MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 6

Arjun wants to invest

$Rs. 15,000$ in two types of bonds. He earns

$12\%$ in the first type and

$15\%$ in the second. His investment in

$15\%$ bond, if he has a total earning of

$Rs. 1,950$ , is

0%
$Rs. 10,000$
0%
$Rs. 5,000$
0%
$Rs. 6,000$
0%
$Rs. 7,000$

Explanation

Let Arjun's investment in

$12\%$ bond be

$x$ and in

$15\%$ bond be

$(15000-x)$

total earnings is

$1950$

earning in

$12\%$ bond is

$=\dfrac { 12 }{ 100 } \times x$

earning in

$15\%$ bond is

$=\dfrac { 15 }{ 100 } \times \left( 15000-x \right)$

As per problem,total earning

$=1950$

$\Rightarrow \left( \dfrac { 12 }{ 100 } \times x \right) +\dfrac { 15 }{ 100 } \times \left( 15000-x \right) =1950$

$\Rightarrow \dfrac { 12x }{ 100 } +\dfrac { 225000-15x }{ 100 } =1950$

$\Rightarrow \dfrac { 12x+225000-15x }{ 100 } =1950$

$\Rightarrow -3x+225000=195000$

$\Rightarrow -3x=-30000$

$\Rightarrow x=10000$

Therefore his investment in

$15\%$ bond is

$(15000-10000)=Rs5000$

Rs 2,000 amounts to Rs 2226.05 in two years at compound interest. The rate of interest is

0%
5.5%
0%
5%
0%
4.5%
0%
4%

Explanation

Rs 226.05 is the interest on Rs 2000 for 2 years at r%.

$\therefore 226.05=2000\left \{\left (1+\frac {r}{100}\right )^2-1\right \}$

$\Rightarrow r=5.5$ %.

The compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly is

0%
Rs 6644.025
0%
Rs 7,600
0%
Rs 8,500
0%
none of these

Explanation

We know formula for compound interest compounded quarterly

$Amount=P{ \left( 1+\frac { r }{ n } \right) }^{ nt }$

$Compound\quad Interest=Amount-Principal$
Given,
principal=Rs64000
Time=1year
Rate=10%

$=64000{ \left( 1+\frac { .1 }{ 4 } \right) }^{ 4 }$

$=64000{ \left( 1.025 \right) }^{ 4 }$

$=64000\times 1.10381$

$=70644.025$
Amount is Rs70644.025
Compound interest will be=(70644.025-64000)=Rs6644.025

The compound interest on

$Rs. 10,000$ in

$\displaystyle2\frac{1}{2}$ years at

$4\%$ per annum is

0%
$Rs. 1,032.32$
0%
$Rs. 1,045.50$
0%
$Rs. 1,100.25$
0%
$Rs. 1,150.70$

Explanation

Given principal

$=Rs.10,000$

Time

$=2.5$ years

Rate of interest

$=4\%$

Compound interest for first year

$=\dfrac{PTR}{100}=Rs.\dfrac { 10000\times 1\times 4 }{ 100 } =Rs.400$

Amount at the end of first year

$=Rs.(10000+400)=Rs.10400$

Amount for first year

$=$ Principal for second year

$=Rs.10400$

Compound interest for second year

$=\dfrac{PTR}{100}=Rs.\dfrac { 10400\times 1\times 4 }{ 100 } =Rs.416$

For next

$6$ months,

Principal

$=Rs.(10400+416)=Rs.10816$

Time

$=6\ months=\dfrac{1}{2}\ years$

Compound interest for next six months will be

$=\dfrac{PTR}{100}=Rs.\dfrac { 10816\times 1\times 4 }{ 100\times 2 } =Rs.216.32$

Total compound interest

$=Rs.(400+416+216.32)=Rs.1032.32$

An article can be sold for 20000 cash or for 12000 down payment and 4 equal monthly instalments of 2200 each. Find the interest paid.

