[3 marks sum]

Question 13 Marks

A sum of money, lent out at simple interest, doubles itself in $8$ years. Find $:(i)$ the rate of interest $(ii)$ in how many years will the sum become triple $($three times$)$ of itself at the same rate percent?

Answer

Let $P= Rs.100, A=R s .200$
$I=\text { Rs. } 200-\text { Rs. } 100=\text { Rs. } 100, T=8 \text { years }$
$ R=\frac{100 \times I}{P \times T}=\frac{100 \times 100}{100 \times 8}$
$ =\frac{100}{8}=\frac{25}{2} \%$
Now again $P= Rs. 100$
$A=\text { Rs. } 300$
$ I=\text { Rs. } 300-\text { Rs. } 100=\text { Rs. } 200$
$ R=\frac{25}{2} \%$
$ T=\frac{100 \times I}{P \times R}=\frac{100 \times 200}{100 \times \frac{25}{2}}=\frac{100 \times 200 \times 2}{100 \times 25}$
$ =16 \text { Years }$
So the given sum of money will become triple in $16$ years.

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Question 23 Marks

The interest on a certain sum of money is $0.24$ times itself in $3$ years. Find the rate of interest.

Answer

Let the sum borrowed $= Rs.100$
Time $=3$ years
Let rate of interest $=r \%$
$\therefore$ Interest $=\frac{100 \times 3 \times \mathrm{r}}{100} \ldots\left[\because \mathrm{S} . \mathrm{I}=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\right]$
$=3 r=(0.24)(100)=24 \ldots \ldots . .($ Given $)$
$\Rightarrow \mathrm{r}=\frac{24}{3}=8$
Hence reqd. rate of interest $=8 \%$

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Question 33 Marks

The simple interest on a certain sum of money is $\frac{3}{8}$ of the sum in $6 \frac{1}{4}$ years. Find the rate percent charged.

Answer

Let $P= Rs. 8$
$\text { S.I. }=\text { Rs. } \frac{3}{8} \times 8$
$ =\text { Rs. } 3$
$\mathrm{T}=6 \frac{1}{4} \text { years }=\frac{25}{4} \text { years }$
We know that:
$R=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{T}}$
$ =\frac{100 \times 3}{8 \times \frac{25}{4}}=\frac{100 \times 3}{8} \times \frac{4}{25}=2 \times 3$
$ =6 \%$

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Question 43 Marks

Find the time in which $Rs.1200$ will amount to $Rs.1536$ at $3.5\%$ per year.

Answer

$

A=R s .1536$
$ P=\text { Rs. } 1200$
$ I=A-P$
$ =\text { Rs. } 1536 \text { - Rs. } 1200$
$ =\text { Rs. } 336$
We know that:
$\mathrm{T}=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{R}}$
$ =\frac{100 \times 336}{1200 \times 3.5}=\frac{100 \times 336 \times 10}{1200 \times 35} \ldots . . .\left[\because \frac{1}{3.5}=\frac{10}{35}\right]$
$ =\frac{28 \times 10}{35}=8 \text { years }$

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Question 53 Marks

In how many years will $Rs.950$ produce $Rs.399$ as a simple interest at $7\%$?

Answer

$P=\text { Rs. } 950$
$\text { S.I. }=\text { Rs. } 399$
$R=7 \%$
We know that:
$\mathrm{T}=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{R}}=\frac{100 \times 399}{950 \times 7}$
$ =\frac{10 \times 21}{5 \times 7}=2 \times 3=6 \text { years }$

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Question 63 Marks

At what rate, percent per year will a sum double itself in $6 \frac{1}{4}$ years?

Answer

Let $P= Rs. 100$
$\therefore$ Amount $=2 \times Rs. 100= Rs. 200$
Interest $=\mathrm{A}-\mathrm{P}$
$= Rs. 200 - Rs. 100$
$= Rs. 100$
$\mathrm{T}=6 \frac{1}{4}$ years $=\frac{25}{4}$ years
$R=\frac{100 \times I}{P \times T}=\frac{100 \times 100}{100 \times \frac{25}{4}} \%=\frac{100 \times 100}{100} \times \frac{4}{25}$
$=16 \%$

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Question 73 Marks

The interest on a certain sum of money is $ Rs. 1,480$ in $2$ years and at $10$ percent per year. Find the sum of money.

Answer

Let $P=Rs.x$
Time $(T)=2$ years
Rate $(R)=10 \%$
$\therefore$ Interest $=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{\mathrm{x} \times 10 \times 2}{100}=\frac{\mathrm{x}}{5}$
$\frac{\mathrm{x}}{5}= Rs.1480 \quad...($Given$)$
$\therefore x=1480 \times 5= Rs. 7400$
Hence the money $Rs. 7400.$

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Question 83 Marks

Rohit borrowed $Rs. 24,000$ at $7.5$ percent per year. How much money will he pay at the end of the $4th$ years to clear his debt?

Answer

Principal $(P)= Rs. 24,000$
Rate $(R)=7.5 \% P.A.$
Time $(T)=4$ years
$\text { S.I. }=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}$
$ =\text { Rs. } \frac{24,000 \times 4 \times 7.5}{100}$
$=\text { Rs. } 240 \times 4 \times 7.5$
$ =240 \times 30$
$=Rs. 7200$
Amount needed to clear the debt at the end of $4th$ year
$=\text { Rs. } 24000+\text { Rs. } 7200=\text { Rs. } 3,1200$

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Question 93 Marks

Find the interest and the amount on: $₹ 4,000$ in $1 \frac{1}{3}$ years at $2$ paise per rupee per month.

Answer

Here $P= Rs. 4,000$ , Time $(T)=1 \frac{1}{3}$ years
$=1$ years $+\frac{12}{3}$ months $=16$ months
Rate $(R)=2$ paise per rupee per month $=2 \%$ per month
$\therefore$ Interest $(\mathrm{I})=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{4,000 \times 2 \times 16}{100}$
$ =40 \times 32$
$ =\text { Rs. } 1280$
$\therefore$ Amount $(A)=P+1$
$=\text { Rs. } 4000+\text { Rs. } 1280$
$=\text { Rs. } 5280$

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Question 103 Marks

Find the interest and the amount on: $₹ 2,600$ in $2$ years $3$ months at $1\%$ per month.

Answer

Here $P=\text {₹} 2600$
Time $(T)=2$ Years $3$ months $=27$ months
Rate $(R)=1 \%$ per month
$\therefore$ Interest $=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{2600 \times 27 \times 1}{100}$
$ =26 \times 27= Rs \ 702$
$\therefore$ Amount $= Rs.(2600+702)= Rs. 3302$

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Question 113 Marks

Find the interest and the amount on: ₹ $750$ in $3$ years $4$ months at $10\%$ per annum.

Answer

Given $P = ₹750$
Time $(T)=3 \frac{4}{12}=3 \frac{1}{3}$
$=\frac{10}{3}$ years Rate $(R) = 10\%$
$\therefore$ Interest $( I )=\frac{ PRT }{100}=\frac{750 \times 10 \times \frac{10}{3}}{100}$
$=\frac{250 \times 10 \times 10}{100}$
$= ₹250$
$\therefore$ Amount $(A) = P + I$
$= ₹750 + ₹250$
$= ₹1000$

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MATHS [3 marks sum]

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