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Applied Maths Chapter 14 Solutions
Compound Interest And Annuity
EXERCISE – 14.2
Q.1 Rohan borrowed Rs.5120 at 12 1/2% compound interest per annum. At the end of the three years, the money was repaid along with the interest. Calculate the amount of interest paid.
Ans. r = 12 1/2% = 25/2% , n = 3
A = P[(1 + r/100)^n]
= 5120[(1 + 25/2 x 100)^3]
= 5120[(1 + 1/8)^3]
= 5120 x (9/8)^3
= 5120 x 729/512
= Rs.7290
Interest = Amount – Principal
= 7290 – 5120
= Rs.2170
Q.2 Calculate the compound interest and amount of Rs.1600 for 1 year at 10% per annum compounded half-yearly.
Ans. r = 10/2 % = 5% , n = 1 x 2 = 2 , P = Rs.1600
A = P[(1 + r/100)^n]
= 1600[(1 + 5/100)^2]
= 1600[(21/20)^2]
= 1600 x 441/400
= Rs.1764
Interest = Amount – Principal
= 1764 – 1600
= Rs.164
Q.3 Find the amount of the compound interest quarterly for 9 months at the rate of 4% per annum.
Ans. r = 4/4 % = 1% , n = 9months x 4 = 4 36months = 3 quarterly year , P = Rs.100000
A = P[(1 + r/100)^n]
= 100000[(1 + 1/100)^3]
= 100000[(101/100)^3]
= 100000 x (101 x 101 x 101)/(100 x 100 x 100)
= Rs.103030.10
Interest = Amount – Principal
= 103030.10 – 100000
= Rs.3030.10
Q.4 Find the difference between compound interest and simple interest on a sum of Rs.4800 for 2 years at 5% per annum payable yearly.
Ans. C.I = P[1 + r/100)^n – 1]
= 4800[(1 + 5/100)^2 – 1]
= 4800[(21/20)^2 – 1]
= 4800[441/400 – 1]
= 4800(41/400)
= Rs.492
S.I = (P x r x T)/100
= (4800 X 5 X 2)/100
= 480
Difference between C.I and S.I = C.I – S.I
= 492 – 480
= Rs.12
Q.5 Find the difference between simple interest and compound interest on Rs.2500 for 2 years at 4% per annum, with compound interest being reckoned semi-annually.
Ans. P = Rs.2500, r = 4/2 = 2%, n = 2 x 2 = 4 years [semi-annually]
S.I = (P x R x T)/100
= (2500 X 4 X 2)/100
= Rs.200
A = P[(1 + r/100)^n]
= 2500(1 + 2/100)^4
= 2500[(51/50)^4
= 2500 x (51/50 x 51/50 x 51/50 X 51/50)
= Rs.2706.08
Interest = Amount – Principal
= 2706.08 – 2500
= Rs.206.08
Difference between C.I and S.I = C.I – S.I
= 206.08 – 200
= Rs.6.08
Q.6 A deposit of Rs.5000 in the state bank savings account allows simple interest at 4 1/2% per year. At the end of the first year, he transfers the entire amount to a fixed deposit account for 2 years at 7% per annum compound annually. Find the total amount of money in the name of the depositor at the end of the third year, nearest to a rupee.
Ans.
Interest for 1st years = (PxRxT)/100
= (5000 x 9 x 1)/(100 x 2)
= Rs.225
Principal at the end of first year = 5000 + 225
= Rs.5225
A = 5225 (1 + 7/100)^2
= 5225 (107/100)^2
= 5225 x 11449/10000
= Rs.5982.1025 = Rs.5982
Q.7 A man borrowed Rs.5000 for 4 years under the following terms: 4% simple interest for the first 2 1/2 years, 4% compound interest for the rest of the period on the amount due after 2 1/2 years, and the interest is compounded semi-annually. How much should he pay to settle the account?
Ans.
Interest for 1 1/2 years = 5/2years
S.I = (P x R x T)/100
= (5000 x 4 x 5)/(100 x 2)
= Rs.500
Principal for next 1 1/2 year = 5000 + 500
= Rs.5500
Now,
n = 1 1/2 x 2 = 3 , r = 4/2 = 2% [compounded half/yearly]
A = 5500 (1 + 2/100)^3
= 5500 (51/50)^3
= 5500 x 132651/125000
= Rs.5836.64
Q.8 Find the amount the compound interest on Rs.2000 in 2 years if the rate is 4% for the first year and 3% for the second year.
