Math, asked by aparnakaser22, 4 months ago

find simple interest of 7800 from 15 November 2001 to 14 February 2002 at 6 % per annum but kinte month ho gaya vo batao n kyo​

Answers

Answered by anmol1383
2

Answer:

(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

Answered by othakur1991
0

Answer:

117

Step-by-step explanation:

p=7800, r=6, t=3 month

formula = p*r*t/100

convert 3 month into year so t=3/12=1/4 year

so (7800*6*1/4)/100 =117

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