Basics of Financial Mathematics Test 7

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Basics of Financial Mathematics Test 7

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Question 1
1 / -0

During a period of two years, a principal of Rs. 100 amounts to Rs. 121 at the annual compound rate of r%. The value of $$r$$ will be

A
$$9$$

B
$$10$$

C
$$\cfrac{21}{2}$$

D
$$11$$

Solution

Given, Sum (P) $$=$$ Rs. 100,
Amount due (A) $$=$$ Rs. 121,
Time (n) $$=$$ 2 years, Rate (r) $$=$$ ?
We know
$$A = P \left ( 1+\cfrac{r}{100} \right )^n$$
$$\therefore 121 = 100 (1+\cfrac{r}{100})^{2}$$
or $$ (1+\cfrac{r}{100})^{2} = \cfrac{121}{100}$$
or $$1+\cfrac{r}{100}=\cfrac{11}{10}=1+\cfrac{1}{10}=1+\cfrac{10}{100}$$
$$\therefore r = 10\%$$
Question 2
1 / -0

Rakesh took a loan for 7 years at the rate of 6% p.a. S.I. If the total interest paid was Rs. 2100, the principal was

A
Rs. 4400

B
Rs. 4800

C
Rs. 5000

D
Rs. 5200

Solution

Given that,
Rakesh took a loan for $$7$$ years at $$6\%$$ per annum simple interest.
He paide $$Rs.\ 2100$$ interest on the amount.

To find out,
The principal amount.

We know that, $$S.I.=\left(\dfrac{P\times R\times T}{100}\right)$$

Here, $$S.I.=Rs.\ 2100,\ R=6\%$$ and $$T=7\ years$$

Hence, $$2100=\left(\dfrac{P\times 6\times 7}{100}\right)$$

$$\Rightarrow P=\left(\dfrac{2100\times 100}{6\times 7}\right)$$

$$\Rightarrow P=\left(\dfrac{2100\times 100}{42}\right)$$

$$\Rightarrow P=50\times 100$$

$$\therefore\ P=Rs.\ 5000$$

Hence, the principal amount was $$Rs.\ 5000$$.
Question 3
1 / -0

The population of a small town is increased by $$\cfrac{15}{2}$$ % per annum for two years. If the present population is $$73, 960$$, then two years before the population was

A
$$68,000$$

B
$$70,000$$

C
$$64,000$$

D
$$60,000$$

Solution

$$A=P(1+\cfrac{R}{100})^n$$
$$73960=P(1+0.075)^2$$
$$\therefore P=64000$$
Question 4
1 / -0

A certain sum of money Q was deposited for 5 years and 4 months at 4.5% simple interest and mounted to Rs. 248. Then the value of Q is

A
Rs. 200

B
Rs. 210

C
Rs. 220

D
Rs. 240

Solution

We know that,
Amount = principle + simple interest
$$A=P+\cfrac{PRT}{100}$$
$$A=P[1+\cfrac{RT}{100}]$$            ...... (1)
Where, $$A$$ is amount.
            $$P$$ is principle.
            $$R$$ is rate of interest.
            $$T$$ is time in year.

Given, $$A=248$$
            $$P=Q$$
            $$R=4.5$$%
            $$T=5+\cfrac{4}{12}=\cfrac{16}{3}$$
Substitute these values in equation (1) and find the value of $$Q$$.
$$248=Q[1+\cfrac{4.5\times\frac{16}{3}}{100}]$$
$$248=Q[1+\cfrac{24}{100}]$$
$$248=1.24Q$$
$$\therefore Q=200$$
Question 5
1 / -0

The compound interest of Rs. 1,000 at the rate of 10% will be Rs. 210 in:

A
one year

B
two years

C
three years

D
four years

Solution

Given Sum (P) $$=$$ $$Rs. 1000$$, Rate $$=$$ $$10\%$$,
$$C.I.$$ $$=$$ $$Rs. 210$$, Time $$(n)$$ $$=$$ ?
Now $$C.I.$$ $$= P [(1+\cfrac{r}{100})^{n}-1]$$
$$\therefore 210 = 1000[(1+\cfrac{10}{100})^{n}-1]$$
or $$(\cfrac{110}{100})^{n}-1 = \cfrac{210}{1000} = 0.21$$
or $$\left ( \cfrac{11}{10} \right )^{n} = 1.21 = \cfrac{121}{100}$$
$$ = \left ( \cfrac{11}{10} \right )^{2}$$
$$\therefore $$ $$n$$ $$= 2$$ years
Question 6
1 / -0

