Question

Find the time in which RS.6000 will produce Rs1986 as compound interest at 10% compounded annually.​

Answer 1

Given :-

• RS.6000 will produce Rs1986 as compound interest at 10% compounded annually.

To find :-

• Time

Solution :-

• Compound interest = Rs.1986

• Principle = Rs6000

As we know that

→ Compound Interest = Amount - Principle

→ C.I = A - P

→ A = C.I + P

→ A = 1986 + 6000

→ A = 7986

Hence, amount is Rs.7986

• Rate of interest = 10%

• A = P(1 + R/100)ⁿ

Where " R " is rate and " n " is time

→ 7986 = 6000(1 + 10/100)ⁿ

→ 7986 = 6000(1 + 1/10)ⁿ

→ 7986 = 6000(10 + 1/10)ⁿ

→ 7986 = 6000(11/10)ⁿ

→ 7986/6000 = (11/10)ⁿ

→ 2662/2000 = (11/10)ⁿ

→ 1331/1000 = (11/10)ⁿ

→ (11/10)³ = (11/10)ⁿ

→ 3 = n

→ n = 3

Hence,

• Time is 3 years

Answer 2

$\sf{\pink{\underline{\underline{\purple{GIVEN:-}}}}}$

• Compound interest (C.I) = Rs. 1986

• Principle (P) = Rs. 6000

• Rate (r) = 10%

$\sf{\pink{\underline{\underline{\purple{TO\: FIND:-}}}}}$

• The time .

$\sf{\pink{\underline{\underline{\purple{SOLUTION:-}}}}}$

We have know that,

☃️ C.I = Amount - Principle

$\orange\star\:\bf{\gray{\boxed{\pink{\boxed{\red{Amount\:=\:C.I\:+\:P\:}}}}}}$

=> Amount (A) = 1986 + 6000

=> Amount (A) = Rs. 7986

$\green\star\:\bf{\gray{\boxed{\pink{\boxed{\red{Amount\:(A)\:=\:P\:(1\:+\:\dfrac{R}{100})^n\:}}}}}}$

Where,

• n = Time

$\bf{\implies\:7986\:=\:6000\:(1\:+\:\dfrac{10}{100})^n\:}$

$\bf{\implies\:\dfrac{7986}{6000}\:=\:(1\:+\:\dfrac{1}{10})^n\:}$

$\bf{\implies\:\dfrac{7986}{6000}\:=\:(\dfrac{11}{10})^n\:}$

$\bf{\implies\:\dfrac{3993}{3000}\:=\:(\dfrac{11}{10})^n\:}$

$\bf{\implies\:\dfrac{1331}{1000}\:=\:(\dfrac{11}{10})^n\:}$

$\bf{\implies\:\dfrac{(11)^3}{(10)^3}\:=\:(\dfrac{11}{10})^n\:}$

$\bf{\implies\:(\dfrac{11}{10})^3\:=\:(\dfrac{11}{10})^n\:}$

$\bf\purple{\implies\:n\:=\:3\:}$

$\rm\red{\therefore}$ The time is “ 3 years ” .