| Book Name | : Ganit Prakash |
| Subject | : Mathematics |
| Class | : 10 (Madhyamik) |
| Publisher | : Prof Nabanita Chatterjee |
| Chapter Name | : Compound Interest And Uniform Rate Of Increase Or Decrease (6th Chapter) |
Application 1.
If I take a loan of ₹ 1400 at 5% compound interest per annum for years, let us write by calculating how much compound interest and total amount I shall pay.
Solution :
When Principal = ₹ 1400
Compound interest = \frac{10.25 \times 1400}{100} = ₹ 143.50
Principal & compound interest are in direct proportion.
Amount = ₹ (1400+143.50) = ₹ 1543.50.
Application 3.
Let us write by calculating what is the amount on ₹ 1000 for 2 ears at the rate of 5% compound interest per annum. [Let me do it myself]
Solution :
Amount = \text { ₹ } 1000\left(1+\frac{5}{100}\right)^2 = \text { ₹ } 1000 \times\left(\frac{22}{20}\right)^2
= \text { ₹ } 1000 \times \frac{441}{400} = \text { ₹ } \frac{2205}{2} = \text { ₹ } 1102.50
Application 4.
If ₹ 4000 is invested in a bank at the rate of 5% compound interest per annum, then let us write by calculating the amount after 3 year.
Solution :
Compound interest = ₹ (4630.50-4000) = ₹ 630.50.
Application 5.
At 5% compound interest per annum, let us find the compound interest on ₹ 10,000 for 3 year [Let me do it myself]
Solution :
Compound interest for 3 years = ₹ 10000\left\{\left(1+\frac{5}{100}\right)^3-1\right\}
\text { ₹ } 10000\left\{\left(1+\frac{1}{20}\right)^3-1\right\}
= \text { ₹ } 10000\left\{\left(\frac{21}{20}\right)^3-1\right\}
= \text { ₹ } 10000\left\{\frac{9261-8000}{8000}\right\}
= ₹ 10000\left(\frac{1261}{8000}\right) = ₹ \frac{6305}{4} = ₹ 1576.25
Application 6.
If interest is compounded half-yearly, let us write by calculating compounded interest and amount on ₹ 8,000 at the rate of 10% compound interest per annum for 1\frac{1}{2} year
Solution :
Compound interest for 1 \frac{1}{2} years = ₹ 1261
Application 7.
Let us find compound interest on ₹ 1000 at the rate of 10% compound interest per annum and the interest being compounded at 6 months interval [Let me do it myself]
Solution :
Compound interest on ₹ 1000 for 1 year at 10% per annum compound interest compound semi-annually.
\text { Amount = ₹ } 1000\left(1+\frac{1 / 2}{100}\right)^{2 \times 1}
= \text { ₹ } 1000\left(1+\frac{10}{200}\right)² = \text { ₹ } 1000\left(\frac{21}{20}\right)² = \text { ₹ } 1000 \times \frac{441}{400}
= \text { ₹ } \frac{2205}{2} = ₹ 1102.50
∴ Compound interest = ₹ (1102.50-1000) = ₹ 102.50
Application 8.
Let us write by calculating compound interest on ₹ 10,000 at the rate of 8% compound interest per annum for 9 months, compounded at an interval 0f 3 months.
Solution :
Compound interest for 9 months = ₹ (1102.50-1000) = ₹ 102.50.
Application 10.
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 ₹ 25000 for 2 year [Let me do myself]
Solution :
Amount after two years on ₹ 25,000 at two rates of 4% for 1st year & 5% for 2nd year
= \text { ₹ } 25,000\left(1+\frac{4}{100}\right) \cdot\left(1+\frac{5}{100}\right)
= \text { ₹ } 25,000 \times \frac{104}{100} \times \frac{105}{100} = ₹ 27300
\therefore \text { Compound interest } = \text { ₹ } 27300-\text { ₹ } 25000
= \text { ₹ } 2300
Application 11.
I lend ₹ 10,000 at the rate of 4% compound interest per annum for 2 \frac{1}{2} years, let us write by calculating how much total money I shall pay.
Solution :
Amount for 2 \frac{1}{2} years = ₹ (10816+216.32) = ₹ 11032.32.
Application 12.
Let us find the amount of ₹ 30,000, at the rate of 6% compound interest per annum for 2 \frac{1}{2} year [Let me do it myself]
Solution :
Amount for 2 \frac{1}{2} years at the rate of 6%.
compound interest
= ₹ \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]
= ₹ \left[30000\left(1+\frac{6}{100}\right)²\left\{1+\frac{6}{12} \times \frac{6}{100}\right\}\right]
= ₹ \left(30000 \times \frac{106 \times 106}{100 \times 100} \times \frac{103}{100}\right)
= ₹ 34719.24
Application 14.
Let us write by calculating what sum of money will amount ₹ 3528 after 2 years at the rate of 5% compound interest per annum. [Let me do it myself]
Solution :
Let the principal = ₹ x
According to the problem,
x\left(1+\frac{5}{100}\right)^2 = 3528
\text { or, } x \times \frac{105}{100} \times \frac{105}{100} = 3528
\therefore x = \frac{2528 \times 100 \times 100}{105 \times 105} = 3200
∴ Principal = ₹ 3200.
Application 17.