If the population of a town is 64,000 and its annual increase is 10%, then its correct population at the end of 3 years will be

A
80,000

B
85,000

C
85,100

D
85,184

Solution

Population (P) $$= 64,000$$
Annual increase rate $$(r) = 10\%$$
Hence, population after $$n=3$$ years
$$= P(1+\cfrac{r}{100})^n$$
$$= 64,000(1+\cfrac{10}{100})^3$$
$$= 64,000\times (\cfrac{11}{10})^3$$
$$= 64,000\times \cfrac{11\times 11\times 11}{1000}$$
$$= 85,184$$
Question 7
1 / -0

A sum of money becomes Rs. 13380 after 3 years and Rs. 20070 after 6 years on compound interest. The sum is

A
Rs. 8800

B
Rs. 8890

C
Rs. 8920

D
Rs. 9040

Solution

Let the sum be Rs. $$x$$. Then ...(i)
$$x (1+\cfrac{R}{100})^3 = 13380$$
and $$x (1+\cfrac{R}{100})^6 = 20070$$ ...(ii)
Dividing equation (ii) by (i), we get
$$(1+\cfrac{R}{100})^3 = (\cfrac{200070}{13380}) = (\cfrac{3}{2})$$
$$\therefore x\cfrac{3}{2} = 13380$$
$$\Rightarrow x = (13380\times \cfrac{2}{3}) = 8920$$
Hence, the sum is Rs. 8920
Question 8
1 / -0

If in two years time a principal of $$Rs. 100$$ amounts to $$Rs. 121$$, when the interest at the rate of $$r\%$$ is compounded annually, then the value of $$r$$ will be

A
10.5

B
10

C
15

D
14

Solution

We know the formula for amount $$A$$ when prinicipal $$P$$ is compounded annually for $$n$$ years, at the rate of $$R\%$$ per annum is :
$$A=P(1+\cfrac{R}{100})^n$$
$$121=100(1+\cfrac{R}{100})^2$$
$$\dfrac{121}{100}= (1+\cfrac{R}{100})^2$$
$$\left( \dfrac{11}{10}\right )^2= (1+\cfrac{R}{100})^2$$
$$1+\cfrac{R}{100}=\dfrac{11}{10}$$
$$\cfrac{R}{100}=\dfrac{1}{10}=0.1$$
$$R=10$$%
Question 9
1 / -0

The cost of a scooter depreciates every year by 15% of its value at beginning of the year. If the present cost of the scooter is Rs. 8,000; find its cost (in Rs.) after 2 years.

A
5780

B
6690

C
7200

D
7530

Solution

Cost after depreciation in 'n' years $$ = \text{Original cost} \times {(1-\dfrac {R}{100})}^{n} $$

So after two years, cost $$ = Rs 8,000 \times {(1-\dfrac {15}{100})}^{2} = 8,000 \times {(\dfrac {85}{100})}^{2} = Rs 5,780 $$
Question 10
1 / -0

A loan was repaid in two annual instalments of Rs. 1210 each. If the rate of interest be 10% per annum, compounded annually, then the sum borrowed was

A
Rs. 2000

B
Rs. 2100

C
Rs. 2300

D
Rs. 1900

Solution

$$\Rightarrow$$ Here we have, $$I=Rs.1210$$ and $$R=10\%$$

$$\Rightarrow$$ The sum borrow = $$[\dfrac{1210}{1+\dfrac{10}{100}}+\dfrac{1210}{(1+\dfrac{10}{100})^2}]$$

$$\Rightarrow$$ The sum borrow = $$[1100+1000]=Rs.2100$$

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Basics of Financial Mathematics Test 7

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Question 1

$$9$$

$$10$$

$$\cfrac{21}{2}$$

$$11$$

Solution

Question 2

Rs. 4400

Rs. 4800

Rs. 5000

Rs. 5200

Solution

Question 3

$$68,000$$

$$70,000$$

$$64,000$$

$$60,000$$

Solution

Question 4

Rs. 200

Rs. 210

Rs. 220

Rs. 240

Solution

Question 5

one year

two years

three years

four years

Solution

Question 6

80,000

85,000

85,100

85,184

Solution

Question 7

Rs. 8800

Rs. 8890

Rs. 8920

Rs. 9040

Solution

Question 8

10.5

10

15

14

Solution

Question 9

5780

6690

7200

7530

Solution

Question 10

Rs. 2000

Rs. 2100

Rs. 2300

Rs. 1900

Solution

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