Basics of Financial Mathematics Test 38

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Basics of Financial Mathematics Test 38

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Question 1
1 / -0

Useful life of fixed asset is _________.

A
The period over which a depreciable asset expected to be used by the enterprises, or

B
The number of production or similar units expected to be obtained from the use of the asset by the enterprise

C
(A) or (B)

D
None of the above

Solution

Useful life of an asset is the estimated economic or commercial life of the asset. Physical life is not important for this purpose because an asset may still exist physically but may not be capable of commercially viable production.

In the allocation of depreciable amount is the ‘expected useful life’ of an asset. It has been described as “either (i) the period over which a depreciable asset is expected to the used by the enterprise, or (ii) the number of production of similar units expected to be obtained from the use of the asset by the enterprise.”
Question 2
1 / -0

John borrowed Rs.1,00,000 at 10 percent per annum simple interest. He immediately lent the whole sum at 10 percent per annum compound interest. At the end of 2 years, he would gain -

A
Rs.1,000

B
Rs.1,800

C
Rs.2,000

D
Rs.100.

Solution

Principal Amount borrowed $$=Rs$$ $$1,00,000$$.
Money is borrowed at S.I.
$$T=2$$ years
$$R=10\%$$
In simple interest, we know that the amount $$(A)=P\left( 1+\dfrac { R\times T }{ 100 } \right) $$
$$\therefore $$ Amount after two years $$=1,00,000\left( 1+\dfrac { 2\times 10 }{ 100 } \right) $$
$$=1,00,000\times \dfrac { 6 }{ 5 } $$
$$=1,20,000$$
The borrowed money is lent by John on C.I. at same rate for same time.
In CI, we know that the amount $$(A)=P{ \left( 1+\dfrac { R }{ 100 } \right) }^{ T }$$

$$\therefore $$ Amount that John will receive $$=1,00,000{ \left( 1+\dfrac { 10 }{ 100 } \right) }^{ 2 }$$
$$=1,00,000{ \left( \dfrac { 11 }{ 10 } \right) }^{ 2 }$$
$$=1,21,000$$

$$\therefore $$ John will get $$1,21,000$$ while be will have to pay $$1,20,000$$
$$\therefore $$ Gain $$=Rs$$ $$(1,21,000-1,20,000)$$
$$=Rs$$ $$1,000$$

Hence, John will gain $$Rs.\ 1000$$.
Question 3
1 / -0

Depreciation is the process of:

A
Asset Valuation

B
Verification of assets

C
Allocation of cost of assets to the period of its life

D
Decreasing the value of asset

Solution

Depreciation accounting is a system of accounting which aims to distribute the cost of tangible capital assets, less salvage value (if any) over the estimated useful life of the unit in a systematic and rational manner. It is a process of allocation, not of valuation
Question 4
1 / -0

A sum of money invested at 14$$\%$$ per annum amounted to Rs. 10400 in 5 years. Calculate the principal.

A
$$Rs.\ 6067.34$$

B
$$Rs.\ 6278.24$$

C
$$Rs.\ 6384.62$$

D
$$Rs.\ 6117.65$$

Solution

We have the formula for amount in simple interest as:

$$A=P(1+rt)$$

Given, $$A=10400$$, $$t=5$$, $$r=14\%$$

$$\therefore \ 10400=P(1+5(0.14))$$

$$10400=P(1+0.7)=P(1.7)$$

$$P=\dfrac{10400}{1.7}$$

$$=6117.647.6$$

Hence, the principal amount is Rs. $$6117.65$$.
Question 5
1 / -0

Which of the following is correct?

A
Depreciation is not the process of valuation of assets- it is the process of allocation of costs to the period of its economic life

B
If cost of machinery is more than book value there is no need of providing depreciation

C
If plant and machinery are in good condition, through repairs etc., there is no need to provide depreciation

D
All the above

Solution

The purpose of recording depreciation as an expense is to spread the initial price of the asset over its useful life. For intangible assets - such as brands and intellectual property - this process of allocating costs over time is called amortization.
Question 6
1 / -0

A TV was bought at a price of Rs. 21,000. After one year the value of the TV was depreciated by 5%. (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year.

