Basics of Financial Mathematics Test 16

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Basics of Financial Mathematics Test 16

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Question 1
1 / -0

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.

A
Rs. 1,162

B
Rs. 1,243

C
Rs. 1,408

D
Rs. 1,591

Solution

P $$=Rs.12800$$

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 $$=Rs.14080$$
R $$=10$$%
T $$=1$$ year

then Interest for second year$$=\dfrac{PRT}{100}$$

$$\Rightarrow \dfrac{14080\times 10\times 1}{100}=Rs.1408$$
Question 2
1 / -0

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.

A
$$Rs. 10,000$$

B
$$Rs. 12,000$$

C
$$Rs. 14,000$$

D
$$Rs. 16,000$$

Solution

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
$$\Rightarrow \frac{121}{1000}-\frac{P}{10}=252$$
$$\Rightarrow \frac{121P-100P}{1000}=252$$
$$\Rightarrow \frac{21P}{1000}=252$$
$$\Rightarrow P=\frac{252\times 1000}{21}=Rs. 12000$$
Question 3
1 / -0

Saurabh invests $$Rs.\ 48,000$$ for $$7$$ year at $$10\%$$ per annum compound interest. Calculate the interest for the first year.

A
$$Rs.\ 4,800$$

B
$$Rs.\ 4,900$$

C
$$Rs.\ 5,000$$

D
$$Rs.\ 5,100$$

Solution

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.
Question 4
1 / -0

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.

A
Rs. 2,106

B
Rs. 2,279

C
Rs. 2,506

D
Rs. 2,792

Solution

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 $$Rs\,2,506$$
Question 5
1 / -0

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.

A
3409

B
3333

C
2365

D
8551

Solution

$$\textbf{Step 1: Calculating the cost for the }\mathbf{1^{st}} \textbf{ year}$$
$$\text{Consider the original value to be x,}$$
$$\text{After the first year the value reduces to}$$
$$=x-\dfrac{12}{100}x=x-0.12x=0.88x$$
$$\textbf{Step 2: Calculating the cost for the }\mathbf{2^{nd}}\textbf{ year}$$
$$\text{The value reduces to}$$
$$=0.88x-\dfrac{12}{100}0.88x=0.88x\times(1-0.12)=0.7744x$$
$$\implies 0.7744x=2640$$
$$\implies x=3409$$
$$\textbf{Hence, The original value is Rs.3409}$$
Question 6
1 / -0

Directions For Questions

Saurabh invests Rs. $$48,000$$ for $$7$$ year at $$10\%$$ per annum compound interest.

...view full instructions

Calculate the amount at the end of the second year.

A
Rs. $$42,856$$

B
Rs. $$48,140$$

C
Rs. $$52,937$$

D
Rs. $$58,080$$

Solution

Given,
Sum $$(P)=Rs.48000$$
Rate $$=10\%$$
Time $$=2\ year$$
$$Amount=P\left(1+\dfrac{R}{100}\right)^t$$
$$= 48000\left(1+\dfrac{10}{100}\right)^2$$
$$= 48000\times \dfrac{110}{100}\times \dfrac{110}{100}\\=Rs.58080$$
Question 7
1 / -0

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 ?

A
$$56000$$

B
$$70000$$

C
$$40000$$

D
$$60000$$

E
None of these

Solution

$$\Rightarrow$$ 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 $$70000$$
Question 8
1 / -0

Naman purchased an old bike for Rs. $$20000$$. If the cost of his bike is depreciated at a rate of $$5\%$$ per annum, then find the cost of the bike after $$2$$ years?

A
Rs. $$18050$$

B
Rs. $$15000$$

C
Rs. $$1900$$

D
Rs. $$18000$$

Solution

Initial price of bike is Rs. $$20000$$, time $$=2$$ years, rate od depreciation $$=5\%$$
Cost of bike after $$2$$ years $$=$$ initial price of bike $$\times\left (1-\dfrac {\text{rate of depreciation} }{100}\right)^{\text{time}}$$
Therefore, cost of bike after $$2$$ years $$=20000\left (1-\dfrac {5}{100}\right)^2$$
$$=20000\left (\dfrac {19}{20}\right)^2$$
$$=18050$$
Thus, cost of bike after $$2$$ years is Rs. $$18050$$.
Question 9
1 / -0

Directions For Questions

The cost of a machine depreciated by Rs. 4,752 during the second year and by Rs. 4,181.76 during the third year. Calculate:

...view full instructions

The rate of depreciation

A
10%

B
12%

C
14%

D
16%

Solution

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.
Question 10
1 / -0

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?

A
$$10350$$

B
$$10395$$

C
$$10500$$

D
$$11500$$

E
$$12705$$

Solution

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. $$10,395$$.

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Basics of Financial Mathematics Test 16

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Basics of Financial Mathematics Test 16

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Question 1

Rs. 1,162

Rs. 1,243

Rs. 1,408

Rs. 1,591

Solution

Question 2

$$Rs. 10,000$$

$$Rs. 12,000$$

$$Rs. 14,000$$

$$Rs. 16,000$$

Solution

Question 3

$$Rs.\ 4,800$$

$$Rs.\ 4,900$$

$$Rs.\ 5,000$$

$$Rs.\ 5,100$$

Solution

Question 4

Rs. 2,106

Rs. 2,279

Rs. 2,506

Rs. 2,792

Solution

Question 5

3409

3333

2365

8551

Solution

Question 6

Rs. $$42,856$$

Rs. $$48,140$$

Rs. $$52,937$$

Rs. $$58,080$$

Solution

Question 7

$$56000$$

$$70000$$

$$40000$$

$$60000$$

None of these

Solution

Question 8

Rs. $$18050$$

Rs. $$15000$$

Rs. $$1900$$

Rs. $$18000$$

Solution

Question 9

10%

12%

14%

16%

Solution

Question 10

$$10350$$

$$10395$$

$$10500$$

$$11500$$

$$12705$$

Solution

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