| Book Name | : Ganit Prakash |
| Subject | : Mathematics |
| Class | : 10 (Madhyamik) |
| Publisher | : Prof Nabanita Chatterjee |
| Chapter Name | : Simple Interest (2nd Chapter) |
Let me do it myself :
Application 2.
| Principal | Time | Rate of simple interest | Total interest |
| ₹ 600 | 1 yr. | 5% | |
| ₹ 1800 | 1 yr. | 4 \frac{1}{2}% |
Solution
Application 3.
(i) Interest on ₹ 100 for 1 year is ₹ 5.
\therefore Interest on ₹ 600 for 1 year ₹ \frac{5}{100} \times 600=30.
\therefore Interest = ₹ 30 Ans
(ii) Solution :
Interest on ₹ 100 for 1 year is ₹ 4 \frac{1}{2}= \text{ ₹ } \frac{9}{2}
Interest on ₹ 1800 for 1 year ₹ \frac{9}{2} \times \frac{1800}{100}= ₹ 81
Application 4.
In one year, if Shraboni would get ₹ 60 as interest at the rate of simple interest 5% per annum in the bank, then how much amount would she deposit, let us calculate and write it.
Solution :
₹ 5 is the interest for 1 year when the principal is ₹ 100.
Re. 1 is the interest for 1 year when the principal is ₹ \frac{100}{5}. ₹ 60 is the interest for 1 year when the principal is ₹ \frac{100}{5} \times 60= ₹ 1200
\therefore Principal = ₹ 1200 Ans
Application 5.
(i) Solution :
₹ 6 is the interest for 1 year when the principle is ₹ 100.
Re. 1 is the interest for 1 year when the principal is ₹ \frac{100}{6}.
₹ 90 is the interest for 1 year when the principal is ₹ \frac{100}{6} \times 90= ₹ 1500.
\therefore Principal = ₹ 1500 . Ans.
(ii) Solution :
₹ 3.5 is the interest for 1 year when the principal is ₹ 100.
₹ 59.50 is the interest for 1 year when the principal = ₹ \frac{100}{3.5} × 59.50 = ₹ 1700.
\therefore Principal = ₹ 1700. Ans.
Application 10.
(i) Solution :
\mathrm{ Interest }(\mathrm{I})=\frac{\text { Principal }(\mathrm{P}) × \text { Rate }(\mathrm{r}) × \text { time }(\mathrm{t})}{100}
=\text { ₹ } \frac{500 × \frac{25}{4} × 3}{100}=5 × \frac{25}{4} × 3 \text { ₹ } 93.75
Amount (P+I) = ₹ (500 + 93.75) = ₹ 593.75 Ans.
(ii) Solution :
Here P = ₹ 146 ; r = 2 \frac{1}{2} \%=\frac{5}{2} \%, t=1 day =\frac{1}{365}
\text { Interest }=\frac{146 × \frac{5}{2} × \frac{1}{365}}{100}=\text { ₹ } \frac{1}{100}=\text { ₹ } 0.01 \\
\text { Amount }=\text { ₹ }(146+0.01)=\text { ₹ } 146.01 \text { Ans. }
(iii) Solution :
Here P = Rs .4565 ; r = 4 \% ; t=2 years \ 6 months =2 \frac{1}{2} years =\frac{5}{2} yea₹
\text { Interest }=\text { ₹ } \frac{4565 × 4 × \frac{5}{2}}{100}=\text { ₹ } \frac{4565 × 10}{100}=\text { ₹ } 456.50 \\
\text { Amount }=\text { ₹ }(4565+456.50)=\text { ₹ } 5021.50 \text { Ans. }
Application 13.
Let us write in the blank by calculating it. [Let me do it myself]
| Principal | Time | Rate of simple interest | Total interest |
| 4 years | 4 \frac{1}{2} \% | ₹ 72 | |
| 1 day. | 5% | ₹ 1 |
(i) Solution :
Here P = \frac{1 × 100}{r × t}
Here I = ₹ 72
r=4 \frac{1}{2} \%=\frac{9}{2} \% \\
t=4 \text { yea₹ } \\
=\text { ₹ } \frac{72 \times 100}{\frac{9}{2} \times 4} \\
=\text { ₹ } \frac{72 \times 100}{18} \\
=\text { ₹ } 400. \text { Ans. }
(ii) Solution :
Here P=\frac{1 \times 100}{r \times t}
=\text { Here } I=\text { Re. } 1 \\
r = 5%
t=1 \text { day }=\frac{1}{365} \\
=\frac{1 \times 100}{5 \times \frac{1}{365}} \\
=\text { ₹ } 73 \times 100
= ₹ 73 \times 100
= ₹ 7300. Ans.
