value of the scooter after $4$ years $=Rs\,\,12913.154$

Hence, this is the answer.

If meena gives an interest of Rs.

$45$ for one year at

$9\%$ rate p.a. What is the has borrowed ?

0%
Rs. $450$
0%
Rs. $480$
0%
Rs. $500$
0%
Rs. $550$

Explanation

Let the amount, Meena has borrowed

$=P=$ Principal

Interest of one year

$=Rs. 45$

Rate,

$r=9$ % per annum

Time,

$t=1$ year

Simple Interest

$=\cfrac{P\times r\times t}{100}$

$45=\cfrac{P\times 9\times 1}{100}$

$9P=4500$

$P=Rs. 500$

What would be the

$C.I$ obtained on an amount of

$12500$ at the rate of

$12$

$p.c.p.a.$ after

$2$ years?

0%
Rs. $3180$
0%
Rs. $3360$
0%
Rs. $3540$
0%
Rs. $3720$

Explanation

The formula of for compound interest is

$C.I. = P\left( 1+\cfrac{r}{100} \right)^{n}-P$

$=12,500 \left( 1+\cfrac{12}{100} \right)^{2}-12500$

$= 12500\times \frac{112}{100}\times \frac{112}{100}-12500$

$=3180$

What would be C.I. on Rs.

$17500$ at the rate of

$12\%$ per annum compounded annually after

$2$ years?

0%
Rs. $4442$
0%
Rs. $4452$
0%
Rs. $4462$
0%
Rs. $4482$

Explanation

As we know,

$A=P \left(1+\dfrac{r}{100} \right)^{n}$

here,

$P=$ Principal amount

$r=$ Rate

$n=$ Number of years

$\implies A=17500 \left(1+\dfrac{12}{100} \right)^{2}$

$=\dfrac{17500 \times 112 \times 112}{ 100 \times 100}$

$\implies A= 7 \times 20 \times 112$

$\implies A= 21952$

$C.I.=A-P=21952-17500$

$~=\text{ Rs }4452$

Ans is

$(B)$

What would be the C.I. obtained on an amount of Rs.

$12000$ at the rate of

$9$ p.c.p.a for

$3$ years?

0%
Rs. $3840$
0%
Rs. $3540.348$
0%
Rs. $3540$
0%
Rs. $3640$

Explanation

Given:

$P=12000$

$R=9\%$

$F=n=3\ year$

$A=P \left (1+\dfrac {R}{100}\right)^n$

$=12000 \left (1+\dfrac {9}{100}\right)^3$

$=12000 \left (\dfrac {109}{100}\right)^3$

$=12000\times \dfrac {109}{100}\times \dfrac {109}{100}\times \dfrac {109}{100}$

$A=15540.348$

C.I.

$=A- P$

C.I.

$=15540.348-12000$

C.I

$=$ Rs.

$3540.348$

Hence option (B) is the correct option

Mark against the correct answer in the following:
A sum amounts to

$Rs.3605$ in

$219$ days at

$5\%$ per annum. The sum is

0%
$Rs. 3250$
0%
$Rs.3500$
0%
$Rs.3400$
0%
$Rs.3550$

Explanation

Let the required sum be

$Rs. x$
Then,

$SI = (P \times R \times T) / 100$

$Amount = P + SI$

$= x + [( x \times R \times T) / 100]$

$= x [ 1 + ((R \times T) / 100)]$

$x = Amount / [1 + ((R \times T) / 100)]$

$= 3605 / [1 + ((5 / 100) \times (219 / 365))]$

$= (3605 \times 36500) / 37595$

$x = Rs. 3500$

The population of a town in

$2009$ was

$125000$ . It increase

$10\%$ per year. What is the population after

$3$ years?

0%
$166375$
0%
$170000$
0%
$125000$
0%
$10000$

Explanation

Taking

$P$ as the population of the town in

$2009$ ,

$n$ is the number of years after which the population is to be calculated and

$R$ is the rate at which the population increases, then:

Population after

$3$ years

$= P \left(1 + \dfrac{R}{100}\right)^{n}$

$= 12500 \left(1 + \dfrac{10}{100}\right)^{3}$

$= 12500 \left(1 + \dfrac{1}{10}\right)^{3}$

$= 12500 \left( \dfrac{11}{10}\right)^{3}$

$= 12500 \times \dfrac{1331}{1000}$

$= 16,637.5$

Aditya invest

$Rs. 50,000$ in a fixed deposit at

$10\%\ C.I.$ for

$2$ years. At the end of

$2$ years he puts the money in another deposit at

$12\%\ S.I.$ for

$3$ years, What was the final value of the initial Investment?

