Basics of Financial Mathematics Test 25

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Basics of Financial Mathematics Test 25

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Question 1
1 / -0

What amount will sum up to Rs. 12100 at 10$$\%$$ per annum, in CI in $$2$$ years?

A
$$Rs.\ 10,000$$

B
$$Rs.\ 11,000$$

C
$$Rs.\ 9,000$$

D
$$Rs.\ 8,000$$

Solution

Given that, $$A=Rs.\ 12100,\,R=10\%$$ and $$T=2\,years$$
To find out: Principal amount, $$P$$.
For compound interest, we know that, $$A=P\left(1+\dfrac{R}{100}\right)^T$$

$$\Rightarrow$$ $$12100=P\times \left(1+\dfrac{10}{100}\right)^2$$

$$\Rightarrow$$ $$12100=P\times \left(\dfrac{11}{10}\right)^2$$

$$\therefore$$ $$P=12100\times \dfrac{100}{121}$$

$$\therefore$$ $$P=Rs.\ 10,000.$$

Hence, $$Rs.\ 10,000$$ will sum up to $$Rs. \ 12,100$$ at $$10\%$$ per annum, in $$2$$ years.
Question 2
1 / -0

Calculate the compound interest for the third year on Rs. $$30,000$$ invested for $$5$$ years at $$10$$% per annum.

A
$$3630$$

B
$$3530$$

C
$$3330$$

D
$$3230$$

Solution

$$\Rightarrow$$ $$P=Rs.30000,\,R=10\%$$
$$\Rightarrow$$ C.P. for third year = $$P\times \dfrac{R}{100}(1+\dfrac{R}{100})^2$$

$$\Rightarrow$$ C.P. for third year = $$30000\times \dfrac{10}{100}\times (1+\dfrac{10}{100})^2$$

$$\Rightarrow$$ C.P. for third year =$$3000\times \dfrac{121}{100}$$

$$\therefore$$ C.P. for third year = $$Rs.3630.$$
Question 3
1 / -0

What amount will sum up to Rs. $$6,655$$ at $$10$$% p.a. in C.I. in $$3$$ years?

A
$$4000$$

B
$$6000$$

C
$$5000$$

D
$$7000$$

Solution

$$\Rightarrow$$ Here, $$A=Rs.6655\,R=10\%$$ and $$T=3years$$
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$

$$\Rightarrow$$ $$6655=P\times (1+\dfrac{10}{100})^3$$

$$\Rightarrow$$ $$6655=P\times (\dfrac{11}{10})^3$$

$$\therefore$$ $$P=\dfrac{6655\times 1000}{1331}=Rs.5000$$
Question 4
1 / -0

In what time will a sum of Rs. $$1600$$ at $$5$$% p.a. $$C.I$$ amounts to Rs. $$1764$$?

A
$$1$$

B
$$1.5$$

C
$$2$$

D
$$3$$

Solution

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

where, $$P= $$ principle
$$A=$$ amount
$$R=$$ rate of interest
$$T=$$ number of years

Given, $$P=Rs.1600\ , R=5\%$$ and $$A=Rs.1764$$

by substituting the values we get ,

$$\Rightarrow$$ $$1764=1600\times (1+\dfrac{5}{100})^T$$

$$\Rightarrow$$ $$\dfrac{1764}{1600}=(\dfrac{21}{20})^T$$

$$\Rightarrow$$ $$\dfrac{441}{400}=(\dfrac{21}{20})^T$$

$$\Rightarrow$$ $$(\dfrac{21}{20})^2=(\dfrac{21}{20})^T$$

$$\therefore$$ $$T=2\,years.$$
Question 5
1 / -0

What sum lent out at CI will amount to Rs. $$1936$$ in $$2$$ years at $$10$$% p.a. interest?

A
$$1500$$

B
$$1600$$

C
$$1700$$

D
$$1800$$

Solution

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

$$\Rightarrow$$ $$1936=P\times (1+\dfrac{10}{100})^2$$

$$\Rightarrow$$ $$1936=P\times (\dfrac{121}{100})$$

$$\Rightarrow$$ $$P=\dfrac{193600}{121}=Rs.1600$$

$$\therefore$$ Sum lent out money is $$Rs.1600.$$
Question 6
1 / -0

Find the compound interest on Rs. $$70,000$$ for $$2$$ years, compounded annually at $$10\%$$ per annum.