Application – 16.
Let us write in the blank by calculation. [Let me do it myself]
| Principal | Time | Rate of simple interest | Principal with interest |
| 5 years | 3% | ₹ 966 | |
| 6 years | 6% | ₹ 13,600 |
(i) Solution :
Principal =\frac{\text { Interest } \times 100}{\text { Rate } \times \text { Time }}
On ₹ 100 interest for 1 year = ₹ 3.
On ₹ 100 interest for 5 years = ₹ 3 × 5= ₹ 15 \therefore Amount = ₹ (100+15) = ₹ 115
When principal & the interest is 115, the principal = ₹ 100.
When principal & the interest is 966, the principal = ₹ \frac{100}{115} \times 966= ₹ 840. Ans.
(ii) Solution :
Interest on ₹ 100 for 1 year = ₹ 6
Interest on ₹ 100 for 6 years = ₹ 6 × 6 = ₹ 36.
Amount = ₹ (100+36) = ₹ 136
When amount is ₹ 136, principal = ₹ 100
When amount is ₹ 13600, principal = ₹ \frac{100}{136} \times 13600
\therefore Principal = ₹ 10,000 Ans.
Application – 19. [Let me do it myself]
| Principal | Time | Rate of simple interest | Total interest |
| 4 \frac{1}{2} \% | ₹ 1008 | ||
| 5% | ₹ 50 |
(i) Solution :
Time =\frac{\text { Interest } \times 100}{\text { Principal } \times \text { rate }}
Here, Interest = ₹ 1008
Rate =4 \frac{1}{2} \%=\frac{9}{2} \%
Principal = Rs, 6,400
=\frac{1008 \times 100}{6400 \times \frac{9}{2}}
Time =\frac{1008}{32 \times 9}=3.5 \text{ years } =3 \frac{1}{2} yea₹ Ans.
(ii) Solution :
Time =\frac{\text { Interest } \times 100}{\text { Principal } \times \text { rate }}
Here, Interest = ₹ 50
Rate = 5%
Principal = ₹ 500
=\frac{50 \times 100}{500 \times 5}
= 2 yea₹
\therefore Time = 2 yea₹ Ans.
Application – 23
(i) Let us determine the rate of simple interest in per cent per annum, if the interest of ₹ 500 in 4 years is 100. [Let me do it myself]
Solution :
Principal = ₹ 500, Interest = ₹ 100, Time = 4 years, Rate =?
Interest on ₹ 500 for 4 years is 100.
Interest on ₹ 100 for 4 years is \frac{100}{500} \times 100.
Interest on ₹ 100 for 1 yr. is \frac{100 \times 100}{500 \times 4 } = 5.
\therefore Rate = 5%
2nd Process :
Rate =\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}
=\frac{100 \times 100}{500 \times 4}=5
\therefore Rate = 5%.
(ii) By calculating let us write the rate of simple interest in percent per annum when the principal with interest of ₹ 910 in 2 years 6 months will be ₹ 955.50. [Let me do myself]
Solution :
Here, Principal = ₹ 910
Interest = ₹ (955.50-910)= ₹ 45.50
\text { Time }=2 \text { years } 6 \text { months }=2 \frac{1}{2} \mathrm{yrs}=\frac{5}{2} \mathrm{yrs}
\therefore \text { Rate }=\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}
=\frac{45.50 \times 100}{910 \times \frac{5}{2}} \\
=\frac{4550 \times 2}{910 \times 5}=2 \%
\therefore Rate = 2% . Ans.
Application 26.
The amount (principal along with interest) of some money becomes 1s. 496 in 3 years and ₹ 560 in 5 years at the same rate of simple interest in percent per annum. By calculating, let us write the principal and the rate of simple interest in percent per annum. [Let me do it myself]
Solution :
Principal + Interest for 5 years = ₹ 560
Principal + Interest for 3 years = ₹ 496
\therefore \text{ Interest for } 2 \text { years }=\text { ₹ } 64
\therefore \text{ Interest for } 1 \text { year }=\text { ₹ } \frac{64}{2}=32
& Interest for 3 \text { years }=\text { ₹ } 32 \times 3=\text { ₹ } 96
\therefore Principal + Interest for 3 years = ₹ 496
Interest for 3 years = ₹ 96
\therefore Principal = ₹ (496-96) = ₹ 400. Ans.