0%
$Rs. 80000$
0%
$Rs. 81280$
0%
$Rs. 82280$
0%
$Rs. 83280$

Explanation

$P=\text{Rs } 50,000$

$A=50,000 \left( 1+ \dfrac{10}{100} \right)^{2}$ as

$A=P \left( 1+ \dfrac{r}{100} \right)^{n}$

$A= 50,000 (1.1)^{2}$

$A= 50,000 (1.21)$

$A=60,500$

After

$2\ yrs. P = 60,500$

$SI= \dfrac{P \times R \times S}{100}$

$\Rightarrow \dfrac{60,500 \times 12 \times 3}{100}$

$= 21, 780$

Total Amount

$\Rightarrow 60,500+21,780$

$= \text{Rs } 82, 280$

Aditya invested certain amount in two different schemes

$A$ and

$B$ . Scheme

$A$ offer

$S.I.@$

$12 \%\ \text{p.a}$ . and scheme

$B$ offer

$C.I. @$

$10 \%\ \text{p.a}$ interest accrued on the amount invested in schemes

$A$ in

$2$ years was

$\text{Rs.}\ 3600$ and the total amount invested was

$\text{Rs.}\ 35000$ . What was the interest accrued on the amount invested in scheme

$B$ ?

0%
$\text{Rs.}\ 4500$
0%
$\text{Rs.}\ 4200$
0%
$\text{Rs.}\ 4000$
0%
Cannot be determined

Explanation

Scheme A: SI @

$12\ pcpa$

Scheme B: CI @

$10\ pcpa$

$T = 2\ \text{years}$

$P = \text{Rs.}\ 35,000$

$SI = \dfrac{P\times R\times T}{100}$

$3600= \dfrac{p\times 12\times 2}{100}$

$P = \text{Rs.} 15,000$

Investment in other scheme

$= \text{Rs.}\ 20,000$

$CI = A-P$

$A= 20,000 \left(1+\dfrac{10}{100} \right)^{2}$

$A = \text{Rs.}\ 24,200$

Interest increased in scheme

$B = CI = A-P$

$= \text{Rs.}\ \left(24,000- 20,000\right)$

$= \text{Rs.}\ 4200$

Decrease

$300$ by

$24$ %

0%
$238$
0%
$220$
0%
$235$
0%
$228$

Explanation

$24$ percent of

$300$ is

$\dfrac{24}{100}\times 300=72$

Decrease 300 by

$24$ percent

$=300-72=228$

The population of Mumbai increase

$5\%$ per annum, its population was

$1000000$ in

$2004$ . What was it's population after

$3$ year?

0%
$1157625$
0%
$123575$
0%
$132557$
0%
$11000$

Explanation

Population after

$3$ years

$= P\left(1 + \dfrac{r}{100}\right)^{n}$

$= 1000000\left(1 + \dfrac{5}{100}\right)^{3}$

$= 1000000\left(1 + \dfrac{1}{20}\right)^{3}$

$= 1000000\left( \dfrac{21}{20}\right)^{3}$

$= 1000000 \times \dfrac{9261}{8000}$

$= 1157625$

A steno-typist deposits Rs.

$10,000$ in a bank for a period of

$3$ years at

$12$ % per annum compound interest. The interest accrued to her after maturity will be

0%
$Rs. 3600$
0%
$Rs. 4029$
0%
$Rs. 4032$
0%
$Rs. 4049.28$

Calculate the amount and compound interest on
(i) INR

$10800$ for

$3$ years at

$12\dfrac{1}{2}\%$ per annum compounded annually .

0%
$4555.25$
0%
$4050.42$
0%
$4577.34$
0%
$4566.20$

Explanation

The amount after

$n$ years compounded annually is given as,

$A=P\left(1+\dfrac{r}{100}\right)^n$

Where,

$P=$ Principal

$r=$ rate of interest

According to the given condition,

Amount After three years

$=10800\left({1}+{\dfrac{12.5}{100}}\right)^3$

$=10800\left (\dfrac{112.5}{100}\right)^3\\$

$=15377.34$

compound interest

$=15377.34-10800$

$=Rs\ 4577.34$

What is CI on INR

$7500$ for

$4$ years if the rate of interest is

$10\%$ p.a. For the first

$2$ years and

$20\%$ p.a. for the next

$2$ years?

