11. A sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself ?

1. 1. 9 years

   
   
   

   2. 27 years

   
   
   

   3. 6 years

   
   
   

   4. 3 years

   
   
   

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1. View Hint View Answer Discuss in Forum

   Suppose P = ₹ 1, then A = ₹ 3 ,
   Using the given formula ,
   

   A = P		1 +	R		T
   			100

   
   According to question ,
   

   ⇒  3 = 1		1 +	R		3
   			100

   
   On squaring both sides, we get
   

   ⇒  9 = 1		1 +	R		6
   			100

   
   ∴  Time = 6 years
   We can find required answer with the help of given formula :
   Here, p = 3, n₁ = 3 , q = 9, n₂ = ?
   Using formula ,
   p^1/n₁ = q^1/n₂
   

   Correct Option: C

   Suppose P = ₹ 1, then A = ₹ 3 ,
   Using the given formula ,
   

   A = P		1 +	R		T
   			100

   
   According to question ,
   

   ⇒  3 = 1		1 +	R		3
   			100

   
   On squaring both sides, we get
   

   ⇒  9 = 1		1 +	R		6
   			100

   
   ∴  Time = 6 years
   We can find required answer with the help of given formula :
   Here, p = 3, n₁ = 3 , q = 9, n₂ = ?
   Using formula ,
   p^1/n₁ = q^1/n₂
   ⇒ (3)^1/3 = (9)^1/n₂
   ⇒ 3^1/3 = (3²)^1/n₂
   ⇒ 3^1/3 = 3^2/n₂
   

   ⇒	1	=	2
   	3		n2

   
   ⇒  n₂ = 6 years

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12. A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?

1. 1. 12 years

   
   
   

   2. 13 years

   
   
   

   3. 8 years

   
   
   

   4. 16 years

   
   
   

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1. View Hint View Answer Discuss in Forum

   As we know that ,
   

   A = P		1 +	R		T
   			100

   
   Let P be ₹ 1, then A = ₹ 2
   Here , T = 4 years , Rate = R%
   

   ⇒  2 = 1		1 +	R		4
   			100

   
   

   ⇒  22 =		1 +	R		8
   			100

   
   ∴  Time = 8 years
   Second Method to solve this question :
   Here, p = 2, n₁ = 4 , q = 4, n₂ = ?
   Using p^1/n₁ = q^1/n₂
   

   Correct Option: C

   As we know that ,
   

   A = P		1 +	R		T
   			100

   
   Let P be ₹ 1, then A = ₹ 2
   Here , T = 4 years , Rate = R%
   

   ⇒  2 = 1		1 +	R		4
   			100

   
   

   ⇒  22 =		1 +	R		8
   			100

   
   ∴  Time = 8 years
   Second Method to solve this question :
   Here, p = 2, n₁ = 4 , q = 4, n₂ = ?
   Using p^1/n₁ = q^1/n₂
   ⇒ (2)^1/4 = (4)^1/n₂
   ⇒ 2^1/4 = (2²)^1/n₂
   ⇒ 2^1/4 = 2^1/n₂
   

   ⇒	1	=	2
   	4		n2

   
   ⇒ n₂ = 8 years

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13. A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?

1. 1. 6 years

   
   
   

   2. 4 years

   
   
   

   3. 8 years

   
   
   

   4. 5 years

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that , Time = 3 years , Rate = R%
   Let the principal be ₹ 1 and Amount = ₹ 8
   

   ∴ A = P		1 +	R		T
   			100

   
   

   ⇒  8 = 1		1 +	R		3
   			100

   
   

   ⇒  23 = 1		1 +	R		3
   			100

   
   

   ⇒  2 =		1 +	R		1
   			100

   
   

   ⇒  24 =		1 +	R		4
   			100

   
   ∴  Time = 4 years
   We can find required answer with the help of given formula :
   Here, p = 8, n₁ = 3 and q = 16, n₂ = ?
   Using p^1/n₁ = q^1/n₂
   