\text { Rate }=\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}=\frac{96 \times 100}{400 \times 3}=8
\therefore Rate = 8% Ans.
Application 30.
Bimalkaku deposited ₹ 1,87,500 for his son of 12 years age and daughter of 14 years age in the bank at the rate of simple interest 5% per annum in such away that both of them will get equal principal along with interest after reaching their 18 years of age. Let us calculate the money he had deposited in the bank for each of his son and daughter. [Let me do it myself]
Solution :
Let Bimal babu deposited ₹ x for his 12 yrs old son & ₹ y for his 14 yrs old daughter.
\therefore x + y = 187500
When his son will be 18 yrs, i.e., after 6 yrs, his son’s amount = ₹ (x+\frac{x \times 6 \times 5}{100})
=\frac{10 x+3 x}{5}=\text { ₹ } \frac{13 x}{10}
Again, when his daughter will be 18 years, i.e., after 4 years
His daughter’s amount =Rs(y+\frac{y \times 4 \times 85}{100})=\frac{5 y+y}{5}=\frac{6 y}{5}
According to the problem,
\frac{13 x}{10}=\frac{6 y}{5}
or, 13x – 12y = 0
x + y= 187500
or, 12x + 12y = 12 × 187500 = 2250000
& 13x – 12y = 0
Adding, 25x = 2250000
x=\frac{2250000}{25}=90000 \\
y = 187500 – 90000 = 97500
\therefore He deposited ₹ 90,000 for his son and ₹ 97500 for his daughter. Ans.
Application 32.
Jayanta deposits ₹ 1000 on the first day of every month in a monthly savings scheme. In the bank, if the rate of simple interest is 5% per annum then let us calculate the amount Jayanta will get at the end of 6 months. [Let me do it myself]
Solution :
Interest on ₹ 1000 at the rate 5% for each month
=\text { ₹ }[\frac{1000 \times 5 \times 6}{100 \times 12}+\frac{1000 \times 5 \times 5}{100 \times 12}+\frac{1000 \times 5 \times 4}{100 \times 12}+\frac{1000 \times 5 \times 3}{100 \times 12}+\frac{1000 \times 5 \times 2}{100 \times 12}+\frac{1000 \times 5 \times 1}{100 \times 12}] \\
=\text { ₹ } \frac{1000 \times 5}{100 \times 12}[6+5+4+3+2+1]=\text { ₹ } \frac{50}{12} \times 21=\text { ₹ } \frac{175}{2}=87.50
Amount = ₹ (1000 × 6 + 87.50)
= ₹ (6000+87.50) = ₹ 6087.50 Ans.
Application – 34.
Soma aunti deposits ₹ 6,20,000 in such a way in three banks a the rate of simple interest of 5% per annum for 2years and 5years respectively so tha the total interests in the 3 banks are equal. Let us calculate the money deposited By Soma Aunti in each of the three banks. [Let me do it myself]
Solution :
Let Soma aunti deposited ₹ p, ₹ x & ₹ y in 3 banks respectively at the rate of 5% S.I.
\therefore p + x + y = 620000
Now, the interest from 3 banks are
\frac{p \times 5 \times 2}{100}=\frac{x \times 5 \times 3}{100}=\frac{y \times 5 \times 5}{100} \\
2p = 3x = 5y = k (let)
\therefore p=\frac{k}{2} ; x=\frac{k}{3} ; y=\frac{k}{5}
As, p + x + y = 620000
or, \frac{k}{2}+\frac{k}{3}+\frac{k}{5}=620000 \text { or, } \frac{15 k+10 k+6 k}{30}=62,0000 \\ \therefore \frac{31 k}{30}=620000 \\
k=\frac{620000 \times 30}{31}=60,0000 \\ \therefore p=\frac{k}{2}=\frac{60,0000}{2}=300,000 \\
x=\frac{k}{3}=\frac{60,0000}{3}=200,000 \\
y=\frac{k}{5}=\frac{60,0000}{5}=120,0000 \therefore She deposited ₹ 300,000; ₹ 200,000; and ₹ 1200,000 respectively in 3 banks. Ans.
LET US WORK OUT – 2.