Let us see by calculating alternatively that Minatididi deposits ₹ 3,00,000 in the bank.
Solution :
Let the principal for 1st year = ₹ x.
Interest for the 1st year at 8% = ₹ \frac{8 x}{100}.
∴ Principal for the 2nd year = \text{ ₹ } \left(x+\frac{8 x}{100}\right) = \text{ ₹ } \frac{108 x}{100}.
Again, the interest for the 2nd year at 8% = \text{ ₹ } \frac{8}{100} \times \frac{108 x}{100} = \text{ ₹ } \frac{864 x}{10000}
∴ Principal for the 3rd year = \text{ ₹ } \left(\frac{108 x}{100}+\frac{864 x}{10000}\right) = \text{ ₹ } \frac{11664 x}{10000}
∴ The interest for the 3rd year (one year) at 8% = ₹ \frac{8 x}{100}+\frac{11664 x}{10000} = \text{ ₹ }\frac{93312 x}{1000000}
∴ After 3 years, amount will be ₹ \left(\frac{11664 x}{10000}+\frac{93312 x}{1000000}\right) = \text{ ₹ } \frac{1259712 x}{1000000}
∴ According to the problem,
\frac{1259712 x}{1000000} = 37791.36
\therefore \text{ x } = \frac{37791.36 \times 1000000}{1259712} = \text { ₹ } 30,000
Ans. Minati Di deposited ₹ 30,000 in the bank.
Application 19.
The simple interest and compound interest of a certain sum of money for 2 years are ₹ 840 and ₹ 869.40 respectively. Let us calculate that sum of money and the rate of interest. [Let me do it myself]
Solution :
Simple interest for 2 years = ₹ 840
∴ Simple interest for 1 year = ₹ \frac{840}{2} = ₹ 420.
Difference between Compound interest & Simple interest for 2 years = ₹ (869.40-840) = ₹ 29.40.
∴ Interest on ₹ 420 for 1 year = ₹ 29.40
∴ Interest on ₹ 100 for 1 year = ₹ \frac{29.40}{420} × 100. = ₹ 7.
∴ Rate of interest = 7%.
Principal = ₹ \frac{420 \times 100}{7} = ₹ 6,000 Ans.
Application 21.
Let us calculate at what rate of compound interest, ₹ 5,000 will amount to ₹ 5832 in 2 year [Let me do it myself]
Solution :
Let the rate of compound interest = r%
P = ₹ 5,000 ; A = ₹ 5832 ; Time = n = 2 year
∴ 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(1+\frac{r}{100}\right)^2 = \left(\frac{27}{25}\right)^2
1+\frac{r}{100} = \frac{27}{25} \text { or, } \frac{r}{100} = \frac{27}{25}-1 = \frac{2}{25} \\
\therefore r = \frac{2 \times 100}{25} = 8
∴ The rate of comound interest = 8%.
Application 23.
Let us write by calculating in how many years ₹ 5000 will compound interest at the rate of 10% per annum amount to ₹ 6050.
Solution :
Let the required time (year) = n.
P = ₹ 5000 ; A = ₹ 6050 , Rate (r) = 10%
∴ A = P\left(1+\frac{r}{100}\right)^n
or, 6050 = 5000\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{1.1}{10}\right)^n
∴ n = 2.
∴ Required Time = 2 years
LET US SEE BY CALCULATING – 6.1
Question 1
I have ₹ 5000 in my hand. I deposited that money in a bank at the rate of 8.5% compound interest per annum for two year Let us write by calculating how much money I shall get at the end of 3 year
Solution :
P = ₹ 5000, R = 8.5% Time (n) = 2. year
∴ Amount A = P\left(1+\frac{R}{100}\right)^n \\
= \text { ₹ } 5,000\left(1+\frac{8.5}{100}\right)^2 = \text { ₹ } 5,000\left(\frac{1085}{1000}\right)^2 \\
= \text { ₹ } \frac{5000 \times 1085 \times 1085}{1000 \times 1000} = \text { ₹ } \frac{5886125}{1000} \\
= ₹ 5886.125 Ans.
Question 2
Let us calculate the amount on ₹ 5000 at the rate of 8% compound interest per annum for 3 year
Solution :
Principal (P) = ₹ 5000, Rate (R) = 8% Time (n) = 3 year
A = P\left(1+\frac{R}{100}\right)^n \\
= \text { ₹ } 5000\left(1+\frac{8}{100}\right)^3 \\
= \text { ₹ } 5000 \times\left(\frac{27}{25}\right)^3 \\
= \text { ₹ } 5000 \times \frac{27 \times 27 \times 27}{25 \times 25 \times 25} \\
= ₹ 6298.56
Question 3
Goutam babu borrowed ₹ 2000 at the rate of 6% compound interest per annum for 2 year Let us write by calculating how much compound interest at the end of 3 years he will pay.
Solution :
P = R s, 2000 ; R = 6%, Time (n) = 2 years, C.I. = ?
A = P\left(1+\frac{R}{100}\right)^n = \text { ₹ } 2000\left(1+\frac{6}{100}\right)^2 \\
= \text { ₹ } 2000 \times \frac{53}{50} \times \frac{53}{50} = ₹ 2247.20
∴ Compound interest = ₹ (2247.20-2000) = ₹ 247.20 Ans.
