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Let the money lent be Rs. X

Now,

Compound Interest = P (1 + R/100)ⁿ - P

And,

SI = (P × r × n)/100

Where, P = Principal Amount

r = Rate of interest (PER annum)

n = NUMBER of years

Hence,

X × [1 + 7/100]³ - X - (X × 7 × 3)/100 = 800

X × (1.07)³ - 1.21X = 800

X = 53181

<P>Let the sum be P and rate be r.

⇒ P (1 + r/100)³ = 2000------ (1)

⇒ P (1 + r/100)⁶ = 5000------ (2)

Dividing eq(2) by eq(1), we GET,

⇒ (1 + r/100)³ = 5/2----- (3)

Putting EQUATION (3) in equation (1)

⇒ P × 5/2 = 2000

∴ P = 2000 × 2/5 = Rs. 800

LET the principle amount be P.

SI for 2 YEARS = 800

⇒ P × 12.5 × 2/100 = 800

⇒ P = 3200

CI for 2 years = 3200(1 + 12.5/100)² - 3200

⇒ 3200(1 + 1/8)² - 3200

⇒ 3200(9/8)² - 3200

CI = 4050 - 3200

∴ CI = Rs. 850

Difference between S.I. and C.I. for 2 years = 2250 - 2000 = 250

S.I. for 1 year = 2000/2 = 1000

∴ r = (250/1000) × 100

= 25% per annum

Now, S.I.(25%) = 1000

P(100%) = 1000/25× 100 = 4000

When RATE is increased by 4% then new S.I = 4000 × 29 × 2/100 = 2320

Rate of C.I. for 2 years = 29 + 29 + (29 × 29)/100

= 58 + 8.41

= 66.41%

∴ C.I. = 4000 × $(\FRAC{{66.41}}{{100}})$ = 2656.4

Now the formula for compound interest can be given as

CI = P(1 + R/100)^t - P

Where CI = Compound Interest

P = Principal

R = Rate of interest

T = Time period

In our CASE the interest EARNED at the end of first YEAR can be given as

CI = 6000(1.08)¹-6000

CI = Rs. 480

Income tax @20% is DEDUCTED on the interest earned

∴ 480 × 0.2 = 96

∴ Amount at the end of first year = 6000 + 480-96 = 6384

Now the amount at the end of first year is TAKEN as principal at the beginning of second year

∴ Interest earned at the end of second year

CI = 6384(1.08) - 6384 = Rs. 510.72

Income tax deducted = 510.72 × 0.2 = 102.144

∴ Amount at the end of second year = 6384 + 510.72 - 102.144 = 6792.576

Interest earned at the end of third year

CI = 6792.576(1.08) - 6792.576 = 543.40

Income tax deducted = 543.40 × 0.2 = 108.68

Principal = 3000

Amount $(= \;5333\frac{1}{3}\; = \;\frac{{16000}}{3})$

Time = 10 years

Let the rate be R%

Using the formula for CI

$(compound\;interest = principal{\left[ {1 + \frac{{rate}}{{100}}} \right]^{time}} - Principal)$

$(5333\frac{1}{3} = 3000{\left( {1 + \frac{R}{{100}}} \right)^{10}})$

$(\RIGHTARROW \frac{{16000}}{{3 \times 3000}} = {\left( {1 + \frac{R}{{100}}} \right)^{10}})$

$(\Rightarrow \frac{{16}}{9} = {\left( {1 + \frac{R}{{100}}} \right)^{10}})$

Taking square ROOT of the above equation

$(\frac{4}{3} = {\left( {1 + \frac{R}{{100}}} \right)^5})$…………………………………….(equation 1)

Now for half the time i.e for T = 5 years

Let the amount be A

Again using the formula for CI

$(A = 3000{\left( {1 + \frac{R}{{100}}} \right)^5})$

From equation (1)

$(\Rightarrow A = 3000\left( {\frac{4}{3}} \right))$

Let the principle amount be P and RATE of interest be r.

From the beginning of 2001 and beginning of 2004, there will be accumulated interest of 3 yrs.

So, we can write,

⇒ P (1 + r/100)³ = P + 10000 ---- (1)

Similarly, from the beginning of 2001 and beginning of 2007, there will be accumulated interest of 6 years.

We can write,

⇒ P (1 + r/100)⁶ = P + 25000 ----- (2)

Let, (1 + r/100)³ = A

We can write, EQUATION (1) as,

⇒ P ? A = P + 10000

⇒ P (A – 1) = 10000 ---- (3)

Similarly, we can write equation (2) as,

⇒ P × A² = P + 25000

⇒ P (A² – 1) = 25000

⇒ P (A – 1) (A + 1) = 25000 ----- (4)

Dividing equation (4) by equation (3), we get,

⇒ A + 1 = 5/2

⇒ A + 1 = 2.5

⇒ A = 1.5

Putting A = 1.5 in eq (3), we get,

⇒ P × (1.5 - 1) = 10000