0%
400
0%
600
0%
800
0%
1000

Explanation

With the downpayment option, total amount paid for article

$= Rs 12000 + 4(2200) = Rs 20800$

But actually SP of article

$= Rs 20000$

So, interest paid

$= Rs 20800 - 20000 = Rs 800$

A sum of money invested at simple interest amounts to 2480 at the end of four years and 4080 at the end of eight years. Find the principal.

0%
2040
0%
1480
0%
1240
0%
880

Explanation

Simple Interest

$SI = \dfrac {PNR}{100}$
Also Amount

$A = P + SI$
So, for four years,

$2480 - P = \dfrac {P \times 4 \times R}{100}$ -- (1)

And, for eight years,

$4080 - P = \dfrac {P \times 8 \times R}{100}$ -- (2)

Subtracting equation (1) from equation (2), we get

$=> 1600 = \dfrac {4PR}{100}$

$=> PR = 40,000$

Substituting it in equation (1), we get

$2480 - P = \dfrac { 4 \times 40,000}{100}$

$=> P = Rs$

$880$

Calculate the amount and the compound interest on Rs.

$12,000$ for

$2$ years at

$5$ % per annum compounded annually.

0%
Rs. $13,500$ and Rs. $1500$
0%
Rs. $13,230$ and Rs. $1,230$
0%
Rs. $14728$ and Rs. $2,728$
0%
None of the above

Explanation

Amount

$A=P(1+r)^{t}$
Here,

$P=12000, r=\dfrac5{100}=0.05, t=2$

$\therefore A=12000(1+0.05)^{2}$

$=12000(1.05)^2$

$=12000(1.1025)$

$=13230$
Compound interest

$=13230-12000=1230$
Option B is correct.

A sum of money at simple interest amounts to 800 in 2 years and to 1200 in 6 years. The sum is

0%
600
0%
1000
0%
400
0%
500

Explanation

Simple Interest

$SI = \frac {PNR}{100}$

Given,

$Amount A = 800$

$=> P + SI = 800$

$=> P + \frac {P \times 2 \times R}{100} = 800$

$=> P(100+2R) = 80,000$ -- (2)

And

$Amount A = 1200$

$=> P + SI = 1200$

$=> P + \frac {P \times 6\times R}{100} = 1200$

$=> P(100+6R) = 120000$ - (2)

Dividing eqn 2 by 1, we get

$\frac {100+6R}{100+2R} = \frac {12}{8}$

$=> R = \frac {100}{6}$ %

Substituting in eqn 2, we get

$P = Rs 600$

A sum was put at SI at a certain rate for 8 years Had it been put at 4% higher rate it would have fetched Rs. 80 more Find the sum___

0%
Rs. 300
0%
Rs. 200
0%
Rs. 250
0%
Rs. 400

Explanation

$\displaystyle \frac{P\times \left ( R+4 \right )\times 8}{100}-\frac{P\times 8\times R}{100}=80$

$\displaystyle \frac{2PR+8P}{25}-\frac{8PR}{100}=80$

$\displaystyle \frac{8PR+32P-8PR}{100}=80$

$\displaystyle \frac{32P}{100}=80$
P =

$\displaystyle \frac{80\times 100}{32}$
= Rs. 250

A sum of Rs.

$13,500$ is invested at

$16\%$ per annum compound interest for

$5$ years. Calculate the interest for the second year, correct to the nearest rupee.

0%
Rs. 2,106
0%
Rs. 2,279
0%
Rs. 2,506
0%
Rs. 2,792

Explanation

Sum=Rs 13,500

Rate=16%

Amount at the end of first year

$=P\left(1+\frac{R}{100}\right)^T$

$\Rightarrow 13500\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 13500\times \dfrac{116}{100}=Rs.15,660$

For the second year

Sum

$=Rs\,15,660$

Then amount after second year

$=15660\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 15,660\times \dfrac{116}{100}=Rs.18165.6$

$\therefore$ Interest for second year

$=18,165.6-15,660$

$=Rs\,2505.6$

Rounding off

$Rs\,2505.6$ we get

$Rs\,2,506$

Therefore interest for the second year is

$Rs\,2,506$

During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates to Rs. 2,640 during the second financial year of its purchase.

