WBBSE Solutions For Class 10 Maths Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.1
Question 1. If I take a loan of Rs. 1400 at 5% compound interest per annum for 2 years, let us write by calculating how much compound interest and the total amount I shall pay.
Solution:
Given
I take a loan of Rs. 1400 at 5% compound interest per annum for 2 years
When Principal = Rs. 1400,
Compound interest =10.25×1400 / 100
= Rs. 143.50
Principal & compound interest are in direct proportion.
Amount = Rs. (1400+ 143.50) Rs. 1543,50.
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Question 2. Let us write by calculating what is the amount of Rs. 1000 for 2 years at the rate of 5% compound interest per annum.
Solution: Amount = Rs. 1000 (1+5/100)²
= Rs. 1000 x (22/20)²
=Rs.1000 x 441/400
=Rs. 2205/2
=Rs. 1102.50
Question 3. At 5% compound interest per annum, let us find the compound inter- est on Rs. 10,000 for 3 years.
Solution: Compound interest for 3 years = Rs. \(10000\left\{\left(1+\frac{5}{100}\right)^3-1\right\}\)
= \(\text { Rs. } 10000\left\{\left(1+\frac{1}{20}\right)^3-1\right\}\)
= \(\text { Rs. } 10000\left\{\left(\frac{21}{20}\right)^3-1\right\}\)
= \(\text { Rs. } 10000\left\{\frac{9261-8000}{8000}\right\}\)
= \(\text { Rs. } 10000\left(\frac{1261}{8000}\right)\)
= \(\text { Rs. } \frac{6305}{4}\)
= Rs. 1576.25
Question 4. Let us find compound interest on Rs. 1000 at the rate of 10% compound interest per annum and the interest being compounded at 6 monthly intervals.
Solution: Compound interest on Rs.1000 for 1 year at 10% per annum compound interest compound semiannually.
Amount = \(\text { Rs. } 1000\left(1+\frac{1 / 2}{100}\right)^{2 \times 1}\)
= \(\text { Rs. } 1000\left(1+\frac{10}{200}\right)^2\)
= \(\text { Rs. } 1000\left(\frac{21}{20}\right)^2\)
= \(\text { Rs. } 1000 \times \frac{441}{400}\)
= \(\text { Rs. } \frac{2205}{2}\)
= Rs. 1102.50
∴ Compound interest = Rs.(1102.50 – 1000) = Rs. 102.50
Question 5. Let us write by calculating compound interest on Rs. 10,000 at the rate of 8% compound interest per annum for 9 months, compounded at an interval of 3 months.
Solution: Compound interest for 9 months = Rs. (1102.50-1000)
= Rs. 102.50.
Question 6. If the rate of compound interest for the first year is 4% and for 2nd year is 5%, let us find the compound interest on Rs. 25000 for 2 years.
Solution: Amount after two years on Rs.25,000 at two rates of 4% for 1st year & 5% for 2nd year
= \(\text { Rs. } 25,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\)
= \(\text { Rs. } 25,000 \times \frac{104}{100} \times \frac{105}{100}\)
= Rs. 27300
∴ Compound interest = Rs. 27300 – Rs. 25000
= Rs. 2300
Question 7. I lend Rs. 10,000 at the rate of 4% compound interest per annum for 2 1/2 years, let us write by calculating how much total money I shall pay.
Solution: Amount for 2 1/2 years = Rs. (10816+216.32) = Rs. 11032.32.
Question 8. Let us find the amount of Rs. 30,000, at the rate of 6% compound interest per annum for 2 1/2 years.
Solution: Amount for 2 1/2 years at the rate of 6%. Compound interest
= Rs. \(\left[30000\left(1+\frac{6}{100}\right)^2+3000\left(1+\frac{6}{100}\right)^2 \times \frac{6}{12} \times \frac{6}{100}\right]\)
= Rs. \(\left[30000\left(1+\frac{6}{100}\right)^2\left(1+\frac{6}{12} \times \frac{6}{100}\right\}\right]\)
= Rs. \(\left(30000 \times \frac{106 \times 106}{100 \times 100} \times \frac{103}{100}\right)\)
= Rs. Ps. 34719.24
Question 9. Let us write by calculating what sum of money will amount to Rs. 3528 after 2 years at the rate of 5% compound interest per annum.
