Basics of Financial Mathematics Test 21

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Basics of Financial Mathematics Test 21

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Question 1
1 / -0

What is true about deferred annuity ?

A
It is an annuity when the payments are made at the end of payment period.

B
It is an annuity when the payments are made at the beginning of payment period.

C
It is an annuity when the payments are made at the middle of payment period.

D
None of the above

Solution

$$\Rightarrow$$ True statement about deferred annuity is,
$$-\,It\,is\,an\,annuity\,when\,the\,payment\,are\,made\,at\,the\,end\,of\,payment\,period.$$
$$\Rightarrow$$ A deferred annuity is an insurance contract designed for long-term savings.
$$\Rightarrow$$ Unlike an immediate annuity, which starts annual or monthly payments almost immediately, investors can delay payments from a deferred annuity indefinitely. During that time, any earnings in the account are tax-deferred.
Question 2
1 / -0

The first year's interest on a sum of money lent at $$8$$% compound interest is $$Rs. 48$$. The second year's amount is

A
$$Rs. 48$$

B
$$Rs. 51.48$$

C
$$Rs. 56.48$$

D
$$Rs. 96$$

Solution
Question 3
1 / -0

The calculation of interest on the interest of principal amount is called

A
simple interest

B
compound interest

C
multiple interest

D
interest quarterly compounded

Solution

The calculation of interest on the interest of principal amount is called $$Compound\,\,interest$$.
Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest.
It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest.
Compound interest is standard in finance and economics.
Question 4
1 / -0

Anagha borrowed Rs.$$ 70,000$$ from her friend at the rate of $$3.5\%$$ p.a.compounded yearly. She returned the amount after three years. So calculate interest of first year and second year .

A
$$Rs.2,450, 85.75$$

B
$$Rs.2,450, 2,450$$

C
$$Rs.2,450, 2,535.75$$

D
$$Rs.2,450, 2100$$

Solution

$$P=70000,R=3.5$$%
Interest in $$1$$st year$$=\cfrac { P\times R\times T }{ 100 } $$
$$=\cfrac { 70000\times 3.5\times 1 }{ 100 } $$
$$=2450$$
Now, $$(2450+P)$$ will be principal amount for $$2$$nd year since interest is compounded yearly
Interest in $$2$$nd year$$=\cfrac { { P }^{ ' }\times R\times T }{ 100 } $$
$$=\cfrac { (2450+70000)\times 3.5\times 1 }{ 100 } $$
$$=724.5\times 3.5=2535.75$$
Question 5
1 / -0

Amar gave Rs.$$50,000 $$ as a loan to Amir at the rate of $$4\%$$ p.a. Amir return the amount after two years. Calculate the interest on the first year's interest.

A
$$200$$

B
$$70$$

C
$$80$$

D
$$60$$

Solution

$$1$$ year interest$$=\cfrac { P\times R\times T }{ 100 } $$
$$=\cfrac { 50000\times 4\times 1 }{ 100 } =2000$$
Interest of $$1$$ year interest$$=\cfrac { \left( { P }^{ ' }\times R\times T \right) }{ 100 } $$
$$=\cfrac { 2000\times 4\times 1 }{ 100 } $$
$$=80$$
Question 6
1 / -0

What principal will amount to Rs. $$9,744$$ in two years, if the rates of interest for successive years are $$16$$% and $$20$$% respectively?

A
$$7000$$

B
$$8000$$

C
$$5000$$

D
$$9000$$

Solution

$$\Rightarrow$$ Let the principal (P) be $$Rs.x$$.
$$\Rightarrow$$ Rate of interests for two successive years are $$(R_1)\,\,16\%$$ and $$(R_2)\,\,20\%$$.
$$\Rightarrow$$ $$A=P(1+\dfrac{R_1}{100})(1+\dfrac{R_2}{100})$$

$$\Rightarrow$$ $$9744=x(1+\dfrac{16}{100})(1+\dfrac{20}{100})$$

$$\Rightarrow$$ $$9744=x\times \dfrac{29}{25}\times {6}{5}$$

$$\Rightarrow$$ $$9744=\dfrac{174}{125}x$$

$$\therefore$$ $$x=\dfrac{9744\times 125}{174}$$

$$=56\times 125=Rs.7000$$
$$\therefore$$ Hence, principal = $$Rs.7000$$.
Question 7
1 / -0

Calculate compound interest for Rs $$15,000$$ for $$1$$ year at $$16$$% compounded semi -annually.

A
Rs.$$3172$$

B
Rs.$$2496$$

C
Rs.$$3000$$

D
Rs.$$2572$$

Solution

Given:
$$\Rightarrow$$ $$P=$$ Rs.$$15000,\,T=1\ year=2$$ and $$R=16\%$$

$$R_{eq}=8\%$$ Since, semi-annually compounded.

$$T_{eq}=2.T= 2.$$ Since, semi-annually compounded.

$$\Rightarrow$$ $$A=P(1+\dfrac{R_{eq}}{100})^{T_{eq}}$$

$$\Rightarrow$$ $$A=15000\times (1+\dfrac{8}{100})^2$$

$$\Rightarrow$$ $$A=15000\times (\dfrac{27}{25})^2$$

$$\Rightarrow$$ $$A=15000\times (\dfrac{729}{625})$$

$$\therefore$$ $$A=$$Rs.$$17496$$

$$\therefore$$ $$C.I.=A-P=$$Rs. $$17496- $$Rs.$$15000=$$Rs.$$2496.$$
Question 8
1 / -0

Megha lended Rs.$$8000$$ as a loan for $$4$$ years at the rate $$4\%$$ compounded annualy. Calculate the interest she will recive by the method of simple interest.