Question 1
Two friends together took a loan amount of ₹ 15,000 to run a business from a bank at the rate of simple interest of 12% per annum. Let us write, by calculating, the interest they have to pay after 4 years
Solution :
Here principal (P) = ₹ 15,000
Rate (r) = 12%
Time (t) = 4 years
Simple Interest (I) = \frac{\text { Prt }}{100}
= ₹ \frac{15000 \times 12 \times 4}{100} = ₹ 7200 Ans
Question 2
Let us determine the interest of ₹ 2000 at the rate of simple interest of 6% per annum from 1st January to 26th May, 2005.
Solution :
Principal (P) = ₹ 2000
Rate (r) = 6%
No. of days from 1st January to 26th May 2005.
Time =31+28+31+30+25
= 146 days = \frac{146}{365} = \frac{2}{5} years
Simple Interest (I) = \frac{\text { Prt }}{100}
= ₹ \frac{2000 \times 6 \times \frac{2}{5}}{100} = ₹ 48
Question 3
Let us determine the amount (principal along with interest) of ₹ 960 at the rate of simple interest of 8 \frac{1}{3} \% annum for 1 yr 3 months.
Solution :
Principal (P) = ₹ 960
Rate (r) = 8 \frac{1}{3} \% = \frac{25}{3} \%
\text { Time }(t)=1 \text { yr } 3 \mathrm{~m}=1 \frac{3}{12}=1 \frac{1}{4}=\frac{5}{4} \mathrm{yrs} \\
Simple Interest (I) = \frac{\text { Prt }}{100}
= ₹ \frac{960 \times \frac{25}{3} \times \frac{5}{4}}{100}
= ₹ 100
Amount = Principal + Interest
= ₹ 960 + ₹ 100 = ₹ 1060
Question 4
Utpalbabu took a loan of ₹ 3200 for 2 years from a Cooperative bank for the cultivation of his land at the rate of simple interest of 6% per annum. Let us write by calculating the money she has deposited in the bank.
Solution :
Here, Principal (P)= ₹ 3200
Rate (r) = 6%
Time (t) = 2 years
Simple Interest (I) = \frac{\text { Prt }}{100}
= ₹ \text{ ₹ } \frac{3200 \times 6 \times 2}{100}
= ₹ 384
∴ Amount = ₹ 3200 + ₹384 = ₹ 3584
Question 5
Sovadebi deposited some amount of money in a bank at the rate of simple interest of 5.25% per annum. After 2 years she has got ₹ 840 as interest. Let us write by calculating the money she has deposited in the bank.
Solution :
Here, Simple Interest (I) = ₹ 840
Rate (r) = 5.25%
Time (t) = 2 years
Principal (P) =\frac{\text { Interest } \times 100}{\text { Rate } \times \text { Time }} \\
=\text { ₹ } \frac{840 \times 100}{5.25 \times 2}=\text { ₹ } \frac{840 \times 100 \times 100}{525 \times 2} \\
= ₹ 40 × 200 = ₹ 8000
∴ She deposited ₹ 8000
Question 6
Goutam took a loan of some money from a Cooperative bank for opening a poultry farm at the rate of simple interest of 12% per annum. Every month he has to repay ₹ 378 as interest. Let us determine the loan amount taken by him.
Solution :
Total Interest = ₹ 378
Rate = 12%
\text { Time }=\frac{1}{12} \text { yrs }
\text { Principal }=\frac{\text { Interest } \times 100}{\text { Rate } \times \text { Time }} \\
=\text { ₹ } \frac{378 \times 100}{12 \times \frac{1}{12}}=\text { ₹ } 37800
\therefore He takes loan of ₹ 37800. Ans.
Question 7
Let us write by calculating the number of years for which an amount becomes twice of its principal having the rate of simple interest of 6% per annum.
Solution :
Let Principal (P) = ₹ x
Rate (r) = 6%
Simple Interest (I) = ₹ x
Time (t) =?
\text { Time }=\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Rate }} \\
=\frac{x \times 100}{x \times 6}=16 \frac{2}{3} \mathrm{yrs} \\
\therefore \text { Time }=16 \frac{2}{3} \mathrm{yrs} .
Question 8
Mannan Miyan observed, after 6 years of taking a loan of some money, that the interest to be paid had become 3/8 th of its principal. Let us determine the rate of simple interest in per cent per annum.
Solution :
Let the principal = ₹ x
Simple Interest (I) = ₹ \frac{3 x}{8} \\
Time (t) = 6 years
Rate (r) = ?