Question 4
Let us write by calculating the amount on ₹ 30,000 at the rate of 9% compound interest per annum for 3 year
Solution :
P = ₹ 30,000 , Rate (R) = 9%, Time (n) = 3 years, C.I. = ?
A = \text { ₹ } 30000 \times\left(1+\frac{9}{100}\right)^3 \\
= \text { Rs } 30,000 \times \frac{109}{100} \times \frac{109}{100} \times \frac{109}{100} = \text { ₹ } 38850.87. \\
∴ Compound interest = ₹ (38850.87-30,000) = ₹ 8850.87 Ans.
Question 5
Let us write by calculating the amount on ₹ 80,000 for 2 \frac{1}{2} years at the rate of 5 % compound interest per annum.
Solution :
P = ₹ 80,000, R = 5% n = 2 \frac{1}{2} year
A = \text { ₹ } 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 { ₹ } 80,000\left(1+\frac{5}{100}\right)^2\left(1+\frac{5}{200}\right) \\
= \text { ₹ } 80,000 \times \frac{105}{100} \times \frac{105}{100} \times \frac{205}{200} = ₹ 90,405
Question 6
Chandadavi borrowed some money for 2 years in the compound interest at the rate of 8% per annum. Let us calculate, if the compound interest is ₹ 2496, then how much money she had lended.
Solution :
Let she look a loan of ₹ x.
According to the problem,
x\left\{\left(1+\frac{8}{100}\right)^2-1\right\} = 2496 \\
\text { or, }\left\{\left(1+\frac{2}{25}\right)^2-1\right\} = 2496 \\
\text { or, } x \times \frac{729-625}{625} = 2496
\therefore x = \frac{2496 × 625}{104}
x = 15000
∴ She took a loan of ₹ 15000.
Question 7
Let us write by calculating the principal which becomes ₹ 2648 after getting 8% compound interest per annum for 3 year
Solution :
Let the principal = ₹ 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 × 331 = 2648 × 1000
or, x = \frac{2648 × 1000}{331}
∴ x = 8000
∴ Required principal = ₹ 8,000.
Question 8
Rahaman chacha deposited some money in a cooperative bank at the rate of 9% compound interest and he received amount ₹ 29702.50 after 2 year Let us calculate how much money Rahaman chacha had deposited in cooperative bank:
Solution :
Let he deposited ₹ x for 2 years at the rate of 9% C.I.
According to the problem,
x\left(1+\frac{9}{100}\right)^2 = 29702.50
or, x × \frac{109}{100} × \frac{109}{100} = \frac{2970250}{100}
or, x = \frac{2970250 × 100 × 100}{109 × 109 × 100}
x = 25000.
∴ He deposited ₹ 25000 in the bank.
Question 9
Let us write by calculating what sum of money at the rate of 8% compound interest per annum for 3 years will amount to ₹ 31492.80.
Solution :
Let the principal = ₹ x
According to the problem,
x\left(1+\frac{8}{100}\right)^3 = 31492.80
or, x × \left(\frac{108}{100}\right)^3 = 31492.80
x= \frac{31492.80 × 100 × 100}{108 × 108} = 25000
∴ Principal = ₹ 25000.
Question 10
Let us calculate the difference between the compound interest and simple interest on ₹ 12,000 for 2 years, at 7.5% interest per annum.
Solution :
Compound interest on ₹ 12,000 for 2 years at the rate of 7.5% C.I.
= ₹ 12,000\left\{\left(1+\frac{7.5}{100}\right)^2-1\right\}
= ₹ 12,000\left\{\left(\frac{1075}{1000}\right)^2-1\right\}
= ₹ 12,000\left\{\left(\frac{43}{40}\right)^2-1\right\} = ₹ 12,000\left(\frac{1849-1600}{1600}\right)
= ₹ 12000 × \frac{249}{1600} = ₹ \frac{3735}{2}
= ₹ 1967.50
Again, simple interest on ₹ 12,000 for 2 years at the rate of 7.5% S.I.
= ₹ 12000 × \frac{7.5}{100} × 2
= ₹ 12000 × \frac{75}{100 × 10} × 2 = ₹ 1800
∴ Difference between C.I. & S.I.
₹ (1867.50-1800) = ₹ 67.50
Question 11
Let us write by calculating the difference between compound interest and simple interest of ₹ 10,000 for 3 years at 5% per annum.
Solution :
Compound interest on ₹ 10,000 for 3 years at the rate of 5%.
= ₹ 10000\left\{\left(1+\frac{5}{100}\right)^3-1\right\}
= ₹ 10000\left\{\left(\frac{21}{20}\right)^3-1\right\}
= ₹ 10000\left(\frac{9261}{8000}-1\right) = ₹ 10000\left(\frac{9261-8000}{8000}\right)
= ₹ 10000 × \frac{1261}{8000} = \frac{12610}{8} = 1576.25
Again, simple interest = ₹ 10000 × \frac{5}{100} × 3 = ₹ 1,500.
∴ Different between C.I & S.I
= ₹ (1576.25-1500) = ₹ 76.25 Ans.
Question 12
Let us write by calculating the sum of money, if the difference between compound interest and simple interest for 2 years at the rate of 9% interest per annum is ₹ 129.60.