Solution: Let the principal = Ra, X
According to the problem,
\(x\left(1+\frac{5}{100}\right)^2=3528\)or, \(\times \frac{105}{100} \times \frac{105}{100}=3528\)
∴ \(x=\frac{2528 \times 100 \times 100}{105 \times 105}=3200\)
∴ Principal = ps 3200.
Question 10. Let us see by calculating alternatively that Minatididi deposits Rs. 3,00,000 in the bank.
Solution: Let the principal for 19t year = Rs. x.
Interest for the 151 year at 8% = Rs. \(\frac{8 x}{100}\)
∴ Principal for the 2nd year = Rs. \(\left(x+\frac{8 x}{100}\right)={Rs} \frac{106 x}{100}\)
Again, the interest for the and year at 8% = Rs. \(\frac{B}{100} \times \frac{106 x}{100}={Rs} \frac{864 x}{10000}\)
∴ Principal for the 3rd year = Rs. \(\left(\frac{106 x}{100}+\frac{864 x}{10000}\right)=\text { Rs. } \frac{11664 x}{10000}\)
∴ The interest for the 3rd year (one year) at 85% Rs.\( \frac{8 \mathrm{x}}{100}+\frac{11664 \mathrm{x}}{10000}=\text { Rs. } \frac{93312 \mathrm{x}}{1000000}\)
∴ After 3 years, amount will be Rs. \(\left(\frac{11664 x}{10000}+\frac{93312 x}{1000000}\right)={Rs} \cdot \frac{1259712 x}{1000000}\)
∴ According to the problem,
\(\frac{1259712 x}{1000000}=37791.36\)∴ \(x=\frac{37791.36 \times 1000000}{1259712}=\mathrm{Rs} \cdot 30,000\)
Minati Di depcaited Rs.30,000 in the bank.
Question 11. The simple interest and compound interest of a certain sum of money for 2 years are Rs. 840 and Rs. 869.40 respectively. Let us calculate that sum of money and the rate of interest.
Solution: Simple interest for 2 years = Rs. 840
∴ Simple interest for 1 year = Rs. \(\frac{840}{2}\) = Rs. 420
Difference between Compound interest & simple interest for 2 years
= Rs. (869.40 – 840) = Rs. 29.40
∴ Interest on Rs. 420 for 1 year = Rs. 29.40
∴ Interest on Rs. 100 for 1 year = \(\text { Rs. } \frac{29.40}{420} \times 100 .=\text { Rs. } 7\)
∴ Rate of interest = 7%.
Principal = \(\text { Rs. } \frac{420 \times 100}{7}\) = Rs. 6,000.
Question 12. Let us calculate at what rate of compound interest, Rs. 5,000 will amount to Rs. 5832 in 2 years.
Solution: Let the rate of compound interest = r%
p = Rs.5,000; A = Rs.5832; Time = n = 2 years.
∴ \(A=P\left(1+\frac{r}{100}\right)^n\)
or, \(5832=5000\left(1+\frac{r}{100}\right)^2\)
or, \(\frac{5832}{5000}=\left(1+\frac{r}{100}\right)^2\)
or, \(\left(i+\frac{r}{100}\right)^2=\left(\frac{27}{25}\right)^2\)
\(1+\frac{r}{100}=\frac{27}{25}\)or, \(\frac{r}{100}=\frac{27}{25}-1=\frac{2}{25}\)
∴ \(r=\frac{2 \times 100}{25}=8\)
∴ the rate of compound interest = 8%.
Question 13. Let us write by calculating in how many years Rs. 5000 will by compound interest at the rate of 10% per annum amount to Rs. 6050.
Solution: Let the required time (year) = n.