   Correct Option: B

   Given that , Time = 3 years , Rate = R%
   Let the principal be ₹ 1 and Amount = ₹ 8
   

   ∴ A = P		1 +	R		T
   			100

   
   

   ⇒  8 = 1		1 +	R		3
   			100

   
   

   ⇒  23 = 1		1 +	R		3
   			100

   
   

   ⇒  2 =		1 +	R		1
   			100

   
   

   ⇒  24 =		1 +	R		4
   			100

   
   ∴  Time = 4 years
   We can find required answer with the help of given formula :
   Here, p = 8, n₁ = 3 and q = 16, n₂ = ?
   Using p^1/n₁ = q^1/n₂
   (8)^1/3 = (16)^1/n₂
   (2³)^1/3 = (2⁴)^1/n₂
   2¹ = 2^4/n₂
   

   ⇒  1 =	4
   	n2

   
   ∴ n₂ = 4 years

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14. A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself ?

1. 1. 4 years

   
   
   

   2. 6 years

   
   
   

   3. 6.4 years

   
   
   

   4. 7.5 years

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given , Time = 3 years , Rate = R%
   Let Principal be ₹ P and Amount = ₹ 2P
   Using the given formula
   

   A = P		1 +	R		T
   			100

   
   According to question ,
   

   ⇒  2 = 1		1 +	R		3
   			100

   
   On squaring both sides and multiplying by P
   

   4P = P		1 +	R		6
   			100

   
   ∴  Time = 6 years
   Second Method to solve this question :
   Here, p = 2, n₁ = 3
   q = 4, n₂ = ?
   ∴ p^1/n₁ = q^1/n₂
   

   Correct Option: B

   Given , Time = 3 years , Rate = R%
   Let Principal be ₹ P and Amount = ₹ 2P
   Using the given formula
   

   A = P		1 +	R		T
   			100

   
   According to question ,
   

   ⇒  2 = 1		1 +	R		3
   			100

   
   On squaring both sides and multiplying by P
   

   4P = P		1 +	R		6
   			100

   
   ∴  Time = 6 years
   Second Method to solve this question :
   Here, p = 2, n₁ = 3
   q = 4, n₂ = ?
   ∴ p^1/n₁ = q^1/n₂
   ⇒ 2^1/3 = 4^1/n₂
   2^1/3 = (2²)^1/n₂
   ⇒ 2^1/3 = 2^2/n₂
   Equating powers of 2 on both sides , we get
   

   1	=	2
   3		n2

   
   ∴  n₂ = 6 Years

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15. At what rate percent per annum of compound interest, will a sum of money become four times of itself in two years ?

1. 1. 100%

   
   
   

   2. 75%

   
   
   

   3. 50%

   
   
   

   4. 20%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the Principal be P and rate of interest be R%.
   According to question ,
   

   4P = P		1 +	R		2
   			100

   
   

   ⇒  4 =		1 +	R		2
   			100

   
   Taking square root both sides , we get
   

   ⇒  1 +	R	= 2
   	100

   
   

   ⇒	R	= 1
   	100

   
   ⇒  R = 100%
   We can find required answer with the help of given formula :
   Here, n = 4, t = 2 years
   

   Correct Option: A

   Let the Principal be P and rate of interest be R%.
   According to question ,
   

   4P = P		1 +	R		2
   			100

   
   

   ⇒  4 =		1 +	R		2
   			100

   
   Taking square root both sides , we get
   

   ⇒  1 +	R	= 2
   	100

   
   

   ⇒	R	= 1
   	100

   
   ⇒  R = 100%
   We can find required answer with the help of given formula :
   Here, n = 4, t = 2 years
   R% = (n^1/t − 1) × 100%
   R% = [(4)^1/2 − 1] × 100%
   R% = [2 - 1] × 100%
   ∴ R% = 100%

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