Solution :
Let the principal = ₹ x.
∴ C.I. on ₹ x for 2 years at the rate of 9%.
= ₹ x\left\{\left(1+\frac{9}{100}\right)^2-1\right\}
= ₹ x\left\{\left(\frac{109}{100}\right)^2-1\right\} = \text{ ₹ x }\left\{\frac{109 × 109-100 × 100}{100 × 100}\right\}
= ₹ x × \left(\frac{11881-10000}{10000}\right)
= ₹ x × \frac{1881}{10000} = \text{ ₹ } \frac{1881 x}{10000}
Again, simple interest on Rs x for 2 years at the rate of 9%.
= ₹ x × \frac{9}{100} × 2 = \frac{18 x}{100}
According to the problem,
₹ \left(\frac{1881 x}{10000}-\frac{18 x}{100}\right) = ₹ 129.60
or, \frac{81 x}{10000} = 129.60
or, 81x = 129.60 × 10000.
x = \frac{1296000}{81} = 16000
∴ Required Principal = ₹ 16,000
Question 13
Let us write by calculating the sum of money if the difference between compound interest and simple interest for 3 years becomes ₹ 930 at the rate of 10% interest per annum.
Solution :
Let the principal = ₹ x
∴ C.I. on ₹ x for 3 years at the rate of 10%
= ₹ x\left\{\left(1+\frac{10}{100}\right)^3-1\right\} = x\left\{\left(\frac{11}{10}\right)^3-1\right\}
= ₹ x ×\left(\frac{331}{1000}-1\right)
= ₹ x ×\frac{331}{1000} = \text{ ₹ }\frac{331 x}{1000}
Again, the S.I on ₹ x for 3 years at the rate of 10%
= ₹ x × \frac{10}{100} × 3 = \text{ ₹ } \frac{30 x}{100} = ₹ \frac{3 x}{10}
According to the problem,
₹ \left(\frac{331 x}{1000}-\frac{3 x}{10}\right) = ₹ 930
or, \frac{331 x-300 x}{1000} = 930
or, \frac{31 x}{1000} = 930
\therefore x = \frac{930 × 1000}{31} = 30,000
∴ Required principal = ₹ 30,000
Question 14
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 ₹ 6000 for 2 year
Solution :
After 2 years amount on ₹ 6,000 at the rate of 7% for 1st year & 8% for 2nd year respectively
= ₹ 6000 ×\left(1+\frac{7}{100}\right)\left(1+\frac{8}{100}\right)
= ₹ 6000 × \frac{107}{100} × \frac{108}{100} = ₹ \frac{69336}{10}
= ₹ 6933.6
∴ Compound interest = ₹ (6933.6-6000) = ₹ 933.6 Ans.
Question 15
If the rate of compound interest for the first and second year are 5% and 6% respectively, let us calculate the compound interest on ₹ 5000 for 2 year
Solution :
After 2 years amount on ₹ 5000 at the rate of 5% for the 1st year & 6% for the 2nd year respectively
= ₹ 5,000 ×\left(1+\frac{5}{100}\right) ×\left(1+\frac{6}{100}\right)
= ₹ 5,000 × \frac{105}{100} × \frac{106}{100} = ₹ 5565
∴ Comound interest = ₹ (5565-5000) = ₹ 565. Ans.
Question 16
If simple interest on a certain sum of money for 1 year is ₹ 50 and compound interest for 2 years is ₹ 102, let us write by calculating the sum of money and the rate of interest.
Solution :
Simple Interest for 1 year = ₹ 50
∴ Simple Interest for 2 years = 2 ₹ 50 = ₹ 100
But compound interest for 2 years = ₹ 102.
∴ Interest on ₹ 50 for 1 year = ₹ (102-100) = ₹ 2
∴ Interest on ₹ 100 for 1 year = \frac{2}{50} × 100 = ₹ 4.
∴ Rate of interest = 4%
∴ Principal = \frac{S . I × 100}{R × t}
= \frac{50 × 100}{4 × 1} = ₹ 1250.
Question 17
If simple interest and compound interest of a certain sum of money for two years are ₹ 8400 and ₹ 8652, then let us write by calculating the sum of money and the rate of interest.
Solution :
Simple interest for 2 years = ₹ 8400
∴ Simple interest for 1 year = ₹ 4200.
Difference between compound interest & simple interest for 2 years = ₹ (8652-8400) = Rs 252.
∴ Interest on ₹ 4200 for 1 year = ₹ 252
∴ Interest on ₹ 100 for 1 year = ₹ \frac{252}{4200} × 100 = ₹ 6.
∴ Rate of interest = 6%
∴ Principal = \frac{S.1 × 100}{R × \text { time }} = \frac{4200 × 100}{6 × 1} = ₹ 70,000. Ans.
Question 18
Let us calculate compound interest on ₹ 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 = P\left\{\left(1+\frac{R}{200}\right)^{2}-1\right\}
= ₹ 6,000\left\{\left(1+\frac{8}{200}\right)^{2}-1\right\}
= ₹ 6,000\left\{\left(\frac{26}{25}\right)^{2}-1\right\} = \text { ₹ } 6,000\left(\frac{676-625}{625}\right)
= ₹ 6,000 × \frac{51}{625} = \text { ₹ } \frac{2488}{5} = ₹ 489.60
Question 19
Let us write by calculating compound interest on ₹ 6250 for 9 months at the rate of 10% compound interest per annum, compounded at the interval of 3 months.
