Mathematics, 26.12.2021 17:40 bwalker411

The difference between CI and SI for 3 years at 10% per annum is rupees 62. find the sum. No spam, please..

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Answers

Answer from: cjdolce9790

» 2,000

Step-by-step explanation:

Given :

CI - SI difference for 3 years = Rs. 62

Rate = 10%

Time = 3 years

Formula used :

CI - SI = $p( \frac{r}{100} ) {}^{2} \times \frac{(300 + r)}{100} \\$

Calculation:

CI - SI = $p( \frac{r}{100} ) {}^{2} \times \frac{(300 + r)}{100} \\$

⇒ 62 = $p( \frac{10}{100} ) {}^{2} \times \frac{(300 + 10)}{100} \\$

⇒ 62 = $p( \frac{100}{10000} ) \times \frac{(310)}{100} \\$

⇒ P = (62) × ( $\frac{10000}{100} )$ × ( $\frac{100}{310}$ )

⇒ P = 2,000

∴ The required sum is Rs. 2,000.

Answer from: xatziry1717

The sum (Principle) is Rs.2000.

Step-by-step explanation:

Given :

➝ C.I - S.I = Rs.62➝ Time = 3 years➝ Rate = 10% per annum

$\begin{gathered}\end{gathered}$

To Find :

➝ Sum = Principal

$\begin{gathered}\end{gathered}$

Using Formulas :

${\star{\small{\underline{\boxed{\sf{\pink{C.I = P \bigg( 1 + \dfrac{R}{100} \bigg)^{T}}}}}}}}$

➝ C.I = Compound Interest ➝ P = Principle ➝ R = Rate ➝ T = Time

${\star{\small{\underline{\boxed{\sf{\pink{S.I = \dfrac{ P \times R \times T}{100}}}}}}}}$

➝ S.I = Simple Interest ➝ P = Principle ➝ R = Rate ➝ T = Time

$\begin{gathered}\end{gathered}$

Solution :

Here, the difference between the compound interest and simple interest is 62. So,

$\longrightarrow{\sf{C.I - S.I = 62}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg( 1 + \dfrac{R}{100} \bigg)^{T} - P\bigg\} - \Bigg\{\dfrac{ P \times R \times T}{100} \Bigg\} = 62}}}$

Substituting all the given values in the formulas.

${\longrightarrow{\sf{ \Bigg\{P \bigg( 1 + \dfrac{10}{100} \bigg)^{ 3} - P\bigg\} - \Bigg\{\dfrac{ P \times 10 \times 3}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg(\dfrac{100 + 10}{100} \bigg)^{ 3} - P\bigg\} - \Bigg\{\dfrac{ P \times 30}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg(\dfrac{110}{100} \bigg)^{ 3} - P\bigg\} - \Bigg\{\dfrac{ P \times 30}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg( \cancel{\dfrac{110}{100}} \bigg)^{ 3} - P\bigg\} - \Bigg\{\dfrac{ P \times 30}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg(\dfrac{11}{10} \bigg)^{3} - P\bigg\} - \Bigg\{\dfrac{ P \times 30}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{P \bigg(\dfrac{1331}{1000} \bigg) - P\bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \bigg(\dfrac{P \times 1331}{1000} \bigg) - P\bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \bigg(\dfrac{1331P}{1000} \bigg) - P\bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{1331P}{1000} - P\bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{1331P - 1000P}{1000} \bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{331P}{1000} \bigg\} - \Bigg\{\dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{331P}{1000} - \dfrac{30P}{100} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{331P - 300P}{1000} \Bigg\} = 62}}}$

${\longrightarrow{\sf{ \Bigg\{ \dfrac{31P}{1000} \Bigg\} = 62}}}$

${\longrightarrow{\sf{\dfrac{31P}{1000} = 62}}}$

${\longrightarrow{\sf{P= 62 \times \dfrac{1000}{31}}}}$

${\longrightarrow{\sf{P= \cancel{62} \times \dfrac{1000}{\cancel{31}}}}}$

${\longrightarrow{\sf{P= 2 \times 1000}}}$

${\longrightarrow{\sf{P= Rs.2000}}}$

${\star \: {\large{\underline{\boxed{\sf{\red{P= Rs.2000}}}}}}}$

Hence, the sum (Principle) is Rs.2000.

$\underline{\rule{220pt}{3.5pt}}$

Answer from: Quest

i think it is 18

step-by-step explanation:

If y varies directly as x, and y is 180b when x is n when x is 5, what is the value of n?

Answer from: Quest

Hello!

the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

why?

the root or roots of a polynomial are the values of the variable that makes the polynomial equal to 0. we can find the roots of a polynomial making the expression equal to 0 and then, isolating the variable (it it has just one variable).

for example, given the following polynomial:

$3x-6$

the root or zero of the polynomial, can be found by making the expression equal to 0, and then, isolating the variable, so, finding the root we have:

$3x-6=0$

$3x=6$

$x=\frac{6}{3}=2$

therefore, we have that the root of the polynomial is equal to 2.

we must remember that the number of roots that a polynomial can have, will depend on the degree of the polynomial, a first degree polynomial have will have one single root, a second degree polynomial (quadratic expression) may have two roots and so.

hence the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

have a nice day!