A
19,750

B
17,950

C
19,950

D
19,000

Solution

$$\textbf{Step -1:Calculate the value to be deduced}.$$
$$\text{We are given that after one year, the value of the TV is depreciated by}$$ $$5\%.$$
$$\text{Value to be deduced =}$$$$5$$$$\text{% of}$$ $$Rs.21,000$$
$$=Rs.\dfrac{21000\times 5}{100}$$
$$=Rs.\: 1050$$
$$\textbf{Step -2:Calculate the value of the TV after one year}.$$
$$\text{Value after one year=}$$$$Rs.(21000-1050)$$
$$=Rs.\: 19,950$$

$$\textbf{Hence, the correct option is C.}$$
Question 7
1 / -0

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.

A
$$12020$$

B
$$9616$$

C
$$6154.24$$

D
$$4923.39$$

Solution

Value after 5 years $$=15025\times(1-(\dfrac{20}{100})^5)$$
$$=15025\times\>0.32768=Rs.4923.39$$
Question 8
1 / -0

The time in which $$Rs.1800$$ amount to $$Rs.2178$$ at $$10 \%$$ per annum, compounded annually is

A
$$3$$ years

B
$$2$$ years

C
$$4$$ years

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

Solution

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
Question 9
1 / -0

Find the compound interest to the nearest rupee on Rs. $$7500$$ for $$2$$ years $$4$$ months at $$12\%$$ per annum reckoned annually.

A
Rs. $$2240$$

B
Rs. $$2250$$

C
Rs. $$2284$$

D
Rs. $$2276$$

Solution

$$P=7500$$,
$$r=12$$,
$$n=2\tfrac{4}{12}=\cfrac{7}{3}$$
$$CI=P[(1+\cfrac{r}{100})^n-1]$$
$$CI=Rs.7500[(1+\cfrac{12}{100})^{7/3}-1]$$
$$=2270.19$$
Value to the nearest integer$$=Rs.2276$$.
Question 10
1 / -0

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.

A
$$Rs\,\,1213.154$$

B
$$Rs\,\,12000.154$$

C
$$Rs\,\,12913.154$$

D
None of these

Solution

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.

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Basics of Financial Mathematics Test 38

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Question 1

The period over which a depreciable asset expected to be used by the enterprises, or

The number of production or similar units expected to be obtained from the use of the asset by the enterprise

(A) or (B)

None of the above

Solution

Question 2

Rs.1,000

Rs.1,800

Rs.2,000

Rs.100.

Solution

Question 3

Asset Valuation

Verification of assets

Allocation of cost of assets to the period of its life

Decreasing the value of asset

Solution

Question 4

$$Rs.\ 6067.34$$

$$Rs.\ 6278.24$$

$$Rs.\ 6384.62$$

$$Rs.\ 6117.65$$

Solution

Question 5

Depreciation is not the process of valuation of assets- it is the process of allocation of costs to the period of its economic life

If cost of machinery is more than book value there is no need of providing depreciation

If plant and machinery are in good condition, through repairs etc., there is no need to provide depreciation

All the above

Solution

Question 6

19,750

17,950

19,950

19,000

Solution

Question 7

$$12020$$

$$9616$$

$$6154.24$$

$$4923.39$$

Solution

Question 8

$$3$$ years

$$2$$ years

$$4$$ years

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

Solution

Question 9

Rs. $$2240$$

Rs. $$2250$$

Rs. $$2284$$

Rs. $$2276$$

Solution

Question 10

$$Rs\,\,1213.154$$

$$Rs\,\,12000.154$$

$$Rs\,\,12913.154$$

None of these

Solution

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