The time for which interest is calculated

The time for which loan is given

Period of conversion of simple interest into compound interest

MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 15

XYZ Ltd. incurs a net loss of Rs. 5, 000 during are accounting years. Depreciation for the relevant year amount to Rs. 1000, preliminary expenses written off during the accounting years is Rs. 3, 000 and a loss of Rs. 4, 000 is due to sale of plant and mach ____________.

0%
Rs. 3, 000
0%
Rs. 2, 000
0%
Rs. 5, 000
0%
Rs. 4, 000

Depreciation means ___________________.

0%
some fixed assets losing price with the passage of time
0%
loss of value by the movable properties over the time of their use
0%
loss in value of the equipment of a plant over time due to their use
0%
gaining of value by a domestic currency in its foreign exchange market

The compound interest on $₹\ 10000$ at $10 \%$ per annum for $3$ years, compounded annually, is

0%
1331
0%
3310
0%
3130
0%
13310

Explanation

Given:

Present value $= ₹\ 10000$

Interest rate $= 10 \%$ per annum

Time

$=3$ years

To find the amount we have the formula,

Amount $(A) = P (1+(r/100))^n$

where

$P$ is present value,

$r$ is rate of interest,

$n$ is time in years

Now substituting the values in above formula we get,

$\therefore A = 10000 (1 +10/100)^3$

$\Rightarrow A = 10000 (11/10)^3$

$\Rightarrow A = 121 (10) (11)$

$\Rightarrow A = 1331 (10)$

$\Rightarrow A = ₹\ 13310$

∴ Compound interest $= A – P$

$= 13310 – 10000= ₹\ 3310$

Radhika bought a car for

$Rs. 2,50,000$ . Next year its price decreased by

$10\%$ and further next year it decreased by

$12\%$ . In the two years overall decrease per cent in the price of the car is

0%
$3.2\%$
0%
$22\%$
0%
$20.8\%$
0%
$8\%$

Explanation

The correct answer is option

$(c)$ .
Cost price of the car

$=Rs. 2,50,000$
It's price decreased next year by

$10\%$
Now, the price

$=Rs.(2,50,000-\dfrac{10}{100}\times 2,50,000)=Rs. 2,25,000$
Next year, further its price decreased by

$12\%$
Now, the price

$=Rs.(2,25,000-2,25,000\times \dfrac{12}{100})=Rs. 1,98,000$
Overall decreased percentage in two years,

$=(\dfrac{2,50,000-1,98,000}{2,50,000}\times 100)\%=(\dfrac{52,000}{2,50,000}\times100)\%=20.8\%$

To calculate the growth of bacteria if the rate of growth is known, the formula for calculation of amount in compound interest can be used.

0%
True
0%
False

The cost of a sewing machine is

$Rs\ 7,000$ . Its value depreciates at

$8\%$\ p.a$ . Then the value of the machine after

$2$ years is

$Rs\ 5,924.80$ .

0%
True
0%
False

Explanation

The given statement is true.

Value of machine after

$1^{st}$ year

$=Rs\ 7000\left(1-\dfrac{8}{100}\right)$

$=Rs\ 7000\left(\dfrac{92}{100}\right)= Rs\ 6,440$

Value of machine after

$2^{nd}$ year

$=Rs\ 6440\left(1-\dfrac{8}{100}\right)$

$=Rs\ 6440\left(\dfrac{92}{100}\right)= Rs\ 5924.80$

The value of a car, bought for

$Rs\ 4,40,000$ depreciates each year by

$10\%$ of its value at the beginning of that year. So its value becomes

$Rs\ 3,08,000$ after three years.

