Question

50,000 amounts to 62000 in 3 years at some rate, then by
the same rate how much 75000 amounts to in 2 years?
(1) 87000
(2) * 12000
(3) 92000
(4) 87500 show step by step​

Answer 1

AnswEr :

$\bf{ Given}\begin{cases}\sf{Principal = Rs. \:50000}\\\sf{Amount = Rs. \:62000}\\ \sf{Time = 3 \:Years}\\\sf{Rate = ?} \end{cases}$

• Let's Head to the Question Now :

$\longrightarrow \tt SI = \dfrac{P \times R \times T}{100} \\ \\\longrightarrow \tt(Amount - Principal) = \dfrac{P \times R \times T}{100} \\ \\\longrightarrow \tt62000 - 50000 = \dfrac{500 \cancel{00} \times R \times 3}{1\cancel{00}} \\ \\\longrightarrow \tt12000 = 500 \times R \times 3 \\ \\\longrightarrow \tt12000 = 1500R \\ \\\longrightarrow \tt \cancel\dfrac{12000}{1500} = R \\ \\\longrightarrow \blue{\tt Rate = 8\% \:p.a.}$

$\rule{300}{1}$

$\bf{ New Case}\begin{cases}\sf{Principal =Rs. \:75000}\\\sf{Rate = 8\%\:p.a.}\\ \sf{Time = 2\:Years}\\\sf{Amount = ?} \end{cases}$

• Calculation of Simple Interest :

$\longrightarrow \tt SI = \dfrac{P \times R \times T}{100} \\ \\\longrightarrow \tt SI = \dfrac{750\cancel{00}\times 8 \times 2}{1 \cancel{00}} \\ \\\longrightarrow \tt SI =750 \times 8 \times 2 \\ \\\longrightarrow \blue{\tt SI =Rs. \:12000}$

$\rule{300}{2}$

• A M O U N T :

↠ Amount = Principal + Simple Interest

↠ Amount = Rs. (75,000 + 12,000)

↠ Amount = Rs. 87,000

⠀

∴ Amount will turn into ( 1) Rs. 87,000.

#answerwithquality #BAL

Answer 2

$\begin{lgathered}\red{\bf{Given}}\begin{cases} \sf \: Principal=Rs.50000 \\ \sf \: Amount=Rs.62000 \\ \sf \: Time=3Years \\ \bf \green{Rate}=?\end{cases}\end{lgathered} \:$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

we know That , in simple interest ,

$\boxed{\bf \blue{Rate}= \orange{\dfrac{Simple\:Interest \times 100}{P \times T}}}$

$\boxed{\bf \: SI=Amount-Principal}$

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Putting values we get,

→ SI = 62000 - 50000 = Rs.12000

Now, putting all values in above Told Formula we get,

$\red\leadsto \sf \: Rate = \dfrac{120 \cancel{00} \times \cancel{100}}{5 \cancel{0000} \times 3} \\ \\ \red\leadsto \large\boxed{\sf \pink{Rate}= \purple{8}\%}$

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Again Now,,,

$\begin{lgathered} \red{\bf{Given}}\begin{cases} \sf \: Principal=Rs.75000 \\ \sf \: Time=2years \\ \sf \: Rate=8\%(same) \\ \bf \orange{Amount=?}\end{cases}\end{lgathered}$

Now, we know that,,,

$\red{\boxed{\bf \: SI=\dfrac{P \times R \times T}{100}}} \\ \\ \\ \blue{\boxed{\bf \: Amount=Principal+SI}}$

__________________________________

Putting values again we get,,,

$\red\longrightarrow \bf \: SI = \dfrac{750 \cancel{00} \times 8 \times 2}{ \cancel{100}} \\ \\ \red\longrightarrow \orange{\bf \: SI \: =rs. 12000}$

So, our Amount will be :----

$\red\longrightarrow \bf \: Amount = 75000 + 12000 \\ \\\red\longrightarrow \: \large\red{\boxed{\tt\blue{Amount}\purple{=} \green {Rs} \orange,\pink{87000}}}$

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Hence , Option A is Right Answer....

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$\rule{200}{4}$

$\Large\bold\star\underline{\underline\textbf{Extra\:Brainly\:Knowledge}}$

For a principal amount of P at the interest rate of r per annum in t years if the interest is I then ...

1) For Simple Interest

$\bf\rightarrow \boxed{I=\frac{P\times r\times t}{100}}$

2)For Compound Interest

$\bf\rightarrow \boxed{I=P(1+\frac{r}{100}){}^{nt}}$

where, n=interest period

Ex. If the principal is compounded

annually(per year) then ,

n=1

If the principal is compounded

half yearly(in every 6 months) then ,

n=2

If the principal is compounded

quarterly (in every 3 months) then ,

n=4....

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#BAL

#answerwithquality