Solution :
9 months = \frac{9}{12} \text{ year }= \frac{3}{4} year.
∴ Compound interest on ₹ 6250 for \frac{3}{4} year at 10%
= \text { ₹ } P\left\{\left(1+\frac{R}{400}\right)^{4 × \frac{3}{4}}-1\right\} = \text { ₹ } 6250\left\{\left(1+\frac{10}{400}\right)^{3}-1\right\}
= \text { ₹ } 6250 ×\left\{\left(\frac{41}{40}\right)^{3}-1\right\} = \text { ₹ } 6250\left\{\frac{68921-64000}{64000}\right\}
= \text { ₹ } 6250 × \frac{4921}{64000} = \text { ₹ } \frac{123025}{256} = ₹ 480.57 Apporx.
Question 20
Let us write by calculating at what rate of interest per annum ₹ 60000 will amount to ₹ 69984 in 2 year
Solution :
P = R s 60,000; A = ₹ 69,984; Time (n) = 2 years; Rate (R) = ?
\therefore P\left(1+\frac{R}{100}\right)^{n} = A
₹ 60000\left(1+\frac{R}{100}\right)^{2} = ₹ 69984
or, \left(1+\frac{R}{100}\right)^{2} = \frac{69984}{60000} = \frac{11664}{10000} = \left(\frac{108}{100}\right)^{2}
\therefore 1+\frac{R}{100} = \frac{108}{100}.
\therefore \frac{R}{100} = \frac{8}{100} \quad \therefore R = 8.
∴ Rate (R) = 8%
Question 21
Let us calculate in how many years ₹ 40000 will amount to ₹ 46656 at the rate of 8% compound interest per annum.
Solution :
P = ₹ 40000; A = ₹ 46656; R = 8%; Time (n) = ?
\therefore \text{ P }\left(1+\frac{R}{100}\right)^{n} = A
or, ₹ 40,000\left(1+\frac{8}{100}\right)^{n} = ₹ 46656
\therefore\left(1+\frac{8}{100}\right)^{n} = \frac{46656}{40000} = \frac{11664}{10000} = \left(\frac{108}{100}\right)^{2}
or, \left(\frac{108}{100}\right)^{n} = \left(\frac{108}{100}\right)^{2}
\quad \therefore n = 2 \therefore Time = 2 year
Question 22
Let us write by calculating at what rate of compound interest per annum, the amount on ₹ 10,000 for 2 years is ₹ 12100.
Solution :
P = ₹ 10,000, A = ₹ 12100 , Time (n) = 2 years, Rate (R) = ?
\therefore \text{ P }\left(1+\frac{R}{100}\right)^{n} = A
₹ 10000\left(1+\frac{R}{100}\right)^{2} = 12100
or, \left(\frac{100+R}{100}\right)^{2} = \frac{12100}{10000} = \left(\frac{11}{10}\right)^{2}
\therefore \frac{100+R}{100} = \frac{11}{10}
∴ 100+R = 110
∴ Rate = 10%.
∴ R = 10
Question 23
Let us calculate in how many years ₹ 50000 will amount to ₹ 60500 at the rate of 10% compound interest per annum.
Solution :
P = ₹ 50,000, R = R s.60,500, Rate (R) = 10% Time (n) = ?
\therefore P\left(1+\frac{R}{100}\right)^{n} = A
₹ 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} \quad \therefore n = 2
∴ Time = 2 year
Question 24
Let us write by calculating in how many years ₹ 30,000 will amount to ₹ 399300 at the rate of 10% compound interest per annum.
Solution :
P = ₹ 300000, A = ₹ 399300 , Rate (R) = 10%, Time (n) = ?
\therefore P\left(1+\frac{R}{100}\right)^{n} = A
₹ 300000\left(1+\frac{10}{100}\right)^{n} = ₹ 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} \quad \therefore n = 3
∴ Time = 3 year
Question 25
Let us calculate the compound interest and amount on ₹ 1600 for 1 \frac{1}{2} years at the rate of 10% compound interest per annum, compounded at an interval of 6 months.
Solution :
P = ₹ 1600, R = 10%. Time (n) = \frac{3}{2} × 2 = 3, A = ?
A = P\left(1+\frac{R}{200}\right)^{n}
= ₹ 1600\left(1+\frac{10}{200}\right)^{3} = ₹ 1600\left(1+\frac{10}{200}\right)^{3} = \text{ ₹ } 1600\left(\frac{21}{20}\right)^{3}
= ₹ 1600 × \frac{21 × 21 × 21}{20 × 20 × 20}
= ₹ \frac{21 × 21 × 21}{5} = \text{ ₹ } \frac{9261}{5} = ₹ 1852.20
∴ Amount = ₹ 1852.20
Compound interest = ( Rs 1852.20 - ₹ 1600)
= ₹ 252.20.
Application 24
At present, 4000 students have been taking training from this training centre. In last 2 years it has been decided that the facility to get a chance for training progreamme in this centre will be increased by 5% in comparison to its previous year. Let us see by calculating how many students will get chance to join this training programme at the end of 2 year
Solution :
After 2 years total candidates
= 4000 ×\left(1+\frac{5}{100}\right)^{2} = 4000 × \frac{105}{100} × \frac{105}{100} = 4410 Ans.