P = Rs. 5000, A = Rs. 6050, Rate (f) = 10%
∴ \(A=P\left(1+\frac{r}{100}\right)^n\)
or, \(6050=6000\left(1+\frac{10}{100}\right)^n\)
\(\frac{6050}{5000}=\left(\frac{11}{10}\right)^n\)or, \(\left(\frac{11}{10}\right)^2=\left(\frac{11}{10}\right)^{\mathrm{n}}\)
∴ n = 2.
∴ Required Time = 2 years.
Question 14. I have Rs. 5000 in my hand. I deposited that money in a bank at the rate of 8.5% compound interest per annum for two years. Let us write by calculating how much money. I shall get it at the end of 3 years.
Solution: P = Rs. 5000, R = 8.5%
Time (n) = 2 years.
∴ Amount \(A=P\left(1+\frac{R}{100}\right)^n\)
= \(\text { Rs. } 5,000\left(1+\frac{8.5}{100}\right)^2=\text { Rs. } 5.000\left(\frac{1065}{1000}\right)^2\)
= \(\text { Rs. } \frac{5000 \times 1085 \times 1005}{1000 \times 1000}=\text { Rs } \cdot \frac{6660125}{1000}\)
= Rs. 5686.125.
Question 15. Let us calculate the amount of Rs. 5000 at the rate of 8% compound interest per annum for 3 years.
Solution: Principal (P) = Rs.5000,
Rate (R) = 8%
Time (n) = 3 years.
\(A=P\left(1+\frac{R}{100}\right)^n\)= \(\text { Rs. } 5000\left(1+\frac{8}{100}\right)^8\)
= \(\text { Rs. } 5000 \times\left(\frac{27}{25}\right)^3\)
= \(\text { Rs. } 6000 \times \frac{27 \times 27 \times 27}{25 \times 25 \times 25}\)
= Rs. 6296.56
Question 16. Goutam babu borrowed Rs. 2000 at the rate of 6% compound interest per annum for 2 years. Let us write by calculating how much compound interest at the end of 3 years he will pay.
Solution: P = Rs. 2000, R = 6%, Time (n) = 2 years, C.I. = ?
\(A=P\left(1+\frac{A}{100}\right)^n=R s .2000\left(1+\frac{6}{100}\right)^2\)= \(\text { Rs. } 2000 \times \frac{53}{50} \times \frac{53}{50}=\text { Rs. } 2247.20\)
∴ Compound interest = Rs. (2247.20 – 2000)
= Rs. 247,20.
Question 17. Let us write by calculating the amount of Rs. 30,000 at the rate of 9% compound interest per annum for 3 years.
Solution: P = Rs. 30,000
Rate (R) = 9%,
Time (n) = 3 yedra,
C.I. ?
\(\mathrm{A}=\mathrm{P}_s \cdot 30000 \times\left(1+\frac{9}{100}\right)^3\)= Rs \(30,000 \times \frac{109}{100} \times \frac{109}{100} \times \frac{109}{100}=\text { Rs. } 38850.87\)
∴ Compound interest = Rs. (38850.87 – 30,000)
= Rs. 8850.87
Question 18. Let us write by calculating the amount of Rs. 80,000 for 2 years at the rate of 5% compound interest per annum.
Solution: P = Rs. 80,000
R = 5%
n = \(2 \frac{1}{2} \text { years }\)
A = \(\text { Rs. } 80000\left(1+\frac{5}{100}\right)^2+80000\left(1+\frac{5}{100}\right)^2 \times \frac{6}{12} \times \frac{5}{100}\)
= \(\text { Rs. } 80,000\left(1+\frac{5}{100}\right)^2\left(1+\frac{5}{200}\right)\)
= \(\text { Rs. } 60,000 \times \frac{105}{100} \times \frac{105}{100} \times \frac{205}{200}=\text { Rs. } 80,405\)
Question 19. Chandadavi borrowed some money for 2 years in compound interest at the rate of 8% per annum. Let us calculate, if the compound interest is Rs. 2496, then how much money she had lent?
Solution: Let she look a loan of Rs. X.