0%
3409
0%
3333
0%
2365
0%
8551

Explanation

$\textbf{Step 1: Calculating the cost for the }\mathbf{1^{st}} \textbf{ year}$

$\text{Consider the original value to be x,}$

$\text{After the first year the value reduces to}$

$=x-\dfrac{12}{100}x=x-0.12x=0.88x$

$\textbf{Step 2: Calculating the cost for the }\mathbf{2^{nd}}\textbf{ year}$

$\text{The value reduces to}$

$=0.88x-\dfrac{12}{100}0.88x=0.88x\times(1-0.12)=0.7744x$

$\implies 0.7744x=2640$

$\implies x=3409$

$\textbf{Hence, The original value is Rs.3409}$

John earned Rs.

$100$ as simple interest on Rs.

$600$ for

$6$ months. Find the annual rate of interest.

0%
$11.11\%$
0%
$22.22\%$
0%
$33.33\%$
0%
$44.44\%$

Explanation

Given, Simple interest

$=$ Rs.

$100$ , Principal

$=$ Rs.

$600$ and period

$6$ months

We know the formula,
Rate of interest,

$R =$

$\dfrac{I\times 100}{PT}$

$R =$

$\dfrac{100\times 100}{600\times 0.5}$

$R = 33.33\%$

Therefore, the annual rate of interest is

$33.33\%$ .

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%
Rs. $18000$
0%
Rs. $19800$
0%
Rs. $20000$
0%
Rs. $22500$

Explanation

Cost price

$=$ Rs.

$18000$
cost on repairing

$=$ Rs.

$1800$
So, Total cost

$= 18000 + 1800 = 19800$
Depreciation

$= \dfrac {100}{11} \%= 9.09 \%$
Gain

$= \dfrac {100}{9 } \%= 11.11 \%$
After depreciation, the cost price would be

$= 19800 - 9.09\%$ of

$19800 =$ Rs.

$18000$
S.P. to gain

$11.11 \% = 18000 + 11.11 \%$ of

$18000 = 18000 + 1999.8 =$ Rs.

$20000$ .

A certain sum is interested at compound interest . The interest occurred in the first two years is Rs.

$272$ and that in the First three years is Rs.

$434$ . Find the rate

$\%$ ?

0%
$33.33$ $\%$
0%
$25$ $\%$
0%
$12.5$ $\%$
0%
$16.67$ $\%$

Explanation

For first two year interest $= P\left [\left (1+\dfrac{R}{100}\right)^2-1\right] = 272$ .......(1)
For first three year interest $= P\left [\left (1+\dfrac{R}{100}\right)^3-1\right] = 434$ .......(2)
On dividing equation (1) by (2), we get
$\Rightarrow\dfrac{ \left (1+\dfrac{R}{100}\right)^2-1 }{ \left (1+\dfrac{R}{100}\right)^3-1 }= \dfrac{272}{434}$
Let $\left (1+\dfrac{R}{100}\right)=x$ ...........(3)

$\Rightarrow \dfrac{(x^2-1)}{ (x^3-1) }= \dfrac{136}{217}$
$\Rightarrow \dfrac{(x+1)}{(x^2+x+1) }= \dfrac{136}{217}$
$136x^2-81x-81 = 0$
After solving $x$ will be $1.1248$
By putting this value in eq (3), we get
$R = 12.48$ approx $12.50$ $\%$

Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find the interest he earns for the second year.

