91. Ramesh invested an amount that is 10% of ₹ 10000 at simple interest. After 3 yr, the amount becomes ₹ 2500. Find out the 4 times of actual interest rate.
    

1. 1. 5000%

   
   
   

   2. 250%

   
   
   

   3. 200%

   
   
   

   4. 600%

   
   
   

   5. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   Investment of Ramesh = 10% of 10000 = ₹ 1000
   After 3 yr = 2500 - 1000 = ₹ 1500
   
   We know.
   SI = (P x R x T)/100

   Correct Option: C

   Investment of Ramesh = 10% of 10000 = ₹ 1000
   After 3 yr = 2500 - 1000 = ₹ 1500
   
   We know.
   SI = (P x R x T)/100
   ⇒ 1500 = (1000 x R x 3)/100
   ∴ R = (1500 x 100) / (1000 x 3) = 50%
   ∴ 4 times of 50% = 200%

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92. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 12 yr and for 18 yr ?

1. 1. 2 : 5

   
   
   

   2. 1 : 3

   
   
   

   3. 2 : 3

   
   
   

   4. 3 : 1

   
   
   

   5. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   If the principal = P and interest = R%
   Then, required ratio = [(P x R x 12)/100] / [(P x R x 18)/100]

   Correct Option: C

   If the principal = P and interest = R%
   Then, required ratio = [(P x R x 12)/100] / [(P x R x 18)/100]
   = 12 / 18
   = 2/3
   = 2 : 3

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93. Divide ₹6800 into two parts so that S.I. on the first part for 3	1	Years at 6% may be equal to
    	3

    the interest on the second part for 3	1	Years at 4% Percent Per annum
    	3

    
    

1. 1. ₹ 2600 ; ₹ 4200
      

   
   
   

   2. ₹ 2800 ; ₹ 4000
      

   
   
   

   3. ₹ 2500 ; ₹ 4300
      

   
   
   

   4. ₹ 2700 ; ₹ 4100

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the first part be p. then ,
   second part = (6800 – p)
   

   Interest on first part for 3	1	years at 6%
   	3

   
   

   =	p × 6 × ( 10/3 )	=	p
   	100		5

   
   

   Interest on second part for 3	1	years at 4%
   	2

   
   

   SI =	(6800 - p) × 4 × ( 7/2 )
   	100

   
   

   SI = ₹	(6800 - p) × 7
   	50

   
   According to the question,
   

   p	=	(6800 - p) × 7
   5		50

   
   

   Correct Option: B

   Let the first part be p. then ,
   second part = (6800 – p)
   

   Interest on first part for 3	1	years at 6%
   	3

   
   

   =	p × 6 × ( 10/3 )	=	p
   	100		5

   
   

   Interest on second part for 3	1	years at 4%
   	2

   
   

   SI =	(6800 - p) × 4 × ( 7/2 )
   	100

   
   

   SI = ₹	(6800 - p) × 7
   	50

   
   According to the question,
   

   p	=	(6800 - p) × 7
   5		50

   
   ⇒ 10p = (6800 – p) 7
   ⇒ 10p = 47600 – 7p
   ⇒ 17p = 47600
   ⇒ p = 2800
   Hence, first part = ₹ 2800
   and second part = ₹ (6800 – 2800) = ₹ 4000.

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94. A certain sum of money amounts to ₹6780 in 2 years and to ₹7360.50 in

    3	1	Years. Find the sum and the rate of interest.
    	3

1. 1. ₹ 6006 ; 6.4 Percent Per annum
      

   
   
   

   2. ₹ 8006 ; 6.4 Percent Per annum
      

   
   
   

   3. ₹ 5006 ; 5 Percent Per annum
      

   
   
   

   4. ₹ 5506 ; 5 Percent Per annum

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question ,
   

   Principal + S.I. for 3	1	years = ₹ 7360.50 ....... (i)
   	2

   
   Principal + S.I. for 2 years = ₹ 6780 ...... (ii)
   On subtracting equation (ii) from (i),
   

   S.I. for 1	1	years = ₹ 580.50
   	2

   
   

   ∴ S.I. for 2 years = ₹		580.50 × 2 × 2		= ₹ 774
   		3

   
   

   Correct Option: A

   According to question ,
   

   Principal + S.I. for 3	1	years = ₹ 7360.50 ....... (i)
   	2

   
   Principal + S.I. for 2 years = ₹ 6780 ...... (ii)
   On subtracting equation (ii) from (i),
   

   S.I. for 1	1	years = ₹ 580.50
   	2

   
   

   ∴ S.I. for 2 years = ₹		580.50 × 2 × 2		= ₹ 774
   		3

   
   ∴ Principal = (6780 – 774 ) = ₹ 6006
   

   And, rate of interest =	774 × 100	= 6.4% per annum.
   	6006 × 2

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95. A borrowed ₹ 1500 at 4% per annum and ₹ 1400 at 5% per annum for the same period. He paid ₹ 390 as total interest. Find the time for which he borrowed the sum.
    

1. 1. 3.5 years
      

   
   
   

   2. 2.5 years
      

   
   
   

   3. 3 years
      

   
   
   

   4. 4 years

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that ,
   SI = SI₁ + SI₂
   

   SI =	P1R1T	+	P2R2T
   	100		100

   
   

   or, T =	100 × SI
   	P1R1 + P2R2

   
   

   T =	100 × 390
   	(1500 × 4) + (1400 × 5)

   
   

   Correct Option: C

   We know that ,
   SI = SI₁ + SI₂
   

   SI =	P1R1T	+	P2R2T
   	100		100

   
   

   or, T =	100 × SI
   	P1R1 + P2R2

   
   

   T =	100 × 390
   	(1500 × 4) + (1400 × 5)

   
   

   T =	39000
   	13000

   
   T = 3 years.

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