Rd Sharma Solutions for Class 8 Math Chapter 14 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 Rd Sharma Book of Class 8 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 14.14:

Question 1:

Compute the amount and the compound interest in each of the following by using the formulae when:
(i) Principal = Rs 3000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3000, Rate = 18%, Time = 2 years
(iii) Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years
(v) Principal = Rs 12800, Rate = 712%, Time = 3 years
(vi) Principal = Rs 10000, Rate 20% per annum compounded half-yearly, Time = 2 years
(vii) Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half-yearly, Time = 2 years.

Answer:

Applying the rule A=P1+R100n on the given situations, we get:(i)A=3,0001+51002=3,0001.052=Rs 3,307.50Now,CI=A-P=Rs 3,307.50-Rs 3,000=Rs 307.50(ii)A=3,0001+181002=3,0001.182=Rs 4,177.20Now,CI=A-P=Rs 4,177.20-Rs 3,000=Rs 1,177.20(iii)A=5,0001+101002=5,0001.102=Rs 6,050Now,CI=A-P=Rs 6,050-Rs 5,000=Rs 1,050(iv)A=2,0001+41003=2,0001.043=Rs 2,249.68Now,CI=A-P=Rs 2,249.68-Rs 2,000=Rs 249.68(v)A=12,8001+7.51003=12,8001.0753=Rs 15,901.40Now,CI=A-P=Rs 15,901.40-Rs 12,800=Rs 3,101.40(vi)A=10,0001+202004=10,0001.14=Rs 14,641Now,CI=A-P=Rs 14,641-Rs 10,000=Rs 4,641(vii)A=16,0001+102004=16,0001.054=Rs 19,448.1Now,CI=A-P=Rs 19,448.1-Rs 16,000=Rs 3,448.1

Page No 14.14:

Question 2:

Find the amount of Rs 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.

Answer:

Given: P=Rs 2,400R=20% p.a.n=3 yearsWe know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by A=P1+R100n. A=2,4001+201003        =2,4001.23        =4,147.20Thus, the required amount is Rs 4,147.20.

Page No 14.14:

Question 3:

Rahman lent Rs 16000 to Rasheed at the rate of 1212% per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.

Answer:

Given: P=Rs 16,000 R=12.5% p.a.n=3 yearsWe know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by A=P1+R100n. A=16,0001+12.51003        =16,0001.1253        =22,781.25Thus, the required amount is Rs 22,781.25.

Page No 14.14:

Question 4:

Meera borrowed a sum of Rs 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back.

Answer:

Given: P=Rs 1,000R=10% p.a. n=2 yearsWe know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by A=P1+R100. A=1,0001+101002=1,0001.12=1,210Thus, the required amount is Rs 1,210.

Page No 14.14:

Question 5:

Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.

Answer:

Given: P=Rs 50,000R=10% p.a. n=2 yearsWe know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by A=P1+R100. A=Rs 50,0001+101002        =Rs 50,0001.12        =Rs 60,500Also,CI=A-P    =Rs 60,500- Rs 50,000    =Rs 10,500We know that:SI=PRT100   =50,000×10×2100   =Rs 10,000 Difference between CI and SI=Rs 10,500-Rs 10,000                                                         =Rs 500



Page No 14.15:

Question 6:

Amit borrowed Rs 16000 at 1712% per annum simple interest. On the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years?

Answer:

Amount to be paid by Amit:SI=PRT100    =16000×17.5×2100    =Rs 5,600Amount gained by Amit:A=P1+R100n   =Rs 16,0001+17.51002   =Rs 16,0001.1752   =Rs 22,090We know that:CI=A-P    =Rs 22,090-Rs 16,000    =Rs 6090Amit's gain in the whole transaction=Rs 6,090-Rs 5,600                                                               =Rs 490

Page No 14.15:

Question 7:

Find the amount of Rs 4096 for 18 months at 1212% per annum, the interest being compounded semi-annually.

Answer:

Given:P=Rs 4,096R=12.5% p.a.n=18 months=1.5 yearsWe have:A=P1+R100nWhen the interest is compounded semi-annually, we have:A=P1+R2002n  =Rs 4,0961+12.52003  =Rs 4,0961.06253  =Rs 4,913Thus, the required amount is Rs 4,913.