Application 27.
The price of a motor car is ₹ 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 year [Let me do it myself]
Solution :
After 3 years, the price of the car
= \text { ₹ } 300000 ×\left(1-\frac{30}{100}\right)^{3}
= \text { ₹ } 30,0000 ×\left(\frac{7}{10}\right)^{3} = \text { ₹ } 300000 × \frac{7 × 7 × 7}{10 × 10 × 10}
= ₹ 102900 Ans.
Application 29.
At present the population of a city is 576000; if it is being increased at the rate of 6 \frac{2}{3}% every year, let us calculate its population 2 years ago. [Let me do it myself]
Solution :
Let 2 years before, population of the city was x.
According to the problem,
\therefore x\left(1+\frac{62 / 3}{100}\right)^{2} = 576000
or, x ×\left(1+\frac{20}{3 × 100}\right)^{2} = 576000
or, x ×\left(1+\frac{1}{15}\right)^{2} = 576000
or, x × \frac{16}{15} × \frac{16}{15} = 576000
\therefore x = \frac{576000 × 15 × 15}{16 × 16}
= \frac{576000 × 225}{256}
= 2250 × 225 = 506250
∴ 2 years before, the population of the city was = 506250.
LET US WORK OUT - 6.2
Question 1
At present the population of the village of Pahalanpur is 10000; if population is being increased at the rate of 3% every year, let us write by calculating its population after 2 year
Solution :
Present population = 10000
After one year, population will be
= 10000+\frac{3}{100} × 10000 = 10000+300 = 10300
After 2nd year, population will be = 10300+\frac{3}{100} × 10300 + 309 = 10609 Ans.
Question 2
Rate of increase in population of a state is 2% in a year. The present population s 80000000; let us calculate the population of the state after 3 year
Solution :
Present population = 80000000
After one year population will be
= 8000000+\frac{2}{100} × 80000000 = 80000000+1600000 = 81600000
After 2nd year the population will be = 83232000+83232000 × \frac{2}{100}
= 81600000+1632000 = 83232000
After 3rd year the population will be
= 83232000+83232000 × \frac{2}{100}
= 83232000+1664640 = 84896640 Ans.
Question 3
The price of a machine in a leather factory depreciates at the rate of 10% every lear. If the present price of the machine be ₹ 100000, let us calculate what will be he price of that machine after 3 year
Solution :
Present price of the machine = ₹ 100000
After 1st year the price will be = ₹ \left(100000-\frac{10}{100} × 100000\right) = ₹ 90,000
After 2nd year the price will be = ₹ \left(90000-\frac{10}{100} × 90000\right) = ₹ 81000
∴ After 3rd year the price will be
= \text { ₹ }\left(81000-\frac{10}{100} × 81000\right) = ₹ (81000-8100) = ₹ 72900 Ans.
Question 4
As a result of Sarba Siksha Abhiyan, the students leaving the school before completion, the students are readmitted, so the number of students in a year increased by 5% in comparison to its previous year. If the number of such readmitted students in a district be 3528 in the present year, let us write by calculating, the number of students readmitted 2 years before in this manner.
Solution :
Let at present no. of students = 3528.
Let two years before, no. of student was x & rate = 5%.
∴ According to the problem,
x\left(1+\frac{5}{100}\right)^{2} = 3528
or, x × \frac{105}{100} × \frac{105}{100} = 3528
\therefore \text{ x } = \frac{3528 × 100 × 100}{105 × 105} = 3200
∴ Two years before no. of students was 3200. Ans.
Question 5
Through the publicity of road-safety programme, the street accidents in Purulia district are decreased by 10% in comparison to its previous year. If the number of street accidents in this year be 8748, let us write by calculating the number of street accidents 3 years before in the district.
Solution :
No. of street accidents decreases by 10% every year in comparison to previous year.
No. of street accidents in this year is 8748.
Let no. of street accidents 3 years before was x.
According to the problem,
x\left(1-\frac{10}{100}\right)^{3} = 8748
or, x ×\left(\frac{9}{10}\right)^{3} = 8748
or, x × \frac{9 × 9 × 9}{10 × 10 × 10} = 8748
\therefore x = \frac{8748 × 10 × 10 × 10}{9 × 9 × 9} = 12000
\therefore. No. of street accidents 3 years before was 12000 Ans.
Question 6
A cooperative society of fishermen implemented such an improved plan for the production of fishes that the production in a year will be increased 10% in comparison son to its previous year. In the present year if the cooperative society can produce 400 quintals of fishes, let us write by calculating cooperative society can produce after 3 year
Solution :
Production in a year will be increased 10% in comparison son to its previous year. In the present year the production of fish is 400 quintals.
∴ Production of fish after 3 years will be
= 400\left(1+\frac{10}{100}\right)^{3} quintals
= 400 × \frac{11}{10} × \frac{11}{10} × \frac{11}{10} = 532.4 quintals (Ans.)
Question 7
The height of a tree increases at the rate of 20% every year. If the present height of the tree is 28.8 metre, let us calculate the height of the tree 2 years before.