Ans. A = P (1 + r1/100) (1 + r2/100)
= 2000(1 + 4/100) (1 + 3/100)
= 2000(1 + 1/25)(103/100)
= 10712/5
= Rs.2142.40
Interest = Amount – Principal
= 2142.40 – 2000
= Rs.142.40
Q.9 Find the compound interest on Rs.3125 for 3 years if the rate of interest for the first, second, and third years are respectively 4%, 5%, and 6% per annum.
Ans.
A = P (1 + r1/100) (1 + r2/100) (1 + r3/100)
= 3125(1 + 4/100) (1 + 5/100) (1 + 6/100)
= 3125 (26/25) (21/20) (53/50)
= Rs.3617.25
Interest = Amount – Principal
= 3617.25 – 3125
= Rs.492.25
Q.10 What sum of money will amount to Rs.1352 in 4 years for 4% compound interest per annum compounded annually?
Ans. A = P[(1 + r/100)^n]
1352 = P[(1 + 4/100)^2]
1352 = P[(26/25)^2]
1352 = P x 676/625
P = (1352 x 625)/676
P = Rs.1250
Q.11 A sum invested for 1 1/2 years compounded half-yearly at the rate of 4%p.a. will amount to Rs.132651?
Ans. P = ?, n = 1 1/2 = 1.5 x 2 = 3 years, r = 4/2 = 2% [Half-yearly]
A = P[(1 + r/100)^n]
132651 = P(1 + 2/100)^3
132651 = P(51/50)^3
P = (132651 x 50 x 50 x 50)/(51 x 51 x 51)
= Rs.125000
Q.12 On what sum will the compound interest for 2 years at 4% per annum be Rs.5712?
Ans. C.I = P[1 + r/100)^n – 1]
5712 = P[(1 + 4/100)^2 – 1]
5712 = P[(26/25)^2 – 1]
5712 = P[676/625 – 1]
5712 = P(51/625)
P = (5712 x 625)/51
= Rs.70000
Q.13 A man invests Rs.1200 for two years at compound interest. After one year the money amounts to Rs.1275. Find the interest for the second year correct to the nearest rupee.
Ans.
Interest for 1st year = 1275 – 1200 = Rs.75
S.I = (P x R x T)/100
75 = (1200 x R x 1)/100
R = 75/12
R = 6.25%
P for the second year = Rs.1275
I = (P x R x T)/100
= (1275 x 25 x 1)/(100 x 4)
= 1275/16
= 79.70
= Rs.80
Q.14 At what rate percent per annum compound interest will Rs.2304 amount to Rs.2500 in 2 years?
Ans. A = P[(1 + r/100)^n]
2500 = 2304[(1 + r/100)^2]
2500/2304 = (1 + r/100)^2
(50/48)^2 = (1 + r/100)^2
50/48 = 1 + r/100
25/24 – 1 = r/100
1/24 = r/100
r = 100/24
r = 4 1/6%
Q.15 A sum compounded annually becomes 25/16 times itself in two years. Determine the rate of interest per annum.
Ans. P = Rs.x, A = 25/16x
A = P[(1 + r/100)^n]
25/16x = x[(1 + r/100)^2]
(5/4)^2 = (1 + r/100)^2
5/4 = 1 + r/100
5/4 – 1 = r/100
1/4 = r/100
r = 100/4
r = 25%
Q.16 Determine the rate of interest for a sum that becomes 216/125 times itself in 1 1/2 years compounded semi-annually.
Ans. A = 216/125x, n = 1 1/2 = 1.5 = 3years
A = P[(1 + r/100)^n]
216/125x = x[(1 + r/100)^3]
(6/5)^3 = (1 + r/100)^3
6/5 = 1 + r/100
6/5 – 1 = r/100
1/5 = r/100
r = 100/5
r = 20% [Half-yearly]
Now, r for yearly
r = 20 x 2 = 40%
Q.17 At what rate percent per annum compound interest would Rs.80000 amount to Rs.88200 in two years, interest is compounded yearly. Also, find the amount after 3 years at the above rate of compound interest.
