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Contact [7]
3 years ago
9

Need help ASAP !!!!!!!!

Mathematics
1 answer:
luda_lava [24]3 years ago
3 0

Answer:

correct amount- 31.63

what you got- 63.31

Step-by-step explanation:

Im not good at explaining but i'll write out a step by step process

100b+.01a-.05=2a+2b

100b+.01a .05=2a+2b

100b-2b=2a- .01a+ .05

98b= 1.99a+ .05

b=  (1.99a+ .05)/98

a=31

b=.63

63.26 is twice as much as 31.63

we go to the same school btw :)

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Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

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P = \dfrac{100}{21} \times 5,460 = 26,000

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

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