Ex. 2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th Feb., 2005 to 18th April, 2005.

Sol. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.

P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a ∴S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

Ex. 3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Sol. Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50 ∴amount = (Rs. x+(27x/50)) = Rs.77x/50 ∴ 77x/50 = 2502.50 ⇔ x = (2502.50 * 50)/77 = 1625

Hence , sum = Rs.1625.

Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere interest rate is increased by 8%, it would amount to bow mucb ? Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs. R = ((100 x 120)/(800*3) ) % = 5%. New rate = (5 + 3)% = 8%. New S.l. = Rs. (800*8*3)/100 = Rs. 192. New amount = Rs.(800+192) = Rs. 992.

Ex. 5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. if he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Ex. 6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.

Sol.. S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156. S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208

Principal = Rs. (1008 - 208) = Rs. 800.

Now, P = 800, T = 2 and S.l. = 208.

Rate =(100* 208)/(800*2)% = 13%

Ex. 7. At what rate percent per annum will a sum of money double in 16 years.

Sol.. Let principal = P. Then, S.l. = P and T = 16 yrs.

∴Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

Ex. 8. The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.

Sol. Let sum = Rs. x. Then, S.l. = Rs. 4x/9

Let rate = R% and time = R years.

Then, (x*R*R)/100=4x/9 or R^{2} =400/9 or R = 20/3 = 6 2/3.

∴Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.

Ex. 9. The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40 less tban the simple interest on the same sum for 3 1/2 years at 10% per annum. Find the sum.

Sol. Let the sum be Rs. x Then, ((x*10*7)/(100*2)) â?? ( (x*12*5)/(100*2)) = 40 ⇔ (7x/20)-(3x/10)=40 ⇔ x = (40 * 20) = 800.

Hence, the sum is Rs. 800.

Ex. 10. 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, [ (P*(R+2)*3)/100] - [ (P*R*3)/100] = 360.

⇔ 3PR + 6P - 3PR = 36000 ⇔ 6P=36000 ⇔ P=6000

Hence, sum = Rs. 6000.

Ex. 11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest? Sol . Let each Instalment be Rs. x Then, ( x+ ((x*12*1)/100)) + (x+ ((x*12*2)/100) ) + x = 1092 ⇔ ((28x/25) + (31x/25) + x) = 1092 ⇔ (28x+31x+25x)=(1092*25) ⇔ x= (1092*25)/84 = Rs.325.

∴ Each instalment = Rs. 325.

Ex. 12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.

Sol. Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x). ⇔ ((x*8*1)/100) + ((1550-x)*6*1)/100=106 ⇔ 8x + 9300 -6x=10600 ⇔ 2x = 1300 ⇔ x = 650. ∴ Money lent at 8% = Rs. 650. Money lent at 6% = Rs. (1550 - 650) = Rs. 900.

Posted by naveen

24 Sep, 2013

In approximately how many years will a certain sum of money triple itself at 22% simple interest?

Ex. 1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.

Sol. P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years.∴ S.I. = (P*R*T)/100 = Rs.(68,000*(50/3)*(3/4)*(1/100))= Rs.8500

Ex. 2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from

Sol. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.4th Feb., 2005 to 18th April, 2005.

P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a

∴S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

Ex. 3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Sol. Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50∴amount = (Rs. x+(27x/50)) = Rs.77x/50

∴ 77x/50 = 2502.50 ⇔ x = (2502.50 * 50)/77 = 1625

Hence , sum = Rs.1625.

Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere

Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs.interest rate is increased by 8%, it would amount to bow mucb ?

R = ((100 x 120)/(800*3) ) % = 5%.

New rate = (5 + 3)% = 8%.

New S.l. = Rs. (800*8*3)/100 = Rs. 192.

New amount = Rs.(800+192) = Rs. 992.

Ex. 5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. if he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Sol. Let the sum borrowed be x. Then,(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400

⇔ (3x/25 + 27x/100 + 14x / 25) = 11400 ⇔ 95x/100 = 11400 ⇔ x = (11400*100)/95 = 12000.

Hence , sum borrowed = Rs.12,000.

Ex. 6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.

Sol.. S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208

Principal = Rs. (1008 - 208) = Rs. 800.

Now, P = 800, T = 2 and S.l. = 208.

Rate =(100* 208)/(800*2)% = 13%

Ex. 7. At what rate percent per annum will a sum of money double in 16 years.

Sol.. Let principal = P. Then, S.l. = P and T = 16 yrs.∴Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

Ex. 8. The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.

Sol. Let sum = Rs. x. Then, S.l. = Rs. 4x/9Let rate = R% and time = R years.

Then, (x*R*R)/100=4x/9 or R

^{2}=400/9 or R = 20/3 = 6 2/3.∴Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.

Ex. 9. The simple interest on a certain sum of money for 2 l/2 years at 12% per

Sol. Let the sum be Rs. x Then, ((x*10*7)/(100*2)) â?? ( (x*12*5)/(100*2)) = 40annum is Rs. 40 less tban the simple interest on the same sum for 3 1/2 years at 10% per annum. Find the sum.

⇔ (7x/20)-(3x/10)=40

⇔ x = (40 * 20) = 800.

Hence, the sum is Rs. 800.

Ex. 10. A sum was put at simple interest at a certain rate for 3 years. Had it been

Sol. Let sum = P and original rate = R.put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.

Then, [ (P*(R+2)*3)/100] - [ (P*R*3)/100] = 360.

⇔ 3PR + 6P - 3PR = 36000 ⇔ 6P=36000 ⇔ P=6000

Hence, sum = Rs. 6000.

Ex. 11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years

Sol . Let each Instalment be Rs. xat 12% simple interest?

Then, ( x+ ((x*12*1)/100)) + (x+ ((x*12*2)/100) ) + x = 1092

⇔ ((28x/25) + (31x/25) + x) = 1092 ⇔ (28x+31x+25x)=(1092*25)

⇔ x= (1092*25)/84 = Rs.325.

∴ Each instalment = Rs. 325.

Ex. 12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at

Sol. Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x).6%. If the total annual income is Rs. 106, find the money lent at each rate.

⇔ ((x*8*1)/100) + ((1550-x)*6*1)/100=106

⇔ 8x + 9300 -6x=10600 ⇔ 2x = 1300 ⇔ x = 650.

∴ Money lent at 8% = Rs. 650. Money lent at 6% = Rs. (1550 - 650) = Rs. 900.

In approximately how many years will a certain sum of money triple itself at 22% simple interest?

## Comments Received:

## Please give your suggestions and feedback:

00At what rate percent per annum will a sum of money double in 16 years.

Sol.. Let principal = P. Then, S.l. = P and T = 16 yrs.∴Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.

i didnt get thz ansr. can anybody clarify my doubt

## Comments Received:

Mahesh.:Hey, Its simple math formula Simple intrest = PTR/100 R = 100 * Simple Intrest / (PT) R = 100 * P / (P*16)Mrinmay:Simple interest(SI) = PTR/100 R = 100 * Simple Interest / Principal * Time So,Rate = 100 * P / (P*16)## Please give your suggestions and feedback:

03Ex. 3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Sol. Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50∴amount = (Rs. x+(27x/50)) = Rs.77x/50

∴ 77x/50 = 2502.50 ⇔ x = (2502.50 * 50)/77 = 1625

Hence , sum = Rs.1625. how we will get 27x/50.plz clarify me

## Comments Received:

Mahesh:Hey its very simple math calculation. x * (27/2) * 4 * (1/100) = (27x*4)/200 = 27x/50## Please give your suggestions and feedback:

00