According to the problem,
\(x\left\{\left(1+\frac{8}{100}\right)^2-1\right\}=2496\)or, \(\left\{\left(1+\frac{2}{25}\right)^2-1\right\}=2496\)
or, \(x \times \frac{729-625}{625}=2496\)
∴ \(x=\frac{2496 \times 625}{104}\)
x = 15000
∴ She took a loan of Rs. 15000
Question 20. Let us write by calculating the principal which becomes Rs. 2648 after getting 8% compound interest per annum for 3 years.
Solution: Let the principal = Rs. x.
According to the problem,
\(x\left\{\left(1+\frac{10}{100}\right)^3-1\right\}=2648\)or, \(x\left\{\left(\frac{11}{10}\right)^3-1\right\}=2648\)
or, \(x\left(\frac{1331-1000}{1000}\right)=2648\)
or, x x 331 = 2648 x 1000
or, \(x=\frac{2648 \times 1000}{331}\)
∴ x = 8000
∴ Required principal = Rs. 8,000.
Question 21. Let us write by calculating what sum of money at the rate of 8% compound – interest per annum for 3 years will amount to Rs. 31492.80.
Solution: Let the pricipal = Rs. x
According to the problem,
\(x\left(1+\frac{8}{100}\right)^3=31492.80\)or, \(x \times\left(\frac{108}{100}\right)^3=31492.80\)
\(x=\frac{31492.80 \times 100 \times 100}{108 \times 108}=25000\)∴ Principal = Rs. 25000.
Question 22. Let us calculate the difference between the compound interest and simple in- terest on Rs. 12,000 for 2 years, at 7.5% interest per annum.
Solution: Compound interest on Rs. 12,000 for 2 years at the rate of 7.5% C.I.
= \(\text { Rs. } 12,000\left\{\left(1+\frac{7.5}{100}\right)^2-1\right\}\)
= \(\text { Rs. } 12,000\left\{\left(\frac{1075}{1000}\right)^2-1\right\}\)
= \(\text { Rs. } 12,000\left\{\left(\frac{43}{40}\right)^2-1\right\}=\text { Rs. } 12,000\left(\frac{1849-1600}{1600}\right)\)
= \(\text { Rs. } 12000 \times \frac{249}{1600}=\text { Rs. } \frac{3735}{2}\)
= Rs.1967.50
Again, simple interest on Rs. 12,000 for 2 years at 7.5% S.I.
= \(\text { Rs. } 12000 \times \frac{7.5}{100} \times 2\)
= \(\text { Rs. } 12000 \times \frac{75}{100 \times 10} \times 2=\text { Rs. } 1800\)
∴ Difference between C.I. & S.I.
= Rs. (1867.50 – 1800) = Rs. 67.50
Question 23. Let us write by calculating the difference between compound interest and simple interest of Rs. 10,000 for 3 years at 5% per annum.
Solution: Compound interest on Rs.10,000 for 3 years at the rate of 5%.
= \(\text { Rs. } 10000\left\{\left(1+\frac{5}{100}\right)^3-1\right\}\)
=\( \text { Rs. } 10000\left\{\left(\frac{21}{20}\right)^3-1\right\}\)
= \(\text { Rs. } 10000\left(\frac{9261}{8000}-1\right)=\text { Rs. } 10000\left(\frac{9261-8000}{8000}\right)\)
= \(\text { Rs, } 10000 \times \frac{1261}{8000}=\frac{12610}{8}=1576.25\)
Again, simple interest = Rs. \(10000 \times \frac{5}{100} \times 3\) = Rs. 1,500.
∴ Different between C.I & S.I
= Rs. (1576.25 – 1500) = Rs. 76.25
Question 24. Let us write by calculating the sum of money, if the difference between com- pound interest and simple interest for 2 years at the rate of 9% interest per annum is Rs. 129.60.
Solution: Let the principal = Rs. x
∴ C.I. on Rs, x for 2 years at the rate of 9%.