0%
INR $5273$
0%
INR $5568$
0%
INR $5554.8$
0%
INR $5325$

Explanation

Given,

$=P\left[\left(1+\dfrac{r}{100}\right)^t-1\right]$

$=CI_1+CI_2$

$CI_1$ (For the first two years)

$=7500\left[\left(1+\dfrac{10}{100}\right)^2-1\right]$

$=7500[(1.1)^2-1]$

$=7500\times 0.21=1575$

For

$CI_2$ ,

$P$ will be

$7500+1575$

$=9075$

$CI_2=9075\left[\left(1+\dfrac{20}{100}\right)^2-1\right]$

$=9075[(1.2)^2-1]$

$=9075\times 0.44$

$=3979.8$

$CI=CI_1+CI_2$

$=1575+3979.8$

$=5554.8$ .

A sum of money doubles in

$3$ years at compound interest, compounded annually. It will become

$4$ times of itself in

0%
$12$ years
0%
$9$ years
0%
$8$ years
0%
$6$ years

Explanation

Let principal

$=p$ , rate

$\%=R\%$

$t=4$ years

$\therefore$ Amount

$=2p$

Case I,

$2P=P\left(1+\dfrac{R}{100}\right)^4$

$2=\left(1+\dfrac{R}{100}\right)^4$ …..

$(1)$

Case II,

Let after t years it will be

$8$ times

$8p=p\left(1+\dfrac{R}{100}\right)^t$

$2^3=\left(1+\dfrac{R}{10}\right)^t$

By using equation

$(1)$ &

$(2)$

$\left(1+\dfrac{R}{100}\right)^{12}=\left(1+\dfrac{R}{100}\right)^t$

By comparing both sides

$t=12$ years.

Find the compound interest on

$R s . 2 8 0 0$ for

$1 \dfrac{1}{2}$ years at

$1 0$ % per annum compound interest annually -

0%
$R s . 4 3 4$
0%
$R s . 5 4 4$
0%
$R s . 5 3 4$
0%
$R s . 4 3 5$

The compound interest on

$Rs. 2800$ for

$1\dfrac{1}{2}$ years at

$10 \%$ p.a. is

0%
$Rs. \,441.35$
0%
$Rs. \,436.75$
0%
$Rs. \,434$
0%
$Rs. \,420$

Explanation

Compound interest for

$1\frac{1}{2}$ year =

C.I. for 1 year + C.I for

$\frac{1}{2}$ year tacking

new amount.

$\therefore A_{1} = P(1+\frac{R}{100})^{n} = 2800(1+\frac{10}{100})^{1}$

$A_{1} = 3080 Rs$

$A_{2} = A_{1}(1+\frac{R/2}{100})^{2n}$ (for half yearly).

$A_{2} = 3080 (1+\frac{10/2}{100})^{\frac{1}{2}\times 2} = 3080 \times (1.05)^{1}$

$A_{2} = 3234 Rs$

$\therefore C.I. = A_{2}-P = 3234-2800 = 434 Rs$

1126214_1271639_ans_70899e6384ef4048bd7397443596746b.JPG

At what rate of interest will

$Rs. 20000$ become

$Rs. 24200$ after

$2$ years when interest is compounded annually?

0%
$5 \%$
0%
$6 \%$
0%
$10 \%$
0%
$15 \%$

Explanation

$P =Rs. 20,000, A = Rs. 24,200. n = 2 years.$

$A = P\left [ 1+\dfrac{r}{100} \right ]^{n}$

$\Rightarrow 24,200 = 20,000(1+\dfrac{r}{100})^{2}$

$\Rightarrow \sqrt{1.21} = 1+\dfrac{r}{100}$

$\Rightarrow 1.1 = 1+\dfrac{r}{100}$

$\Rightarrow \dfrac{r}{100} = 1.1-1 = 0.1 \Rightarrow r = 10$

The rate of interest is

$10$ %

Amit invested an amount of Rs.25000 in fixed deposit @C.I. 8% per annum for 2 years. What amount Amit will get?

0%
Rs. 28240
0%
Rs. 28540
0%
Rs. 29240
0%
Rs. 29160

The compound interest on

$Rs.5,000$ at

$4\%$ per annum for

$2$ years compounded annually will be

0%
$Rs.806$
0%
$Rs.708$
0%
$Rs.5,408$
0%
$Rs.408$

Explanation

Given,

$P=5000$

$R=4\%$

$T=2\ years$

Amount

$A=P\Big(1+\dfrac{R}{100}\Big)^T$

$A=5000\Big(1+\dfrac{4}{100}\Big)^2$

$=5000\times\Big(\dfrac{104}{100}\Big)^2$

$=Rs. 5408$

Compound interest

$=A-P$

$=5408-5000=Rs. 408$

The population of a particular area of a city is

$5000$ . It increases by

$10 \%$ in

$1^{st}$ year. It decreases by

$20 \%$ in the

$2^{nd}$ year because of some reason. In the

$3^{rd}$ year, the population increases by

$30 \%$ . What will be the population of area at the end of

$3$ years?

