[5 marks sum]

Question 15 Marks

A sum of money was invested for $3$ years, interest being compounded annually. The rates for successive years were $10\%, 15\%$ and $18\%$ respectively. If the compound interest for the second year amounted to $Rs. 4,950$, find the sum invested.

Answer

Given : $C.I.$ for the $2^{\text {nd }}$ year $= Rs. 4,950$ and rate $=15 \%$
Then, $\text { C.I. }=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$ \Rightarrow 4,950=P\left[\left(1+\frac{15}{100}\right)^1-1\right]$
$ \Rightarrow 4,950=P\left[\frac{3}{20}\right]$
$ \Rightarrow P=\frac{4,950 \times 20}{3}$
$ \Rightarrow P=\text { Rs. } 33,000 .$
Then amount at the end of $2^{\text {nd }}$ year$= Rs. 33,000$
For first $2$ years
$A=\text { Rs. } 33,000 ; r_1=10 \%$
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r_1}{100}\right)$
$\Rightarrow 33,000=P\left(1+\frac{10}{100}\right)$
$ \Rightarrow 33,000=P\left(\frac{11}{10}\right)$
$\Rightarrow \mathrm{P}=\frac{33,000 \times 10}{11}=Rs. 30,000$
The sum invested is $Rs. 30,000 .$

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Question 25 Marks

The value of a machine, purchased two years ago, depreciates at the annual rate of $10\%$. If its present value is $Rs.97,200$, find:

Its value after $2$ years.
Its value when it was purchased.

Answer

$(i)$ Present value of machine $(\mathrm{P})=\mathrm{Rs} .97,200$
Depreciation rate $=10 \%$
$\therefore$ Value of machine after $2$ years $=\mathrm{P}\left(1-\frac{r}{100}\right)^n$
$=97,200\left(1-\frac{10}{100}\right)^2$
$ =97,200\left(\frac{9}{10}\right)^2$
$ =\text { Rs. } 78732 .$
$(ii)$ Present value of machine $(A)= Rs. 97,200$
Depreciation rate $=10 \%$ and time $=2$ years
To calculate the cost $2$ years ago
$\therefore A=P\left(1-\frac{r}{100}\right)^n$
$ \Rightarrow 97,200=P\left(1-\frac{10}{100}\right)^2$
$ \Rightarrow 97,200=P\left(\frac{9}{10}\right)^2$
$ \Rightarrow P=\text { Rs. } 97,200 \times\left(\frac{10}{9}\right)^2=1,20,000$

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Question 35 Marks

A sum of money, invested at compound interest, amounts to $Rs. 16,500$ in $1$ year and to $Rs. 19,965$ in $3$ years. Find the rate per cent and the original sum of money invested.

Answer

Let sum of money be $Rs. \mathrm{P}$ and rate of interest $=\mathrm{r} \%$
Money after $1$ year $= Rs. 16,500$
Money after $3$ years $= Rs. 19,965$
For $1$ year
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 16,500=\mathrm{P}\left(1+\frac{r}{100}\right)^1 \dots...(1)$
For $3$ years
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 19,965=\mathrm{P}\left(1+\frac{r}{100}\right)^3 \dots...(2)$
Divide eq ${ }^{\mathrm{n}}(2)$ by eq ${ }^{\mathrm{n}}(1)$
$ \frac{19,965}{16,500}=\frac{P\left(1+\frac{r}{100}\right)^3}{P\left(1+\frac{r}{100}\right)^1}$
$ \Rightarrow \frac{121}{100}=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow\left(\frac{11}{10}\right)^2=\left(1+\frac{r}{100}\right)^2$
On comparing, we get
$\Rightarrow \frac{11}{10}=1+\frac{r}{100}$
$ \Rightarrow r=10 \%$
Put value of $r$ in eq ${ }^n(1)$
$16,500=P\left(1+\frac{10}{100}\right)$
$ \Rightarrow P=\frac{16,500 \times 10}{11}=\text { Rs. } 15,000 .$

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Question 45 Marks

Ashok invests a certain sum of money at $20\%$ per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half$-$yearly. If Geeta gets $Rs. 33$ more than Ashok in $18$ months, calculate the money invested.

