MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 3

The compound interest on Rs. 50,000 at 4% per annum for two years compounded anually is :

0%
$4000$
0%
$4080$
0%
$4280$
0%
$4050$

Explanation

C.I. = Amount - Principle

$P((1+\dfrac{r}{100})^{T}_{} - 1)$

CI =

$50,000(1+\dfrac{4}{100})^{2}_{} - 1)$

C.I.=

$4080.$

Tax money is used by ________________.

0%
the common people to invest in technology and education
0%
the government to invest in technology and education
0%
not used
0%
none of these

What rate percent of S.I. will a sum of money double itself

0%
1/2%
0%
8.33%
0%
both 1 &2
0%
none of these

Explanation

In case of simple Interest, total amount is

$A = P \left( 1 + \dfrac{rt}{100} \right)$

Where,

P = Principal

R = Rate

T = time

After 12 year sum of money double itself ?

that is A = 2 P

Now apply the formula

$A = P \left( 1 + \dfrac{r \times 12}{100} \right)$

$2P = P \left( 1 + \dfrac{r \times 12}{100} \right)$

$2 = 1 + \dfrac{ 12 r}{100}$

$1 = \dfrac{ 12 r}{100} \\ \therefore r = \dfrac{100}{12} = 8 \dfrac{1}{3} \%$

A specialized agency of the UNO can claim refund of tax paid on ___________________.

0%
Intra-State supply of goods and/or services
0%
Inter-state supply of goods and/or services
0%
Inward supply of goods and/or services
0%
All of the above

Compound interest is the interest calculated on the previous year's amount.

0%
True
0%
False

If the population of a town is 64,000 and its annual increase is 10%, then its correct population at the end of 3 years will be

0%
80,000
0%
85,000
0%
85,100
0%
85,184

Explanation

Population (P)

$= 64,000$
Annual increase rate

$(r) = 10\%$
Hence, population after

$n=3$ years

$= P(1+\cfrac{r}{100})^n$

$= 64,000(1+\cfrac{10}{100})^3$

$= 64,000\times (\cfrac{11}{10})^3$

$= 64,000\times \cfrac{11\times 11\times 11}{1000}$

$= 85,184$

If a sum of money placed at compound interest doubles itself in 5 years, then the same amount of money will be 8 times of itself in

0%
25 years
0%
20 years
0%
15 years
0%
10 years

Explanation

Let the sum be Rs.

$x$ .

Then

$x$ becomes

$2x$ in 5 years,

$2x$ becomes

$4x$ in next 5 years,

$4x$ becomes

$8x$ in next 5 years.
Hence,

$x$ becomes

$8x$ in 15 years.

How much should money lender lend at the simple rate of interest of

$15\%$ in order to have

$\text{Rs. }3234$ at the end of

$1\frac{1}{2}$ years?

0%
$\text{Rs. }1640$
0%
$\text{Rs. } 2620$
0%
$\text{Rs. }2610$
0%
$\text{Rs. } 2640$

Explanation

Let the required money be

$x$ .

Then,

Total amount

$=$ initial amo

$(x+\cfrac{x\times 15}{100}\times \cfrac{3}{2}) = 3234$

$\Rightarrow \cfrac{49x}{40} = 3234$

$\Rightarrow x = (\cfrac{3234\times 40}{49}) = 2640$

The compound interest of Rs. 1,000 at the rate of 10% will be Rs. 210 in:

0%
one year
0%
two years
0%
three years
0%
four years

Explanation

Given Sum (P)

$=$

$Rs. 1000$ , Rate

$=$

$10\%$ ,

$C.I.$

$=$

$Rs. 210$ , Time

$(n)$

$=$ ?

Now

$C.I.$

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

$\therefore 210 = 1000[(1+\cfrac{10}{100})^{n}-1]$

$(\cfrac{110}{100})^{n}-1 = \cfrac{210}{1000} = 0.21$

$\left ( \cfrac{11}{10} \right )^{n} = 1.21 = \cfrac{121}{100}$

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

$\therefore$

$n$

$= 2$ years

A certain sum of money Q was deposited for 5 years and 4 months at 4.5% simple interest and mounted to Rs.Then the value of Q is

0%
Rs. 200
0%
Rs. 210
0%
Rs. 220
0%
Rs. 240

Explanation

We know that,
Amount = principle + simple interest

$A=P+\cfrac{PRT}{100}$

$A=P[1+\cfrac{RT}{100}]$ ...... (1)
Where,

$A$ is amount.

$P$ is principle.

$R$ is rate of interest.

$T$ is time in year.

Given,

$A=248$

$P=Q$

$R=4.5$ %

$T=5+\cfrac{4}{12}=\cfrac{16}{3}$
Substitute these values in equation (1) and find the value of

$Q$ .

