Question

A SUM OF MONEY PUT AT 11% PER ANNUM SIMPLE INTEREST AMOUNTS TO RS.10370 IN 2 YEARS . WHAT WILL IT AMOUNT TO IN 3 YEARS AT THE SAME RATE​

Answer 1

Given ;

Amount = ₹10,370

Rate = 11%

Time = 2 years

To Find :-

Amount for 3 years on interest rate being same.

Let Principal be ₹x

Formula :-

SI = ${\dfrac{P \times R \times T}{100}}$

On Inserting the value in the formula

we get ,

SI = ${\dfrac{x \times 11 \times 2}{100}}$

SI = ${\dfrac{22x}{100}}$

SI = ${\dfrac{11x}{50}}$

A = P + I

= ₹x + ${\dfrac{11x}{50}}$

= ${\dfrac{50x + 11x}{50}}$

= ${\dfrac{61x}{50}}$

Also given that, A = ₹10,370

Acc. to question,

₹10,370 = ${\dfrac{61x}{50}}$

=> x = $\dfrac{10370 \times 50}{61}$

=> x = ₹8500

Therefore, Principal = ₹x = ₹8500

For 3 years :-

P = ₹8500

R = 11%

T = 3 years

Formula :-

SI = ${\dfrac{P \times R \times T}{100}}$

On Inserting the value in the formula

We get,

SI = ${\dfrac{8500 \times 11 \times 3}{100}}$

SI = 85 × 11 × 3

SI = ₹ 2805

Amount = Principal + Interest

= ₹8500 + ₹2805

= ₹11,305

Therefore, amount for 3 years = ₹11,305

Short-hands Used :-

• SI = Simple Interest.
• P = Principal.
• R = Rate.
• T = Time.
• A = Amount.
• Acc = According.

$\bf{ \underline{Some\: Formulae}} \\ \\ \rightarrow \quad \sf P = \dfrac{100 \times S.I}{R \times T} \\ \\ \\ \rightarrow \quad T = \dfrac{100 \times S.I}{P \times R} \\ \\ \\ \rightarrow \quad R = \dfrac{100 \times S.I}{P \times T} \\ \\ \\ \rightarrow \quad A = P + S.I \\ \\ \\ \rightarrow \quad S.I = \dfrac{P \times R \times T}{100}$

Answer 2

so answer is ₹11305

Formula

P× R×T /100

Steps

Rs x× 11×2/100

S.I =22/50

S.I = 11/50

x+50x+11/50

61/50

so ₹10370×50/61

8500

S.I =8500×11×3/100

2805

A=p+s.i

=₹8500-2805

=₹11305