Page No 14.15:

Question 8:

Find the amount and the compound interest on Rs 8000 for 112 years at 10% per annum, compounded half-yearly.

Answer:

Given: P=Rs 8,000R=10% p.a. n=1.5 yearsWhen compounded half-yearly, we have:A=P1+R2002n  =Rs 8,0001+102003  =Rs 8,0001.053  =Rs 9,261Also,CI=A-P    =Rs 9,261-Rs 8,000    =Rs 1,261

Page No 14.15:

Question 9:

Kamal borrowed Rs 57600 from LIC against her policy at 1212% per annum to build a house. Find the amount that she pays to the LIC after 112 years if the interest is calculated half-yearly.

Answer:

Given: P=Rs 57,600R=12.5% p.a.n=1.5 yearsWhen the interest is compounded half-yearly, we have: A=P1+R2002n  =Rs 57,6001+12.52003  =Rs 57,6001.06253  =Rs 69,089.06Thus, the required amount is Rs 69,089.06.

Page No 14.15:

Question 10:

Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.

Answer:

Given:P=Rs 64,000R=5% p.a. n=1.5 yearsWhen the interest is compounded half-yearly, we have:A=P1+R2002n  =Rs 64,0001+52003  =Rs 64,0001.0253  =Rs 68,921Also,CI=A-P    =Rs 68,921-Rs 64,000    =Rs 4,921Thus, the required interest is Rs 4,921.

Page No 14.15:

Question 11:

Rakesh lent out Rs 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?

Answer:

Given:P=Rs 10,000R=20% p.a.n=2 yearsA=P1+R100n  = Rs 10,0001+201002  =Rs 10,0001.22  =Rs 14,400When the interest is compounded half-yearly, we have:A=P1+R2002n  =Rs 10,0001+202004  =Rs 10,0001.14  =Rs 14,641Difference=Rs 14,641-Rs 14,400                  =Rs 241

Page No 14.15:

Question 12:

Romesh borrowed a sum of Rs 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years.

Answer:

Given: P=Rs 245,760R=12.5% p.a.n=2 yearsWhen compounded annually, we have:A=P1+R100n   =Rs 245,7601+12.51002   =Rs 311,040When compounded semi-annually, we have:A=P1+R2002n   =Rs 245,7601+12.52004   =Rs 245,7601.06254   =Rs 313,203.75Romesh's gain=Rs 313,203.75-Rs 311,040                            =Rs 2,163.75

Page No 14.15:

Question 13:

Find the amount that David would receive if he invests Rs 8192 for 18 months at 1212% per annum, the interest being compounded half-yearly.

Answer:

Given:P=Rs 8,192R=12.5% p.a.n=1.5 yearsWhen the interest is compounded half-yearly, we have:A=P1+R2002n   =Rs 8,1921+12.52003   =Rs 8,1921.06253   =Rs 9,826Thus, the required amount is Rs 9,826.

Page No 14.15:

Question 14:

Find the compound interest on Rs 15625 for 9 months, at 16% per annum, compounded quarterly.

Answer:

Given: P=Rs 15,625R=16%=164=4% quarterlyn=9 months=3 quartersWe know that:A=P1+R100n   =Rs 15,6251+41003   =Rs 15,6251.043   =Rs 17,576Also,CI=A-P    =Rs 17,576-Rs 15,625    =Rs 1,951Thus, the required compound interest is Rs 1,951.

Page No 14.15:

Question 15:

Rekha deposited Rs 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.

Answer:

Given:P=Rs 16,000R=20% p.a.n=1 yearWe know that:A=P1+R100nWhen compounded quarterly, we have:A=P1+R4004n   =Rs 16,0001+204004   =Rs 16,0001.054   =Rs 19,448.10Also,CI=A-P    =Rs 19,448.1-Rs 16,000    =Rs 3,448.10Thus, the interest received by Rekha after one year is Rs 3,448.10.

Page No 14.15:

Question 16:

Find the amount of Rs 12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.

Answer:

Given: P=Rs 12,500R1=15% p.a.R2=16% p.a. Amount after two years=P1+R11001+R2100=Rs 12,5001+151001+16100=Rs 12,5001.151.16=Rs 16,675Thus, the required amount is Rs 16,675.

Page No 14.15:

Question 17:

Ramu borrowed Rs 15625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after 214 years?