Ans. A = P[(1 + r/100)^n]
88200 = 80000[(1 + r/100)^2]
882/800 = (1 + r/100)^2
441/400 = (1 + r/100)^2
(21/20)^2 = (1 + r/100)^2
21/20 – 1 = r/100
1/20 = r/100
r = 100/20
r = 5%
Interest for 3rd year
S.I = (P x R x T)/100
= (88200 x 5 x 1)/100
=Rs.4410
Amount = 88200 + 4410
= Rs.92610
Q.18 A certain sum amounts to Rs.798.60 after 3 years and Rs.878.46 after 4 years. Find the interest rate and sum.
Ans. Interest for 4th year =
878.46 – 798.60
= Rs.79.60
Principal for 4th year = 798.60
S.I = (P x R x T)/100
79.86 = (798.60 x R x 1)/100
7986/100 = (79860 x R x 1)/(100 x 100)
R = 10%
A = P(1 + r/100)^n
798.60 = P (1 + 10/100)^3
79860/100 = P (11/10)^3
79860/100 = 798.6 P x 1331/100
P = (7986 x 100)/1331
= Rs.600
Q.19 In what time will Rs.15625 amount to Rs.17576 at 4% per annum compound interest?
Ans. A = P[(1 + r/100)^n]
17576 = 15625(1 + 4/100)^n
17576/15625 = (26/25)^n
(26/25)^3 = (26/25)^n
n = 3
Q.20 Rs.6000 invested at 10% per annum, compounded semi-annually, amounts to Rs.18522. Find the period of investment.
Ans. A = P (1 + r1/100) (1 + r2/100)
2782.50 = P(1 + 5/100) (1 + 6/100)
278250/100 = P(1 + 1/20)(1 + 3/50)
278250/100 = P x 21/20 x 53/50
(278250 x 2 x 5)/(21 x 53) = P
P = Rs.2500
Q.21 What sum will amount to Rs.2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive year?
Ans. A = P (1 + r1/100) (1 + r2/100)
2782.50 = P(1 + 5/100) (1 + 6/100)
278250/100 = P(1 + 1/20)(1 + 3/50)
278250/100 = P(21/20) (53/50)
P = (278250 x 2 x 5)/(21 x 53)
P = Rs.2500
Q.22 A sum of money is invested at compound interest payable annually. The interest in two successive years is RS.225 and Rs.240. Find the rate of interest.
Ans. Interest for 1st year
= 240 – 225 = Rs.15
S.I = (P x R x T)/100
15 = (225 x r x 1)/100
r = (15 x 100)/225
r = 6 2/3%
Q.23 The difference between compound interest for a year payable half-yearly and simple interest on a sum of money lent out at 10% for a year is Rs.15. Find the sum of money lent out.
Ans.
n = 1 x 2 = 2 2 , r = 10/2 = 5% [half-yearly]
C.I – S.I = 15
[P(1 + 5/100)^2 – 1] – (P x 5 x 2)/100 = 15
P(21/20)^2 – 1] – 1/10 = 15
P(441/400 – 1 – 1/10] = 15
P(441 – 400 – 40)/400 = 15
P (1/400) = 15
P = Rs.6000
Q.24 The simple interest on a certain sum for 3 years is Rs.225 and the compound interest on the same sum at the sum rate for 2 years is Rs.153. Find the rate of interest and principal.
Ans. S.I = (P x R x T)/100
75 = (P x R x 3)/100
PR = 7500 ……….(i)
C.P = P[1 + r/100)^n – 1]
153 = P[(1 + R/100)^2 – 1]………(ii)
Dividing eq (i) & (ii)
R/[(1 + R/100)^2 – 1] = 7500/153
153R = [(1 + R/100)^2 – 1] x 7500
153R = (1 + R^2/10000 + R/50 – 1) x 7500
153R = [(R^2 + 200R)/10000] x 7500
153R = [(R^2 + 200R)] x 3
612R = 3R^2 + 600R
612R – 600R = 3R^2
12R = 3R^2
R = 4
From eq (i)
P x 4 = 7500
P = 7500/4
P = Rs.1875
Ans. For annually,
S.I = (P x R x T)/100
= (8000 x 10 x 1)/100
= Rs.800
Principal for next 1 1/2 year
= 8000 + 800
= Rs.8800
S.I = (P x R x T)/100
= (8800 x 10 x 1)/(100 x 2)
= Rs.440
A = 8800 + 440
= Rs.9240
For semi-annually,
n = 1 1/2 x 2 = 3years , r = 4/2 = 2% [compounded half-yearly]
A = P (1 + r/100)^n
A = 8000 (1 + 5/100)^3
= 8000 (1 + 1/20)^3
= 8000 (21/20)^3
= 8000 x 9261/8000
= Rs.9261
Difference = 9261 – 9240
= Rs.21
Q.26 A sum of money is lent out at compound interest for two years at 20%p.a., C. I am being reckoned yearly. If the same sum of money is let out at compound interest at the same rate percent per annum, C. I being reckoned half-yearly. It would have fetched Rs.482 more by way of interest. Calculate the sum of money lent out.