Rate = \frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}=\frac{\frac{3}{8} \times 100}{x \times 6} \\
=\frac{3}{8} \times \frac{100}{6}=\frac{25}{4}=6 \frac{1}{4} \\
∴ Rate =6 \frac{1}{4} \\ %
Question 9
An agricultural Cooperative society gives agricultural loan to its members at the rate of simple interest of 4% per annum. But interest is to be given at the rate of simple interest of 7.4% per annum for a loan taken from the bank. If a farmer being a member of the Co-operative Society takes a loan of ₹ 5000 from it instead of taking loan from the bank, then let us write, by calculating, the money to be saved as interest per annum.
Solution :
Principal = ₹ 5,000
Time = 1 yr
Rate (1st case )= 4 %
Rate (2nd case )= 7.4 %
Simple interest in the 1st case =\frac{\text { Prt }}{100}
= ₹ \frac{\text { 5000×4×1 }}{100}
= ₹ 200
Simple interest in the 1st case =\frac{\text { Prt }}{100}
= ₹ \frac{\text { 5000×7.4×1 }}{100}
= ₹ 370
Interest saved = ₹ 370 – ₹ 200 = ₹ 170
Question 10
If the interest of ₹ 292 in 1 day is 5 paise, then let us write by calculating the rate of simple interest in per cent per annum.
Solution :
Principal (P) = ₹ 292
Simple Interest (I) = 5 p
Time (t) = 1 day = \frac{1}{365} years
Rate = \frac{Interest×100}{Principal×Time}
Rate = \frac{5×100}{292×{\frac{1}{365}}}
∴ Rate = \frac{25}{4} = 6\frac{1}{4}%
Question 11
Let us write by calculating the number of years for which the interest of ₹ 600 at the rate of simple interest in percent per annum.
Solution :
Principal (P) = ₹ 600
Rate (r) = 8 %
Interest (I) = ₹ 168
Time =?
Time = \frac{\text { Interest } × 100}{\text { Principal } × \text { Rate }} \\
=\frac{168 × 100}{600 × 8}=\frac{21}{6}=\frac{7}{2} yrs=3 \frac{1}{2} \\yrs
∴ Time = 3\frac{1}{2}yrs
Question 12
If I get ₹ 1200 return as the amount (principal along with interest) by depositing ₹ 800 in the bank at the rate of simple interest of 10% per annum, then let us write by calculating the time for which the money was deposited in the bank.
Solution :
Amount = ₹ 120
Principal = ₹ 800
Simple Interest = ₹ (1200-800) = ₹ 400
Rate =10%
Time =?
Time = \frac{\text { Interest } × 100}{\text { Principal } × \text { Rate }}
= \frac{400 × 100}{800 × 10}
= 5 \text { years }
∴ Time = 5 years
Question 13
At the same rate of simple interest in per cent per annum, if a principal becomes the amount of ₹ 7100 in 7 years and of ₹ 6200 in 4 years, let us determine the principal and rate of simple interest in per cent per annum.
Solution :
Principal + Interest for 7 years = ₹ 7100
Principal + Interest for 4 years = ₹ 6200
∴ Interest for 3 years = ₹ 900
∴ Interest for 1 years = ₹ 300
Interest for 4 years interest = ₹ 300 × 4 = ₹ 1200
According to question
Principal + 4 years interest = ₹ 6200
or, Principal + ₹ 1200 = ₹ 6200
∴ Principal = ₹ 6200 – ₹ 1200 = ₹ 5000
Calculation of Rate
Rate = \frac{\text { Interest } × 100}{\text { Principal } × \text { Time }}
= \frac{1200 × 100}{5000 × 4} = 6
∴ Rate = 6 % and Principal = ₹ 5000
Question 14
Amal Roy deposits ₹ 2000 in the bank and Pashupati Ghosh deposits ₹ 2000 in the post office at the same time. After 3 years: they get the return amounts ₹ 2360 and ₹ 2480 respectively. Let us write by calculating the ratio of the rate of simple interest in per cent per annum in the bank and that in the post office.
Solution :
In the case of Amal Roy :
Principal = ₹ 2000
Interest = ₹ (2360-2000) = ₹ 360
Time = 3 yrs
\text { Rate }=\frac{\text { Interest } × 100}{\text { Principal } × \text { Time }} \\
\text { Rate }=\frac{360 × 100}{2000 × 3}=6
In case of Pasupati Ghosh :
Principal = ₹ 2000
Interest = ₹ (2480-2000)= Rs 480
Time = 3 years
\text { Rate }=\frac{\text { Interest } × 100}{\text { Principal } × \text { Time }} \\
=\frac{480 × 100}{2000 × 3}=8
\therefore Ratio of Rates = 6 : 8 = 3 : 4
Question 15
A weaver cooperative society takes a loan of ₹ 15,000 at the time of buying a power loom. After 5 years the society has to repay ₹ 22125 for recovering the loan. Let us determine the rate of simple interest in per cent per annum.