Answer:

Given: P=Rs 15,625R=16% p.a.n=214 years Amount after 214 years=P1+R10021+14(R)100=Rs 15,6251+1610021+164100=Rs 15,6251.1621.04=Rs 21,866Thus, the required amount is Rs 21,866.

Page No 14.15:

Question 18:

What will Rs 125000 amount to at the rate of 6%, if the interest is calculated after every 3 months?

Answer:

Because interest is calculated after every 3 months, it is compounded quarterly.Given:P=Rs 125,000R=6% p.a.=64% quarterly=1.5% quarterlyn=4So,A=P1+R100n  =125,0001+1.51004  =125,0001.0154  =132,670 approxThus, the required amount is Rs 132,670.

Page No 14.15:

Question 19:

Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs 12000 as simple interest.

Answer:

P=SI×100RTAccording to the given values, we have: =12,000×1005×3=80,000The principal is to be compounded annually. So, A=P1+R100n    =80,0001+51003    =80,0001.053    =92,610Now,CI=A-P    =92,610-80,000    =12,610Thus, the required compound interest is Rs 12,610.  

Page No 14.15:

Question 20:

A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.

Answer:

A=P1+R100nAlso, P=A-CILet the sum of money be Rs x.If the interest is compounded annually, then:A1=x1+201002    =1.44x CI=1.44x-x         =0.44x         ...(1)If the interest is compounded half-yearly, then:A2=x1+101004    =1.4641x CI=1.4641x-x         =0.4641x    ...(2)It is given that if interest is compounded half-yearly, then it will be Rs 482 more.0.4641x=0.44x+482    [From (1) and (2)]0.4641x-0.44x=4820.0241x=482x=4820.0241  =20,000Thus, the required sum is Rs 20,000.  

Page No 14.15:

Question 21:

Simple interest on a sum of money for 2 years at 612% per annum is Rs 5200. What will be the compound interest on the sum at the same rate for the same period?

Answer:

P=SI×100RT P=5,200×1006.5×2        =40,000Now,A=P1+R100n   =40,0001+6.51002   =40,0001.0652   =45,369Also,CI=A-P    =45,369-40,000    =5,369Thus, the required compound interest is Rs 5,369.  

Page No 14.15:

Question 22:

Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1200 as simple interest.

Answer:

We know that:P=SI×100RT P=1200×1005×3        =8,000Now,A=P1+R100n   =8,0001+51003   =8,0001.053   =9,261Now,CI=A-P    =9,261-8,000    =1,261Thus, the required compound interest is Rs 1,261.  



Page No 14.20:

Question 1:

On what sum will the compound interest at 5% per annum for 2 years compounded annually be Rs 164?

Answer:

Let the sum be Rs x.We know that:CI=A-P    =P1+R100n-P    =P1+R100n-1164= x1+51002-1164=x1.052-1x=1640.1025  =1,600Thus, the required sum is Rs 1,600.

Page No 14.20:

Question 2:

Find the principal if the interest compounded annually at the rate of 10% for two years is Rs 210.

Answer:

Let the sum be Rs x.We know that:CI=A-P    =P1+R100n-P    =P1+R100n-1210=x1+101002-1210=x1.102-1x=2100.21  =1,000Thus, the required sum is Rs 1,000.

Page No 14.20:

Question 3:

A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Answer:

Let the sum be Rs x.Then,A= P1+R100n   =P1+R100n756.25=x1+101002756.25=x1.102x=756.251.21  =625Thus, the required sum is Rs 625.

Page No 14.20:

Question 4:

What sum will amount to Rs 4913 in 18 months, if the rate of interest is 1212% per annum, compounded half-yearly?

Answer:

Let the sum be Rs x.Given: A=Rs 4913R=12.5% n=18 months=1.5 yearsWe know that:A=P1+R2002n4,913=P1+R2002n4,913=x1+12.520034,913=x1.06253x=4,9131.1995  =4,096Thus, the required sum is Rs 4,096.

Page No 14.20:

Question 5:

The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.

Answer:

Given:CI-SI=Rs 283.50R=15%n=3 yearsLet the sum be Rs x.We know that:A = P(1+R100)n    = P(1+R100)n    =x(1+15100)3     =x1.153                     ...(1)Also,SI = PRT100 = x(15)(3)100= 0.45 xA = SI + P = 1.45x      ...(2)Thus, we have:x1.153-1.45x=283.50   [From (1) and (2)]1.523x-1.45x=283.500.070875x=283.50x=283.500.070875  =4,000Thus, the sum is Rs 4,000.