$248=Q[1+\cfrac{4.5\times\frac{16}{3}}{100}]$

$248=Q[1+\cfrac{24}{100}]$

$248=1.24Q$

$\therefore Q=200$

The population of a small town is increased by

$\cfrac{15}{2}$ % per annum for two years. If the present population is

$73, 960$ , then two years before the population was

0%
$68,000$
0%
$70,000$
0%
$64,000$
0%
$60,000$

Explanation

$A=P(1+\cfrac{R}{100})^n$

$73960=P(1+0.075)^2$

$\therefore P=64000$

Rajan borrowed Rs. 50,000 from Rakesh at simple interest. After 3 years, Rakesh got Rs. 3000 more than what he had given to Rajan. What was the rate of interest per annum?

0%
2%
0%
5%
0%
8%
0%
10%

Explanation

$\Rightarrow$ Here, we have

$P=Rs.50,000,\, T=3\, years$ and

$S.I.=Rs.3000$

$\Rightarrow$

$Simple\, Interest=\dfrac{P\times R\times T}{100}$

$\Rightarrow$

$3000=\dfrac{50000\times R\times 3}{100}$

$\Rightarrow$

$R=\dfrac{3000\times 100}{50000\times 3}$

$\therefore$

$R=2\%$

A certain sum of money lent out at S.I. amounts to Rs. 6900 in 3 years and Rs. 7500 in 5 years. The sum lent is

0%
Rs. 4000
0%
Rs. 4500
0%
Rs. 5000
0%
Rs. 6000

Explanation

S.I. for 2 years

$= (7500 - 6900) = Rs. 600$
S.I. for 3 years

$= Rs. (\cfrac{600}{2}\times 3)$

$= Rs. 900$

$\therefore$ Principal

$= Rs. (6900 - 900)$

$= Rs. 6000$

Calculate the amount and the compound interest on Rs.

$12,000$ in

$2$ years and at

$10 \%$ per year.

0%
$Rs.\ 14,520$ and $Rs.\ 2,520$
0%
$Rs.\ 14,000$ and $Rs.\ 2,000$
0%
$Rs.\ 13,520$ and $Rs.\ 2,820$
0%
$Rs.\ 17,520$ and $Rs.\ 2,120$

Explanation

Amount after

$n$ of compound interest annually

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

So, required Amount after

$2$ years

$A = Rs\ 12,000 \times \left( 1+\dfrac { 10 }{ 100 } \right) ^{ 2} = Rs\ 12,000 \times {1.1}^{2} = Rs\ 14,520$

And

$C.I$ =

$A - P = Rs\ 14,520 - Rs\ 12,000 = Rs\ 2,520$

During a period of two years, a principal of Rs. 100 amounts to Rs. 121 at the annual compound rate of r%. The value of

$r$ will be

0%
$9$
0%
$10$
0%
$\cfrac{21}{2}$
0%
$11$

Explanation

Given, Sum (P)

$=$ Rs. 100,
Amount due (A)

$=$ Rs. 121,
Time (n)

$=$ 2 years, Rate (r)

$=$ ?
We know

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

$\therefore 121 = 100 (1+\cfrac{r}{100})^{2}$
or

$(1+\cfrac{r}{100})^{2} = \cfrac{121}{100}$
or

$1+\cfrac{r}{100}=\cfrac{11}{10}=1+\cfrac{1}{10}=1+\cfrac{10}{100}$

$\therefore r = 10\%$

Rakesh took a loan for 7 years at the rate of 6% p.a. S.I. If the total interest paid was Rs. 2100, the principal was

0%
Rs. 4400
0%
Rs. 4800
0%
Rs. 5000
0%
Rs. 5200

Explanation

Given that,

Rakesh took a loan for

$7$ years at

$6\%$ per annum simple interest.

He paide

$Rs.\ 2100$ interest on the amount.

To find out,

The principal amount.

We know that,

$S.I.=\left(\dfrac{P\times R\times T}{100}\right)$

Here,

$S.I.=Rs.\ 2100,\ R=6\%$ and

$T=7\ years$

Hence,

$2100=\left(\dfrac{P\times 6\times 7}{100}\right)$

$\Rightarrow P=\left(\dfrac{2100\times 100}{6\times 7}\right)$

$\Rightarrow P=\left(\dfrac{2100\times 100}{42}\right)$

$\Rightarrow P=50\times 100$

$\therefore\ P=Rs.\ 5000$

Hence, the principal amount was

$Rs.\ 5000$ .