Solution :
Let 2 years before the height of the tree was xm.
Now the height of the tree = x ×\left(1+\frac{20}{100}\right)^{2} m = 28.8m
x ×\left(\frac{6}{5}\right)^{2} = 28.8
x × \frac{36}{25} = 28.8
\therefore \text{ x } = \frac{28.8 × 25}{36} = 20m.
∴ 2 years before the height of the tree was = 20m. Ans.
Question 8
Three years before from today a family had planned to reduce the expenditure of electric bill by 5% in comparison to its previous year. 3 years before, that family had p spend ₹ 4000 in a year for electric bill. Let us write by calculating how much amount the family will have to spend to pay the electric bill in the present year.
Solution :
The expenditure for electric bill reduced by 5% in comparison to the previous ear.
3 years before the expenditure for electric bill was ₹ 4000.
∴ The present electric expenditure will be
= ₹ 4000\left(1-\frac{5}{100}\right)^{3}
= ₹ 4000 × \frac{95}{100} × \frac{95}{100} × \frac{95}{100}
= ₹ 3429.50. Ans.
Question 9
The weight of Savan Babu is 80 kg. In order to reduce his weight, he started regular morning walks. He decided to reduce his weight every year by 10%. Let us write by calculating his weight after 3 year
Solution :
Present mass of Savan Babu is 80 kg.
He decided to reduce his mass every year by 10%.
∴ After 3 years his mass will be
= 80 ×\left(1-\frac{10}{100}\right)^{3}
= 80 ×\left(\frac{9}{10}\right)^{3} kg \quad = 80 × \frac{9}{10} × \frac{9}{10} × \frac{9}{10} kg
= 58.32 kg. Ans.
Question 10
At present the sum of the number of students in all M.S.K in a district is 399. If number of students increased in a year was 10% of its previous year, let us calculate the sum of the number of students 3 years before in all the M.S.K in the district.
Solution :
If the number of students increased in a year by 10% in comparison to the previous year.
The present no. of students = 3993 and let the no. of students 3 years was x.
\therefore \text{ x } ×\left(1+\frac{10}{100}\right)^{3} = 3993
or, x × \frac{11}{10} × \frac{11}{10} × \frac{11}{10} = 3993
\therefore \text{ x } = \frac{3993 × 10 × 10 × 10}{11 × 11 × 11}
x = 3 × 1000 = 3000 Ans.
Question 11
As the farmers are becoming more alert of the harmful effects of applying only the chemical fertilisers and insecticides in agricultural lands, the number of farmers using fertilisers and insecticides in the village of Rasulpur decreases by 20% in a year in comparison to its previous year. Three years before, the number of such farmers was 3000; let us calculate the number of such farmers in that village now.
Solution :
No. of farmers using chemical fertiliser decreases by 20% in current year comparison to its previous year.
3 years before no. of such farmers was 3000.
∴ No. of farmers will be
= 3000 ×\left(1-\frac{20}{100}\right)^{3}
= 3000 × \frac{4}{5} × \frac{4}{5} × \frac{4}{5} = 1536. Ans.
Question 12
The price of a machine of a factory is ₹ 18000. The price of that machine decreases by 10% in each year. Let us calculate its price after 3 year
Solution :
Value of a machine in a factory is ₹ 180000.
The value of the machine depreciates at 10% every year.
∴ The value of the machine after 3 years will be
= \text { ₹ } 180000 ×\left(1-\frac{10}{100}\right)^{3}
= \text { ₹ } 18000 ×\left(\frac{9}{10}\right)^{3}
= \text { ₹ } 18000 × \frac{729}{1000} = ₹ 131220 Ans.
Question 13
For the families having no electricity in their house, a Panchayat samity village Bakultala accepted a plan to offer electricity connections. 1200 families this village have no electric connection in their house. In comparison to its previous year, it is possible to arrange electricity every year for 75% of the families having no electricity, let us write by calculating the number of families without electricity after 2 year
Solution :
No. of family = 1200
In comparison to previous year it is possible to arrange electricity every year for 75% of the family having no electricity.
∴ No. of families without electricity after 2 years
= 1200 ×\left(1-\frac{75}{100}\right)^{2} = 1200 ×\left(1-\frac{3}{4}\right)^{2}
= 1200 × \frac{1}{4} × \frac{1}{4} = 75 Ans.
Question 14
As a result of continuous publicity on harmful reactions to the use of cold drinks filled, bottles, the number of users of cold drinks has decreased by 25% every year in comparison to the previous year. 3 years before the number of users of cold drinks in a town was 80000. Let us write by calculating the number of users of cold drinks in the present year.
Solution :
No. of users of cold drinks decreases by 25% every year in comparison to the previous year.
3 years before number of users of cold drinks in a town was 80000.
∴ No. of users of cold drinks in the present year
= 80000 × \left(1-\frac{25}{100}\right)^{3} = 80000 ×\left(\frac{3}{4}\right)^{3}
= 80000 × \frac{27}{64} = 33750 Ans.
Question 15
As a result of publicity on smoking, the number of smokers is decreased by 6 \frac{1}{4}% every year in comparison to its previous year. If the number of smokers at present in a city is 33750, let us write by calculating the number of smokers in that city 3 years before.
