In how many years 2500 rupees will amount to 2704 rupees at a compound interest rate of 4 pcpa

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Mathematics Solutions Solutions for Class 8 Math Chapter 15 Compound Interest are provided here with simple step-by-step explanations. These solutions for Compound Interest are extremely popular among Class 8 students for Math Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 90:

Question 1:

Find the amount and the compound interest.

Answer:

(1) Here, P = ₹ 2000; R = 5 % ; N = 2 years

A=P1+R100N   =2000 1+51002   =20001051002   =200021202   =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=P1+R100N   =50001+81003   =5000 1081003   =500027253   =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=P1+R100 N   =40001+7.51002   =40001+751000 2   =4000107510002   =400043402   =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=P1+R100N   = 125001+121003   =125001+3253   =125002825 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=P1+R100 N   =80001+10121002   =80001+21 2002   =80002212002   =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=P1+R100N    =3201+251002   =3201+142   =32054 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=P1+R100N   =200 1+81003   =2001+2253   =20027253    =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=P1+R100 N   =400001+51003   =400001+1203   =4000021203   =46305

Hence, the expected number of trees after 3 years is 46,305.

Page No 93:

Question 4:

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.

Answer:

Here, P = Cost price of the machine = 2,50,000
A = Cost price after 2 years
I = Depreciation in price after 2 years
R = Rate of depreciation = 10 %
N = 2 years

A=P1+R 100N   =2500001+-101002   =2500001-110 2   =2500009102   =202500

Also,
I = P − A
  = 250000 − 202500
  = 47500

Hence, the depreciation in price of the machine after two years is Rs 47,500.

Page No 93:

Question 5:

Find the compound interest if the amount of a certain principal after two years is ₹4036.80 at the rate of 16 p.c.p.a.

Answer:

Here, P = Principal
A = ₹ 4036.80
I = Compound Interest
R = 16 %
N = 2 years

A=P1+ R100N⇒4036.80=P1+161002⇒4036.80=P1 +4252⇒4036.80=P29252⇒P=4036.80×25×2529×29 ⇒P=3000

Also,
I = A − P
  = 4036.80 − 3000
  = 1036.80

Hence, the compound interest is ₹ 1036.80

Page No 93:

Question 6:

A loan of ₹15000 was taken on compound interest. If the rate of compound interest s 12 p.c.p.a. find the amount to settle the loan after 3 years.

Answer:

Here, P = Principal = ₹ 15000
A = Amount
R = 12 %
N = 3 years

A=P1+R100N   =150001 +121003   =150001+3253   =1500028253   =21073.92

Hence, the amount is ₹ 21073.92

Page No 93:

Question 7:

A principal amounts to ₹13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

Answer:

Here, P = Principal
A = ₹ 13924
R = 18 %
N = 2 years

A=P1+R100N⇒13924=P1+181002⇒13924=P1+9502⇒13924=P59502⇒P=13924 ×50×5059×59⇒P=10000

Hence, the principal is ₹ 10000.

Page No 93:

Question 8:

The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.

Answer:

Here, P = Population of a suburb = 16000
A = Population after two years =17640
R = R %
N = 2 years

A=P1+R100N⇒17640=160001 +R1002⇒1764016000=1+R1002⇒441400=1+R1002⇒21202=1+R1002⇒1+R100=21 20⇒R100=2120-1⇒R100=120⇒R=5

Hence, the rate of increase in the population is 5 p.c.p.a.

Page No 93:

Question 9:

In how many years ₹700 will amount to ₹847 at a compound interest rate of 10 p.c.p.a.

Answer:

Here, P = Principal = ₹ 700
A = Amount = ₹ 847
R = 10 %
N = N years

A=P1+R 100N⇒847=7001+10100N⇒847700=1+110N⇒121100=1110N⇒11102=1110N ⇒N=2

Hence, the number of years is 2 years.

Page No 93:

Question 10:

Find the difference between simple interest and compound interest on ₹20000 at 8 p.c.p.a.

Answer:

Disclaimer: In the question "Time" is not given. So the question is solved taking time as 2 years, because simple interest and compound interest will be same for one year.

Here, P = Principal = ₹ 20000
R = 8 %
N = 2 years

Simple interest=P×R×N100                           =20000×8×2100                           =3200Amount=P1 +R100N              =200001+81002               =200001+2252               =2000027252              =23328 Compound interest=23328-20000                                   =3328Compound interest-Simple interest=3328-3000                                                                =128

Hence, the difference between simple interest and compound interest is ₹ 128.

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