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’p’ is considered the principal AMOUNT

’r’ is considered the rate of interest

’t’ is considered the number of years

C.I is the compound interest

‘n’ is the number of TIMES compound interest is calculated in a year

‘A’ is the final amount after the compound interest

For the first year:

We know that

$({\rm{A\;}} = {\rm{\;p\;}} \times {\rm{\;}}(1{\rm{\;}} + \frac{r}{{100\; \times \;n}})$)^{(n × t)}

 $(\therefore {\rm{\;A\;}} = {\rm{\;}}21,000{\rm{\;}} \times {\rm{\;}}\left( {1{\rm{\;}} + {\rm{\;}}\frac{{25}}{{100\; \times \;2}}} \right))$^{(1 × 2)}

∴ A = 26578.125

Thus the MONEY Mr.Chopra got after year 1 is RS 26578.125

This will be the principal amount for the 2^nd year

For the second year:

$({\rm{A\;}} = {\rm{\;p\;}} \times {\rm{\;}}(1{\rm{\;}} + {\rm{\;}}\frac{r}{{100\; \times \;n}})$)^{(n × t)}

∴ A = 26578.125 × (1 + 25/100)

∴ A = 33222.65

LET the amount deposited by Adolf be Rs. x.

We know the formula for compound interest:-

$(\Rightarrow {\bf{CI}} = \left[ {{\bf{P}}\left\{ {{{\left( {1 + \frac{{\bf{R}}}{{100}}} \right)}^t} - 1} \right\}} \right])$

Where,

CI = Compound interest

P = PRINCIPAL

R = Rate of interest

t = Time period

(? Interest = Amount – Principal)

$(\Rightarrow {\rm{\;}}9408 - {\rm{x}} = {\rm{}}\left[ {{\rm{x}}\left\{ {{{\left( {1{\rm{}} + {\rm{}}\frac{{12}}{{100}}} \right)}^2} - 1} \right\}} \right])$

⇒ x(1.2544-1) = 0.2544x

⇒ 9408 = 0.2544x + x

⇒ x = 9408/1.2544 = Rs. 7,500

Quantity 1:

Let the sum be RS. 100. Then,

⇒ S.I. for first six MONTHS = [(100 × 10 × 1)/100 × 2] = Rs. 5

⇒ S.I. for NEXT six months = [(105 × 10 × 1)/100 × 2] = Rs. 5.25

⇒ So, amount at the end of 1 YEAR = Rs. (100 + 5 + 5.25) = Rs. 110.25

⇒ EFFECTIVE rate = (110.25 – 100) = 10.25%

Quantity 2:

let the original rate be r%

Then the new rate will be 2r%

⇒ (725 × r × 1)/100 + (362.50 × 2r × 1)/100 × 3 = 33.50

⇒ (2175 + 725) r = 33.50 × 100 × 3 = 10050

⇒ r = 10050/2900 = 3.46%

⇒ Original rate = 3.46%

Quantity 3:

Let the interest be r%

Difference between SI – CI

⇒ [15000 × (1 + r/100)² – 15000] – [15000 × r × 2/100] = 96

⇒15000 × [(1 + r/100)² – 1 – (r × 2/100)] = 96

⇒ 15000 × [(r + 100)² – 10000 – 200 r)/10000] = 96

⇒ r² = 96 × 2/3 = 64

⇒ r = 8%

Comparing all three quantities

∴ Quantity 1 > Quantity 3 > Quantity 2

Now the formula for simple interest can be given as

SI = P × R × T/100

Where SI = Simple Interest, P = PRINCIPAL, R = Rate of interest, T = Time period

∴ 600 = P × R × 2/100

∴ P × R = 30000-------- equation 1

Now the formula for compound interest can be given as

CI = P(1 + R/100)^t - P

Where CI = Compound Interest

∴ 630 = P(1 + R/100)² - P-----------equation 2

Put the value of P from equation 1 in equation 2

∴ 630 = 30000/R(1 + R/100)² - 30000/R

∴ 630 = 30000/R(1 + 2R/100 + R²/10000) - 30000/R (Using the formula (a + b)² = a² + 2ab + b²)

∴ 630 = 30000/R + 600 + 3R - 30000/R

∴ 630 = 600 + 3R

∴ 3R = 30

Quantity A:

Price of mobile = Rs. 26040 and rate of interest = 12.5% = 1/8

Let the amount of each installment is Rs. x and principals for three years are P₁, P₂ and P₃.

First year:

x = P₁ (1 + 1/8)

⇒ P₁ = 8x/9

Second year:

x = P₂ (1 + 1/8)²

⇒ P₂ = 64x/81

Third year:

x = P₃ (1 + 1/8)³

⇒ P₃ = 512x/729

Sum of principle for 3 years = 26040.

 ∴ 8x/9 + 64x/81 + 512x/729 = 26040

⇒ 1736x/729 = 26040

⇒ x = 15 × 729

⇒ x = Rs. 10935

∴ Amount of each installment = Rs. 10935

Quantity B:

Man borrowed a sum of Rs. 25000 from BANK at 10% interest PER annum.

∴ Amount after 1 year = 25000 × 1.1 = Rs. 27500

Since he pays Rs. 10000 at the end of each year.

∴ Principal for second year = 27500 – 10000 = Rs. 17500

∴ Amount after 2 year = 17500 × 1.1 = Rs. 19250

Principal for third year = 19250 – 10000 = Rs. 9250

∴ Amount after 3 year = 9250 × 1.1 = Rs. 10175

∴ He will have to pay Rs. 10175 at the end of 3^rd year to clear his dues.