Question

T = 6 months
(a) P=37200
(b) P= 9000
R = 7.5% per annum, ,
R = 8% per annum,
T= 2 years 6 months​

Answer 1

Answer:

hope it is helpful to you

Step-by-step explanation:

P denotes the principal ($), R denotes the rate (percentage p.a.) and T denotes time (years), then:-

S.I = (P × R × T)/100

R = (S.I × 100)/(P × T)

P = (S.I × 100)/(R × T)

T = (S.I × 100)/(P × R)

If the denotes the amount, then A = P + S.I

Note:

● When we calculated the time period between two dates, we do not could the day on which money is deposited but we count the day on which money is retuned.

● Time is always taken according to the per cent rat.

● For converting time in days into years, divide th number of days by 365 (for ordering or lap year.)

● For converting time in month into years, divide th number of month by 12 (for ordering or lap year.)

Examples to find or calculate simple interest when principal, rate and time are known

Calculate Simple Interest

Find the simple interest on:

(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.

Solution:

P = $ 900,

R = 5% p.a.

T = 3 years 4 months = 40/12 years = 10/3 years

Therefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150

Amount = P + S.I = $ 900 + $ 150 = $ 1050

(b) $ 1000 for 6 months at 4% per annum. Find the amount also.

Solution:

P = $ 1000,

R = 4% p.a.

T = 6 months = 6/12 years

S.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20

Therefore, A = P + I = $( 1000 + 20) = $ 1020

(c) $ 5000 for 146 days at 15¹/₂% per annum.

Solution:

P = $ 5000, R = 151/2% p.a. T = 146 days

S.I = ( 5000 × 31 × 146)/(100 × 2 × 365)

= $ 10 × 31 = $ 310

(d) $ 1200 from 9ᵗʰ April to 21ˢᵗ June at 10% per annum.

Solution:

P = $ 1200, R = 10% p.a. T = 9th April to 21st June

= 73 days [April = 21, May = 31, Jun = 21, 73 days]

= 73/365 years

S.I = (1200 × 10 × 73)/(100 × 365) = $ 24

Examples to find or calculate Time when Principal, S.I and Rate are known

Calculate Simple Interest

1. In how much time dose $ 500 invested at the rate of 8% p.a. simple interest amounts to $ 580.

Solution:

Here P = $ 500, R = 8% p.a A = $ 580

Therefore S.I = A - P = $ (580 - 500) = $ 80

Therefore T = (100 × S.I)/(P × R) = (100 × 80)/(500 × 3) = 2 years

2. In how many years will a sum of $ 400 yield an interest of $ 132 at 11% per annum?

Solution:

P = $ 400, R = 11% S.I = $ 132

T = (100 × S.I)/(P × R) = (132 × 100)/(400 × 11) = 3 years