Basics of Financial Mathematics Test 23

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Basics of Financial Mathematics Test 23

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Question 1
1 / -0

At what rate percent per annum will a sum of Rs. 2000 amount to Rs. 2205 in 2 years at compound interest?

A
$$3$$%

B
$$2$$%

C
$$5$$%

D
$$4$$%

Solution

Given that, a sum of $$Rs.\ 2000$$ amounts to $$Rs.\ 2205$$ in $$2$$ years.
To find out: The rate of interest

For compound interest, we know that,
$$A=P\left(1+\dfrac{R}{100}\right)^T$$

Here, $$P=Rs.\ 2000,\ A=Rs.\ 2205$$ and $$T=2\,$$years.

$$\therefore \ 2205=2000\times \left(1+\dfrac{R}{100}\right)^2$$

$$\Rightarrow$$ $$\dfrac{441}{400}=\left(1+\dfrac{R}{100}\right)^2$$

$$\Rightarrow$$ $$\dfrac{21}{20}=\left(1+\dfrac{R}{100}\right)$$

$$\Rightarrow$$ $$R=\dfrac{1}{20}\times 100=5\%$$

Hence, the required rate of interest is $$5\%$$.
Question 2
1 / -0

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 % ?

A
$$10$$%

B
$$12.5$$%

C
$$20$$%

D
$$22.5$$%

Solution

$$\Rightarrow$$ First two years compound interest $$=P\left[\left(1+\dfrac{R}{100}\right)^2-1\right]=272$$ ..... $$(1)$$

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

$$\Rightarrow$$ $$136x^2-81x-81=0$$
On solving we get $$x=\dfrac{9}{8}$$
$$\therefore$$ $$1+\dfrac{R}{100}=\dfrac{9}{8}$$

$$\therefore$$ $$R=12.5\%$$
Question 3
1 / -0

$$A$$ borrowed Rs. $$2,500$$ from $$B$$ at $$12$$% per annum compound interest. After $$2$$ years, $$A$$ gave Rs. $$2,936$$ and a watch to $$B$$ to clear the account. Find the cost of the watch.

A
Rs. $$100$$

B
Rs. $$200$$

C
Rs. $$300$$

D
Rs. $$400$$

Solution

$$\Rightarrow$$ $$P=$$Rs. $$2500,\,R=12\%$$ and $$T=2\,$$years
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$

$$\Rightarrow$$ $$A=2500\times (1+\dfrac{12}{100})^2$$

$$\Rightarrow$$ $$A=2500\times \dfrac{28}{25}\times \dfrac{28}{25}$$

$$\Rightarrow$$ $$A=$$Rs. $$3136.$$
$$\Rightarrow$$ Cost of Watch $$= $$Rs. $$(3136-2936) =$$ Rs. $$200$$
Question 4
1 / -0

Calculate the compound interest for the third year on Rs. $$15,000$$ invested for $$5$$ years at $$10\%$$ per annum.

A
Rs. $$1815$$

B
Rs. $$1825$$

C
Rs. $$1845$$

D
Rs. $$1875$$

Solution

Here $$P=$$ Rs. $$15,000,\,R=10\%$$

C.I. for 2 years $$=$$ $$P\times \left (1+\dfrac{R}{100}\right)^2 - P$$ ......(1)

C.I. for 3 years $$=$$ $$P\times \left (1+\dfrac{R}{100}\right)^3-P$$ ......(2)

By subtracting (1) from (2):

C.I. for third year $$=$$ C.I. for three years $$-$$ C.I. for two years $$= P\times \dfrac{R}{100}\times \left (1+\dfrac{R}{100}\right)^2$$

C.I. for third year $$=$$ $$15000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)^2$$

C.I. for third year $$=$$ $$1500\times \dfrac{121}{100}$$

C.I. for third year $$=$$ Rs. $$1815$$
Question 5
1 / -0

Calculate the compound interest for the second year on Rs. $$16,000/-$$ invested for $$3$$ years at $$10$$% per annnum.

A
Rs. $$1760$$

B
Rs. $$1560$$

C
Rs. $$1660$$

D
Rs. $$1860$$

Solution

$$\Rightarrow$$ $$P=Rs.16000,\,R=10\%$$
$$\Rightarrow$$ C.I. for second year = $$P\times \dfrac{R}{100}\times (1+\dfrac{R}{100})$$

$$\Rightarrow$$ = $$16000\times \dfrac {10}{100}\times (1+\dfrac{10}{100})$$

$$\Rightarrow$$ = $$1600\times \dfrac{11}{10}$$

$$\Rightarrow$$ C.I. for second year = $$Rs.1760.$$
Question 6
1 / -0

Calculate the principal when time $$= 10$$ years, interest = Rs. $$3000$$; rate $$= 5\%$$ p.a.