= \(\text { Rs. } x\left\{\left(1+\frac{9}{100}\right)^2-1\right\}\)
= \(\text { Rs. } x\left\{\left(\frac{109}{100}\right)^2-1\right\}=\text { Rs. } x\left\{\frac{109 \times 109-100 \times 100}{100 \times 100}\right\}\)
= \(\text { Rs. } x \times\left(\frac{11881-10000}{10000}\right)\)
= \(\text { Rs. } x \times \frac{1881}{10000}=\text { Rs. } \frac{1881 \mathrm{x}}{10000}\)
Again, simple interest on Rs x for 2 years at the rate of 9%.
= \(\text { Rs. } x \times \frac{9}{100} \times 2=\frac{18 x}{100}\)
According to the problem,
⇒ \(\text { Rs. }\left(\frac{1881 \mathrm{x}}{10000}-\frac{18 \mathrm{x}}{100}\right)=\text { Rs. } 129.60\)
or, \(\frac{81 x}{10000}=129.60\)
or, 81x = 129.60 x 10000
x= \(\frac{1296000}{81}=16000\)
∴ Required principal = Rs. 16,000
Question 25. If the rates of compound interest for the first and the second year are 7% and 8% respectively, let us write by calculating compound interest on Rs. 6000 for 2 years.
Solution: After 2 years amount on Rs. 6,000 at the rate of 7% for 1st year & 8% for 2nd year respectively
= \(\text { Rs. } 6000 \times\left(1+\frac{7}{100}\right)\left(1+\frac{8}{100}\right)\)
= \(\text { Rs. } 6000 \times \frac{107}{100} \times \frac{108}{100}=\text { Rs. } \frac{69336}{10}\)
= Rs. 6933.6
∴ Compound interest = Rs.(6933.6 – 6000) = Rs. 933.6
Question 26. If the rate of compound interest for the first and second year are 5% and 6% respectively, let us calculate the compound interest on Rs. 5000 for 2 years.
Solution: After 2 years amount on Rs. 5000 at the rate of 5% for the 1st year & 6% for the 2nd year respectively.
= \(\text { Rs. } 5,000 \times\left(1+\frac{5}{100}\right) \times\left(1+\frac{6}{100}\right)\)
= \(\text { Rs. } 5,000 \times \frac{105}{100} \times \frac{106}{100}=\text { Rs. } 5565\)
∴ Compound interest = Rs. (5565 – 5000) = Rs. 565
Question 27. If the simple interest on a certain sum of money for 1 year is Rs. 50 and compound interest for 2 years is Rs. 102, let us write by calculating the sum of money and the rate of interest.
Solution: Simple Interest for 1 year = Rs. 50
∴ Simple Interest for 2 years = 2 Rs. 50 = Rs. 100
but compound interest for 2 years = Rs. 102
∴ Interest on Rs. 50 for 1 year = Rs. (102 – 100) = Rs. 2
∴ Interest on Rs. 100 for 1 year = \frac{2}{50} \times 100 = Rs. 4
∴ Rate of interest = 4%
∴ Principal = \(\frac{\mathrm{Sl} \times 100}{\mathrm{R} \times \mathrm{t}}\)
= \(\frac{50 \times 100}{4 \times 1}\)
= Rs. 1250.
Question 28. If simple interest and compound interest of a certain sum of money for two years are Rs. 8400 and Rs. 8652, then let us write by calculating the sum of money and the rate of interest.
Solution:
Given
If simple interest and compound interest of a certain sum of money for two years are Rs. 8400 and Rs. 8652
Simple interest for 2 years = Rs. 8400
∴ Simple interest for 1 year = Rs. 4200.
Difference between compound interest & simple interest for 2 years = Rs. (8652-8400) = Rs 252.
∴ Interest on Rs. 4200 for 1 year = Rs. 252
∴ Interest on Rs. 100 for 1 year = Rs. 252 / 4200 x 100 = Rs. 6.
∴ Rate of interest = 6%.
∴ Principal S.I x 100 / Rx time
= 4200 x 100 / 6×1
= Rs. 70,000.
Question 29. Let us calculate compound interest on Rs. 6000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months.
