Chapter – 2 : Simple Interest
1. Two friends together took a loan of ₹15,000 from a bank at the rate of simple interest of 12% per annum. Find the interest after 4 years.
Solution :
Principal (P) = ₹15000
Rate of Interest (R) = 12% per annum
Time (T) = 4 years
We know,
Simple Interest =
P × R × T
100
=
15000 × 12 × 4
100
= 150 × 12 × 4
= 1800 × 4
= ₹7200
∴ Required interest = ₹7200
2. Find the simple interest on ₹2000 at 6% per annum from 1st January to 26th May, 2005.
Solution :
Principal (P) = ₹2000
Rate (R) = 6%
Time from 1st January to 26th May
= 145 days
T =
145
365
year
We know,
S.I. =
P × R × T
100
=
2000 × 6 × 145
100 × 365
= ₹47.67
∴ Required simple interest = ₹47.67
3. Determine the amount on ₹960 at the rate of simple interest of 8¼% per annum for 1 year 3 months.
Solution :
Principal (P) = ₹960
Rate (R) = 8¼%
=
33
4
%
Time = 1 year 3 months
=
5
4
years
S.I. =
960 × 33 × 5
100 × 4 × 4
= ₹99
Amount = Principal + Interest
= 960 + 99
= ₹1059
∴ Required amount = ₹1059
4. Utpalbabu took a loan of ₹3200 for 2 years at 6% simple interest per annum. Find the amount to be repaid.
Solution :
Principal (P) = ₹3200
Rate (R) = 6%
Time (T) = 2 years
S.I. =
3200 × 6 × 2
100
= ₹384
Amount = Principal + Interest
= 3200 + 384
= ₹3584
∴ Required amount = ₹3584
5. Sovadebi deposited money at 5.25% simple interest per annum. After 2 years she got ₹840 as interest. Find the deposited money.
Solution :
Simple Interest = ₹840
Rate = 5.25%
Time = 2 years
P =
S.I. × 100
R × T
=
840 × 100
5.25 × 2
= ₹8000
∴ Required deposited money = ₹8000
6. Goutam took a loan at 12% simple interest per annum. Every month he has to repay ₹378 as interest. Determine the loan amount.
Solution :
Interest for 1 month = ₹378
Interest for 1 year = 378 × 12
= ₹4536
P =
4536 × 100
12
= ₹37800
∴ Required loan amount = ₹37800
7. Find the time in which a sum becomes double at 6% simple interest per annum.
Solution :
If amount becomes double,
Interest = Principal
P =
P × 6 × T
100
100 = 6T
T =
100
6
=
50
3
years
= 16
2
3
years
∴ Required time = 16⅔ years
8. After 6 years, the interest became ⅜ of the principal. Determine the rate of simple interest.
Solution :
S.I. =
3P
8
3P
8
=
P × R × 6
100
300 = 48R
R = 6.25%
∴ Required rate = 6.25% per annum
9. A Co-operative society gives loan at 4% while bank gives loan at 7.4%. Find the saving in interest on ₹5000 for 1 year.
Solution :
Difference of rate = 7.4% − 4%
= 3.4%
Saving =
5000 × 3.4 × 1
100
= ₹170
∴ Required saving = ₹170
10. If the interest on ₹292 for 1 day is 8 paisa, determine the rate percent per annum.
Solution :
Principal = ₹292
Interest = 8 paisa = ₹0.08
T =
1
365
year
0.08 =
292 × R × 1
100 × 365
R = 10%
∴ Required rate = 10% per annum
11. Let us calculate the number of years for which the interest on ₹600 at the rate of simple interest of 8% per annum will be ₹168.
Solution :
Principal (P) = ₹600
Rate (R) = 8%
Simple Interest (S.I.) = ₹168
We know,
S.I. =
P × R × T
100
168 =
600 × 8 × T
100
168 =
4800T
100
168 = 48T
T =
168
48
T = 3.5 years
∴ Required time = 3.5 years
12. If ₹800 amounts to ₹1200 at the rate of simple interest of 10% per annum, find the time for which the money was deposited.
Solution :
Principal (P) = ₹800
Amount (A) = ₹1200
Simple Interest = Amount − Principal
= 1200 − 800
= ₹400
Rate (R) = 10%
We know,
S.I. =
P × R × T
100
400 =
800 × 10 × T
100
400 = 80T
T =
400
80
T = 5 years
∴ Required time = 5 years
13. At the same rate of simple interest per annum, if a principal yields the amount ₹7100 in 7 years and ₹6200 in 4 years, determine the principal and rate of interest.
