Math, asked by vipinbatham02, 14 hours ago

can someone clear how to factorize any question ​

Answers

Answered by ffkpgamer30
0

Answer:

Principal, P = Rs 20000

Rate of interest, r = 10 % per annum

Time, n = 1 year

We know,

Simple interest (SI) received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

\begin{gathered}\boxed{ \rm{ \:SI \: = \: \frac{P \times r \times n}{100} \: }} \\ \end{gathered}

SI=

100

P×r×n

So, on substituting the values, we get

\begin{gathered}\rm \: SI_1 \: = \: \dfrac{20000 \times 10 \times 1}{100} \\ \end{gathered}

SI

1

=

100

20000×10×1

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:SI_1 \: = \: Rs \: 2000 \: \: }} - - - (1) \\ \end{gathered}

SI

1

=Rs2000

−−−(1)

Case :- 2

Principal, P = Rs x

Rate of interest, r = 5 % per annum

Time, n = 1 year

So,

\begin{gathered}\rm \: SI_2 = \dfrac{x \times 5 \times 1}{100} \\ \end{gathered}

SI

2

=

100

x×5×1

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:SI_2 \: = \: \frac{5x}{100} \: \: }} - - - (2) \\ \end{gathered}

SI

2

=

100

5x

−−−(2)

Case :- 3

Principal, P = Rs (20000 + x)

Rate of interest, r = 7 % per annum

Time, n = 1 year

So,

\begin{gathered}\rm \: SI_3 \: = \: \dfrac{(20000 + x) \times 7 \times 1}{100} \\ \end{gathered}

SI

3

=

100

(20000+x)×7×1

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:SI_3 \: = \: \frac{140000 + 7x}{100} \: }} - - - (3)\\ \end{gathered}

SI

3

=

100

140000+7x

−−−(3)

Now, According to statement

\begin{gathered}\rm \: SI_3 = SI_1 + SI_2 \\ \end{gathered}

SI

3

=SI

1

+SI

2

On substituting the values from equation (1), (2) and (3), we get

\begin{gathered}\rm \: \dfrac{140000 + 7x}{100} = 2000 + \dfrac{5x}{100} \\ \end{gathered}

100

140000+7x

=2000+

100

5x

\begin{gathered}\rm \: \dfrac{140000 + 7x}{100} = \dfrac{200000 + 5x}{100} \\ \end{gathered}

100

140000+7x

=

100

200000+5x

\begin{gathered}\rm \: 140000 + 7x = 200000 + 5x \\ \end{gathered}

140000+7x=200000+5x

\begin{gathered}\rm \: 7x - 5x = 200000 - 140000 \\ \end{gathered}

7x−5x=200000−140000

\begin{gathered}\rm \: 2x = 60000 \\ \end{gathered}

2x=60000

\begin{gathered}\rm\implies \:x = 30000 \\ \end{gathered}

⟹x=30000

So,

\begin{gathered}\rm \:Total\:investment \: = \: 20000 +x \\ \end{gathered}

Totalinvestment=20000+x

\begin{gathered}\rm \:Total\:investment \: = \: 20000 +30000\\ \end{gathered}

Totalinvestment=20000+30000

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