Solution :
Principal = ₹ 15,000
Interest = ₹ 22125 – ₹ 15,000 = ₹ 7125
Time = 5 years
Rate = \frac{\text { Interest } × 100}{\text { Principal } × \text { Time }}
= \frac{7125 × 100}{15000 × 5}
= \frac{1425}{150}
= \frac{95}{10} = 9.5
∴ Rate = 9.5 %
Question 16
Aslamchacha got ₹ 100000 when he retired from his service. He deposited some of that money in the post office and got ₹ 5400 in total per year as interest. If the rates of simple interest per annum in the bank and in the post office are 5% and 6% respectively, then let us write by calculating the money he had deposited in the bank and post office.
Solution :
Let Aslamchacha deposit ₹ x at the rate of 5 % for 1 year in the bank and he deposited ₹ (100000 – x) at the rate of 6 % for 1 year in the post office.
1st case (In Bank) :
Simple Interest = \frac{\text { Principal } × \text { rate } × \text { time }}{100}
= ₹ \frac{x × 5 × 1}{100} = ₹ \frac{5 x}{100}
2nd case (In Post office) :
Simple Interest = \frac{\text { Principal } × \text { rate } × \text { time }}{100}
= ₹ \frac{(100000 - x) × 6 × 1}{100}
= ₹ \frac{(100000 - x) × 6}{100}
According to the problem,
\frac{5 x}{100}+\frac{(10000-x) 6}{100}=5400 \\
or, \frac{5 x+600000-6 x}{100}=5400 \\
or, -x + 600000 = 540000
or, 60,0000 – 540000 = x
or, x = 60000
∴ He deposited ₹ 60,000 in bank and ₹ (100000-60000) = ₹ 40,000 in post office
Question 17
Rekhadidi deposited ₹ 10,000 of her savings in two separate banks at the same time. The rate of simple interest per annum is 6% in one bank and 7% in another bank; After 2 years, if she gets ₹1280 in total as interest, then let us write by calculating, the money she had deposited separately in each of the two banks.
Solution :
Let she deposit ₹ x in the 1st bank at the rate of 6% and Rs (10000-x) in the 2nd bank at the rate of 7 % for 2 years
According to the problem,
₹ {\frac{x × 6 × 2}{100}+\frac{(10000-x) × 7 × 2}{100}} = ₹ 1280
or, \frac{12x}{100}+\frac{14(10000-x)}{100} = 1280
or, frac{12x+140000-14 x}{100} = 1280
or, -2x + 140000 = 1280 × 100
or, 140000 – 128000 = 2x
∴ x = =\frac{12000}{2}=6000
∴ She deposited ₹ 6000 in the 1st bank & ₹ 4000 in the 2nd bank.
Question 18
A bank gives 5% simple interest per annum. In that bank, Dipubabu deposits ₹ 15,000 at the beginning of the year, but withdraws ₹ 3000 after 3 months and then again, after 3 months he deposits ₹ 8000. Let us determine the amount (principal along with interest) Dipubabu will get at the end of the year.
Solution :
Dipu Babu’s total interest for 1 year
= (\frac{15000 × 5}{100} × \frac{3}{12}+\frac{12000 × 5}{100} × \frac{3}{12}+\frac{20000 × 5}{100} × \frac{6}{12}) \\
= ₹ (\frac{150 × 5}{4}+150+500)
= ₹ 187.50 +₹ 150 +₹ 500 = ₹ 837.50
∴ His amount will be ₹ [20000+837.50]= ₹ 20837.50
Question 19
Rahamatchacha takes a loan amount of ₹ 2,40,000 from a bank for constructing a building at the rate of simple interest of 12% per annum. After 1yr of taking the loan, he rents the house at the rate of ₹ 5200 per month. Let us determine the number of years he would take to repay his loan along with interest from the income of the house rent.
Solution :
Let after x years he will repayment the amount.