Solution :
Amount in 7 years = ₹7100
Amount in 4 years = ₹6200
Interest for 3 years = 7100 − 6200
= ₹900
Interest for 1 year =
900
3
= ₹300
Interest for 4 years =
300 × 4 = ₹1200
Principal = 6200 − 1200
= ₹5000
Now,
S.I. =
P × R × T
100
300 =
5000 × R × 1
100
300 =
5000R
100
300 = 50R
R = 6%
∴ Principal = ₹5000
∴ Rate of interest = 6% per annum
14. Amal Roy deposited ₹2000 in a bank and Poshupoti Ghosh deposited ₹2000 in a post office at the same time. After 3 years they received ₹2360 and ₹2480 respectively. Find the ratio of rates of simple interest.
Solution :
For Amal Roy :
Interest = 2360 − 2000
= ₹360
R₁ =
360 × 100
2000 × 3
=
36000
6000
= 6%
For Poshupoti Ghosh :
Interest = 2480 − 2000
= ₹480
R₂ =
480 × 100
2000 × 3
=
48000
6000
= 8%
Required ratio = 6 : 8
= 3 : 4
∴ Ratio of rates of interest = 3 : 4
15. A weaver Cooperative society takes a loan of ₹15,000 for buying a power loom. After 5 years the society has to repay ₹22,125. Find the rate of simple interest per annum.
Solution :
Principal (P) = ₹15000
Amount (A) = ₹22125
Time (T) = 5 years
Simple Interest = Amount − Principal
= 22125 − 15000
= ₹7125
We know,
S.I. =
P × R × T
100
7125 =
15000 × R × 5
100
7125 = 750R
R =
7125
750
R = 9.5%
∴ Required rate of simple interest = 9.5% per annum
16. Aslamchacha got ₹1,00,000 after retirement. He deposited a part in a bank at 5% and the rest in a post office at 6%. If the total yearly interest is ₹5400, find the amounts deposited in the bank and post office.
Solution :
Let the money deposited in bank = ₹x
Then money deposited in post office = ₹(100000 − x)
Interest from bank =
x × 5 × 1
100
=
x
20
Interest from post office =
(100000 − x) × 6
100
According to the question,
x
20
+
6(100000 − x)
100
= 5400
5x + 6(100000 − x) = 540000
5x + 600000 − 6x = 540000
−x = −60000
x = 60000
Money deposited in bank = ₹60000
Money deposited in post office =
100000 − 60000 = ₹40000
∴ Required amounts are ₹60000 and ₹40000
17. Rekhadidi deposited ₹10,000 in two banks at 6% and 7% simple interest. After 2 years she got ₹1280 as total interest. Find the amounts deposited in each bank.
Solution :
Let the amount deposited at 6% = ₹x
Then amount deposited at 7% = ₹(10000 − x)
Interest from first bank =
x × 6 × 2
100
Interest from second bank =
(10000 − x) × 7 × 2
100
According to the question,
12x
100
+
14(10000 − x)
100
= 1280
12x + 140000 − 14x = 128000
−2x = −12000
x = 6000
Amount at 6% = ₹6000
Amount at 7% =
10000 − 6000 = ₹4000
∴ Required amounts are ₹6000 and ₹4000
18. A bank gives 5% simple interest per annum. Dipubabu deposited ₹15,000 at the beginning of the year, withdrew ₹3000 after 3 months and deposited ₹8000 again after another 3 months. Find the amount at the end of the year.
Solution :
Interest on ₹15000 for first 3 months :
S.I. =
15000 × 5 × 3
100 × 12
= ₹187.50
Remaining amount after withdrawal :
15000 − 3000 = ₹12000
Interest on ₹12000 for next 3 months :
S.I. =
12000 × 5 × 3
100 × 12
= ₹150
New deposit after 6 months :
12000 + 8000 = ₹20000
Interest on ₹20000 for last 6 months :
S.I. =
20000 × 5 × 6
100 × 12
= ₹500
Total interest =
187.50 + 150 + 500
= ₹837.50
Final principal = ₹20000
Amount = Principal + Interest
= 20000 + 837.50
= ₹20837.50
∴ Required amount = ₹20837.50
19. Rahamatchacha takes a loan of ₹2,40,000 at 12% simple interest per annum. After 1 year he rents the house at ₹5200 per month. Find the number of years required to repay the loan with interest from the rent income.
Solution :
Principal (P) = ₹240000
Rate (R) = 12%
Interest for 1 year =
240000 × 12 × 1
100
= ₹28800
Total amount after 1 year =
240000 + 28800
= ₹268800
Monthly rent = ₹5200
Time required =
268800
5200
= 51.69 months
≈ 52 months
=
52
12
years
≈ 4 years 4 months
∴ Required time ≈ 4 years 4 months
20. Rothinbabu deposited money for his two daughters so that each gets ₹1,20,000 at age 18 years. The rate of simple interest is 10% per annum. Present ages are 13 years and 8 years. Find the deposited amounts.