A
Rs. $$5000$$

B
Rs. $$6000$$

C
Rs. $$7000$$

D
Rs. $$8000$$

Solution

simple interest formula is given by:

$$S.I.=\dfrac{P\times R\times T}{100}$$

where $$S.I = $$ simple interest
$$P=$$ principle
$$R=$$ rate of interest
$$T=$$ number of years

Given, $$T=10$$ years, $$S.I=$$ Rs. $$3000$$, $$R=5\%$$

by substituting the values in the formula we get:

$$\Rightarrow$$ $$3000=\dfrac{P\times 5\times 10}{100}$$

$$\Rightarrow$$ $$P=\dfrac{3000\times 100}{5\times 10}=$$ Rs. $$6000$$

Therefore, the principle is Rs. $$6000$$.
Question 7
1 / -0

If the compound interest on sum of Rs. $$29000$$ for $$2$$ years is Rs. $$9352.5$$ Find the rate of interest.

A
$$10$$

B
$$12$$

C
$$15$$

D
$$18$$

Solution

Given:
$$\Rightarrow$$ $$P=29000,\,T=2\,$$years and $$C.I.=$$Rs.$$9352.5$$
$$\Rightarrow$$ $$A=P+C.I.$$
According to the given condition,

$$\Rightarrow$$ $$29000\left(1+\dfrac{R}{100}\right)^T=29000+9352.5$$

$$\Rightarrow$$ $$29000\times \left(1+\dfrac{R}{100}\right)^2=38352.5$$

$$\Rightarrow$$ $$\left(1+\dfrac{R}{100}\right)^2=\dfrac{38352.5}{29000}$$

$$\Rightarrow$$ $$\left(1+\dfrac{R}{100}\right)^2=\dfrac{529}{400}$$

$$\Rightarrow$$ $$1+\dfrac{R}{100}=\dfrac{23}{20}$$

$$\Rightarrow$$ $$\dfrac{R}{100}=\dfrac{23}{20}-1$$

$$\Rightarrow R=\dfrac{3}{20}\times 100\\=15\%$$
Question 8
1 / -0

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?

A
$$10,000$$

B
$$20,000$$

C
$$30,000$$

D
$$40,000$$

Solution

$$C.P$$ of the scooter=Rs.$$18000$$
Gain=$$\dfrac{100}{9}\%$$

$$SP=\dfrac{(100+\dfrac{100}{9})}{100}\times 18000$$

$$=\dfrac{100}{9}\times 180=Rs.20000$$
Question 9
1 / -0

The annual production in cement industry is subject to business cycles. The production increases for two consecutive years consistently by 18?

A
$$24$$ % increase

B
$$25$$ % increase

C
$$42$$ % increase

D
$$45$$ % increase

Solution

After every $$2$$ years $$18$$% increase/year third year $$12$$% decrease
Base year is $$2008$$
$$2008+2=2010$$
In $$2010,$$ effect$$=18+18=36$$
$$2010+1=2011$$
In $$2011$$, effect$$=36-12=24$$
$$2012$$ effect$$=24+18=42$$
$$\therefore$$ In $$2012$$, total effect will be $$42$$% increase.
Question 10
1 / -0

Calculate the compound interest for the final year on Rs. $$15,000$$ invested for $$4$$ years at $$10$$% per annum.

A
$$1815$$

B
$$1996.5$$

C
$$1500$$

D
$$1650$$

Solution

$$\Rightarrow$$ $$P=Rs.15000$$ and $$R=10\%$$.
$$\Rightarrow$$  C.I. for final year = $$P\times R\times (1+\dfrac{R}{100})^3$$

$$\Rightarrow$$  C.I. for final year = $$15000\times \dfrac{10}{100}\times (1+\dfrac{10}{100})^3$$

$$\Rightarrow$$  C.I. for final year = $$1500\times (\dfrac{11}{10})^3$$

$$\Rightarrow$$  C.I. for final year = $$\dfrac{1500\times 1331}{1000}=Rs.1996.5$$

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Basics of Financial Mathematics Test 23

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Basics of Financial Mathematics Test 23

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SHARING IS CARING

Question 1

$$3$$%

$$2$$%

$$5$$%

$$4$$%

Solution

Question 2

$$10$$%

$$12.5$$%

$$20$$%

$$22.5$$%

Solution

Question 3

Rs. $$100$$

Rs. $$200$$

Rs. $$300$$

Rs. $$400$$

Solution

Question 4

Rs. $$1815$$

Rs. $$1825$$

Rs. $$1845$$

Rs. $$1875$$

Solution

Question 5

Rs. $$1760$$

Rs. $$1560$$

Rs. $$1660$$

Rs. $$1860$$

Solution

Question 6

Rs. $$5000$$

Rs. $$6000$$

Rs. $$7000$$

Rs. $$8000$$

Solution

Question 7

$$10$$

$$12$$

$$15$$

$$18$$

Solution

Question 8

$$10,000$$

$$20,000$$

$$30,000$$

$$40,000$$

Solution

Question 9

$$24$$ % increase

$$25$$ % increase

$$42$$ % increase

$$45$$ % increase

Solution

Question 10

$$1815$$

$$1996.5$$

$$1500$$

$$1650$$

Solution

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