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.
Read and Learn More WBBSE Solutions For Class 10 Maths
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
“WBBSE Class 10 Maths Compound Interest and Uniform Rate of Increase or Decrease Exercise 6.1 solutions”
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
“West Bengal Board Class 10 Maths Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease Exercise 6.1 solutions”
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.
“WBBSE Class 10 Compound Interest and Uniform Rate of Increase or Decrease Exercise 6.1 solutions explained”
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
“WBBSE Class 10 Maths Exercise 6.1 Compound Interest and Uniform Rate of Increase or Decrease problem solutions”
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.
“Class 10 WBBSE Maths Exercise 6.1 Compound Interest and Uniform Rate of Increase or Decrease step-by-step solutions”
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
“WBBSE Class 10 Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease Exercise 6.1 solution guide”
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 = \(\mathrm{P}\left\{\left\{1+\frac{\mathrm{H}}{200}\right)^2-1\right\}\)
= \(\text { Rs. } 6,000\left\{\left(1+\frac{8}{200}\right)^2-1\right\}\)
= \(\text { Rs. 6,000 }\left\{\left(\frac{26}{26}\right)^2-1\right\}=\text { Rs. 6,000 }\left(\frac{676-625}{625}\right)\)
= \(\text { Rs. } 6,000 \times \frac{51}{625}=\text { Rs. } \frac{2488}{5}=\text { Rs. } 499.60\)
“West Bengal Board Class 10 Maths Exercise 6.1 Compound Interest and Uniform Rate of Increase or Decrease solutions”
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.
“Class 10 WBBSE Maths Exercise 6.1 solutions for Compound Interest and Uniform Rate of Increase or Decrease”
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.
“WBBSE Class 10 Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease Exercise 6.1 problem-solving steps”
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.