A
$$14,400$$

B
$$14,700$$

C
$$14,300$$

D
$$14,100$$

Solution

Here,$$P=$$ Rs $$70000$$ , $$R=10\%\ p.a$$
interest for the first year $$=\cfrac{P\times R\times T}{100}$$
$$=\cfrac{70000\times 10\times 1}{100}$$
$$=$$ Rs $$7000$$
The amount after the first year $$=$$ Rs $$77000$$
Principal for the second year $$=$$ Rs $$77000$$
Interest for the second year $$=$$ Rs $$\cfrac{77000\times 10\times 1}{100}$$
$$=$$ Rs $$7700$$
The final amount $$=$$ Rs $$77000$$ + Rs $$7700$$
$$=$$ Rs $$84700$$
Compound interest $$= $$ Rs $$84700 -70000$$
$$=$$ Rs $$14700$$
Question 7
1 / -0

The amount on Rs. $$20,500$$ at $$7%$$ per annum compunded annually for $$2$$ years, is:

A
$$22470$$

B
$$23470$$

C
$$24470$$

D
$$25470$$

Solution

$$\Rightarrow$$ $$P=Rs.20,500,\,R=7\%$$ and $$T=2\,years$$

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

$$\Rightarrow$$ $$A=20500\times (1+\dfrac{7}{100})^2$$

$$\Rightarrow$$ $$A=20500\times (\dfrac{107}{100})^2$$

$$\Rightarrow$$ $$A=20500\times (1.07)^2$$

$$\Rightarrow$$ $$A=20500\times 1.1449$$

$$\therefore$$ $$A=Rs.23470.$$
Question 8
1 / -0

A sum of Rs $$15,000$$ is invested for $$3$$ years at $$13$$ % per annum compound interest. Calculate the approx interest for the second year.

A
$$2100$$

B
$$2200$$

C
$$2300$$

D
$$2400$$

Solution

Interst for the first year
$$=Rs \cfrac{15000\times 13\times 1}{100}$$
$$=Rs1950$$
Amount after the first year
$$=Rs15000+Rs1950$$
$$Rs16950$$
Interest for the second year
$$=Rs\cfrac{16950\times 13\times 1}{100}$$
$$=Rs2203.5$$
$$=Rs2200$$(approx)
Question 9
1 / -0

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

A
$$1500$$

B
$$1650$$

C
$$1800$$

D
$$2000$$

Solution

$$\Rightarrow$$ $$P=Rs.15,000,\,R=10\%$$
$$\Rightarrow$$ C.I. for second year = $$15000\times \dfrac{10}{100}\times (1+\dfrac{10}{100})$$

$$\Rightarrow$$ C.I. for second year = $$1500\times \dfrac{11}{10}$$

$$\therefore$$ C.I. for second year = $$Rs.1650.$$
Question 10
1 / -0

A certain sum is to be invested for a certain time at Compound interest at $$12$$% p.a., such that there is an increase of $$40.4$$% in the sum. Find the time.

A
$$2.5$$ years

B
$$3$$ years

C
$$3.5$$ years

D
$$2.75$$ years

Solution

$$A=140.4\%\times P=1.404P$$
$$A=P(1+\cfrac{r}{100})^n$$
$$\implies 1.404P=P(1+\cfrac{12}{100})^n\\ \implies n\approx3$$
Time$$=3$$years.

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Basics of Financial Mathematics Test 25

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Basics of Financial Mathematics Test 25

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Question 1

$$Rs.\ 10,000$$

$$Rs.\ 11,000$$

$$Rs.\ 9,000$$

$$Rs.\ 8,000$$

Solution

Question 2

$$3630$$

$$3530$$

$$3330$$

$$3230$$

Solution

Question 3

$$4000$$

$$6000$$

$$5000$$

$$7000$$

Solution

Question 4

$$1$$

$$1.5$$

$$2$$

$$3$$

Solution

Question 5

$$1500$$

$$1600$$

$$1700$$

$$1800$$

Solution

Question 6

$$14,400$$

$$14,700$$

$$14,300$$

$$14,100$$

Solution

Question 7

$$22470$$

$$23470$$

$$24470$$

$$25470$$

Solution

Question 8

$$2100$$

$$2200$$

$$2300$$

$$2400$$

Solution

Question 9

$$1500$$

$$1650$$

$$1800$$

$$2000$$

Solution

Question 10

$$2.5$$ years

$$3$$ years

$$3.5$$ years

$$2.75$$ years

Solution

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