Solution :
For first daughter :
Time = 18 − 13 = 5 years
Amount = ₹120000
120000 = P +
P × 10 × 5
100
120000 = P + 0.5P
120000 = 1.5P
P = ₹80000
For second daughter :
Time = 18 − 8 = 10 years
120000 = P +
P × 10 × 10
100
120000 = 2P
P = ₹60000
∴ Deposited amounts are ₹80000 and ₹60000
21(i). If the interest of principal p at the rate r% per annum in t years is I, then —
Solution :
I =
p × r × t
100
prt = 100I
∴ Correct option = (c)
21(ii). A principal becomes double in 20 years at simple interest. In how many years will it become three times?
Solution :
If amount becomes double,
Interest = Principal
This takes 20 years.
To become triple,
Interest = 2 × Principal
Required time =
20 × 2 = 40 years
∴ Correct option = (c)
21(iii). If a principal becomes double in 10 years, find the rate of simple interest.
Solution :
If amount becomes double,
Interest = Principal
P =
P × R × 10
100
100 = 10R
R = 10%
∴ Correct option = (b)
21(iv). If the total interest becomes x for a principal at x% simple interest in x years, then the principal is —
Solution :
x =
P × x × x
100
100x = Px²
P =
100
x
∴ Correct option = (d)
21(v). The total interest of a principal in n years at the rate of simple interest of r% per annum is
pnr
25
. Then the principal will be —
Solution :
We know,
S.I. =
P × R × T
100
Given,
S.I. =
pnr
25
P × r × n
100
=
pnr
25
P = 4p
∴ Correct option = (b)
(B)(i). A man who takes a loan is called debtor.
Solution :
The statement is True.
(B)(ii). If the principal and the rate of simple interest per annum are constant, then the total interest and the time are in inverse relation.
Solution :
We know,
S.I. =
P × R × T
100
If P and R are constant, then S.I. is directly proportional to time.
∴ The statement is False.
(C)(i). A man who gives loan is called ______.
Solution :
A man who gives loan is called creditor.
(C)(ii). The amount of 2p in t years at the rate of simple interest of
r
2
% per annum is (2p + ____).
Solution :
S.I. =
2p × r × t
2 × 100
=
prt
100
Amount = Principal + Interest
= 2p +
prt
100
∴ Blank =
prt
100
(C)(iii). The ratio of the principal and the amount in 1 year is 8 : 9. Find the rate of simple interest per annum.
Solution :
Let principal = ₹8
Amount after 1 year = ₹9
Interest = 9 − 8 = ₹1
R =
1 × 100
8 × 1
= 12.5%
∴ Required rate = 12.5% per annum
22(i). Write the number of years in which the amount becomes twice of the principal at the rate of simple interest of 6¼% per annum.
Solution :
If amount becomes twice,
Interest = Principal
P =
P × 6.25 × T
100
100 = 6.25T
T =
100
6.25
T = 16 years
∴ Required time = 16 years
22(ii). The rate of simple interest per annum reduces to 3¾% from 4% and Amal babu's annual income decreases by ₹60. Determine the principal.
Solution :
Difference in rate =
4% − 3.75% = 0.25%
Decrease in yearly interest = ₹60
60 =
P × 0.25 × 1
100
60 =
P
400
P = 60 × 400
= ₹24000
∴ Required principal = ₹24000
22(iii). Determine the rate of simple interest per annum when the interest of a principal in 4 years becomes
8
25
part of the principal.
Solution :
S.I. =
8P
25
8P
25
=
P × R × 4
100
8
25
=
4R
100
800 = 100R
R = 8%
∴ Required rate = 8% per annum
22(iv). Determine the rate of simple interest per annum when the interest of a sum in 10 years becomes
2
5
part of the amount.
Solution :
Let principal = ₹P
S.I. =
P × R × 10
100
=
PR
10
Amount =
P +
PR
10
According to the question,
PR
10
=
2
5
(
P +
PR
10
)
5PR = 20P + 2PR
3PR = 20P
3R = 20
R =
20
3
%
= 6
2
3
%
∴ Required rate = 6⅔% per annum
22(v). Calculate the principal whose monthly interest is ₹1 at the rate of simple interest of 5% per annum.
Solution :
Monthly interest = ₹1
Yearly interest = 1 × 12 = ₹12
12 =
P × 5 × 1
100
12 =
P
20
P = 12 × 20
= ₹240
∴ Required principal = ₹240