0%
Rs. 1,162
0%
Rs. 1,243
0%
Rs. 1,408
0%
Rs. 1,591

Explanation

$=Rs.12800$

Rate of interest

$=10$ %

Interest for the first year

$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{12800\times 10\times 1}{100}=Rs.1280$

Amount after first year

$=12800+1280=Rs.14080$

For Second year

Principal

$=Rs.14080$

$=10$ %

$=1$ year

then Interest for second year

$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{14080\times 10\times 1}{100}=Rs.1408$

Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs. 252.

0%
$Rs. 10,000$
0%
$Rs. 12,000$
0%
$Rs. 14,000$
0%
$Rs. 16,000$

Explanation

let the sum=Rs.P

rate=10%

C.I for 3 year=Amount at the end of 3rd year-Amount at the end of 2nd year

$=P\left(1+\frac{10}{100}\right)^3-P\left(1+\frac{10}{100}\right)^2$

$=P\left[1+\left(\frac{11}{10}\right)^3-\left(\frac{11}{10}\right)^2\right]$

$=P(\frac{11}{10})^2-[\frac{11}{10}-1]$

$=P\times \frac{121}{100}\times \frac{1}{10}$

$=\frac{121}{1000}P$

C.P for 1 st year

$=\frac{P\times 10\times 1}{100}=\frac{P}{10}$

Then according to the question

$\Rightarrow \frac{121}{1000}-\frac{P}{10}=252$

$\Rightarrow \frac{121P-100P}{1000}=252$

$\Rightarrow \frac{21P}{1000}=252$

$\Rightarrow P=\frac{252\times 1000}{21}=Rs. 12000$

Saurabh invests

$Rs.\ 48,000$ for

$7$ year at

$10\%$ per annum compound interest. Calculate the interest for the first year.

0%
$Rs.\ 4,800$
0%
$Rs.\ 4,900$
0%
$Rs.\ 5,000$
0%
$Rs.\ 5,100$

Explanation

Given:

$P=Rs.\ 48000$

$r=10\%$

$t=1$ year

$CI$ for

$1^{st}$ year

$=$

$SI$ for

$1^{st}$ year

$=\dfrac{PRT}{100}$

$=\dfrac{48000\times 10\times 1}{100}$

$=Rs.\ 4800$

Hence, option

$A$ is correct.

Calculate the amount at the end of the second year.

0%
Rs. $42,856$
0%
Rs. $48,140$
0%
Rs. $52,937$
0%
Rs. $58,080$

Explanation

Given,

Sum

$(P)=Rs.48000$

Rate

$=10\%$

Time

$=2\ year$

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

$= 48000\left(1+\dfrac{10}{100}\right)^2$

$= 48000\times \dfrac{110}{100}\times \dfrac{110}{100}\\=Rs.58080$

In a city

$20 \%$ of the total population is student community which is not employed , if the number of students who are unemployed is

$14000$ , then find the population of the city ?

0%
$56000$
0%
$70000$
0%
$40000$
0%
$60000$
0%
None of these

Explanation

$\Rightarrow$ In this question we have given, 20% of student are not employed. Total student population is 100%.

$\Rightarrow$ Total unemployed student is 14000.

$\Rightarrow$ Let

$x$ be the total population

$\Rightarrow$

$\dfrac{Number\, of\, unemployed\, student}{Total\, number\, of\, student}=\dfrac{Number\, of\, unemployed\, student\, \%}{Total\, number\, of\, student\, \%}$

$\Rightarrow$

$\dfrac{14000}{x}=\dfrac{20}{100}$

$\therefore$

$x=70000$

$\therefore$ Population of city is

$70000$

A sum of

$Rs.\ 9,600$ is invested for

$3$ years at

$10\%$ per annum at compound interest.

What is the sum due at the end of the first year?

