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Basics of Financial Mathematics Test 34

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Basics of Financial Mathematics Test 34
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  • Question 1
    1 / -0
    An annuity is?
    Solution
    An annuity is 
    $$(1)$$A fixed sum
    $$(2)$$ Paid at regular intervals
    $$(3)$$Under certain stated conditions.
  • Question 2
    1 / -0
    Mrs. S deposited Rs. $$1,00,000$$ in a nationalized bank for $$3$$ years. If the rate of interest is $$7\%$$ p.a., Interest is compounded annually. Calculate the amount at the end of third year (i.e., Principal $$+$$ compound interest)
    Solution
    Interest for the$$1^{ st }$$ year $$= Rs.\dfrac { 100000\times 1\times 7 }{ 100 } =Rs. 7000$$

    Amount after the$$1^{ st }$$ year $$= Rs. 100000 + Rs.7000= Rs. 107000$$

    Interest for the $$2^{nd}$$ year $$= Rs.\dfrac{107000\times 1\times 7}{100} = Rs.7490$$

    Amount after the $$2^{nd}$$ year $$= Rs. 107000 + Rs. 7490 = Rs. 114490$$

    Interest for the$$3^{ rd }$$ year $$= Rs. \dfrac { 114490\times 1\times 7 }{ 100 } = Rs. 8014.3$$

    $$\therefore$$ Amount at the end of the $$3^{rd}$$ year $$= Rs. 114490+ Rs. 8014.3= Rs. 122504.30$$
  • Question 3
    1 / -0
    A sum of Rs. $$1,200$$ becomes Rs. $$1,323$$ in $$2$$ years at compound interest compounded annually. Find the rate per cent.
    Solution
    Let r% be rate per annum, then,
    $$1323 = 1200(1 + \cfrac{r}{100})^2$$
    $$\implies \cfrac{1323}{1200} = (1+\cfrac{r}{100})^2$$
    $$\implies \sqrt{\cfrac{1323}{1200}} = (1+\cfrac{r}{100})$$
    $$\implies 1.05 = (1+\cfrac{r}{100})$$
    $$\implies \cfrac{r}{100} = 1.05 - 1 = 0.05$$
    $$\implies r= 100 \times 0.05 = 5$$%
  • Question 4
    1 / -0
    Find the compound interest for Rs. $$1,000$$ for $$4$$ years at $$5\%$$ p.a. When the interest is compounded annually.
    Solution
    Interest for first year $$ =\cfrac{1000 \times 5 \times 1}{100} = ₹50$$
    Amount after first year $$= 1000+50=₹1050$$
    Interest for second year $$= \cfrac{1050 \times 5 \times 1}{100} = ₹52.5$$
    Amount after second year  $$= 1050+52.5=₹1102.5$$
    Interest for third year $$= \cfrac{1102.5 \times 5 \times 1}{100} = ₹55.125$$
    Amount after third year  $$= 1102.5+55.125=₹1157.625$$
    Interest for fourth year $$= \cfrac{1157.625 \times 5 \times 1}{100} = ₹57.88125$$
    Amount after fourth year $$= 1157.625+57.88125=₹1215.50625$$
    Compound interest $$=1215.50625 - 1000 = 215.50625 = ₹215$$
  • Question 5
    1 / -0
    Tanishka deposited Rs. $$50,000$$ in a bank for two years with the interest rate of $$5.5\%$$ p.a. How much interest would she earn?
    Solution
    $$ Interest = \cfrac{ 50000 \times 2 \times 5.5 }{100} = 5500$$
  • Question 6
    1 / -0
    The S.I. accrued on an amount of Rs. $$25,000$$ at the end of three years is Rs. $$7,500$$. What would be the C.I. accrued on the same amount at the same rate in the same period?
    Solution
    Apply the formula for simple interest

    $$7500=\dfrac{25000\times R\times3}{100}$$

    $$\implies R=\dfrac{7500}{250\times3}=10\%$$

    Interest for the $$1$$st year$$=\dfrac{25000\times10\times1}{100}$$
    $$=2500$$

    Amount After the $$1$$st year$$=25000+$$$$2500$$
    $$=27500$$
    Interest for the $$2$$nd year$$=\dfrac{27500\times10\times1}{100}$$
    $$=2750$$

    Amount After the $$2$$nd year$$=27500+$$$$2750$$
    $$=30250$$
    Interest for the $$3$$rd year$$=\dfrac{30250\times10\times1}{100}$$
    $$=3025$$

    Amount After the $$3$$rd year$$=30250+$$$$3025$$
    $$=33275$$

    $$\therefore$$Compound Interest(CI)$$=33275-$$$$25000$$
    $$=Rs.8275$$
  • Question 7
    1 / -0
    What will be the C.I. on a sum of Rs. $$25,000$$/- after $$3$$ years at the rate of $$12\%$$p.a?
    Solution
    $$P=25000$$
    $$T=3years$$
    $$R=12\%pa$$
    $$A=P(1+\frac{R}{100})^T$$
    $$=>A=25000(1+\frac{12}{100})^3$$
    $$=25000(\frac{28}{25})^3$$
    $$=35123.2$$
    $$I=A-P$$
    $$=35123.2-25000$$
    $$I$$$$=Rs.10123.2$$
  • Question 8
    1 / -0
    Which of the following is not an effect of simplification of procedural aspects in Taxation?
    Solution

  • Question 9
    1 / -0
    If $$P=Rs. 1,000$$, R$$=5\%$$, p.a., $$n=4$$; What is amount and CI?
    Solution
    $$P=Rs$$ $$1000$$, $$R=5\%$$ $$p.a.$$, $$n=4$$
    Interest for the $$1^{st}$$ year $$= Rs$$ $$\cfrac{1000\times 5\times 1}{100}= Rs$$ $$50$$
    Amount after $$1^{st}$$ year $$= Rs$$ $$1000+ Rs$$ $$50=Rs$$ $$1050$$
    Interest for the $$2^{nd}$$ year $$= Rs$$ $$\cfrac{1050\times 5\times 1}{100}= Rs$$ $$52.5$$
    Amount after $$2^{nd}$$ year $$= Rs$$ $$1050+ Rs$$ $$52.5=Rs$$ $$1102.5$$
    Interest for the $$3^{rd}$$ year $$= Rs$$ $$\cfrac{1102.5\times 5\times 1}{100}= Rs$$ $$55.125$$
    Amount after $$3^{rd}$$ year $$= Rs$$ $$1102.5+ Rs$$ $$55.125=Rs$$ $$1157.625$$
    Interest for the $$4^{th}$$ year $$= Rs$$ $$\cfrac{1157.625\times 5\times 1}{100}= Rs$$ $$57.88125$$
    Amount after $$1^{st}$$ year $$= Rs$$ $$1157.625+ Rs$$ $$57.88125=Rs$$ $$1215.50$$
    $$\therefore$$ Compound Interest $$=Rs$$ $$1215.50-Rs$$ $$1000=Rs$$ $$215.50$$
  • Question 10
    1 / -0
    Which of the following is not part of the Taxation Reforms in 1991?
    Solution

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