Page No 14.20:

Question 6:

Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1290 as interest compounded annually, find the sum she borrowed.

Answer:

Let the money borrowed by Rachana be Rs x.Then, we have:CI=P1+R100n-P1,290=x1+151002-11,290=x0.3225x=1,2900.3225  =4,000Thus, Rachana borrowed Rs 4,000.

Page No 14.20:

Question 7:

The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.

Answer:

Let the time period be n years.Then, we have:CI=P1+R100n-P163.20=2,0001+4100n-2,0002,163.20=2,0001.04n1.04n=2,163.202,0001.04n=1.08161.04n=1.042On comparing both the sides, we get:n=2Thus, the required time is two years.

Page No 14.20:

Question 8:

In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest?

Answer:

Let the time period be n years.Thus, we have:CI=P1+R100n-P6,655=5,0001+10100n-5,00011,655=5,0001.10n1.1n=11,6555,0001.1n=2.3311.1n=1.13On comparing both the sides, we get:n=3Thus, the required time is three years.

Page No 14.20:

Question 9:

In what time will Rs 4400 become Rs 4576 at 8% per annum interest compounded half-yearly?

Answer:

Let the time period be n years.R=8%=4% (Half-yearly)Thus, we have:A=P1+R100n4,576=4,4001+4100n4,576=4,4001.04n1.04n=4,5764,0001.04n=1.041.04n=1.041On comparing both the sides, we get: n=1Thus, the required time is half a year.

Page No 14.20:

Question 10:

The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.

Answer:

Given:CI-SI=Rs 20P1+41002-P-P×4×2100=20P1.042-1-0.08P=200.0816 P-0.08 P=200.0016 P=20P=200.0016  =12,500Thus, the required sum is Rs 12,500.

Page No 14.20:

Question 11:

In what time will Rs 1000 amount to Rs 1331 at 10% per annum, compound interest?

Answer:

Let the time be n years.Then,A=P1+10100n1,331=1,0001+10100n1.1n=1,3311,0001.1n=1.3311.1n=1.13On comparing both the sides, we get:n=3Thus, the required time is three years.

Page No 14.20:

Question 12:

At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years?

Answer:

Let the rate of interest be R%.Then,A=P1+R100n774.40=6401+R10021+R1002=774.406401+R1002=1.211+R1002=1.121+R100=1.1R100=0.1R=10Thus, the required rate of interest is 10% per annum.

Page No 14.20:

Question 13:

Find the rate percent per annum if Rs 2000 amount to Rs 2662 in 112 years, interest being compounded half-yearly?

Answer:

Let the rate of interest be R%.Then,A=P1+R100n2,662=2,0001+R10031+R1003=2,6622,0001+R1003=1.3311+R1003=1.131+R100=1.1R100=0.1R=10Because the interest rate is being compounded half-yearly, it is 20% per annum.

Page No 14.20:

Question 14:

Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.

Answer:

Let the sum be Rs P and the rate of interest be R%.We know that Kamla paid Rs 200 as simple interest. 200 = PR(2)100PR=10,000    ...(1)Also, Kamla received Rs 210 as compound interest. 210=P(1+R100)2-1 210(10,000) = P(R2+200R) 210R =  R2+200R         [from (1)]R=10% p.a.Putting the equation in (1), we get:P = 1,000Thus, the required sum is Rs 1,000 and the rate of interest is 10%

Page No 14.20:

Question 15:

Find the rate percent per annum, if Rs 2000 amount to Rs 2315.25 in an year and a half, interest being compounded six monthly.

Answer:

Let the rate percent per annum be R.Because interest is compounded every six months, n will be 3 for 1.5 years.Now,A=P1+R200n2,315.25=2,0001+R20031+R2003=2,315.252,0001+R2003=1.1576251+R2003=1.0531+R200=1.05R200=0.05=10Thus, the required rate is 10% per annum.

Page No 14.20:

Question 16:

Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

Answer:

Let the rate percent per annum be R.Then,A=P1+R100n2P=P1+R10031+R1003=21+R100=1.2599R100=0.2599R=25.99Thus, the required rate is 25.99% per annum.