Solution :
No. of smokers decreases by 6 \frac{1}{4}%\left(\frac{25}{4}%\right) every year in comparison to its previous year.
Let the no. of smokers 3 years before was x.
According to the problem,
x ×\left(1-\frac{\frac{25}{4}}{100}\right)^{3} = 33750
or, x ×\left(1-\frac{25}{100}\right)^{3} = 33750
or, x ×\left(\frac{15}{16}\right)^{3} = 33750
or, x × \frac{15 × 15 × 15}{16 × 16 × 16} = 33750
\therefore \text{ x }= \frac{33750 × 16 × 16 × 16}{15 × 15 × 15} = 40960
∴ 3 years before no. of smokers was 40960.
A. M.C.Q.
Question 1
In case of compound interest, the rate of compound interest per annum is-
- Equal
- Unequal
- Both equal or unequal
- None of these
Solution :
(a) Equal
Question 2
In case of compound interest
- The principal remains unchanged each year
- Principal changes in each year
- Principal may be equal or unequal in each year
Solution :
(c) Principal changes in each year
Question 3
At present the population of a village is p and if the rate of increase of po tion per year be 2r%, the population will be after n year
- p\left(1+\frac{r}{100}\right)^{n}
- p\left(1+\frac{r}{50}\right)^{n}
- p\left(1+\frac{r}{50}\right)^{2 n}
- p\left(1-\frac{r}{50}\right)^{n}
Solution :
(b) p\left(1+\frac{r}{50}\right)^{n}
Question 4
Present price of a machine is ₹ 2 p and if price of the machine decreases b 2r% in each year, the price of machine will be
- ₹ p\left(1-\frac{r}{50}\right)^{n}
- ₹ 2 p\left(1-\frac{r}{50}\right)^{n}
- ₹ p\left(1-\frac{r}{100}\right)^{2 n}
- ₹ 2 p\left(1-\frac{r}{100}\right)^{2 n}
Solution :
(d) ₹ 2 p\left(1-\frac{r}{100}\right)^{2 n}
Question 5
A person deposited ₹ 100 in a bank and got the amount ₹ 121 for two years
- 10%
- 20%
- 5%
- 10 \frac{1}{2}%
Solution :
(a) 10%
B. Let us write true or false for the following statements:
Question 1
The compound interest will be always less than simple interest for some money at fixed rate of interest for fixed time.
Solution :
False
Question 2
In case of compound interest, interest is to be added to the principal at the fixed time interval, i.e., the amount of principal increases continuously.
Solution :
True
C. Let us fill in the blanks:
Question 1
The compound interest and simple interest for one year at the fixed rate of interest on fixed sum of money are _________________.
Solution :
Equal.
Question 2
If some things are increased by fixed rate with respect to time, that is _________________________.
Solution :
Same rate.
Question 3
If some things are decreased by fixed rate with respect to time, this is uniform rate of ________________________.
Solution :
Decrease
Short answer (S.A.):
Question 1
Let us write the rate of compound interest per annum, so that the amount on ₹ 400 for 2 years becomes ₹ 441.
Solution :
Let the rate of compound interest = r%
\therefore 400 ×\left(1+\frac{r}{100}\right)^{2} = 441
\left(1+\frac{r}{100}\right)^{2} = \frac{441}{400}
\left(1+\frac{r}{100}\right)^{2} = \left(\frac{21}{20}\right)^{2}
\therefore 1+\frac{r}{100} = \frac{21}{20}
or, \frac{r}{100} = \frac{21}{20}-1
\therefore \frac{r}{100} = \frac{1}{20}
\therefore r = \frac{100}{20} = 5
∴ Rate = 5%
Question 2
If a sum of money doubles itself at compound interest in n years, let us write in how many years it will become four times.
Solution :
Let the principal = ₹ p & rate of C.I = R
\therefore p\left(1+\frac{R}{100}\right)^{n} = 2p
or, \left(1+\frac{R}{100}\right)^{n} = 2
Now in 't' years, amount will be 4p
\therefore\left(1+\frac{R}{100}\right)^{t}= 4p
\left(1+\frac{R}{100}\right)^{t} = 4 = 2^{2} = \left(1+\frac{R}{100}\right)^{2 n}
∴ Time = 2n.
Question 3
Let us calculate the principal that at the rate of 5% compound interest per annum becomes ₹ 615 after two year
Solution :
Let the principal = ₹ p.
\therefore P\left\{\left(1+\frac{5}{100}\right)^{2}-1\right\} = 615
or, P ×\left\{\left(\frac{21}{20}\right)^{2}-1\right\} = 615
or, P ×\left(\frac{441-400}{400}\right) = 615
\therefore P = \frac{615 × 400}{41} = 15 × 400
P = ₹ 6000
Question 4
The price of a machine depreciates at the rate of r% per annum, let us find the price of the machine that was n years before.
Solution :
The value of the machine n years before was = ₹ \frac{v}{\left(1-\frac{r}{100}\right)^{n}} = v ×\left(1-\frac{r}{100}\right)^{-n}
Question 5
If the rate of increase in population is r% per year, the population after n years p; let us find the population that was n years before.
Solution :
n years before, no. of population was = \frac{P}{\left(1+\frac{r}{100}\right)^{n}} = P ×\left(1+\frac{r}{100}\right)^{-n}