0%
$5120$
0%
$5300$
0%
$5720$
0%
$5620$

Explanation

Population of a city at the starting of first year

$=5000$

It increases by

$10\%$ in

$1^{st}$ year.

i.e., Population at the end of

$1^{st}$ year

$=$ Population at the starting of

$2^{nd}$ year

$=5000+10\%\space of\space5000\\=5000+\dfrac{10}{100}\times5000\\=5500$

It decreases by

$20\%$ in

$2^{nd}$ year.

i.e., Population at the end of

$2^{nd}$ year

$=$ Population at the starting of

$3^{rd}$ year

$=5500-20\%\space of\space5500\\=5500-\dfrac{20}{100}\times5500\\=5500-1100\\=4400$

It increases by

$30\%$ in

$3^{rd}$ year.

i.e., Population at the end of

$3^{rd}$ year

$=4400+30\%\space of\space4400\\=4400+\dfrac{30}{100}\times4400\\=4400+1320\\=5720$

So,

$\text{C}$ is the correct option.

The cost price of a car is

$Rs. \,4,00,000$ . If its price decreases by

$10 \%$ every year, then when will be the cost of car after

$3$ years?

0%
$Rs. \,3,00,000$
0%
$Rs. \,2,91,700$
0%
$Rs. \,2,91,600$
0%
$Rs. \,2,50,000$

Explanation

Cost price of car

$(P)=\mathrm{Rs.}\space4,00,000$

Rate of depreciation

$(R)=10\%$

Time

$(T)=3\space\text{years}$

So, Cost of car after

$3\space\text{years}=P(1-\frac{R}{100})^T$

$=4,00,000(1-\frac{10}{100})^3$

$=4,00,000(1-0.1)^3$

$=4,00,000(0.9)^3$

$=4,00,000\times0.729$

$=\mathrm{Rs.}\space2,91,600$

So,

$\text{C}$ is the correct option.

A four year Indira Vikas certificate with a maturity value of Rs. 700 is purcahsed for Rs.The rate

$\%$ p.a. is :

0%
$9 \dfrac { 1 } { 11 } \%$
0%
$10\%$
0%
$11\%$
0%
$21 \dfrac { 9 } { 11 } \%$

Explanation

Interest

$= 200.....(700-500=200)$

$SI=\dfrac{P\times N\times R}{100}$

$200=\dfrac{500 \times R \times 4}{100}\implies R=10\%$

The population of a country is

$10$ crore and it is the possibility that the population will become

$13.31$ crore in

$3$ years. What will be the annual rate per cent of this growth?

0%
$8 \%$
0%
$12.7 \%$
0%
$10 \%$
0%
$15 \%$

Rohit received an interest of Rs 2520 at the rate 7 % after 3 years. Find out the principal for which he received this interest.

0%
Rs 15,000
0%
Rs 21,000
0%
Rs 1,500
0%
Rs 12,000

Due to reduction of interest rate from 8% to 7.5% by bank, one man got Rs. 600 less as compared to early annual receipts. The amount deposited in the bank is _______________.

0%
Rs. $1, 00, 000$
0%
Rs. $1, 50, 000$
0%
Rs. $1, 20, 000$
0%
Rs. $2, 00, 000$

The present population of a city is 3,02,If the population increases by 10% every year, what was the population of the city two years ago?

0%
$2,80,000$
0%
$2,75,000$
0%
$2,48,000$
0%
$2,50,000$

Explanation

Let the population before two years ago be

$x$

Given the population increases by

$10\%$

So increase in population after one year is

$\dfrac{10}{100}\times x=0.1 x$

So population after one year is

$x+0.1 x=1.1 x$

Increase in population after another one year is

$\dfrac{10}{100} \times 1.1 x=0.11 x$

So population at present is

$1.1 x+0.11 x=1.21 x$

But given population at present is

$3,02,500$

$\implies 1.21 x=302500$

$\implies x=\dfrac{302500}{1.21}=2,50,000$

Population of city two years ago is

$2,50,000$

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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