Page No 14.21:

Question 17:

Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly.

Answer:

Let the rate percent per annum be R.Then,A=P1+R2n4P=P1+R20041+R2004=41+R200=1.4142R200=0.4142R=82.84Thus, the required rate is 82.84%.

Page No 14.21:

Question 18:

A certain sum amounts to Rs 5832 in 2 years at 8% compounded interest. Find the sum.

Answer:

Let the sum be P.Thus, we have:A=P1+R100n5,832=P1+810025,832=1.1664PP=5,8321.1664   =5,000Thus, the required sum is Rs 5,000.

Page No 14.21:

Question 19:

The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.

Answer:

Let the sum be P.Thus, we have:CI-SI=360P1+R100n-P-P×7.5×2100=360P1+7.51002-1-P×7.5×2100=360P1.155625-1-0.15P=3600.155625P-0.15P=3600.005625P=360P=3600.005625P=64000 Thus, the required sum is Rs 64,000.

Page No 14.21:

Question 20:

The difference in simple interest and compound interest on a certain sum of money at 623% per annum for 3 years is Rs 46. Determine the sum.

Answer:

Given:CI-SI=46P1+R100n-1-PRT100=46P1+203003-1-P×20×33×100=464,0963,375P - P5-P=46(4,096-3,375-675)P3,375=46P=46×3,37546  =3,375Thus, the required sum is Rs 3,375.

Page No 14.21:

Question 21:

Ishita invested a sum of Rs 12000 at 5% per annum compound interest. She received an amount of Rs 13230 after n years. Find the value of n.

Answer:

A=P1+R100n13,230=12,0001+5100n1.05n=13,23012,0001.05n=1.10251.05n=1.052On comparing both the sides, we get:n=2Thus, the value of n is two years.

Page No 14.21:

Question 22:

At what rate percent per annum will a sum of Rs 4000 yield compound interest of Rs 410 in 2 years?

Answer:

Let the rate percent be R.We know that: CI=P1+R100n-P410=4,0001+R1002-4,0004,410=4,0001+R10021+R1002=4,4104,0001+R1002=1.10251+R1002=1.0521+R100=1.05R100=0.05R=5Thus, the required rate percent is 5.

Page No 14.21:

Question 23:

A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.

Answer:

A=P1+R100n10,404=P1+2100210,404=P1.022P=10,4041.0404P=10,000Thus, the required sum is Rs 10,000.

Page No 14.21:

Question 24:

In how much time will a sum of Rs 1600 amount to Rs 1852.20 at 5% per annum compound interest?

Answer:

A=P1+R100n1852.20=16001+5100n1852.201600=1.05n1.05n=1.1576251.05n=1.053On comparing both the sides, we get:n=3Thus, the required time is three years.

Page No 14.21:

Question 25:

At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest?

Answer:

A=P1+R100n1102.50=10001+R10021102.501000=1+0.01R21+0.01R2=1.10251+0.01R2=1.052On comparing both the sides, we get:1+0.01R=1.050.01R=0.05R=5Thus, the required rate percent is 5.

Page No 14.21:

Question 26:

The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.

Answer:

CI=P1+R100n-P378=1,8001+10100n-1,8001,8001+10100n=2,1781+10100n=2,1781,8001.1n=1.211.1n=1.12On comparing both the sides, we get: n=2Thus, the required time is two years.

Page No 14.21:

Question 27:

What sum of money will amount to Rs 45582.25 at 634% per annum in two years, interest being compounded annually?

Answer:

A=P1+R100n45,582.25=P1+274002P1.06752=45,582.25P=45,582.251.13955625P=40,000Thus, the required sum is Rs 40,000.

Page No 14.21:

Question 28:

Sum of money amounts to Rs 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.

Answer:

A=P1+R100n453,690=P1+6.51002P1.0652=453,690P=453,6901.134225P=400,000Thus, the required sum is Rs 400,000.



Page No 14.27:

Question 1:

The present population of a town is 28000. If it increases at the rate of 5% per annum, what will be its population after 2 years?

Answer:

Here,P=Initial population=28,000R=Rate of growth of population=5% per annumn=Number of years=2 Population after two years=P1+R100n=28,0001+51002=28,0001.052=30,870Hence, the population after two years will be 30,870.

