Math, asked by answer78, 11 months ago

compound interest practice set 14.1 compound interest practice set 14.1 Standard 8

Answers

Answered by knjroopa
7

Step-by-step explanation:

Given Compound interest practice set 14.1 compound interest practice set 14.1 Standard 8

  • Find the amount and compound interest for P = 2000, R = 5% p.a and time = 2 years
  • So A = P( 1 + R / 100)^n
  •         = 2000 (1 + 5/100)^2
  •         = 2000 (105 / 100)^2
  •         = 2000 (21 / 20)^2
  •        = 5 x 441
  • A = Rs 2205
  • I = A – P
  •   = 2205 – 2000
  • I = Rs 205
  • Similarly we can do for Rs 5000 and Rs 4000
  • So A = P( 1 + R / 100)^n
  •         = 5000 (1 + 8/100)^3
  •         = 5000 (108 / 100)^3
  •         = 5000 (27 / 25)^3
  •        = 200 x 27 x 27 / 25 x 27 / 25
  • A = Rs 6,298.56
  • I = A – P
  •   = 6298.56 – 5000
  • I = Rs 1298.56
  • Sameer rao has taken a loan of Rs 12,500 at the rate of 12% p.a for 3 years, if the interest is compounded annually then how many rupees should he pay to clear his loan?
  • A = p(1 + r/100)^n
  •    = 12,500 (1 + 12 / 100)^3
  •   = 12,500 (112 / 100)^3
  •   = 12,500 (28 / 25)^3
  •  = 12,500 x 28 / 25 x 28/25 x 28/25
  •   = Rs 17,561.60

Sameer rao should pay Rs 17,561.60 to clear his loan

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