Basics of Financial Mathematics Test 24

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Basics of Financial Mathematics Test 24

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Weekly Quiz Competition

Question 1
1 / -0

In what time will the amount Rs $$2,000$$ will turn to Rs $$2,420$$ compounded yearly?

A
$$1$$year

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

C
$$2$$years

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

Solution

$$\Rightarrow$$ $$A=$$Rs.$$2420,\,P=$$Rs.$$2000$$ and Let $$R=10\%$$
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$

$$\Rightarrow$$ $$2420=2000\times (1+\dfrac{10}{100})^T$$

$$\Rightarrow$$ $$\dfrac{2420}{2000}=(\dfrac{11}{10})^T$$

$$\Rightarrow$$ $$\dfrac{121}{100}=(\dfrac{11}{10})^T$$

$$\Rightarrow$$ $$(\dfrac{11}{10})^2=(\dfrac{11}{10})^T$$

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

Ekansh borrowed Rs.$$50,000$$ for $$2$$ years at $$20$$ % per year compound interest. Calculate the final amount at the end of the second year.

A
$$70,000$$

B
$$71,000$$

C
$$72,000$$

D
$$75,000$$

Solution

$$\Rightarrow$$ Here, $$P=$$Rs.50,000,\,T=2\,years and $$R=20\%$$.
$$\Rightarrow$$ $$A=P(1+\dfrac{R}{100})^T$$

$$\rightarrow$$ $$A=50,000\times (1+\dfrac{20}{100})^2$$

$$\Rightarrow$$ $$A=50000\times \dfrac{36}{25}$$

$$\therefore A=$$Rs.$$72,000.$$
Question 3
1 / -0

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

A
$$1815$$

B
$$1996.5$$

C
$$1500$$

D
$$1650$$

Solution

Here $$P=$$ Rs. $$15000$$ and $$R=10\%$$
C.I. for second year $$=$$ $$P\times R\times \left (1+\dfrac{R}{100}\right)$$
$$=$$ $$15000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)$$
$$=$$ $$1500\times \dfrac{11}{10}$$
Therefore, C.I. for second year $$=$$ Rs. $$1650.$$
Question 4
1 / -0

Principal $$= 2500$$, rate $$= 6$$%, time$$= 4$$ years. Calculate the interest.

A
Rs.$$500$$

B
Rs.$$600$$

C
Rs.$$300$$

D
Rs.$$400$$

Solution

Simple interest $$=\dfrac{P \times R \times T}{100}$$
$$=\dfrac{2500 \times 6 \times 4}{100} =600$$
Therefor the interest is $$Rs \ 600$$
Question 5
1 / -0

Find the compound interest approximately rate applied to a principal over $$5$$ years if the total interest paid equals the borrowed principal.

A
$$14$$%

B
$$24$$%

C
$$34$$%

D
$$7$$%

Solution

We know,
$$CI=P(1+\cfrac{r}{100})^n-P$$
$$CI=P$$ ....... given,

solving the above equations
$$\left(1+\cfrac{r}{100}\right)^5=2$$

$$r=(2^{\tfrac 1 5}-1)\times 100$$

$$\implies r=14.86\%\approx 14\%$$
Question 6
1 / -0

A sum of money at simple interest amounts to Rs. $$600$$ in $$2$$ years and to Rs. $$800$$ in $$4$$ years. The sum is:

A
$$100$$

B
$$200$$

C
$$300$$

D
$$400$$
Question 7
1 / -0

Principal + Amount - $$2 \times $$ principle =

A
Principle

B
Interest

C
Amount

D
Rate

Solution

Principal + Amount - $$2 \times $$ principle
$$= \,P+A-2P$$
$$=P+(P+I)-2P$$
$$=P+P+I-2P$$
$$=I$$
Question 8
1 / -0

In what time will Rs. $$1000$$ amount to Rs. $$1210$$ at $$10$$% p.a. in CI?

A
$$2$$

B
$$1$$

C
$$1.5$$

D
$$2.1$$

Solution

$$\Rightarrow$$ $$P=Rs.1000,\,A=Rs.1210$$ and $$R=10\%$$

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

$$\Rightarrow$$ $$1210=1000\times (1+\dfrac{10}{100})^T$$

$$\Rightarrow$$ $$\dfrac{121}{100}=(\dfrac{11}{10})^T$$

$$\Rightarrow$$ $$(\dfrac{11}{10})^2=(\dfrac{11}{10})^T$$

$$\therefore$$ $$T=2\,years$$
Question 9
1 / -0

Jenn borrowed Rs. $$5,000$$ for $$5$$ years and had to pay Rs. $$1,500$$ simple interest at the end of that time. What rate of interest did she pay?

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

$$\Rightarrow$$ $$P=Rs.5000,\,T=5\,years $$ and $$S.I.=Rs.1500$$
$$\Rightarrow$$ $$S.I.=\dfrac{P\times R\times T}{100}$$

$$\Rightarrow$$ $$1500=\dfrac{5000\times R\times 5}{100}$$

$$\Rightarrow$$ $$R=\dfrac{1500}{250}$$

$$\therefore$$ $$R=6\%$$
Question 10
1 / -0

Joshita borrowed Rs. $$3,000$$ for $$3$$ years and had to pay Rs.$$ 1,000$$ simple interest at the end of that time. What rate of interest did she pay?

A
$$11.11$$

B
$$22.22$$

C
$$33.33$$

D
$$44.44$$

Solution

$$\Rightarrow$$ $$P=Rs.3000,\,T=3\,years$$ and $$S.I.=Rs.1000$$.
$$\Rightarrow$$ $$S.I.=\dfrac{P\times R\times T}{100}$$

$$\Rightarrow$$ $$1000=\dfrac{3000\times R\times 3}{100}$$

$$\Rightarrow$$ $$R=\dfrac{1000}{90}$$

$$\therefore$$ $$R=11.11\%$$

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Basics of Financial Mathematics Test 24

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Basics of Financial Mathematics Test 24

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Question 1

$$1$$year

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

$$2$$years

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

Solution

Question 2

$$70,000$$

$$71,000$$

$$72,000$$

$$75,000$$

Solution

Question 3

$$1815$$

$$1996.5$$

$$1500$$

$$1650$$

Solution

Question 4

Rs.$$500$$

Rs.$$600$$

Rs.$$300$$

Rs.$$400$$

Solution

Question 5

$$14$$%

$$24$$%

$$34$$%

$$7$$%

Solution

Question 6

$$100$$

$$200$$

$$300$$

$$400$$

Question 7

Principle

Interest

Amount

Rate

Solution

Question 8

$$2$$

$$1$$

$$1.5$$

$$2.1$$

Solution

Question 9

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 10

$$11.11$$

$$22.22$$

$$33.33$$

$$44.44$$

Solution

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