Page No 14.27:

Question 2:

The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.

Answer:

Here, P=Initial population=125,000Annual birth rate=R1=5.5%Annual death rate=R2=3.5%Net growth rate, R=R1-R2=2%n=Number of years=3 Population after three years=P1+R100n=125,0001+21003=125,0001.023=132,651Hence, the population after three years will be 132,651.

Page No 14.27:

Question 3:

The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.

Answer:

Here,P=Initial population=25,000R1=4%R2=5%R3=8%n=Number of years=3 Population after three years=P1+R11001+R21001+R3100=25,0001+41001+51001+8100=25,0001.041.051.08=29,484Hence, the population after three years will be 29,484.

Page No 14.27:

Question 4:

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.

Answer:

Here,P=Initial population=50,000R1=4%R2=5%R3=3%n=Number of years=3 Population after three years=P1+R11001+R21001+R3100=50,0001+41001+51001+3100=50,0001.041.051.03=56,238Hence, the population after three years is 56,238.

Page No 14.27:

Question 5:

There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago?

Answer:

Population after three years=P1+R100n9,261=P1+510039,261=P1.053P=9,2611.157625=8,000Thus, the population three years ago was 8,000.

Page No 14.27:

Question 6:

In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.

Answer:

Let the annual rate of growth be R. Production of scooters after three years= P1+R100n46,305=4,0001+R10031+0.01R3=46,30540,0001+0.01R3=1.1576251+0.01R3=1.0531+0.01R=1.050.01R=0.05R=5Thus, the annual rate of growth is 5%.

Page No 14.27:

Question 7:

The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago?

Answer:

Population after three years=P1+R100n196,830=P1+81003196,830=P1.083P=196,8301.259712=156,250Thus, the population three years ago was 156,250.

Page No 14.27:

Question 8:

The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.

Answer:

Population after two years=P1+R100n22,050=P1+501000222,050=P1.052P=22,0501.1025=20,000Thus, the population two years ago was 20,000.

Page No 14.27:

Question 9:

The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?

Answer:

Given: R1=10%R2=-8%R3=12% P=Original count of bacteria=13,125,000We know that:P1+R11001-R21001+R3100 Bacteria count after three hours=13,125,0001+101001-81001+12100=13,125,0001.100.921.12=14,876,400Thus, the bacteria count after three hours will be 14,876,400.

Page No 14.27:

Question 10:

The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000?

Answer:

Population at the end of the year 2000=P1+R11001-R2100=72,0001+71001-10100=72,0001.070.9=69,336Thus, the population at the end of the year 2000 was 69,336.

Page No 14.27:

Question 11:

6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year?

Answer:

Number of workers=6,400At the end of the first year, 25% of the workers were retrenched. 25% of 6,400=1,600Number of workers at the end of the first year= 6,400-1,600=4,800At the end of the second year, 25% of those working were retrenched. 25% of 4,800=1,200Number of workers at the end of the second year=4,800-1,200=3,600At the end of the third year, 25% of those working increased. 25% of 3,600=900Number of workers at the end of the third year=3,600+900=4,500Thus, the number of workers during the fourth year was 4,500.

Page No 14.27:

Question 12:

Aman started a factory with an initial investment of Rs 100000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

Answer:

Aman's profit for three years=P1-R11001+R21001+R3100=100,0001-51001+101001+12100=100,0000.951.101.12=117,040 Net profit=Rs 117,040-Rs 100,000                     =Rs 17,040



Page No 14.28:

Question 13:

The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.

Answer:

Population after three years=P1+R1003175,760=P1+4010003175,760=P1.043P=175,7601.124864=156,250Thus, the population three years ago was 156,250.

Page No 14.28:

Question 14:

The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?

Answer:

Production after three years=P1+R110021-R2100=8,0001+151,00021-5100=8,0001.1520.95=10,051Thus, the production after three years will be 10,051.

Page No 14.28:

Question 15:

The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.

Answer:

(i)Population of the city in 2001=P1+R1002=67600001+41002=67600001.042=7311616Thus, Population of the city in 2001 is 7311616.(ii)Population of the city in 1997=P1+R100-2=67600001+4100-2=67600001.04-2=6250000Thus,Population of the city in 1997 is 6250000.