Interest of ₹ 240000 at 12% for x years
= \frac{240000 × 12 × x}{100}=28800 x
Amount = Principal + Interest
= ₹ (240000+28800x)
Now, house rent for 1 year 12m = ₹ 5200 × 12
\therefore House rent for (x – 1) yrs = ₹ 5200 × 12x(x – 1)
= 62400 (x – 1)
According to the problem,
62400 (x – 1) = 240000 + 28800 x
or, 62400 x – 62400 = 240000 + 28800 x
or, 62400 x – 28800 x = 240000+62400
or, 33600 x = 302400
∴ x =\frac{302400}{33600}=9
∴ After 9 years he will repay his loan with interest.
Question 20
Rothinbabu deposits the money for each of his two daughters in such a way that when the age of each of his daughters will be 18 years, each one will get ₹ 1,20,000. The rate of simple interest per annum in the bank is 10 % and the present ages of his daughters are 13 years and 8 years respectively. Let us determine the money he had deposited separately in the bank for each of his daughters
Solution :
Let Rathin babu deposited ₹ x for his 1st daughter (13 yrs old) and he deposited ₹ y for his 2nd daughter ( 8 yrs old).
When his 1st daughter will be 18 yrs, old i.e., after (18 – 13) =5 years
Her amount = x+\frac{x × 5 × 10}{100}=120000
or, \frac{10 x × 5 x}{10}=120000
or, 15x = 120000 × 10
or, x=\frac{1200000}{15}=80,000
When his 2nd daughter will be 18 years old, i.e., after (18 – 8) = 10 years
\text { Her amount }=y+\frac{y × 10 × 10}{100}=2 y \\
\therefore 2y = 120000
y = 60000
M.C.Q
Question 1
If the interest of ₹ p at the rate of simple interest of r% per annum in t years is I, then
(a) I = prt
(b) prt = 100 × I
(c) prt=100 × 1
(d) None of these
Solution :
I=\frac{\text { prt }}{100} or prt = 100 × I ……….(b)
Question 2
A principal becomes twice of its amount in 20 yrs at a certain rate of simple interest. At the same rate of simple interest, that principal becomes thrice of its amount in
(a) 30 years
(b) 35 years
(c) 40 years
(d) 45 years
Solution :
(c) 40 years
Explanation
Simple Interest (I) = \frac{\text{Prt}}{100}
x = \frac{\text{x × r × 20}}{100}
or, r = \frac{\text{x × 100}}{x × 20}
or, r = 5 %
Now, 2x = \frac{\text{x × 5 × t}}{100}
or, t = \frac{\text{2x × 100}}{x × 5}
or, t = 40 years
Question 3
If a principal becomes twice of its amount in 10years, the rate of simple interest per annum is
(a) 5%
(b) 10%
(c) 15%
(d) 20%
Solution :
(b) 10 %
Explanation
Principal (P) = x
Amount (A) = 2x
Simple Interest (I) = 2x – x = x
Simple Interest (I) = \frac{\text{P × r × t}}{100}
or, x = \frac{\text{x × r ×10}}{100}
or, r = \frac{\text{x ×100}}{x × 10} = 10 %
Question 4
The simple interest at x% for x years will be Rs x for a principal amount of
(a) ₹ x
(b) ₹ 100x
(c) \frac{100}{x}
(d) ₹ \frac{100}{x^2}
Solution :
(c) P = 100\over x
Explanation
Simple Interest (I) = \frac{\text{P × r × t}}{100}
or, x = \frac{\text{P × x × x}}{100}
or, P = \frac{\text{ x × 100}}{x × x}
or, P = 100\over x
Question 5
The total interest of a principal in n years at the rate of simple interest of r% per annum is \frac{p n r}{25}, the principal will be
(a) ₹ 2p
(b) ₹ 4p
(c) ₹ \frac{p}{2}
(d) ₹ \frac{p}{4}
Solution :
P=\frac{1 × 100}{r × t}=\frac{p n r × 100}{25 × r × n}=4 p .......... \text { (b) }
Let us write whether the following statements are true or false :
Question 1
A man who takes a loan is called a debtor.
Solution :
True.
Explanation
A man who takes a loan is indeed called a debtor. In financial terms, a debtor is an individual or entity that owes money to another party, typically in the form of a loan or credit. The debtor is obligated to repay the borrowed amount, and they are often referred to as the borrower or debtor in the context of financial transactions.
Question 2
If the principal and the rate of simple interest in percent per annum be constants, then the total interest and the time are in inverse relation.
Solution :
False.
