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Question 15 Marks
(a) Anil bought a TV costing ₹ 13000 by making a down payment of ₹ 3000 and agreeing to make equal annual payment for three years. How much would be each payment if the interest on unpaid amount be 14% compounded annually ?
(b) Experts say that the baby boom generation (Indians born between 1946 and 1960) cannot count on a company pension or social security to provide comfortable retirement, as their parents did. It is recommended that they start to save early and regularly. Mahesh, a baby boomer, has deposit ₹ 200 each month for 20 years in an account that pays interest of 7.2% compounded monthly.
(i) How much will be in the account at the end of 20 years?
(ii) Mahesh believes he needs to accumulate ₹ 130,000 in the 20 years period to have enough for retirement. if he can not get higher rate to produce ₹ 130,000 in 20 years. To meet the goal, he must increase her monthly payment. What payment should he make each month? Given that (1.006) 240 = 4.2026
Answer
(a) In the present case, we have present value of annuity ie., ₹ 10000 (13000-3000) and we have to calculate equal annual payment over the period of 3 years.
We know that,
$ \begin{array}{l} \text {Present value }=\text { C.F. } \times \frac{\left[(1+i)^n-1\right]}{i(1+i)^n} \\ \Rightarrow \quad \text { C.F. }=\frac{\text { Present value }}{\left[(1+i)^n-1\right]} \times i(1+i)^n \end{array} $
Here, $n=3, i=14 \%=\frac{14}{100}=0.14$
$ \begin{aligned} \therefore \quad \text { C.F. } & =\frac{10,000}{\left[(1+0.14)^3-1\right]} \times 0.14 \times(1+0.14)^3 \\ & =\frac{10,000}{\left[(1.14)^3-1\right]} \times 0.14 \times(1.14)^3 \end{aligned} $
$\begin{array}{l}=\frac{10,000}{(1.4815-1)} \times 0.14 \times 1.4815 \\ =\frac{10,000}{0.4815} \times 0.207416 \\ =\frac{2074.16}{0.4815}\end{array}$
=₹ 4307.71
Therefore each payment would be ₹ 4307.71.
(b) (i) The saving plan is an annuity with periodic payments PMT $=200, i=7.2 \%$ monthly $=$ $\frac{7.2}{100} \times \frac{1}{12}$ annually $=0.006$ and $n=12(20)=$ 240 The future value is $ \begin{aligned} FV & =\operatorname{PMT} \frac{(1+i)^n-1}{i} \\ \therefore \quad FV & =200 \times\left\{\frac{(1+0.006)^{220}-1}{0.006}\right\} \\ & =200 \times\left\{\frac{(1.006)^{240}-1}{0.006}\right\} \\ & =200 \times\left(\frac{4.2026-1}{0.006}\right) \end{aligned} $
$\begin{array}{l}=200 \times \frac{3.2026}{0.006} \\ =200 \times 533.7623\end{array}$
= ₹ 106, 752.468
He will accumulate ₹ 106, 252.47 in the account at the end of 20 years.
(ii) Mahesh's goal is to accumulate ₹ 130,000 in 20 years at 7.2% compounded monthly. Therefore the future value is F .V.= ₹ 130,000, the monthly interest rate is $i=\frac{0.072}{12}=0.006$ and the number of periods is n = 12(20) = 240 and the Using the sinking fund payment formula to find the payment PMT.
$\begin{aligned} \text { PMT } & =\frac{i \times F . V .}{(1+i)^n-1} \\ & =\frac{0.006 \times 130,000}{(1+0.006)^{240}-1} \\ & =\frac{780}{4.2026-1} \\ & =\frac{780}{3.2026}\end{aligned}$
= ₹ 243.5521
Mahesh will need payments of ₹ 243.55 each month for 20 years to accumulate at atleast ₹ 130,000.