Page No 14.28:

Question 16:

Jitendra set up a factory by investing Rs 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, compute his total profit.

Answer:

Profit at the end of the first year=P1+R100=2,500,0001+5100=2,500,0001.05=2,625,000Profit at the end of the second year=P1+R100=2,625,0001+10100=2,625,0001.10=2,887,500Total profit = Rs 2,887,500-Rs 2,500,000                    =Rs 387,500



Page No 14.31:

Question 1:

Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.

Answer:

Value of the boat after two years=P1-R100n16,0001-51002=16,0000.952=14,440Thus, the value of the boat after two years will be Rs 14,440.

Page No 14.31:

Question 2:

The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000? Also, find the total depreciation during this period.

Answer:

Value of the machine after two years=P1-R100n100,0001-101002=100,0000.902=81,000Thus, the value of the machine after two years will be Rs 81,000.Depreciation=Rs 100,000-Rs 81,000                       =Rs 19,000

Page No 14.31:

Question 3:

Pritam bought a plot of land for Rs 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years?

Answer:

Given: P= Rs 64,000R= 5% for every six monthsValue of the plot after two years = P1+R100n64,0001+52004=64,0001.0254=706,440.25Thus, the value of the plot after two years will be Rs 706,440.25.

Page No 14.31:

Question 4:

Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.

Answer:

Value of the house after three years=P1-R100n30,0001-251003=30,0000.753=12,656.25Thus, the value of the house after three years will be Rs 12,656.25.

Page No 14.31:

Question 5:

The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.

Answer:

Purchase price=P1-R100-n43,7401-10100-3=43,7400.90-3=60,000Thus, the purchase price of the machine was Rs 60,000.

Page No 14.31:

Question 6:

The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?

Answer:

Purchase price=P1-R100-n9,6801-12100-2=9,6800.88-2=12,500Thus, the purchase price of the refrigerator was Rs 12,500.

Page No 14.31:

Question 7:

The cost of a T.V. set was quoted Rs 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What was the cost of the T.V. set in 2001?

Answer:

Cost of the TV=P1+R1001-R10017,0001+51001-4100=17,0001.050.96=17,136Thus, the cost the TV in 2001 was Rs 17,136.

Page No 14.31:

Question 8:

Ashish started the business with an initial investment of Rs 500000. In the first year he incurred a loss of 4%. However during the second year he earned a profit of 5% which in third year rose to 10%. Calculate the net profit for the entire period of 3 years.

Answer:

Profit for three years=P1-R11001+R21001+R3100500,0001-41001+51001+10100=500,0000.961.051.10=554,400Thus, the net profit is Rs 554,400.



Page No 14.4:

Question 1:

Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.

Answer:

Principal for the first year=Rs 3,000Interest for the first year=Rs3,000×5×1100                                         =Rs 150Amount at the end of the first year=Rs 3,000+Rs 150                                                            =Rs 3,150Principal for the second year=Rs 3,150Interest for the second year=Rs3,150×5×1100                                               =Rs 157.50Amount at the end of the second year=Rs 3,150+Rs 157.50                                                                   =Rs 3307.50 Compound interest=Rs3,307.50-3,000                                       =Rs 307.50

Page No 14.4:

Question 2:

What will be the compound interest on Rs 4000 in two years when rate of interest is 5% per annum?

Answer:

We know that amount A at the end of n years at the rate of R% per annum is given by A=P1+R100n.Given:P=Rs 4,000R=5% p.a.n=2 yearsNow,A=4,0001+51002=4,0001.052=Rs 4,410And,CI=A-P=Rs 4,410-Rs 4,000=Rs 410

Page No 14.4:

Question 3:

Rohit deposited Rs 8000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years?

Answer:

We know that amount A at the end of n years at the rate of R% per annum is given by A=P1+R100n.Given:P=Rs 8,000R=15% p.a.n=3 yearsNow,A=8,0001+151003=8,0001.153=Rs 12,167And,CI=A-P=Rs 12,167-Rs 8,000=Rs 4,167

Page No 14.4:

Question 4:

Find the compound interest on Rs 1000 at the rate of 8% per annum for 112 years when interest is compounded half-yearly.

