The difference of compound interest & simple interest of the first year on 8,00,000 at 16% is 0.
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Answer:
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Step-by-step explanation:
=2000(105100)2 =2000(2120)2 =2205 Rupees∴ Compound Interest after 2 years,I = Amount − Principal =2205−2000 =205 Rupees
Hence, Amount = ₹ 2205 and Compound interest = ₹ 205.
(2) Here, P = ₹ 5000; R = 8 % ; N = 3 years
A=P(1+R100)N =5000(1+8100)3 =5000(108100)3 =5000(2725)3 =6298.56 Rupees∴ Compound Interest after 3 years,I = Amount − Principal =6298.56−5000 =1298.56 Rupees
Hence, Amount = ₹ 6298.56 and Compound interest = ₹ 1298.56
(3) Here, P = ₹ 4000; R = 7.5 % ; N = 2 years
A=P(1+R100)N =4000(1+7.5100)2 =4000(1+751000)2 =4000(10751000)2 =4000(4340)2 =4622.50 Rupees∴ Compound Interest after 2 years,I = Amount − Principal =4622.50−4000 =622.50 Rupees
Hence, Amount = ₹ 4622.50 and Compound interest = ₹ 622.50
Page No 90:
Question 2:
Sameerrao has taken a loan of ₹12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
ANSWER:
Here, P = ₹ 12500; R = 12 % ; N = 3 years
A=P(1+R100)N =12500(1+12100)3 =12500(1+325)3 =12500(2825)3 =17561.60 Rupees
Hence, he should pay an amount of ₹ 17561.60 to clear his loan.
Page No 90:
Question 3:
To start a business Shalaka has taken a loan of ₹8000 at a rate of 1012 p.c.p.a. After two years how much compound interest will she have to pay?
ANSWER:
Here, P = ₹ 8000; R = 1012 % ; N = 2 years
A=P(1+R100)N =8000(1+1012100)2 =8000(1+21200)2 =8000(221200)2 =9768.20 Rupees∴ Compound Interest after 2 years,I = Amount − Principal =9768.20−8000 =1768.20 Rupees
Hence, she will have to pay a Compound interest of ₹ 1768.20 after 2 years.
Page No 93:
Question 1:
On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
ANSWER:
Here, P = Number of workers initially = 320
A = Number of workers after 2 years
R = Rate of increase of number of workers per year = 25 %
N = 2 years
A=P(1+R100)N =320(1+25100)2 =320(1+14)2 =320(54)2 =500
Hence, the number of workers after 2 years is 500.
Page No 93:
Question 2:
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
ANSWER:
Here, P = Number of sheeps initially = 200
A = Number of sheeps after 3 years
R = Rate of increase of number of sheeps per year = 8 %
N = 3 years
A=P(1+R100)N =200(1+8100)3 =200(1+225)3 =200(2725)3 =251.94 =252 (approx)
Hence, the number of sheeps after 3 years is 252.
Page No 93:
Question 3:
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
ANSWER:
Here, P = Number of trees initially = 40,000
A = Number of trees after 3 years
R = Rate of increase of number of trees per year = 5 %
N = 3 years
A=P(1+R100)N =40000(1+5100)3 =40000(1+120)3 =40000(2120)3 =46305
Hence, the expected number of trees after 3 years is 46,305.
Answer:
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