0%
True
0%
False

Explanation

The given statement is false.
As pre given data, the value of the care after depreciation in

$3$ years is:

$A=P\left(1-\dfrac{R}{100}\right)^{T}=440000\left(1-\dfrac{10}{100}\right)^{3}=440\times 729=Rs\ 320760$

Four alternative options are given for each of the following statements.
Select the correct option.
(o) The principal that yields a simple interest of

$Rs. 1280$ at

$16$ % per annum for

$8$ months is

0%
$Rs. 10,000$
0%
$Rs. 12,000$
0%
$Rs. 12,800$
0%
$Rs. 14,000$

How many years will it take for equipment with an initial value of Rs.

$2000$ to reach its scrap value of Rs.

$1000$ if depreciation is reducing-balance at

$15\%$ p.a. of the current value?

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

Explanation

New Value

$=1000$

Rate

$=15\%p.a.$

Original value

$=2000$

Time

$=t$

$2000-\cfrac{2000\times 15\times t}{100}=1000$

On solving , we get

$t\approx 3.34\,years$

Therefore, we can say that

$t=3$ years.

The clan of Rs. 50,000 at 10% p.a. Compounded annually for a certain time period is Rs. 10,What is the time period?

0%
2 years
0%
5 years
0%
7 years
0%
4 years

A man gave

$50\%$ of his savings of

$84100$ to his wife and divided the remaining sum among his two sons

$A$ and

$B$ of

$15$ and

$13$ years of age respectively. He divided it in such a way that each of his sons, when they attain the age of

$20$ years would receive the same amount at

$5\%$ compound interest per annum, the share of

$B$ was -

0%
$20000$
0%
$20500$
0%
$22000$
0%
$22500$

Explanation

Total savings

$=$ Rs.

$84100$
Wife got

$= 50\%$ of

$84100$

$=$ Rs.

$42050$
Let

$B$ gets

$X$ , then

$A$ will have

$( 42050 - X)$
For

$A$ ,
Rate

$= 5\%$
Time (n) = 5 years
Amount

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

$= (42050 - X) \times \left [1 + (\dfrac{5}{100})\right]^5$
Amount

$= (42050 - X) \times 1.27$ .........(i)
For

$B$ ,
Time

$= 7$ years
Amount

$= X \left [1 + \left (\dfrac{5}{100}\right)\right]^7$
Amount

$= 1.40X$ ...........(ii)
According to question, we have
Amount of

$A =$ Amount of

$B$

$(42050 - X)\times 1.27 = 1.40X$

$53403 - 1.27X = 1.40X$

$1.40X + 1.27X = 53403$

$2.67X = 53403$

$X = 20000$
So,

$B$ got Rs.

$20000$ .

A person borrowed Rs.

$7,500$ at

$16$ % compound interest. How much does he have to pay at the end of two years to clear the loan?

0%
Rs. $9,900$
0%
Rs. $10,092$
0%
Rs. $11,000$
0%
Rs. $11,052$

A biology class at Central High School predicted that a local population of animals will double in size every 12 years. The population at the beginning of 2014 was estimated to be 50 animals. If P represents the population n years after 2014, then which of the following equations represents the classs model of the population over time?

0%
$\displaystyle P=12+50n$
0%
$\displaystyle P=50+12n$
0%
$\displaystyle P={ 50\left( 2 \right) }^{ 12n }$
0%
$\displaystyle P={ 50\left( 2 \right) }^{ \frac { n }{ 12 } }$

A person invested part of Rs.

$45000$ at

$4$ % and the rest at

$6$ %. If his annual income from both are equal, then what is the average cost of interest?

0%
$4.6$ %
0%
$4.8$ %
0%
$5.0$ %
0%
$5.2$ %

The principal which yields a simple interest of

$Rs. 90$ at

$6$ % per annum in

$3$ years is

0%
$Rs. 270$
0%
$Rs. 500$
0%
$Rs. 540$
0%
$Rs. 720$

The population of a State increased from

$100$ million to

$169$ million in two decades. What is the average increase in population per decade?

0%
$20$ %
0%
$34.5$ %
0%
$69$ %
0%
$30$ %

Depreciation, as the term is used in accounting, means ___________.