Answer:

Given:P=Rs 1,000R=8% p.a.n=1.5 yearsWe know that:A=P1+R2002n=1,0001+82003=1,0001.043=Rs 1,124.86Now,CI=A-P=Rs 1,124.86-Rs 1,000=Rs 124.86

Page No 14.4:

Question 5:

Find the compound interest on Rs 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.

Answer:

Given:P=Rs 16,000R=20% p.a.n=1 yearWe know that:A=P1+R4004n=16,0001+204004=16,0001.054=Rs 19,448.1Now,CI=A-P=Rs 19,448.1-Rs 16,000=Rs 3,448.1



Page No 14.5:

Question 6:

Swati took a loan of Rs 16000 against her insurance policy at the rate of 1212% per annum. Calculate the total compound interest payable by Swati after 3 years.

Answer:

Given:P=Rs 16,000R=12.5% p.a.n=3 yearsWe know that:A=P1+R100n=16,0001+12.51003=16,0001.1253=Rs 22,781.25Now, CI=A-P=Rs 22,781.25-Rs 16,000=Rs 6,781.25

Page No 14.5:

Question 7:

Roma borrowed Rs 64000 from a bank for 112 years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 112 years, if the interest is compounded half-yearly.

Answer:

Given:P=Rs 64,000R=10% p.a.n=1.5 yearsAmount after n years:A=P1+R2002n=64,0001+102003=64,0001.053=Rs 74,088Now,CI=A-P=Rs 74,088-Rs 64,000=Rs 10,088

Page No 14.5:

Question 8:

Mewa Lal borrowed Rs 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.

Answer:

SI for Mewa Lal=PRT100=20,000×18×2100=Rs 7,200Thus, he has to pay Rs 7,200 as interest after borrowing.CI for Mewa Lal = A-P=20,0001+181002-20,000=20,0001.182-20,000=27,848-20,000=Rs 7,848He gained Rs 7,848  as interest after lending.    His gain in the whole transaction=Rs 7,848-Rs 7,200                                                          =Rs 648

Page No 14.5:

Question 9:

Find the compound interest on Rs 8000 for 9 months at 20% per annum compounded quarterly.

Answer:

P=Rs 8,000T=9 months=3 quartersR=20% per annum=5% per quarterA=8,0001+51003=8,0001.053=9,261The required amount is Rs 9,261.Now,CI=A-P=Rs 9,261-Rs 8,000=Rs 1,261

Page No 14.5:

Question 10:

Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs 200 as simple interest.

Answer:

SI=PRT100 P=SI×100RT=200×10010×2=Rs 1,000A=P1+R100n=1,0001+101002=1,0001.102=Rs 1,210Now,CI=A-P=Rs 1,210-Rs 1,000=Rs 210

Page No 14.5:

Question 11:

Find the compound interest on Rs 64000 for 1 year at the rate of 10% per annum compounded quarterly.

Answer:

To calculate the interest compounded quarterly, we have:A=P1+R4004n=64,0001+104004×1=64,0001.0254=70,644.03Thus, the required amount is Rs 70,644.03.Now,CI=A-P=Rs 70,644.025-Rs 64,000=Rs 6,644.03

Page No 14.5:

Question 12:

Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months.

Answer:

Given:P=Rs 7,500R=12% p.a.=3% quarterlyT=9 months=3 quartersWe know that:A=P1+R100nA=7,5001+31003=7,5001.033=8,195.45Thus, the required amount is Rs 8,195.45.

Page No 14.5:

Question 13:

Anil borrowed a sum of Rs 9600 to install a handpump in his dairy. If the rate of interest is 512% per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.

Answer:

A=P1+R100n=9,6001+5.51003=9,6001.0553=Rs 11,272.72Now,CI=A-P=Rs 11,272.72-Rs 9,600=Rs 1,672.72

Page No 14.5:

Question 14:

Surabhi borrowed a sum of Rs 12000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.

Answer:

A=P1+R100n=12,0001+51003=12,0001.053=13,891.50Thus, the required amount is Rs 13,891.50.Now,CI=A-P=Rs 13,891.50-Rs 12,000=Rs 1,891.50

Page No 14.5:

Question 15:

Daljit received a sum of Rs. 40000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.

Answer:

A=P1+R100n=40,0001+71002=40,0001.072=45,796Thus, the required amount is Rs 45,796.Now,CI=A-P=Rs 45,796-Rs 40,000=Rs 5,796



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