0%
$Rs.\ 10,560$
0%
$Rs.\ 15,630$
0%
$Rs.\ 20,360$
0%
$Rs.\ 30,860$

Explanation

Given:

$P=Rs.\ 9,600$

$r=10\%$

Sum due at the end of the first year will be,

$\begin{aligned}{}A &= 9600{\left( {1 + \frac{r}{{100}}} \right)^n}\\ &= 9600{\left( {1 + \frac{{10}}{{100}}} \right)^1}\\ &= 9600 \times \frac{{110}}{{100}}\\ &= Rs.\ 10,560\end{aligned}$

Hence, option

$A$ is correct.

You invested Rs.

$1500$ and received Rs.

$5000$ after three years. What had been the interest rate?

0%
$111.11\%$
0%
$222.22\%$
0%
$333.33\%$
0%
$77.77\%$

Explanation

Amount invested

$= 1500$

Amount received after three years

$= 5000$

$\Rightarrow$ Interest earned

$= 5000-1500 = 3500$

Let interest rate

$\%$ be

$r$

$\Rightarrow$

$3500 = \dfrac{1500 \times 3 \times r}{100}$

$\Rightarrow$

$r = 77.77\%$

In a culture, the present bacteria count was found as

$67,60,000$ . It was found that the number was increased by

$4\%$ per hour. Find the number of hours before which the bacteria count was

$62,50,000$ .

0%
$1$ hr
0%
$2$ hr
0%
$3$ hr
0%
$4$ hr

Explanation

Let the number of years be

$n$

$\Rightarrow 67,60,000=62,50,000 \left [1+\dfrac{4}{100}\right]^n$

$\Rightarrow \dfrac{676}{625}=[1+\dfrac{1}{25}]^n$

$\Rightarrow \left [\dfrac{26}{25}\right]^2=\left [\dfrac{26}{25}\right]^n$

$n=2$ hrs

Thus, the number of hours are

$2$ .

Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w?

0%
$\displaystyle \frac { w }{ 1+1.08 }$
0%
$\displaystyle \frac { w }{ 1.08+1.16 }$
0%
$\displaystyle \frac { w }{ 1.16+1.24 }$
0%
$\displaystyle \frac { w }{ 1.08+{ \left( 1.08 \right) }^{ 2 } }$
0%
$\displaystyle \frac { w }{ { \left( 1.08 \right) }^{ 2 }+{ \left( 1.08 \right) }^{ 3 } }$

Explanation

Amount Alex deposited in account =

$x$ dollars.

$\Rightarrow$ Annual interest earned after one year =

$8\%\,of\,x = 0.08x$

$\Rightarrow$ Total amount in account after one year =

$(x + 0.08x) = 1.08x$

$\Rightarrow$ After one year, Alex deposited additional

$x$ dollars.

$\Rightarrow$ Total amount after depositing

$x$ dollars =

$1.08x + x$

$\Rightarrow$ Interest after second year =

$8\%\,of\,(1.08x + x) = 0.08(1.08x + x)$

$\Rightarrow$ Total amount after second year :

$\Rightarrow$

$(1.08x+x)+0.08(1.08x+x)$

$\Rightarrow$

$(1.08x+x)(1+0.08)$

$\Rightarrow$

$(1.08x+x)(1.08)$

$\Rightarrow$

$x[1.08+(1.08)^2]$

$\Rightarrow$ Let

$w$ be the total amount in the account, then

$\Rightarrow$

$w=x[1.08+(1.08)^2]$

$\therefore$

$x=\dfrac{w}{1.08+(1.08)^2}$

Find the compound interest on Rs.

$7500$ at

$4\%$ per annum for

$2$ years, compounded annually.

0%
Rs. $612$
0%
Rs. $412$
0%
Rs. $782$
0%
Rs. $112$

Explanation

Given,

$P=$ Rs.

$7500, R=4\%, n=2$ years

We know, Amount

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

Therefore, amount

$=$ Rs

$\begin{bmatrix}7500\times\begin{Bmatrix}1+\begin{pmatrix}\dfrac{4}{100}\end{pmatrix}\end{Bmatrix}^2\end{bmatrix}=$ Rs.

