More solved problems on compound interest using formula are shown below.
Solution:
Given, SI = $ 6750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.
sum = 100 × SI / R × T
= $ (100 × 6750 × ³/₂₀ × 1/3 ) = $ 33750.
Now, P = $ 33750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.
Therefore, amount after 3 years
= $ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)ᵀ]
= $ (33750 × 16/15 × 16/15 × 16/15) = $ 40960.
Thus, amount = $ 40960.
Hence, compound interest = $ (40960  33750) = $ 7210.
Solution:
Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)²  100}
= $ {(100 × 53/50 × 53/50)  100} = $ (2809/25  100) = $ 309/25
Therefore, (CI)  (SI) = $ (309/25 – 100) = $ 9/25
If the difference between the CI and SI is $ 9/25, then the sum = $ 100.
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18 )
= $ 5000.
Hence, the required sum is $ 5000.
Alternative method
Let the sum be $ P.
Then, SI = $ (P × 6/100 × 2) = $ 3P/25
And, CI = $ {P × (1 + 6/100)²  P}
= $ {(P × 53/50 × 53/50)  P} = $ (\(\frac{2809}{2500}\)P  P) = $ (309P/2500)
(CI)  (SI) = $ (309P/2500 – 3P/25) = $ (9P/2500)
Therefore, 9P/2500 = 18
⇔ P = 2500 × 18/9
⇔ P = 5000.
Hence, the required sum is $ 5000.
Solution:
Let the sum be $ 100. Then,
amount = $ {100 × (1 + 8/100)²}
= $ (100 × 27/25 × 27/25) = $ (2916/25)
If the amount is $ 2916/25 then the sum = $ 100.
If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500.
Hence, the required sum is $ 62500.
Alternative method
Let the sum be $ P. Then,
amount = $ {P × (1 + 8/100)²}
= $ {P × 27/25 × 27/25} = $ (729P/625)
Therefore, 729P/625 = 72900
⇔ P = (72900 × 625)/729
⇔ P = 62500.
Hence, the required sum is $ 62500.
Solution:
Let the required rate be R% per annum.
Here, A = $ 2205, P = $ 2000 and n = 2 years.
Using the formula A = P(1 + R/100)ⁿ,
2205 = 2000 × ( 1 + R/100)²
⇒ (1 + R/100)² = 2205/2000 = 441/400 = (21/20)²
⇒ ( 1 + R/100) = 21/20
⇒ R/100 = (21/20 – 1) = 1/20
⇒ R = (100 × 1/20) = 5
Hence, the required rate of interest is 5% per annum.
Solution:
Let the required time be n years. Then,
amount = $ {1000 × (1 + 10/100)ⁿ}
= $ {1000 × (11/10)ⁿ}
Therefore, 1000 × (11/10)ⁿ = 1331 [since, amount = $ 1331 (given)]
⇒ (11/10)ⁿ = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)³
⇒ (11/10)ⁿ = (11/10)³
⇒ n = 3.
Thus, n = 3.
Hence, the required time is 3 years.
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● Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded Yearly
Compound Interest when Interest is Compounded HalfYearly
Compound Interest when Interest is Compounded Quarterly
Variable Rate of Compound Interest
Difference of Compound Interest and Simple Interest
Practice Test on Compound Interest
Uniform Rate of Growth and Depreciation
● Compound Interest  Worksheet
Worksheet on Compound Interest
Worksheet on Compound Interest when Interest is Compounded HalfYearly
Worksheet on Compound Interest with Growing Principal
Worksheet on Compound Interest with Periodic Deductions
Worksheet on Variable Rate of Compound Interest
Worksheet on Difference of Compound Interest and Simple Interest
Worksheet on Uniform Rate of Growth
Worksheet on Uniform Rate of Depreciation
Worksheet on Uniform Rate of Growth and Depreciation
8th Grade Math Practice
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