Ans. A1 = P (1 + r/100)^n, A2 = P (1 + r/100)^n
A1 = P (1 + 20/100)^n , A2 = P(1 + 10/100)^N
Now, according to the question
A2 – A1 = 482
P(11/10)^4 – P(6/5)^2 = 482
P [(14641/10000) – (36/25)] = 482
P [(14641 – 14400)/10000] = 482
P x 241/10000 = 0.0241 482
P = (482 x 10000)/241
P = Rs.20000
Q.27 A sum of money amounts to Rs.13230 in one year and to Rs.13891.50 1 1/2 in years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.
Ans. A = P (1 + r/100)^n
13230 = P [1 + r/(2 x 100)]^2……(i) [semi-annually]
13891.50 = P [1 + r/(2 x 100)]^3……(ii) [ semi-annually]
Dividing eq. (i) & (ii)
13230/13891.50 = [P(1 + r/200)^2]/[P(1+r/200)^3]
1323000/1389150 = 1/[(200 + r)/200]
980/1029 = 200/(200 + r)
200 + r = 210
r = 210 – 200
r = 10%
From eq (i)
13230 = P(1 + 10/200)^2
13230 = P(21/20)^2
13230 = P (442/400)
P = (13230 x 400)/441
P = Rs.12000
Q.28 A man borrows Rs.5800 at 12% p.a, compound interest. He repays Rs.1800 at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest rupee.
Ans. S.I = (P x R x T)/100
= (5800 x 12 x 1)(100 x 2)
= Rs.348
Principal for next 6 month = 5800 + 348 – 1800
= 6148 – 1800
= Rs.4348
Interest for the next 6 month = (4348 x 12 x1)/(100 x 2)
= 26088/100
= Rs.260.88
Principal for next 6 months = 4348 + 260.88 – 1800
= 4608.88 – 1800
= Rs.2808.88
Interest for next 6 month = (2808.88 x 12 x1)/(100 x 2)
= 16853.28/100
= Rs.168.53
Principal for next 6 month = 2808.88 + 168.53 – 1800
= 2977.41 – 1800
= Rs.1177.41
Q.29 Divide Rs.21866 in two parts such that the amount of one in 3 years is the same as the amount of the second in 5 years, and the rate of compound interest is 5% per annum.
Ans. Let,
One amount = P
Another amount = 21866 – P
P(1 + r/100)^n = P(1 + r/100)^n
P(1 + 5/100)^3 = (21866 – P) (1 + 5/100)^5
P = 21866 – P (21/20)^2
P = (21866 – P) x 441/400
400P = 21866 x 441 – 441P
400P + 441P = 21866 x 441
841P = 21866 x 441
P = (21866 x 441)/841
P = Rss.11466
First part = Rs.11466
Second part = 21866 – 11466
= Rs.10400
Q.30 A certain sum is invested at compound interest at 5% per annum. In how many years will the amount will be doubled?
Ans. A = P(1 + r/100)^n
2P = P (1 + 5/100)^n
2 = 2 (1 + 1/20)^n
2 = 2 (21/20)^n
Taking logs on both sides
log 2 = log(21/20)^n
log 2 = n log 21/20
log 2 = n [log21 – log20]
0.3010 = n (1.3222 – 1.201)
0.3010 = n x 0.0212
0.3010/0.0212 = n
n = 3010/212
= 14.2 years
Q.31 Find how many years will it take the sum of Rs.35000 to double when invested at 4%, compounded half-yearly.