Solution: Compound interest for 1 year = \(\mathrm{P}\left\{\left\{1+\frac{\mathrm{H}}{200}\right)^2-1\right\}\)
= \(\text { Rs. } 6,000\left\{\left(1+\frac{8}{200}\right)^2-1\right\}\)
= \(\text { Rs. 6,000 }\left\{\left(\frac{26}{26}\right)^2-1\right\}=\text { Rs. 6,000 }\left(\frac{676-625}{625}\right)\)
= \(\text { Rs. } 6,000 \times \frac{51}{625}=\text { Rs. } \frac{2488}{5}=\text { Rs. } 499.60\)
Question 30. Let us write by calculating compound interest on Rs. 6250 at the rate of 10% compound interest per annum, compounded at the interval of 3 months.
Solution: 9 months = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year.
∴ Compound interest on Rs. 6250 for \(\frac{3}{4}\) year at 10%
= \(\text { Rs. } P\left\{\left(1+\frac{\mathrm{F}}{400}\right)^{4 \times \frac{3}{4}}-1\right\}=\operatorname{Pss} 6250\left\{\left(1+\frac{10}{400}\right)^3-1\right\}\)
= \(\text { Rs. } 6250 \times\left\{\left(\frac{41}{40}\right)^3-1\right\}=\text { Rs. } 6250\left\{\frac{68921-64000}{64000}\right\}\)
= \(\text { Rs. } 6260 \times \frac{4921}{64000}=\text { Rs. } \frac{123025}{256}=\text { Rs. } 480.57 \text { Apporx. }\)
Question 31. Let us write by calculating at what rate of interest per annum Rs. 6000 will amount to Rs. 69984 in 2 years.
Solution: P = Rs. 6,000
A = Rs. 69984
Time (n) = 2 years,
Rate (R) = ?
∴ \(P\left(1+\frac{A}{100}\right)^n=A\)
\(\text { Rs. } 60000\left(1+\frac{A}{100}\right)^2=\text { Rs. } 69984\)or, \(\left(1+\frac{A}{100}\right)^2-\frac{60984}{60000}=\frac{11664}{10000}=\left(\frac{108}{100}\right)^2\)
∴ \(1+\frac{R}{100}=\frac{106}{100}\)
∴ \(\frac{A}{100}=\frac{8}{100}\)
∴ A = 8
∴ Rate(R) = 8%
Question 32. Let us calculate in how many years Rs. 4000 will amount to Rs. 46656 at the rate of 8% compound interest per annum.
Solution: P = Rs. 4000;
A = pa.46656,
R = 8%
Time (n) = ?
∴ \(P\left(1+\frac{A}{100}\right)^n=A\)
or, \text { Rs. } 40,000\left(1+\frac{B}{100}\right)^n=\text { Rs. } 46656
∴ \left(1+\frac{8}{100}\right)^n-\frac{46856}{40000}=\frac{11664}{10000}=\left(\frac{108}{100}\right)^2
or, \left(\frac{108}{100}\right)^n=\left(\frac{108}{100}\right)^2
∴ n = 2
Time = 2 years.
Question 33. Let us write by calculating at what rate of compound interest per annum, the amount on Rs. 10,000 for 2 years is Rs. 12100.
Solution: P = Rs. 10,000
A = Rs. 12100
Time (n) = 2 years,
Rate (R) = ?
∴ \(P\left(1+\frac{A}{100}\right)^n=A\)
\(\text { Rs. } 10000\left(1+\frac{\mathrm{R}}{100}\right)^2=12100\)or, \(\left(\frac{100+\mathrm{R}}{100}\right)^2=\frac{12100}{10000}=\left(\frac{11}{10}\right)^2\)
∴ \(\frac{100+R}{100}=\frac{11}{10}\)
∴ 100 + A = 110
∴ A = 10
∴ Rate = 10%
Question 34. Let us calculate in how many years Rs. 50000 will amount to Rs. 60500 at the rate of 10% compound interest per annum.
Solution: P = Rs 50,000
R = Rs. 60,500
Rate (R) = 10%
Time (n) = ?