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Question 25 Marks
(a) ₹ 200 is invested at the end of each month in an account paying interest 6% per year compounded monthly. What is the future value of this annuity after 10th payment? Given that (1.005)10 = 1.0511 deg
(b) Find the present value of ₹ 10,000 to be required after 5 years if the interest rate be 9%. Given that (1.09)5 = 1.5386
(c) Sneha borrows ₹ 500000 to buy a house. If she pays equal installments for 20 years and 10% interest on outstanding balance what will be the equal annual installment ? Given that ( 1.10)20 =6.7274
Answer
Here, C.F = ₹ 200
n = 10
$ \begin{aligned} i & =6 \% \text { per annum } \\ & =\frac{6}{12} \% \text { per month } \\ & =0.005 \end{aligned} $ Future value of annuity after 10 months is given by $ \begin{aligned} EV . & =C . F \cdot\left[\frac{(1+i)^n-1}{i}\right] \\ & =200 \times\left[\frac{(1+0.005)^{10}-1}{0.005}\right] \\ & =200 \times\left[\frac{(1.005)^{10}-1}{0.005}\right] \\ & =200 \times \frac{(1.0511-1)}{0.005} \\ & =200 \times \frac{0.0511}{0.005} \\ & =200 \times 10.22 \end{aligned} $
= ₹ 2044
(b) Here, $i=9 \%=0.09, n=5, C . E =10,000$
$ \begin{aligned} \text {Present value } & =\frac{A_n}{(1+i)^n} \\ & =\frac{10,000}{(1+0.09)^5}=\frac{10,000}{(1.09)^5} \end{aligned} $
$=\frac{10,000}{1.5386}$
= ₹ 6499.42
(c) We known that, $ C . E=\frac{\text { Present value (P.V.) }}{P(n, i)} $
Here, P.V = ₹ 500000
$\begin{aligned} n & =20 \\ i & =10 \% \text { p.a. }=0.10 \\ \text { C.F } & =\frac{500000}{P(20,0.10)}\end{aligned}$
$P(n, i)=\frac{(1+i)^n-1}{i(1+i)^n}$
$\begin{array}{l}=\frac{(1+0.10)^{20}-1}{0.10(1+0.10)^{20}} \\ =\frac{6.7274-1}{0.10 \times 6.7274} \\ =\frac{5.7274}{0.67274} \\ =8.51356\end{array}$
Therefore, $\quad$ C.F $=\frac{500000}{8.51356}$
=₹ 58729.84
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Question 35 Marks
Leela is an athlete who believes that her playing career will last 3 years.
(a) To prepared for future, she deposits ₹ 24,000 at the end of each year for 3 years in an account paying 6% compounded annually. How much will she have on deposit after 3 years ? Also, find the value of interest earned.
(b) Instead of investing ₹ 24,000 at the end of each year, suppose Leela deposits ₹ 80,000 at the end of each year for 3 years in an account paying 5% compounded monthly. How much will she have on deposit after 3 years? Also, find the value of interest earned.
Answer
(a) Her yearly payments form an annuity regular with C.F.= ₹ 24000 , Since, the interest is compounded annually, no. of conversion periods, n = 3 and the interest rate per period dot $i=6 \%=\frac{6}{100}$= 0.06
Using the formula for the future value of an annuity
$\begin{aligned} \text { F.V. } & =\text { C.F. } \times\left[\frac{(1+i)^n-1}{i}\right] \\ & =24,000\left[\frac{(1+0.06)^3-1}{0.06}\right] \\ & =24,000\left[\frac{(1.06)^3-1}{0.06}\right] \\ & =24,000\left[\frac{1.1910-1}{0.06}\right]\end{aligned}$
$\begin{array}{l}=24,000 \times \frac{0.1910}{0.06} \\ =24,000 \times 3.1836\end{array}$
= ₹ 76, 406.40
the interest earned = F.V. -3 * C.F.
$\begin{array}{l}=76406.40-3 \times 24,000 \\ =76406.40-72,000 \\ =4,406.40\end{array}$
(b) Interest is compounded semi-annually, so no. of conversion periods $n=3 \times 1=3$ and $i=5 \%$ $=\frac{5}{100}=0.05$
C.F. = ₹ 80, 000
$\begin{array}{l} \therefore \\ =80,000\left[\frac{(1+0.05)^3-1}{0.05}\right] \\ =80,000\left[\frac{(1.05)^3-1}{0.05}\right] \\ =80,000\left(\frac{1.1576-1}{0.05}\right) \\ =80,000\left(\frac{0.1576}{0.05}\right) \\ =80,000 \times 3.1525\end{array}$
= ₹ 252, 200
$\begin{aligned} \text { The interest earned } & =\text { F.V. }-3 \times \text { C.F. } \\ & =252,200-3 \times 80,000 \\ & =252,200-240,000\end{aligned}$
= ₹ 12, 200
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Question 45 Marks
(a) A man purchased a house valued at ₹ 300000. He paid ₹ 200000 at the time of purchased and agreed to pay the balance with the interest at 12% per annum compounded half yearly in 20 equal half yearly installments. If first installment is paid after six months from the date of purchase than find the amount of each installment. [Given that (1.06)2020 = 3.2071 ]
(b) A person invests ₹ 500 at the end of each year with a bank which pays interest at 10% p.a. C.I. annually. Find the amount standing to his credit one year after he has made his yearly investment for the 12th time.
