Simple & Compound Interest
Practice aptitude questions
Q1. A money lender borrows money at 4% P.a and pays interest at the end of the year. He lends it at 6% P.a compound interest compounded half-yearly and receives the interest at the end of the year. Thus be gains Rs 104.50 a year. The amount of money he borrows is
Solution
Let the money borrowed be Rs x
Interest paid by the money lender = Rs (x × 4/100 × 1) =Rs x/25
Interest received by the money lender
= Rs [x × (1 + 3/100)2 - x]
= Rs(x × 103/100 × 103/100 - x)
Gain = Rs (609x/1000 – x /25)
= Rs 209x/1000 = Rs 609x/10000
Therefore 209x/10000 = 104.50
=> 209x = 1045000
=> x = 5000
Hence money borrowed = Rs 5000
Q2. A man deposited Rs 6000 in a bank at 5% P.a simple interest another man deposits Rs 5000 at 8% P.a Compounded interest. After 2 years the difference of their interest will be
Solution
S.I = Rs (6000 × 5/100 × 2) = Rs 600
C.I = Rs [5000 × (1 + 8/100)2 - 5000]
= Rs (5000 × 27/25 × 27/25 - 5000)
= Rs(5832 - 5000)
= Rs 832
(C.I) – (S.I) = Rs (832 - 600) = Rs 232
Q3. The difference between simple and compound interest (compounded-annually) on a sum of money for 2 years at 10% per annum is Rs 65. The sum is
Solution
Let the sum be Rs x.
Then [x × (1 +10/100)2 - x] – (x × 10/100 × 2) = 65
=> (x × 11/10 × 11/10 - x) – x/5 = 65
=> (121x/100 - x) – x/5 = 65
=> (21x/100 – x/5) = 65
=> (21x – 20x) = 6500
=> X = 6500
Q4. The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half yearly is:
Solution
Let the sum be Rs 100.
Then P = Rs 100, R = 3 %
per half – year, t = 2 half – years
Amount = Rs [100 × (1 + 3/100)2]
= Rs (100 × 103/100 × 103/100)
= Rs 10609/100 = Rs 106.09
Effective Annual Rate = 6.09%
Q5. At what rate of interest will be Rs 20000 becomes Rs 24200 after 2 years when interest is compounded annually?
Solution
P = Rs 20000, A =Rs 24200,
t = 2 years 20000 × (1 + R/100)2 = 24200
=> (1 + R/100) 2 = 24200/20000 = 121/100 = (11/10) 2
=> 1 + R/100 = 11/10
=> R/100 = (11/10 - 1) = 1/10
=> R = (100 × 1/10) % p.a = 10 % p.a
Hence, Rate = 10 % p.a
Q6. The difference between compound interest and simple interest at the same rate on Rs 5000 for 2 years is Rs 72. The rate of interest per annum is:
Solution
Let the rate be r % p.a,
Then S. I = Rs (5000 × r/100 × 2) = Rs (100r)
C. I = Rs[5000 × (1 + r/100) 2 - 5000]
= Rs {5000 × [(1 + r/100) 2 – 1]}
= Rs [5000 × (r2/2 + r/50)]
= Rs (r2 /2 + 100r)
(C.I) – (S.I) = (r2/2 + 100r) – (100r) = r2/2
Therefore, r2/2 = 72
=> r2 = 144
=> r = 12
Hence, Rate = 12 % p.a
Q7. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:
Solution
NO Solution
Q8. The difference between compound interest and simple interest on a sum for 2 years at 8% P.a is Rs 786. The sum is:
Solution
Let the sum be Rs x.
Then S. I = Rs (x × 8/100 × 2) = Rs 4x/25
C. I = Rs [x × (1 + 8/100)2 - x]
= Rs (x × 27/25 × 27/25 - x)
=Rs 104x/625 (C.I) – (S.I)
= Rs(104x/625 – 4x /25)
= Rs 4x /625
Therefore 4x/625 = 768
=> x = ((768 ×625)/4) = 120000
Therefore sum = Rs 120000
Q9. A sum of money placed at compound interest doubles itself in 5 years in how many years it would to 8 times of itself at the same rate of interest?
Solution
Let principal be Rs x and the rate is R % p.a.
Then X × (1 + R/100)5 = 2x
=> (1 + R/100) 5 = 2
Let x × (1 + R/100) t = 8x
=> (1 + R/100) t = 8 = 23
={(1 + R/100) 5 }3
=> (1 + R/100) t = (1 + R/100) 15
=> T = 15 Years
Q10. If the rate of interest be 4% per annum for first year 5% per annum for the second year and 6% per annum from the third year then the compound interest of Rs 10000 for 3 years will be
Solution
Amount = Rs[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)]
= Rs(1000 × 26/25 × 21/20 × 53/50)
= Rs (57876/5) = Rs 11575.20
C.I = Rs (11575.20 - 10000) = Rs 1575.20