$8112$
Compound Interest

$=$ Rs.

$\begin{pmatrix}8112-7500\end{pmatrix}=$ Rs.

$612$

If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college?

0%
$$ 20,000$
0%
$$ 15,000$
0%
$$ 12,000$
0%
$$ 10,000$
0%
$$ 9,000$

Explanation

$\Rightarrow$ We have given that

$r$ = Percent interest.

$\Rightarrow$ We are also given that an investment with double in approximately

$70/r$ years.

$\Rightarrow$ We are told to invest

$\$5,000$ at

$8\%$ for 18 years.

$\Rightarrow$ Put

$r=8$

$\Rightarrow$ So,

$\dfrac{70}{8}$ is about

$9\,years$ , meaning investment will double in

$9\,years$ .

$\Rightarrow$ In the first 9 years,

$\$5,000$ doubles to

$\$10,000$

$\Rightarrow$ In the next 9 years,

$\$10,000$ doubles to

$\$20,000$

$\therefore$ The approximate total amount of investment 18 years later, when Pat is ready for college is

$\$20,000$ .

The rate of depreciation

0%
10%
0%
12%
0%
14%
0%
16%

Explanation

Difference between depreciations of 2nd year and 3rd year

$= Rs. 4,752 - Rs. 4,181.76 = Rs. 570.24$

$\Rightarrow$ Depreciation of one year on Rs. 4,752

$=$ Rs. 570.24

$\Rightarrow$ Rate of depreciation

$=\displaystyle \frac{Rs. 570.24}{Rs. 4,752}\times 100\%=12\%$ Ans.

Lucy invested

$10,000$ in a new mutual fund account exactly three years ago. The value of the account increased by

$10$ percent during the first year,increased by

$5$ percent during the second year, and decreased by

$10$ percent during the third year. What is the value of the account today?

0%
$10350$
0%
$10395$
0%
$10500$
0%
$11500$
0%
$12705$

Explanation

The first year's increase of

$10$ percent can be expressed as

$1.10$ , the second year's increase can be expressed as

$1.05$ and third year's decrease can be expressed as

$0.90$ .

Multiply the original value accounts by each of these years changes,

$10,000(1.10)(1.05)(0.90)=10,395$

So, the value of the account today is Rs.

$10,395$ .

The present population of town is

$12500$ and it is increasing at the rate of

$8\%$ per annum. What will be the population of town after two years?

0%
$14000$
0%
$12500$
0%
$14580$
0%
$25000$

Explanation

Given, present population of town

$=12500, R=8\%$

Therefore, population after

$2$ years

$=12500\times\begin{Bmatrix}1+\dfrac{8}{100}\end{Bmatrix}^2$

$=12500\times\begin{Bmatrix}\dfrac{108}{100}\end{Bmatrix}^2$

$=14580$

Find the interest for the principal Rs.

$4110$ at the rate of

$6\%$ for

$2$ years.

0%
$343.50$
0%
$493.20$
0%
$460.19$
0%
$531.12$

Explanation

Simple interest,

$I =$

$\dfrac{PRT}{100}$
Where,

$P$ is the principal

$=4110$

$R$ is the rate

$=6\%$

$T$ is the time period

$=2$ years

$I =$

$\dfrac{4110\times 6\times 2}{100}$

$I =$ Rs.

$493.20$

Find rate, when principal

$=$ Rs.

$3000$ ; interest

$=$ Rs.

$400$ ; time

$= 3$ years.

0%
$1.11\%$
0%
$2.22\%$
0%
$3.33\%$
0%
$4.44\%$

Explanation

Given, Principal

$=$ Rs.

$3000$ , interest

$=$ Rs.

$400$ , time

$=3$ years

We know the formula,
Rate of interest,

$R =$

$\dfrac{I\times 100}{PT}$

$R =$

$\dfrac{400\times 100}{3000\times 3}$

$R = 4.44\%$

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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