Explanation
If the principal amount and the rate of simple interest per annum are constants, the relationship between the total interest and the time is direct, not inverse.
Let us fill in the blanks :
Question 1
A man who gives a loan is called _________.
Solution :
The man who gives a loan is called a money lender or creditor.
Question 2
The amount of ₹ 2p in t years at the rate of simple interest of \frac{r}{2}\% per annum is Rs .(2p + _________).
Solution :
Interest =\frac{2 p × t × \frac{r}{2}}{1.00}=\frac{p r t}{100}
Amount =2 p+\frac{p r t}{100}
Question 3
The ratio of the principal and the amount (principal along with interest) in 1 yr is 8 : 9; the rate of simple interest per annum is _________.
Solution :
Let the principal = ₹ 8x and the amount = ₹ 9x.
∴ Interest = ₹ 9x – ₹ 8x
Rate = \frac{\text { Interest } × 100}{\text { Principal } × \text { Time }}
= \frac{x × 100}{8 × 1}=\frac{25}{2}
∴ Rate = 12 \frac{1}{2}%
Short Answer Type :
Question 1
Let us write the number of years for which a principal becomes twice of its amount having the rate of simple interest of 6 \frac{1}{4} % per annum.
Solution :
Let the principal = ₹ 100
Amount = ₹ 200 – ₹ 100 = ₹ 100
Rate = 6 \frac{1}{4} %
∴ Time = \frac{\text { Interest } × 100}{\text { Principal } × \text { Rate }}
= \frac{100 × 100}{100 × \frac{25}{4}}
= \frac{100 × 4}{25} = 16 years
∴ Required time = 16 years
Question 2
The rate of simple interest per annum reduces 4% to 3 \frac{3}{4} %, and for this, Amal Babu’s annual income decreases by ₹ 60. Let us determine Amal Babu’s principal.
Solution :
On ₹ 100, income (Interest) decreased from ₹ 4 to ₹ 3 \frac{3}{4}. = \text{ ₹ } 4-3 \frac{3}{4}=\frac{1}{4} in 1 year.
\therefore His income decreases by ₹ \frac{1}{4} in 1 year on ₹ 100 .
\therefore His income decreases by ₹ 60 in 1 year on ₹ \frac{100}{1 / 4} × 60
= ₹ 100 × 4 × 60= ₹ 24000 .
\therefore Amal babu’s capital = ₹ 24000. Ans.
Question 3
What is the rate of simple interest per annum, when the interest of some money in 4 years will be \frac{8}{25} part of its principal – let us determine it.
Solution :
Let the principal = ₹ x
\therefore \text { Interest }=\text { ₹ } \frac{8}{25} x . \\
\text { Time }=4 \text { yea₹ } \\
\text { Rate }=\frac{\text { Interest } × 100}{\text { Principal } × \text { Time }}=\frac{\frac{8}{25} × x × 100}{x × 4}=8 \\
\therefore \text { Rate }=8 % \text {. Ans. }
Question 4
What is the rate of simple interest per annum, when the interest of some money in 10 years will be \frac{2}{5} part of its amount (principal along with interest) – Let us determine it.
Solution :
Let the amount = ₹ x.
\therefore \text { Interest }=\frac{2}{5} × \text { ₹ } x=\text { ₹ } \frac{2 x}{5} \\
\therefore \text { Principal }=\text { ₹ }\left(x-\frac{2 x}{5}\right)=\text { ₹ } \frac{5 x-2 x}{5}=\text { ₹ } \frac{3 x}{5} \\
\text { Time }=10 \text { years } \\
\therefore \text { Rate }=\frac{\text { Interest } × 100}{\text { Principal } × \text { Time }} \\
=\frac{\frac{2 x}{5} × 100}{\frac{3 x}{5} × 10} \\
=\frac{20 × 2}{2 × 3}=\frac{20}{3}=6 \frac{2}{3} \\
\therefore \text { Rate }=6 \frac{2}{3} % . \text { Ans. }
Question 5
Let us determine the money for which monthly interest is Re. 1 having the rate of simple interest of 5% per annum.
Solution :
Here let principal = ₹ x.
\text { Rate }=5 %
\text { Interest }=\text { Re. } 1
\therefore \text { Principal }=\frac{\text { Interest } × 100}{\text { Rate } × \text { Time }} \\
=\frac{1 × 100}{5 × \frac{1}{12}}=\text { ₹ } 20 × 12 \\
=240. \text { Ans. }