[Given that (1.1) 12 = 3.1348 ]
Answer
Here, Present value = 300000 - 200000 = ₹ 100000
$\begin{aligned} i & =12 \% \text { half yearly } \\ & =\frac{12}{100} \times \frac{1}{2}=0.06 . \\ n & =20 \\ \text { P.V. } & =\text { C.F. } \times\left[\frac{(1+i)^n-1}{i(1+i)^n}\right] \\ \text { C.F } & =\frac{\text { P.V. } \times i(1+i)^n}{(1+i)^n-1}\end{aligned}$
$\begin{array}{l}=100000 \times\left[\frac{0.06(1+0.06)^{20}}{(1+0.06)^{20}-1}\right] \\ =100000 \times\left[\frac{0.06 \times(1.06)^{20}}{(1.06)^{20}-1}\right] \\ =100000 \times \frac{0.06 \times 3.2071}{3.2071-1} \\ =100000 \times \frac{0.1924}{2.2071} \\ =100000 \times 0.0871845 \\ =8718.45\end{array}$
= ₹ 8719 ( approx )
Hence,₹ 8719 is the amount of each installment
(b) We know that, F.V. $=$ C.F. $\frac{\left[(1+i)^n-1\right]}{i}$
Here, C.F. $=500, n=12, i=10 \%=\frac{10}{100}=0.1$
$ \therefore \quad \text { FV. }=\frac{500\left[(1+0.1)^{12}-1\right]}{0.1} $
$ \begin{array}{l} =\frac{500\left[(1.1)^{12}-1\right]}{0.1} \\ =\frac{500 \times[3.138-1]}{0.1} \\ =\frac{1069.214}{0.1} \\ =10692.14 \end{array} $
Amount after one year i.e., after $12^{\text {th }}$ installment
$ \begin{array}{l} =FV . \times(1+i) \\ =10692.14 \times(1.1) \end{array} $
= ₹ 11761
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Question 55 Marks
(a) A sum of money doubles itself in 4 years compound interest. It will amount to 8 times itself at the same rate of interest in how many years?
(b) Compound interest on a sum of money you 2 years at 4% per annum is 2448. Find simple interest on the same sum of money at the same rate of interest for 2 years.
Answer
Let $\quad$ Principal $=P$
$ \text { Rate }=i \%=\frac{i}{100} $
$ t=4 \text { years } $
$ \therefore \quad \text { Amount }=2 P $
Case 1 :
$ \begin{aligned} 2 P & =P\left(1+\frac{i}{100}\right)^4 \\ 2 & =\left(1+\frac{i}{100}\right)^4 \end{aligned} $
Case 2 : Let after $t$ years it will be 8 time
$ \begin{array}{l} 8 P=P\left(1+\frac{i}{100}\right)^t \\ (2)^3=\left(1+\frac{i}{100}\right)^t \end{array} $
$ \begin{array}{l} {\left[\left(1+\frac{i}{100}\right)^4\right]^3=\left(1+\frac{i}{100}\right)^t} \\ \left(1+\frac{i}{100}\right)^{12}=\left(1+\frac{i}{100}\right)^t \end{array} $
By comparing both sides,
$ \begin{aligned} t & =12 \text { years } \\ \text { Time }(t) & =2 \text { years } \\ \text { Rate, } i & =4 \%=0.04 \end{aligned} $
(b)
Effective interest rate of compound interest $4 \%$ of 2 years
$ \begin{array}{l} =(1+i)^n-1 \\ =(1+0.04)^2-1 \\ =(1.04)^2-1 \\ =1.0816-1 \\ =0.0816 \approx 8.16 \% \end{array} $
Effective rate of interest of simple interest for 2 years $=8 \%$
According to the question,
8.16% of sum = ₹ 2448
1% of sum = ₹ $\frac{2448}{8.16}$
8% of sum = ₹ $\frac{2448}{8.16} \times 8$
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Question 65 Marks
(a) There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of ₹ 12,000 after 3 years at the same rate ?
(b) What is the difference between the compound interest on ₹ 5000 for $1 \frac{1}{2}$ years at $4 \%$ per annum compounded yearly and half-yearly.
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Question 75 Marks
(a) In what time will 8000 amount to ₹ 8820 at 10% per annum interest compounded half-yearly?
(b) Find the rate percent per annum if ₹ 200000 amount to ₹ 231525 in $1 \frac{1}{2}$ year interest being compounded half yeariy.