Answer

$(i)$ For Ashok $($interest is compounded yearly$) :$
Let $P= Rs. y ; n=18$ months $=1 \frac{1}{2}$ year and $r=20\ \%\ p.a.$
For $1$ year
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n=y\left(1+\frac{20}{100}\right)^1=\left(\frac{6}{5}\right) y$
For $\frac{1}{2}$ year
$\mathrm{P}=$ Rs. $\left(\frac{6}{5}\right) y ; n=\frac{1}{2}$ year and $\mathrm{r}=20 \%$
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=R s .\left(\frac{6}{5}\right) y\left(1+\frac{20}{2 \times 100}\right)^{\frac{1}{2} \times 2}=R s .\left(\frac{66}{50}\right) y$
$(ii)$ For Geeta $($ interest is compounded half$-$yearly $)$
$P=$ Rs. $y ; n=1 \frac{1}{2}$ year and $r=20 \% p.a.$
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=y\left(1+\frac{20}{2 \times 100}\right)^{\frac{3}{2} \times 2}=y\left(\frac{11}{10}\right)^3$
$ =\text { Rs. } \frac{1,331}{1,000} y$
According to question
$\therefore \frac{1,331}{1,000} y-\left(\frac{66}{50}\right) y=R s .33$
$ =\left(\frac{11}{1,000}\right) y=R s .33$
$ =\mathrm{y}=\text { Rs. } \frac{33 \times 1,000}{11}=R s .3,000$
Money invested by each person$=Rs. 3,000.$

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MCQ 55 Marks

Find the difference between the compound interest compounded yearly and half$-$yearly on $Rs. 10,000$ for $18$ months at $10\%$ per annum.

Answer

(i) When interest is compounded yearly:
Given : $P=$ Rs. 10,$000 ; n=18$ months $=1 \frac{1}{2}$ year and $r=10 \%$ p.a.
For 1 year
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n\left(1+\frac{r}{200}\right)^{\frac{1}{\mathcal{Z}}} \times \not 2$
$ =10,000 \times \frac{11}{10} \times \frac{21}{20}$
$=1020 \times 11$
$ =11,550$
$\text { When interest is compounded half-yearly }$
$A=P\left(1+\frac{r}{200}\right)^{n \times 2}$
$=10,000\left(1+\frac{10}{200}\right)^{\frac{3}{\nexists} \times \not 2}$
$ =10,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$ =\frac{105 \times 441}{4}$
$ =\frac{46305}{4}$
$=11576.25$
Therefore, the difference between both $C.I$
$₹11,576.25 - ₹11,550$
$= ₹26.25$

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Question 65 Marks

Kamal and Anand each lent the same sum of money for $2$ years at $5\%$ at simple interest and compound interst respectively. Anand recived $Rs. 15$ more than Kamal. Find the amount of money lent by each and the interest received.

Answer

Let principal $= Rs. 100, R=5 \% T=2$ years
For Kamal, $SI =\frac{100 \times 5 \times 2}{100}= Rs. 10$
For Anand, $ A=P\left[1+\frac{R}{100}\right]^T$
$=100\left[1+\frac{5}{100}\right]^2$
$ =100 \times \frac{21}{20} \times \frac{21}{20}$
$=\frac{441}{4}$
$\mathrm{Cl}=\frac{441}{4}-100=\frac{41}{4}$
Difference of $\mathrm{Cl}$ and $\mathrm{SI}=\frac{41}{4}-10$
$=\frac{41-40}{4}$
$ =\text { Rs. } \frac{1}{4}$
When difference is $Rs. \frac{1}{4}$, then principal = Rs. 100 .
If difference is $1$ , principal $=100 \times 4$
If difference is $Rs. 15$ , principal $=100 \times 4 \times 15= Rs. 6000$
For kamal, interest $=\frac{6000 \times 5 \times 2}{100}= Rs. 600 .$
For Anand, interest $=6000\left[1+\frac{5}{100}\right]^2-6000$
$=6000\left[\left(\frac{21}{20}\right)^2-1\right]$
$ =6000\left[\frac{441}{400}-1\right]$
$ =6000 \times \frac{41}{400}$
$ =\text { Rs. } 615$

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Question 75 Marks

A sum of money, invested at compound interest, amounts to $Rs. 19,360$ in $2$ years and to $Rs. 23,425.60$ in $4$ years. Find the rate per cent and the original sum of money.