Ans. Let the time to double the money = n years
Half yearly interest = 4/2 = 2% = 0.02
Period of interest = 2 x n = 2n
A = P (1 + R/100)^nt
35000 x 2 = 35000 (1 + 0.02)^2n
70000 = 35000 (1.02)^2n
70000/35000 = (1.02)^2n
2 = (1.02)^2n
Taking logarithm on both sides,
log 2 = log (1.02)^2n
Log 2 = 2n log(1.02)
n = log2/[2 x log(1.02)]
= 17.5 years
Q.32 In what time will Rs.25000 amount to Rs.35000 at 6% compound quarterly?
Ans. A = P(1 + r/100)^n
35000 = 25000 (1 + 6/4 x 100)^n
35000/25000 = (203/200)^4n
Taking logs on both sides
log 7/5 = log(203/200)^4n
log 7/5 = 4n log 203/200
log7 – log5 = 4n [log203 – log200]
0.8450 – 0.6989 = 4n (2.3074 -2.3010)
0.1461 = 4n x 0.0064
n = 1461/(4×0.0064)
n = 5.7
Q.33 Niti owes Rs.12500 without interest 3 years from today. What sum would the creditor be willing to accept today if he can invest the money at 4% compounded semi-annually?
Ans. F.V = Rs.12500, r = 4% = 0.04, n = 2, t = 3 years
P.V = F.V/(1 + r/n)^nt
= 12500/(1 + 0.04/2)^6
= 12500/(1.04)^6
= 12500/1.125515
= Rs.11108.53
Q.34 A sum of Rs.20000 is borrowed at 18% compound interest and paid back in four equal installments, what is the value of each installment?
Ans. Loan amount = P[(100/100+r) + (100/100+r)^2 + (100/100+r)^3 + (100/100+r)^4]
2000 = P[(100/100+18) + (100/100+18)^2 + (100/100+18)^3 + (100/100+18)^4]
2000 = P[50/59 + (50/59)^2 + (50/59)^3 + (50/50)^4]
2000 = 50/59P[1 + 50/59 + (50/59)^2 + (50/59)^3]
(2000 x 59)/50 = P[1 + 50/59 + 2500/3481 + 125000/205379]
400 x 59 = P[651929/205379]
P = (400 x 59 x 205379)/651929
= Rs.7434.77
Q.35 If the nominal rate of interest is 5.25% compound quarterly, find the corresponding effective rate. Given(1.013125)^4 = 1.0535
Ans. E = (1 + r/100p)^p – 1
= (1 + 5.25/100×4)^4 – 1
= (1621/1600) – 1
= (1.013)^4 – 1
= 1.0535 – 1
= 0.0535
= 5.35%
Q.36 Find the nominal rate compounded semi-annually equivalent to a 4.2% effective rate.(Given √1.042 = 1.02078)
Ans. E = (1 + r/100p)^p – 1
4.2/100 = (1 + r/100×2)^2 – 1
0.042 = (1 + r/200) – 1
0.042 + 1 = (1 + r/200)^2
√1.042 = 1 + r/200
1.02078 = 1 + r/200
1.02078 – 1 = r/200
0.02078 x 200 = 4.156 r
r = 4.15600%
= 4.156%
FAQ’s related to Applied Maths Chapter 14 on compound interest and annuity:
Q.1 What is compound interest, and how does it differ from simple interest?
Ans. Compound interest is the interest calculated on the initial principal and also on the accumulated interest from previous periods. Unlike simple interest, which is calculated only on the principal amount, compound interest takes into account the interest earned over time, resulting in exponential growth of the investment.
Q.2 What are some common mistakes to avoid when solving compound interest and annuity problems?
Ans. Common mistakes include misinterpreting the problem statement, using incorrect formulas, neglecting to convert interest rates to decimal form, and misunderstanding the frequency of compounding or payment periods. It’s essential to carefully read the problem and double-check calculations for accuracy.
Q.3 How do I calculate compound interest for a given principal, interest rate, and time period?
Ans. The formula to calculate compound interest is A = P(1 + r/n)^(nt), where A is the future value, P is the principal amount, r is the annual interest rate (in decimal form), n is the number of times interest is compounded per year, and t is the time in years.
These are few Frequently Asked Questions relating to Applied Maths Chapter 14
In Applied Maths chapter 14, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 14 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 14, we delve deep into advanced mathematical concepts that are crucial for understanding.
Applied Maths Chapter 14 EXCERCISES:
