Question 1 1 / -0
A person invests Rs. 1.53 lakh in fixed deposit at 20% simple interest. Find his monthly income.
Solution
P = Rs. 1,53,000, r = 20%, t = 1/12 years
Simple interest =
= Rs. 2550
Question 2 1 / -0
If I get Rs. 50 by investing Rs. 5,000 for a year, what is the rate of interest?
Solution
P = Rs. 5,000, t = 1 year, S.I. = Rs. 50
Simple interest =
50 =
r = 1%
Question 3 1 / -0
At what rate (percent per annum) does Rs. 500 amount to Rs. 600 in 4 years, interest being simple?
Solution
P = Rs. 500, A = Rs. 600, t = 4 years
Simple interest = Amount - Principal = 600 - 500 = Rs. 100
Simple interest =
100 =
r = 5%
Question 4 1 / -0
A certain sum doubles in 30 years on simple interest. In how many years will it become 4 times?
Solution
Let Principal = P, Amount = 2P, t = 30 years
Simple interest = Amount - Principal = 2P - P = P
Simple interest =
P =
r =
%
Now, Simple interest = Amount - Principal = 4P - P = 3P
Simple interest =
3P =
3 × 100 =
t = 90 years
Question 5 1 / -0
The simple interest will be Rs. 810 after 6 years on the money lent at 9% per annum. What will be the sum?
Solution
Let principal be P, SI be Rs. 810, r be 9% and t be 6 years.
Simple interest =
810 =
810 × 100 = 54P
P =
= Rs. 1500
Question 6 1 / -0
The sum of Rs. 450 will amount to Rs. 540 at the rate of 5% per annum at simple interest in
Solution
P = Rs. 450, A = Rs. 540, and r = 5%.
Simple interest = Amount - Principal = 540 - 450 = Rs. 90
Simple interest =
90 =
90 × 100 = 450 × 5 × t
t = 4 years
Question 7 1 / -0
A sum of Rs. 5000 was lent at simple interest and at the end of 2 years, the total amount was Rs. 5,800. Find the rate.
Solution
P = Rs. 5000, A = Rs. 5800, t = 2 years
Simple interest = Amount - Principal = 5800 - 5000 = 800
Simple interest =
, 800 =
800 x 100 = 5000 x r x 2
r = 8%
Question 8 1 / -0
Simple interest on Rs. 5,000 for 5 years at 10% per annum is equal to
Solution
P = Rs. 5000, t = 5 years, r = 10%.
Simple interest =
S.I. =
= Rs. 2500
Question 9 1 / -0
A certain sum of money lent at simple interest amounts to Rs. 690 in 3 years and Rs. 750 in 5 years. Find the sum.
Solution
Let the principal = PAmount after 3 years = 690Amount after 5 years = 750Simple interest for 2 years = 750 - 690 = 60Simple interest for 1 year = 60/2 = 30Simple interest for 3 years = 3 x 30 = 90Principal = Amount - S.I.P = 690 - 90 = Rs.600
Question 10 1 / -0
A certain sum of money lent at simple interest amounts to Rs. 1,300 in 4 years and Rs. 1,525 in 7 years. Find the sum.
Solution
Let principal = P
Question 11 1 / -0
Calculate the C.I. on Rs. 10,000 for 3 years at 10% per annum compounded annually.
Solution
P = 10,000, r = 10%, n = 3 years compounded annually
Amount A = P(1 + r/100)
n A = 10,000 (1 + 10/100)
3 = 10,000 ×
= 13,310
Compound interest = A - P
= 13,310 - 10,000
= Rs. 3310
Question 12 1 / -0
Calculate the C.I. on Rs. 8000 for 1
years at 20% per annum compounded half-yearly.
Solution
P = 8000, r = 20%, n = 3/2 years compound half-yearly.
Amount A = P(1 + r/200)
2n ,
A = 8000 (1 + 20/200)
2 x 3/2 = 8000 x
= 8 x 1331 = 10,648
Compound interest = A - P = 10,648 - 8000 = Rs. 2648
Question 13 1 / -0
Calculate the C.I. on Rs. 40,000 for 9 months at 20% per annum compounded quarterly.
Solution
P = 40,000, r = 20%, n = 9 months = 3/4 years compounded quarterly.
Amount A = P (1 + r/400)
4n , A = 40,000 (1 + 20/400)
4 × 3/4 = 4000 ×
= 5 × 9261 = 46,305.
Compound interest = A - P = 46,305 - 40,000 = Rs. 6305
Question 14 1 / -0
A sum of Rs. 8000 becomes Rs. 9261 in 3 years compounded annually at a certain rate. Find the rate percent.
Solution
P = 8000, A = 9261, n = 3 years compounded annually
Amount, A = P(1 + r/100)
n 9261 = 8000(1 + r/100)
3 = (1 + r/100)
3 r = 5%
Question 15 1 / -0
Find the compound interest on Rs. 21,600 for 3 months at 20% per annum, compounded monthly.
Solution
P = 21,600, r = 20% and n = 3 months = ¼ year, compounded monthly
Amount A = P(1 + r/1200)
12n ,
A = 21,600(1 + 20/1200)
12*1/4 = 21,600 ×
= 22,698.1
Compound interest = A - P = 22,698.1 - 21,600 = Rs. 1098.10
Question 16 1 / -0
A sum becomes 8 times its original amount in 3 years, compounded annually at a certain rate. Find the rate of interest.
Solution
Let principle be P, amount be 8P and time(t) be 3 years compounded annually.
Amount A = P(1 +
)
t 8P = P(1 +
)
3 8 = (1 +
)
3 (2)
3 =
2 =
1 =
r = 100%
Question 17 1 / -0
A sum becomes 6¼ times in n years, compounded annually at 150%. Find the value of n.
Solution
P(1 + r/100)n = An = 25/4 Pn = 25/4 Pn = 25/4
Question 18 1 / -0
A sum of Rs. 4000 becomes Rs. 4410 in 'n' years compounded annually at 5% per annum. Find the number of years.
Solution
P = 4000, A = 4410 and r = 5% compounded annually
Amount, A = P(1 + (r/100))
n 4410 = 4000(1 + (5/100))
n = (1 + (5/100))
n n = 2
Thus, the number of required years is 2.
Question 19 1 / -0
A sum becomes Rs. 6,760 in 2 years compounded annually at 4% per annum. Find the sum.
Solution
A = 6760, n = 2 years, r = 4% compounded annually.
Amount A = P(1 + r/100)
n ,
6760 = P(1 + 4/100)
2 ,
6760 = P
6760 = P ×
P =
P = Rs. 6250
Question 20 1 / -0
Find the difference between the CI and SI on Rs. 500 at 5% after 1 year.
Solution
P = 500, r = 5%, n = 1 year compounded annually
Amount, A = P
A = 500
A = 500 ×
= 525
Compound interest, CI = 525 - 500 = Rs. 25
Simple interest, SI =
=
= Rs. 25
Difference = 25 - 25 = Rs. 0
Question 21 1 / -0
If the difference between CI and SI on a sum of money at 5% for 3 years is Rs. 122, find the sum.
Solution
P = P, r = 5%, n = 3 years compounded annually
A = P(1 + r/100)
n A = P(1 + 5/100)
3 = P × (21/20)
3 =
P
Compound interest =
Simple interest =
=
=
Difference =
(Given)
P ×
= 122
P =
= Rs. 16,000
Question 22 1 / -0
The population of a town increases at 10% every year. If the present population is 20,000, find the population after 4 years.
Solution
P = 20,000, r = 10% and n = 4 years (increases annually)n A = 20,000(1 + 10/100)4 = 20,000 × (11/10)4
Question 23 1 / -0
The population of a town decreases by 5% every year. If the present population of the town is 10, 000, find the population after 2 years.
Solution
P = 10,000, r = 5%, n = 2 years, decreases annually. Population after 3 years
= P(1 - r/100)
n (Note: The minus sign)
A = 10,000(1 - 5/100)
2 = 10,000 ×
= 10,000 ×
= 9025
Question 24 1 / -0
The population of a town increases by 5% in the 1st year, decreases by 4% in the 2nd year and again increases by 3% in the 3rd year. If the present population of the town is 2,00,000, then find the population after 3 years.
Solution
P = 2,00,000
r
1 = 5%
r
2 = 4%
r
3 = 3%
A = P(1 +
)(1 –
)(1 +
)
A = 2,00,000(1 +
)(1 –
)(1 +
)
= 2,00,000 ×
= 2,07,648
Question 25 1 / -0
The population of a town doubles in 2 years. Find the growth rate.
Solution
Let P = P, A = 2P, n = 2 years
A = P(1 + r/100)
n 2P = P(1 + r/100)
2 2 = (1 + r/100)
2 = (1 + r/100)
2 = (1 + r/100)
1.414 = 1 + r/1000
r = 41.4%
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