SIMPLE INTEREST

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EXERCISE PROBLEM . Ex.l Find: S.l. on Rs 68000 at 16 2/3 % per annum for 9 months. Sol. P = 68000, R = 50/3 % p.a. & T = 9/12 year = 3/4 years. .. S.I. = ( PxRxT/100) = Rs. [6800 x 50/3 x ¾ x 1/100] = Rs. 8500 Ex.2 Find: S.l. on Rs 6250 at 14% per annum for 146 days. Sol. P = Rs. 6250, R = 14 % p.a. & T = (146/365) year = 2/5 year . :. S.l. = Rs (6250 x 14 x 2/5 x 1/100)= Rs 350. Ex.3 Find: S.l. on Rs 3000 at 18% per annum for the period from 4th Sol. Time = (24 + 31 + 18) days = 73 days = 1/5 year . P = Rs 3000 and R = 18 % p.a. :. S.I. = Rs (3000 x 18 x 1/5 x 1/100)= Rs 108 . Remark: The day on which money is deposited is not counted while the day on which money is withdrawn is counted. Ex. 4. A sum at simple interest at 13 i % per annum amounts to Rs. 2502.50 after 4 years. Find the sum. Sol. Let sum be x. Then S.I. = (X x 27/2 x 4 x 1/100) = 27X/50
.. Amount = ( X + 27X/50) = 77X/50 .. 77X/50 = 2502.50 or X= (2502.50 X 50)/77 = 1625 Hence Sum = Rs.1625 Ex. 5. A certain sum of money amounts to Rs 1008 in 2 years and to Rs 1164 in 3 i years. Find the sum and (be rate of interest. Sol. S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 . S.I. for 2 years = Rs (156 x 2/3 x 2)= Rs 208. :. Principal = Rs (1008 - 208) = Rs 800. Now, P = 800, T= 2 and S.I. = 208. .. Rate = (100 x S.I.) / (P x T) =[ (100x208)/(800x2)]% =13% Ex.6. At what rate percent per annum will a sum of money double in 8 years Sol. Let principal = P, Then, S.I.=P and Time=8 years .. Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.. Ex. 7. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum. Sol. Let, sum = P and original rate = R. Then, Px (R+2}x 3 P xRx 3 = 360 100 100 or 3PR + 6P - 3PR = 36000 or 6P = 36000 or P = 6000 . Hence, sum = Rs 6000 . Ex. 8. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, If both are numerically equal. Sol. Let sum = x. Then, S.I. = 16x / 25 Let rate = R % and time = R years. .. ( X x R x R)/(100) = 16x/25 or R 2 =1600/25 R= 40/5 =8
Page 1 and 2: SIMPLE INTEREST
Page 3 and 4: (i). Simple Intereest = P x R x T 1
Page 5 and 6: 5. Reena took a loan of Rs. 1200 wi
Page 7 and 8: 10.A sum of Rs. 725 is lent in the
Page 9 and 10: Since the principal is not given, s
Page 11: II gives, T = 10 years. Sum = 100 x
Page 15 and 16: 4% p.a. interest on principal for n
Page 17 and 18: P x T Now, I gives, S.I. = Rs. 4000
Page 19 and 20: COMPOUND INTEREST Compound interest
Page 21 and 22: Simple interest on a certain sum of
Page 23 and 24: Solution S.I C.I = Rs.(1200x10x1/10
Page 25 and 26: If the simple interest on a sum of
Page 27 and 28: 40 40 40 = Rs. 1600 x 41 41 + 1 40
Page 29 and 30: Amount = Rs. (30000 + 4347) = Rs. 3
Page 31 and 32: Amount = Rs. P 1 + R 4 100 C.I. =P
Page 33 and 34: Answer: Option D Explanation: Amoun
Page 35 and 36: 15000 (100 + R)2 - 10000 - (200 x R
Page 37 and 38: 11.The population of a town was 360
Page 39 and 40: ond as the interest that he receive
Page 41 and 42: Difference between C.I. and S.I. ma
Page 43 and 44: For this sum, C.I. and hence amount

SIMPLE INTEREST

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