A
$$8963.26$$

B
$$9358.86$$

C
$$1358.86$$

D
$$9863.25$$

Solution

$$P=8000,R=4$$%, $$T=4$$
Interest in $$1$$st year$$=\cfrac { P\times R\times T }{ 100 } $$
$$=\cfrac { 8000\times 4\times 1 }{ 100 } $$
$$=320$$
Now, Principal amount will be $$(P+320)$$ since the interest in compounded annually.
Interest in $$2$$nd year$$=\cfrac { { P }^{ ' }\times R\times T }{ 100 } $$
$$=\cfrac { 8320\times 4\times 1 }{ 100 } $$
$$=332.8$$
Now, $$P''=332.8+P'$$
Interest in $$3$$rd year$$=\cfrac { 8652.8\times 4\times 1 }{ 100 } =346.112$$
Interest in $$4$$th year$$=\cfrac { (346.112+8652.8)\times 4\times 1 }{ 100 } =359.956$$
Total interest$$=320+332.8+346.112+359.95$$
$$=1358.86$$
Question 9
1 / -0

Ranjana borrowed the money Rs$$1,00,000$$ for $$3$$ years at the rate $$3.5\%$$ p.a. compounded annually. Calculate interest to be paid.

A
$$10871.78$$

B
$$1000.8$$

C
$$8710.68$$

D
$$245$$

Solution

$$\Rightarrow$$ Here, Principal amount for first year = $$Rs.1,00,000$$
$$\Rightarrow$$ First year Interest = $$\dfrac{100000\times 3.5}{100}=Rs.3500$$

$$\Rightarrow$$ Principal amount for second year = $$Rs.1,00,000+Rs.3500=Rs.1,03,500.$$
$$\Rightarrow$$ Second year interest = $$\dfrac{103500\times 3.5}{100}=Rs.3622.5$$

$$\Rightarrow$$ Principal amount for third year = $$Rs.1,03,500+Rs.3622.5=Rs.1,07,122.5$$
$$\Rightarrow$$ Third year interest = $$\dfrac{107122.5\times 3.5}{100}=Rs.3749.28$$

$$\therefore$$ Total Interest = $$Rs.3500+Rs.3622.5+Rs.3749.28=Rs.10871.78$$
Question 10
1 / -0

An industrialist borrowed the $$Rs\ 75,000$$ for two years at the rate $$2.5\%$$ p.a.compounded annually. Calculate the total amount compound interest by simple interest method.

A
$$2750$$

B
$$3750$$

C
$$78,795$$

D
$$78,796.87$$

Solution

$$P=75000,R=2.5$$%, Time$$=2$$yrs

Interest in $$1$$st year$$=\cfrac { P\times R\times T }{ 100 } $$

$$=\cfrac { 75000\times 2.5\times 1 }{ 100 } $$

$$=1875$$

$$P'=(1875+P)$$

Interest in $$2$$nd year$$=\cfrac { { P }^{ ' }\times R\times T }{ 100 } $$

$$=\cfrac { 76875\times 2.5\times 1 }{ 100 } $$

$$=1921.875$$

Total interest$$=1875+1921.875$$

$$=3796.875$$

Total amount=Principal$$+$$Total interest

$$=75000+3796.875$$

$$=78796.875$$

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Basics of Financial Mathematics Test 21

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Question 1

It is an annuity when the payments are made at the end of payment period.

It is an annuity when the payments are made at the beginning of payment period.

It is an annuity when the payments are made at the middle of payment period.

None of the above

Solution

Question 2

$$Rs. 48$$

$$Rs. 51.48$$

$$Rs. 56.48$$

$$Rs. 96$$

Solution

Question 3

simple interest

compound interest

multiple interest

interest quarterly compounded

Solution

Question 4

$$Rs.2,450, 85.75$$

$$Rs.2,450, 2,450$$

$$Rs.2,450, 2,535.75$$

$$Rs.2,450, 2100$$

Solution

Question 5

$$200$$

$$70$$

$$80$$

$$60$$

Solution

Question 6

$$7000$$

$$8000$$

$$5000$$

$$9000$$

Solution

Question 7

Rs.$$3172$$

Rs.$$2496$$

Rs.$$3000$$

Rs.$$2572$$

Solution

Question 8

$$8963.26$$

$$9358.86$$

$$1358.86$$

$$9863.25$$

Solution

Question 9

$$10871.78$$

$$1000.8$$

$$8710.68$$

$$245$$

Solution

Question 10

$$2750$$

$$3750$$

$$78,795$$

$$78,796.87$$

Solution

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