Answer

$($for $2$ years$) A = Rs. 19360$
$T=2$ years
Let $P=X$
$\mathrm{X}\left(1+\frac{\mathrm{R}}{100}\right)^2=19,360 \dots......(1)$
$A ($for $4$ years$) = Rs. 23425.60$
$X\left(1+\frac{R}{100}\right)^4=23425.60\dots......(2)$
$ 2 \div 1$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{23425.60}{19360}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{2342560}{1936000}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{14641}{12100}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\left[\frac{121}{110}\right]^2}$
$ 1+\frac{\mathrm{R}}{100}=\frac{121}{110}$
$ \frac{\mathrm{R}}{100}=\frac{121}{110}-1$
$R=10 \%$
From $(1) \times\left[1+\frac{10}{100}\right]^2=19360$
$X=\frac{19360 \times 10 \times 10}{11 \times 11}$
$X= Rs. 16,000$
Thus, sum$ = Rs. 16000$

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Question 85 Marks

Mohit borrowed a certain sum at $5\%$ per annum compound interest and cleared this loan by paying $Rs. 12,600$ at the end of the first year and $Rs. 17,640$ at the end of the second year. Find the sum borrowed.

Answer

For the payment of $Rs. 12,600$ at the end of first year : $A= Rs. 12,600 ; n=1$ year and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 12,600=\mathrm{P}\left(1+\frac{5}{100}\right)^1$
$ \Rightarrow 12,600=\mathrm{P}\left(\frac{21}{20}\right)$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times 12,600=\text { Rs. } 12,000$
For the payment of $Rs. 17,640$ at the end of second year : $A= Rs. 17,640, n=2$ years and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 17,640=\mathrm{P}\left(1+\frac{5}{100}\right)^2$
$ \Rightarrow 17,640=\mathrm{P}\left(\frac{21}{20}\right)^2$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times \frac{20}{21} \times 17,640=\text { Rs. } 16,000$
$\therefore$ Sum borrowed $= Rs. (12,000+16,000)= Rs. 28,000$.

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Question 95 Marks

A sum of $Rs. 44,200$ is divided between John and Smith, $12$ years and $14$ years old respectively, in such a way that if their portions be invested at $10\%$ per annum compound interest, they will receive equal amounts on reaching $16$ years of age.
$(i)$What is the share of each out of $Rs.44,200$ ?
$(ii)$ What will each receive, when $16.years$ old ?

Answer

$(i)$ Let share of John $=\text { Rs y }$
share of Smith $=\text { Rs }(44,200-y)$
rate of interest $=10 \%$
According to question,
Amount of John in $4$ years $=$ Amount of Smith in $2$ years
$\Rightarrow y\left(1+\frac{10}{100}\right)^4=(44,200-y)\left(1+\frac{10}{100}\right)^2$
$ \Rightarrow y\left(1+\frac{10}{100}\right)^2=(44,200-y)$
$ \Rightarrow y\left(\frac{11}{10}\right)^2=(44,200-y)$
$ \Rightarrow 121 y=100(44,200-y)$
$ \Rightarrow 121 y=100 \times 44,200-100 y$
$ \Rightarrow 121 y+100 y=100 \times 44,200$
$ \Rightarrow 221 y=100 \times 44,200$
$ \Rightarrow y=100 \times 44, \frac{200}{221}=\text { Rs. } 20,000 .$
Therefore share of John $=Rs. 20,000$
Share of Smith $= Rs. 44,200- Rs. 20,000= Rs. 24,200$
$(ii)$ Amount that each will receive
$=20,000\left(1+\frac{10}{100}\right)^4$
$ =20,000\left(\frac{11}{10}\right)^4$
$= Rs. 29,282$

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MATHEMATICS [5 marks sum]

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