∴ \(P\left(1+\frac{A}{100}\right)^n=A\)
\text { Rs. } 50,000\left(1+\frac{10}{100}\right)^n=60,500
or, \left(\frac{110}{100}\right)^n=\frac{60500}{50000}=\frac{605}{500}=\left(\frac{11}{10}\right)^2
or, \left(\frac{11}{10}\right)^n=\left(\frac{11}{10}\right)^2
∴ n = 2
∴ Time = 2 years.
Question 35. Let us write by calculating in how many years Rs. 30,000 will amount to Rs. 399300 at the rate of 10% compound interest per annum.
Solution: P = Rs. 300000,
A = Rs. 399300,
Rate (R) = 10%
Time (n) = ?
∴ \(P\left(1+\frac{A}{100}\right)^n=A\)
\(\text { Rs. } 300000\left(1+\frac{10}{100}\right)^{\text {n }}=\text { Rs. } 399300\)or, \(\left(1+\frac{1}{10}\right)^n=\frac{399300}{300000}\)
or, \(\left(\frac{11}{10}\right)^n=\left(\frac{11}{10}\right)^3\)
∴ n = 3
∴ Time = 3 years.
Question 36. Let us calculate the compound interest and amount on Rs. 1600 for 1 the rate of 10% compound interest per annum, compounded at an interval of 6 months.
Solution: P = Rs.1600, R = 10%
Time (n) = \(\frac{3}{2} \times 2\) = 3, A = ?
\(A=P\left(1+\frac{R}{200}\right)^n\)= \(R s \cdot\left(600\left(1+\frac{10}{200}\right)^3\right.\)
= \(\text { Rs. } 1600\left(1+\frac{10}{200}\right)^3\)
= \(\text { Rs. } 1600\left(\frac{21}{20}\right)^3\)
= \(\text { Rs. } 1600 \times \frac{21 \times 21 \times 21}{20 \times 20 \times 20}\)
= \(\frac{21 \times 21 \times 21}{5}\)
= \(\text { Rs. } \frac{9261}{5}\)
= Rs. 1852.20
∴ Amount = Rs. 1852.20
Compound interest = (Rs 1852.20 – Rs.1600)
= Rs. 25220.
Question 37. At present, 4000 students have been taking training from this train- ing centre. In the last 2 years, it has been decided that the facility to get a chance for a training programme in this centre will be increased by 5% in comparison to its previous year. Let us see by calculating how many students will get a chance to join this training programme at the end of 2 years.
Solution:
Given
At present, 4000 students have been taking training from this train- ing centre. In the last 2 years, it has been decided that the facility to get a chance for a training programme in this centre will be increased by 5% in comparison to its previous year.
After 2 years the total number of candidates
= 4000 x (1+ 5 / 100)
= 4000 x105 /100 x 105 100
= 4410. Ans.
After 2 years the total number of candidates = 4410.
Question 38. The price of a motor car is Rs. 3 lakhs. If the price of the car depreciates at the rate of 30% every year, let us write by calculating the price of the car after 3 years.
Solution:
Given
The price of a motor car is Rs. 3 lakhs. If the price of the car depreciates at the rate of 30% every year
After 3 years, the price of the car
= Rs. 300000 X (1- 30/100)3
= Rs. 30,0000 X (7/10)3
Rs. 300000 X (7x7x7 / 10×10×10)
= Rs. 102900.
After 3 years, the price of the car = Rs. 102900.
Question 39. At present the population of a city is 576000; if it is being increased at the rate of 6 2/3 % every year, let us calculate its population 2 years ago.
Solution: Let 2 years before, population of the city was x.
According to the problem,
∴ \(x\left(1+\frac{62 / 3}{100}\right)^2\)=576000
Or, \(x \times\left(1+\frac{20}{3 \times 100}\right)^2\) = 576000
Or, \(\mathrm{x} \times\left(1+\frac{1}{15}\right)^2\) = 576000
Or, \(x \times \frac{16}{15} \times \frac{16}{15}\) = 576000
∴ x = \(\frac{576000 \times 15 \times 15}{16 \times 16}\)
= \( \frac{576000 \times 225}{256}\)
= 2250 × 225 = 506250
∴ 2 years before, the population of the city was = 506205.