The sum of money at compound interest amounts to

$Rs. 10580$ in

$2$ years and to

$Rs. 12167$ in

$3$ years. The rate of interest per annum is

0%
$12\%$
0%
$14\%$
0%
$15\%$
0%
$16\dfrac{2}{3}$ %

Explanation

Amount in

$2$ years =

$Rs.10580$

Amount in

$3$ years =

$Rs.12167$

Principal of

$3rd$ year=Amount of

$2nd$ year=

$Rs.10580$

So, Interest on

$Rs. 10580$ for

$1$ year

$= Rs. (12167 - 10580)$

$= Rs. 1587$

$\therefore Rate = (\dfrac{100\times Interest}{principal\times time})\%=(\cfrac{100\times 1587}{10580\times1})\% = 15\%$

A sum of money becomes Rs. 13380 after 3 years and Rs. 20070 after 6 years on compound interest. The sum is

0%
Rs. 8800
0%
Rs. 8890
0%
Rs. 8920
0%
Rs. 9040

Explanation

Let the sum be Rs.

$x$ . Then ...(i)

$x (1+\cfrac{R}{100})^3 = 13380$
and

$x (1+\cfrac{R}{100})^6 = 20070$ ...(ii)
Dividing equation (ii) by (i), we get

$(1+\cfrac{R}{100})^3 = (\cfrac{200070}{13380}) = (\cfrac{3}{2})$

$\therefore x\cfrac{3}{2} = 13380$

$\Rightarrow x = (13380\times \cfrac{2}{3}) = 8920$
Hence, the sum is Rs. 8920

The cost of a scooter depreciates every year by 15% of its value at beginning of the year. If the present cost of the scooter is Rs. 8,000; find its cost (in Rs.) after 2 years.

0%
5780
0%
6690
0%
7200
0%
7530

Explanation

Cost after depreciation in 'n' years

$= \text{Original cost} \times {(1-\dfrac {R}{100})}^{n}$

So after two years, cost

$= Rs 8,000 \times {(1-\dfrac {15}{100})}^{2} = 8,000 \times {(\dfrac {85}{100})}^{2} = Rs 5,780$

A sum of money placed at compound interest doubles itself in

$5$ years. It will amount to eight times itself in

0%
$10$ years
0%
$12$ years
0%
$15$ years
0%
$20$ years

Explanation

Use formula for compound ineterest

$P\left(1+\cfrac{R}{100}\right)^5 = 2P$

$\Rightarrow \left(1+\cfrac{R}{100}\right)^5 = 2$

Let in

$n$ years the sum would be eight times

$\Rightarrow P \left(1+\cfrac{R}{100}\right)^n = 8P$

$\Rightarrow \left(1+\cfrac{R}{100}\right)^n = 8 \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2^3$

$=$

$\left(\left(1+\cfrac{R}{100}\right)^5\right)^3$

$= \left(1+\cfrac{R}{100}\right)^{15}$

$\therefore n = 15$ years

A tree increases annually by one-fourth of its height. What will be it's height after

$2$ years, if it stands

$64$ cm high today?

0%
$76$ cm
0%
$80$ cm
0%
$100$ cm
0%
$105$ cm

Explanation

Initial height

$=64\ cm$

Every year it increases one-fourth of it's height.

Increase

$\%= (\cfrac{1}{4}\times 100\%)=25\%$

Height after

$2$ years

$= 64\times \left(1+\cfrac{25}{100}\right)^2\ cm$

$= 64\times \left(1+\cfrac{1}{4}\right)^2\ cm$

$= 64\times \left(\cfrac{5}{4}\right)^2\ cm$

$= (64\times \cfrac{5}{4}\times \cfrac{5}{4})\ cm$

$= 100\ cm$

Hence, the height of the tree after

$2$ years will be

$100\ cm$

The current population of a town is 10,If the population increases by

$10\%$ every year, then the population of the town after three years will be

0%
13,310
0%
13,500
0%
14,000
0%
14,500

Explanation

$A=P(1+\cfrac{R}{100})^n$

$A=10000(1+0.1)^3=13310$

If Rs.3,750 amount to Rs.4,620 in 3 years at simple interest, then find the rate of interest.

0%
$\displaystyle 7\frac{11}{15}$ %
0%
$\displaystyle 8\frac{11}{15}$ %
0%
$\displaystyle 6\frac{11}{15}$ %
0%
$\displaystyle 5\frac{11}{15}$ %

Explanation

Simple Interest

$I = \cfrac {PNR}{100}$
Given,

Amount

$A = Rs. 4620$

$P = Rs. 3750$

$I = A - P = 870$

$N = 3$

So,

$870 = \cfrac {3750 \times 3 \times R}{100}$

$=> R = 7 \cfrac {11}{15} \%$

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 12000, the compound interest on the same sum for the same period at the same rate, is

0%
Rs. 12600
0%
Rs. 12610
0%
Rs. 12640
0%
Rs. 12650

Explanation

Clearly, Rate =

$5\; \%$ , Time =

$3$ years and S.I. =

$\text{Rs.}\;12000$

Sum

$= \text{Rs.} \left (\cfrac{100\times 12000}{3\times 5}\right ) = \text{Rs.}\; 80000$
Amount

$= \text{Rs. } \left [80000\times \left (1+\cfrac{5}{100}\right )^3\right ]$

$= \text{Rs. } 92610$

We know that C.I. = Amount - Sum
So,

C.I.

$= 92610-80000$

$= \text{Rs. } 12610$

Which of the following asset does not depreciate?

0%
Machinery and equipment
0%
Motor Vehicles
0%
Land
0%
Furniture

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. Find the sum.

0%
Rs. $698$
0%
Rs. $798$
0%
Rs. $658$
0%
Rs. $758$

Explanation

$\textbf{Step 1: Find the simple interest in one year and 3 years.}$

$\text{Given,}$

$\text{Sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years.}$

$\text{So, Amount after 4 years= Amount after 3 years }+\text{ Simple interest in 1 year}$

$\therefore \text{Simple interest in one year}=854-815$

$=\text{Rs. } 39$

$\text{Simple interest for 3 years}=39\times 3$

$=\text{Rs. }117$

$\textbf{Step 2: Find the principle value.}$

$\text{Principle}=815-117$

$[\because\textbf{Principle}=\textbf{Amount}-\textbf{Interest}]$

$=\text{Rs. }698$

$\textbf{Hence, Option A is correct.}$

State true or false:

$\bigtriangleup ABC$ ,

$AD$ is the median and

$DE$ is parallel to

$BA$ , where

$E$ is a point in

$AC$ and hence

$BE$ is parallel to

$BC$ .

0%
True
0%
False

Explanation

Given:

$AD$ is the median of

$\triangle ABC$ .
Hence,

$BD = DC$ .
Also,

$DE \parallel AB$ and

$DE$ is drawn from the mid point of

$BC$ , i.e.

$D$ .

Thus, by converse of mid-point theorem,

$DE$ bisects the third side, which is

$AC$ .
Then,

$E$ is the mid point of

$AC$ .

Hence,

$BE$ is the median of

$\triangle ABC$ .

That is, the given statement is false and option

$B$ is correct.

Calculate the compound interest(I) on :

Rs.

$8,000$ in

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

$15\%$ per annum.

0%
$8000$
0%
$11000$
0%
$11373.50$
0%
$3373.50$

Explanation

$Amount =8000\times \left(1+\cfrac{15}{100} \right)^2\left(1+\cfrac{7.5}{100} \right)$

$\Rightarrow 8000\times \dfrac{115}{100}\times \dfrac{115}{100}\times \dfrac{107.5}{100}$

$\Rightarrow \dfrac{11373500}{1000}\Rightarrow 11373.50 Rs.$

$C.I=11373.50-8000=3373.50 Rs.$

A man borrows

$Rs.\ 1,200$ at rate

$10$ percent per annum compound interest. If he repays

$Rs.\ 250$ at the end of each year, find the amount outstanding at the beginning of the fourth year.

0%
$Rs.\ 769.70$
0%
$Rs. 796$
0%
$Rs. 799.70$
0%
$Rs.\ 777.70$

Explanation

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

$A=1200\left ( 1+\dfrac{10}{100} \right )\Rightarrow 1200\times \dfrac{110}{100}\Rightarrow 1320$
According to question, borrower paid

$Rs\ 250$ per year
Then borrowed amount for second year is

$Rs.\ 1320-250=Rs.\ 1070$

$A=1070\left ( 1+\dfrac{10}{100} \right )\Rightarrow 1070\times \dfrac{110}{100}\Rightarrow 1177$
Then borrowed amount for third year is

$Rs.\ 1177-250=Rs.\ 927$

$A=927\left ( 1+\dfrac{10}{100} \right )\Rightarrow 927\times \dfrac{110}{100}\Rightarrow Rs.\ 1019.70$
Borrowed amount for fourth year is

$Rs.\ 1019.70-250=Rs.\ 769.70$

Hence, option A.

Calculate the amount and the compound interest on :

Rs.

$3,500$ at

$10\%$ per annum in

$2$ years

0%
Rs. $4,935$ and Rs. $529$ .
0%
Rs. $4,235$ and Rs. $735$ .
0%
Rs. $3,506$ and Rs. $825$ .
0%
Rs. $3,120$ and Rs. $900$ .

Explanation

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

$P=3500, R=10\%, n=2$ years

$A=3500\left(1+\dfrac{10}{100} \right)^2$

$A=3500\times \dfrac{110}{100}\times \dfrac{110}{100}$

$A=35\times 121=4235 Rs.$

$C.I=A-P=4235-3500=735 Rs.$

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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