(c) What is the monthly interest rate equivalent to an annual rate of 8% capitalized quarterly ?
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Question 85 Marks
A person opened an account on April, 2019 with a deposit of 800. The account paid 6% interest compounded quarterly. On October 1, 2019 he closed the account and added enough additional money to invest in a 6 month time-deposit for ₹ 1000, earning 6% compounded monthly.
(a) How much addition amount did the person invest on October 1?
(b) What was the maturity value of his time deposit on April 1 2002 ?
(c) How much total interest was earned ?
Given that $(1+i)^n$ is 1.03022500 for $i=1 \frac{1}{2}, n=2$
and $(1+i)^n$ is 1.03037751 for $i=\frac{1}{2} \%, n=6$.
Answer
(a) The initial investment earned for April-june and uly-September quarter i.e., two quarters. in this case $i=\frac{6}{4}=1 \frac{1}{2} \%=0.015, n=\frac{6}{12} \times 4=2$ and compounded amount
$ \begin{array}{l} =800(1+0.015)^2 \\ =800 \times 1.03022500 \end{array} $
= ₹ 824.18
The additional amount invested
= ₹ (1000 - 824.18 )
= ₹ 175.82
(b) In this case the time-deposit earned interest compounded monthly for six-month
Here,
$i=\frac{6}{12}=\frac{1}{2} \%=0.005$
n= 6
p = ₹ 1000
$\begin{aligned} \text { Maturity value } & =1000(1+0.005)^6 \\ & =1000 \times 1.03037751\end{aligned}$
= ₹ 1030.38
(c) Total interest earned
= ₹ ( 824.18 - 800 ) + (1030.38 - 1000 )
= ₹ ( 24.18 + 30.38 )
= ₹ 54.56
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Question 95 Marks
(a) In what time will at 4.5% p.a? 85000 amount to 157675 (b) A sum of 46875 was lent out at simple interest and at the end of 1 year 8 months the total amount was 50,000. Find the rate of interest percent per annum.
Answer
(a) We know,
A = P(1 + it)
$\Rightarrow \quad 157675=85000\left(1+\frac{4.5}{100} \times t\right)$
$\begin{array}{ll}\Rightarrow & \frac{157675}{85000}=\frac{100+4.5 t}{100} \\ \Rightarrow & 4.5 t=\left[\frac{157675}{85000} \times 100\right]-100 \\ \Rightarrow & 4.5 t=\frac{85.5}{4.5}=19\end{array}$
In 19 years ₹ 85000 will amount to ₹ 157675 at 4.5% p.a. simple interest rate.
(b) We know,
$A=P(1+i t)$
$\Rightarrow$ $50000=46875\left(1+i 1 \frac{8}{12}\right)$
$\Rightarrow$ $\frac{50000}{46875}=1+\frac{5}{3} i$
$\Rightarrow$ $(1.067-1) \times \frac{3}{5}=i$
$\Rightarrow$ $i=0.04$
$\Rightarrow$ $i= 4$%.
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Question 105 Marks
₹ 2000 is invested at annual rate of interest of 10%. What is the amount after two years if compound-ing is done (a) Annually (b) semi-annually (c) Quarterly (d) monthly.
Answer
Here, P = ₹ 2000,$i=10 \%=\frac{10}{100}=0.1$
(a) Since, interest is compounded yearly
n = 2
Since,$ A_n=P(1+i)^n $
$\therefore \quad A_2=2000(1+0.1)^2$
$\begin{array}{l}=2000(1.1)^2 \\ =2000 \times 1.21\end{array}$
₹ 2420
(b) For semi-annual compounding
$ \begin{array}{l} n=2 \times 2=4 \\ i=\frac{0.1}{2}=0.05 \\ \therefore \quad A=2000(1+0.5)^4 \\ =2000 \times(1.05)^4 \\ =2000 \times 1.2155 \end{array} $
= ₹ 2431
(c) For quarterly compounding
n = 4 * 2 = 8
$\begin{array}{ll}\text { Since, } & i=\frac{0.1}{4}=0.025 \\ \therefore & A_4=2000 \times(1+0.025)^8\end{array}$
$\begin{array}{l}=2000 \times(1.025)^8 \\ =2000 \times 1.2184\end{array}$
= ₹ 2436.80
(d) For monthly compounding
n = 12 * 2 = 24
$\begin{array}{lrl}\text { Since, } & & i=\frac{0.1}{12}=0.00833 \\ \therefore & A_{24} & =2000 \times(1+0.00833)^{24} \\ & & =2000 \times 1.22029\end{array}$
= ₹ 2440.58
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5 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip