Basics of Financial Mathematics Test 40

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Basics of Financial Mathematics Test 40

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Question 1
1 / -0

Amit invested an amount of Rs.25000 in fixed deposit @C.I. 8% per annum for 2 years. What amount Amit will get?

A
Rs. 28240

B
Rs. 28540

C
Rs. 29240

D
Rs. 29160

Solution
Question 2
1 / -0

What is the compound interest on Rs 6, 250 at 4% p. a compoundded annually for 3 years ?

A
Rs. 780.00

B
Rs. 789.00

C
Rs. 788.80

D
780.40

Solution

\begin{array}{l} A=P{ \left( { 1+\frac { r }{ { 100 } } } \right) ^{ n } } \\ A=6250{ \left( { 1+\frac { 4 }{ { 100 } } } \right) ^{ 3 } } \\ A=6250{ \left( { \frac { { 104 } }{ { 100 } } } \right) ^{ 3 } } \\ A=6250{ \left( { 1.04 } \right) ^{ 3 } } \\ A=7030.40 \end{array}

Hence,
Compound Interest $$=$$ $$7030.40-6250$$
$$ = Rs. 780.40 $$
Question 3
1 / -0

The compound interest on $$Rs.5,000$$ at $$4\%$$ per annum for $$2$$ years compounded annually will be

A
$$Rs.806$$

B
$$Rs.708$$

C
$$Rs.5,408$$

D
$$Rs.408$$

Solution

Given, $$P=5000$$
$$R=4\%$$
$$T=2\ years$$

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

$$A=5000\Big(1+\dfrac{4}{100}\Big)^2$$

$$=5000\times\Big(\dfrac{104}{100}\Big)^2$$

$$=Rs. 5408$$

Compound interest $$=A-P$$

$$=5408-5000=Rs. 408$$
Question 4
1 / -0

The compound interest on Rs 4000 at 10% per annum for 2 years 3 months, compounded annually is

A
Rs$$916$$

B
Rs $$900$$

C
Rs $$961$$

D
Rs$$896$$

Solution

Principal $$(P)=Rs.4000$$
Rate of Interest $$(r)=10\%$$
Time $$=2$$ years and $$3$$ months
Here first we take $$n=2$$ years.
$$\Rightarrow$$ Amount for first $$2$$ years $$(A)=P\left(1+\dfrac{r}{100}\right)^n$$

$$=4000\left(1+\dfrac{10}{100}\right)^2$$

$$=4000\left(\dfrac{11}{10}\right)^2$$

$$=4000\times\dfrac{121}{100}$$

$$=Rs.4840$$
$$\Rightarrow$$ The amount after two years $$=Rs.4840$$
$$\Rightarrow$$ Now, Principal $$=Rs.4840$$
Simple interest for last $$3$$ months i.e. $$\dfrac{1}{4}years$$ $$=\dfrac{PRT}{100}$$

$$=\dfrac{4840\times 10\times 1}{100\times 4}$$

$$=Rs.121$$

$$\Rightarrow$$ Amount after $$2$$ years and $$3$$ months $$=Rs.4840+Rs.121=Rs.4961$$
$$\Rightarrow$$ $$C.I.=A-P$$
$$=Rs.4961-Rs.4000$$
$$=Rs.961$$
$$\therefore$$ The compound interest is $$Rs.961$$
Question 5
1 / -0

The population of a particular area of a city is $$5000$$. It increases by $$10 \%$$ in $$1^{st}$$ year. It decreases by $$20 \%$$ in the $$2^{nd}$$ year because of some reason. In the $$3^{rd}$$ year, the population increases by $$30 \%$$. What will be the population of area at the end of $$3$$ years?

A
$$5120$$

B
$$5300$$

C
$$5720$$

D
$$5620$$

Solution

Population of a city at the starting of first year $$=5000$$
It increases by $$10\%$$ in $$1^{st}$$ year.
i.e., Population at the end of $$1^{st}$$ year $$=$$ Population at the starting of $$2^{nd}$$ year
$$=5000+10\%\space of\space5000\\=5000+\dfrac{10}{100}\times5000\\=5500$$
It decreases by $$20\%$$ in $$2^{nd}$$ year.
i.e., Population at the end of $$2^{nd}$$ year $$=$$ Population at the starting of $$3^{rd}$$ year
$$=5500-20\%\space of\space5500\\=5500-\dfrac{20}{100}\times5500\\=5500-1100\\=4400$$
It increases by $$30\%$$ in $$3^{rd}$$ year.
i.e., Population at the end of $$3^{rd}$$ year
$$=4400+30\%\space of\space4400\\=4400+\dfrac{30}{100}\times4400\\=4400+1320\\=5720$$

So, $$\text{C}$$ is the correct option.
Question 6
1 / -0

The cost price of a car is $$Rs. \,4,00,000$$. If its price decreases by $$10 \%$$ every year, then when will be the cost of car after $$3$$ years?

A
$$Rs. \,3,00,000$$

B
$$Rs. \,2,91,700$$

C
$$Rs. \,2,91,600$$

D
$$Rs. \,2,50,000$$

Solution

Cost price of car $$(P)=\mathrm{Rs.}\space4,00,000$$
Rate of depreciation $$(R)=10\%$$
Time $$(T)=3\space\text{years}$$
So, Cost of car after $$3\space\text{years}=P(1-\frac{R}{100})^T$$
$$=4,00,000(1-\frac{10}{100})^3$$
$$=4,00,000(1-0.1)^3$$
$$=4,00,000(0.9)^3$$
$$=4,00,000\times0.729$$
$$=\mathrm{Rs.}\space2,91,600$$
So, $$\text{C}$$ is the correct option.
Question 7
1 / -0

The compound interest on $$Rs \ 2000$$ at $$15\%$$ for $$2$$ years is Rs ........

A
$$645$$

B
$$600$$

C
$$615$$

D
$$315$$

Solution

Given,
Principal, $$P=Rs.2000$$
Rate of interest, $$r=15\%$$
Time, $$t=2$$ years

Let,amount $$=A$$
so,$$A=P(1+\dfrac{r}{100})^t$$
$$=(2000)\left(1+\dfrac{15}{100}\right)^2$$
$$=(2000)(1.15)^2$$
$$=2645$$

CI$$=A-P$$
$$=Rs.(2645-2000)$$
$$=Rs.645$$
Question 8
1 / -0

A four year Indira Vikas certificate with a maturity value of Rs. 700 is purcahsed for Rs. 500. The rate $$\%$$ p.a. is :

A
$$

9 \dfrac { 1 } { 11 } \%

$$

B
$$10\%$$

C
$$11\%$$

D
$$

21 \dfrac { 9 } { 11 } \%

$$

Solution

Interest $$= 200.....(700-500=200)$$

$$SI=\dfrac{P\times N\times R}{100}$$

$$200=\dfrac{500 \times R \times 4}{100}\implies R=10\%$$
Question 9
1 / -0

The population of a country is $$10$$ crore and it is the possibility that the population will become $$13.31$$ crore in $$3$$ years. What will be the annual rate per cent of this growth?

A
$$8 \%$$

B
$$12.7 \%$$

C
$$10 \%$$

D
$$15 \%$$

Solution
Question 10
1 / -0

Rohit received an interest of Rs 2520 at the rate 7 % after 3 years. Find out the principal for which he received this interest.

A
Rs 15,000

B
Rs 21,000

C
Rs 1,500

D
Rs 12,000

Solution

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Basics of Financial Mathematics Test 40

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Basics of Financial Mathematics Test 40

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Question 1

Rs. 28240

Rs. 28540

Rs. 29240

Rs. 29160

Solution

Question 2

Rs. 780.00

Rs. 789.00

Rs. 788.80

780.40

Solution

Question 3

$$Rs.806$$

$$Rs.708$$

$$Rs.5,408$$

$$Rs.408$$

Solution

Question 4

Rs$$916$$

Rs $$900$$

Rs $$961$$

Rs$$896$$

Solution

Question 5

$$5120$$

$$5300$$

$$5720$$

$$5620$$

Solution

Question 6

$$Rs. \,3,00,000$$

$$Rs. \,2,91,700$$

$$Rs. \,2,91,600$$

$$Rs. \,2,50,000$$

Solution

Question 7

$$645$$

$$600$$

$$615$$

$$315$$

Solution

Question 8

$$9 \dfrac { 1 } { 11 } \%$$

$$10\%$$

$$11\%$$

$$21 \dfrac { 9 } { 11 } \%$$

Solution

Question 9

$$8 \%$$

$$12.7 \%$$

$$10 \%$$

$$15 \%$$

Solution

Question 10

Rs 15,000

Rs 21,000

Rs 1,500

Rs 12,000

Solution

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$$

9 \dfrac { 1 } { 11 } \%

$$

$$

21 \dfrac { 9 } { 11 } \%

$$