Question

Find the time period (n) for which money is borrowed for the following cases.(a) P = 2,000, r = 2%, I =500. Solution​

Answer 1

Answer:

Given :-

• Principal is Rs 2000, rate of interest is 2% and the simple interest is Rs 500.

To Find :-

• What is the time period (n).

Formula Used :-

$\clubsuit$ Time Period (n) Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{n =\: \dfrac{S.I \times 100}{P \times r}}}}\: \: \: \bigstar$

where,

• n = Time Period
• S.I = Simple Interest
• P = Principal
• r = Rate of Interest

Solution :-

Given :

• Principal = Rs 2000
• Rate of Interest = 2%
• Simple Interest = Rs 500

According to the question by using the formula we get,

$\implies \bf n =\: \dfrac{S.I \times 100}{P \times r}$

$\implies \sf n =\: \dfrac{500 \times 100}{2000 \times 2}$

$\implies \sf n =\: \dfrac{50\cancel{000}}{4\cancel{000}}$

$\implies \sf n =\: \dfrac{50}{4}$

$\implies \sf\bold{\red{n =\: 12.5\: years}}$

$\therefore$ The time period (n) for which the money is borrowed is 12.5 years .

Answer 2

Answer:

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Class 11

>>Applied Mathematics

>>Basics of financial mathematics

>>Accumulation with simple and compound interest

>>I borrowed Rs. 12,000 from Jamshed at 6%

Question

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I borrowed Rs.12,000 from Jamshed at 6% per annum simple interest for 2 years. If I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay

Medium

Solution

verified

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Given P=Rs.12000N=2R=6%

S.I.=

100

PNR

SI =

100

12000×6×2

=1440

A=SI+P=1440+12000= Rs. 13440

Compound interest:

A=P(1+

100

r

)

n

=12000(1+

100

6

)

(2)

=12000(1.1236)

=Rs.13483.2

Amount =Rs.13483.20

CI=A−P=13483.20−12000

=Rs.1483.20

Extra interest =13483.20−13440=Rs.43.20