0%
Physical deterioration of a fixed asset
0%
Decline in the market value of an asset
0%
Allocation of the cost of fixed asset over its useful life
0%
A business expense for acquiring asset

The population of a City decreases by 3% of the initial at the beginning of every year. If the present population is 1,25,000, what will be after 2 years?

0%
1,17,600
0%
1,17,610
0%
17,610
0%
1,17,613

A town's population increased by

$1200$ people, and then this new population decreased

$11$ %. The town now had

$32$ less people than it did before the

$1200$ increase. Find the original population.

0%
$10000$
0%
$11000$
0%
$12000$
0%
$13000$

Explanation

Let the population be $P$ .

So, the increased population will be $P + 1200$ .

The population after $11\%$ decrease in population becomes $P - 32$ .

Therefore,

${{P}} - 32 = \left( {{{P}} + 1200} \right) - \dfrac{{11}}{{100}}\left( {{{P}} + 1200} \right)$

${{P}} = 10000$

So, the total population is $10000$ .

Depreciation is a measure of the wearing out, consumption or other loss of value of a depreciable asset arising from -
I. Use
II. Effluxion of time
III. Obsolescence through technology and market changes
Select the correct answer from the options given below

0%
I and II.
0%
II and III.
0%
I and III.
0%
Any of the above.

The difference between

$C.I.$ (Compound Interest) and

$S.I.$ (Simple Interest) on a sum of Rs.

$4,000$ for

$2$ years at

$5$ % p.a. payable is:

0%
Rs. $20$
0%
Rs. $10$
0%
Rs. $50$
0%
Rs. $60$

Explanation

For the first year, Simple Interest = Compound Interest

$=\dfrac{PTR}{100}$

$I_1=\dfrac{4000\times 1\times 5}{100}= 200$
For the second year, simple interest remains same and compund interest will be the simple interest on new amount

$=4000+200=4200$
Compound Interest for second year

$= \dfrac{4200\times 1 \times 5}{100}= 210$
Simple Interest for second year

$=I_1 =200$
Difference

$220-210 = Rs.10$

Which of these is not an accepted method of depreciation

0%
Straight line method
0%
Sinking fund method
0%
Written down method
0%
Market value

The SI on a sum of money for 3 years is Rs.

$240$ and the CI on the sum at same rate for

$2$ years is Rs.

$170$ . The rate % p.a. is:

0%
$16 \%$
0%
$12 \displaystyle \frac{1}{2} \%$
0%
$8 \%$
0%
$8 \displaystyle \frac{1}{3} \%$

Conversion period is

0%
The time for which loan is given
0%
The time for which interest is calculated
0%
Period of conversion of simple interest into compound interest
0%
None of the above

Interest rate per conversion period is calculated by

0%
Rate of interest
0%
Conversion period/annual interest rate
0%
Annual interest rate/conversion period
0%
None of the above

How many years will it take to pay of a Rs.

$11,000$ loan with a Rs.

$1241.08$ annual payment and a

$5\%$ interest rate?

0%
$6$ years
0%
$12$ years
0%
$24$ years
0%
$48$ years

Which one of the following terms refer to the risk arises for bond owners from fluctuating interest rates?

0%
Fluctuation Risk
0%
Interest rate Risk
0%
Real-Time Risk
0%
Inflation Risk

If you plan to save Rs.

$5,000$ with a bank an interest rate of

$8\%$ what will be the worth of your amount after

$4$ years if interest is compounded annually?

0%
$Rs.5,400$
0%
$Rs.5,900$
0%
$Rs.6,600$
0%
$Rs.6,802$

The allocation of the cost of a tangible plant asset to expense in the periods, in which services are received from the asset, is termed as _______.

0%
Appreciation
0%
Depreciation
0%
Fluctuation
0%
None of these

Rate of interest is decided by

0%
Borrower
0%
Lender
0%
Bank
0%
Both a & b

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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