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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
At 5% compound interest per annum the compound interest on Rs. 10000 for 3 years isA. Rs. 1567.25B. Rs. 1567.52C. Rs. 1657.25D. Rs. 1576.25 |
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Answer» Correct Answer - D Rs. 1576.25 The required compound interest `=Rs. 10000(1+5/100)^(3)-Rs. 10000` `=Rs. 10000(1+1/20)^(3)-Rs. 10000` `=Rs. 10000{(21/20)^(3)-1}` `=Rs. 1000(9261/8000)-1` `=Rs. 10000xx1261/8000` `=Rs. (10xx1261)/8=Rs. 176.25` `:.` d is the correct answer. |
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| 2. |
Calculate the amount on Rs. 20000 at the rate of 5% compound interest per annum for 2 years. |
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Answer» The required amount `=Rs. 20000(1+5/100)^(2)` `=Rs. 20000(1+1/20)^(2)` `=Rs. 20000xx21/20xx21/20=Rs. 22050` |
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| 3. |
The amount on Rs. 1000 for 2 years at the rate of 5% compound interest per annum isA. Rs. 1102.50B. Rs. 1120.50C. Rs. 1021.50D. Rs. 1202.50 |
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Answer» Correct Answer - A Rs. 1102.50 The required amount `=Rs. 1000(1+5/100)^(2)` `=Rs. 1000(1+1/20)^(2)` `=Rs. 1000xx21/20xx21/20=Rs. 1102.50` Hence a is correct. |
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| 4. |
The population of a village is 20000. If the rate of birth be 4% per annum and the of death be 2% per annum, then find the population of the village after 2 years. |
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Answer» Rate of birth `=4%` per annum. Rate of death `=2%` per annum. `:.` Rate of growth of population `=(4%-2%)` per annum `=2%` per annum The present population of the village is 20000. `:.` The population of the village after 2 years will be `20000xx(1+2/100)^(2)=20000xx(1+1/50)^(2)` `=20000xx(51/50)^(2)=20000xx(51xx51)/(50xx50)` `=20808`. Hence the required population of the village after 2 years will be 20808. |
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| 5. |
Divide Rs. 3903 between A and B in such a way that at the rate of compound interest of 4% pr annum, the amount obtained by A after 7 years is equal to the amount obtained by B after 9 years. |
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Answer» Let the sum of money given to A be Rs.`x` and that given to B be Rs.`y`. Then the amount obtained by A after 7 years at the rate of compound interet of 4% per annum `Rs.x xx(1+4/100)^(7)=Rs.x xx(26/25)^(7)` Similarly, the amount obtained by B after 9 years at the same rate of interest. `=Rs. yxx(1+4/100)^(9)` `impliesRs.yxx(26/25)^(9)` As per question `x xx(26/25)^(7)=yxx(226/25)^(9)` `impliesx=yxx(26/25)^(9-7)` `impliesx=yxx(26/25)^(2)` `impliesx/y=676/625` `:.x:y=676:625` `:.` Sum of money of `A=Rs. 3903xx676/(676+625)` `=Rs. 3903xx676/1301` `=Rs. 3xx676=Rs. 2028` and sum of money of B`=Rs. 3903xx625/(676+625)` `=Rs. 3903xx625/1301` `=Rs. 3xx625=Rs. 1875` Hence the required sum of money is Rs. 2028 obtained by A and the sum of money is Rs. 1875 obtained by B. |
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| 6. |
Find the sum of money if the difference between compound interest and simple interest for 2 years at the rate of 9% interest per annum is Rs. 129.60 |
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Answer» Let the sum of money be Rs. x. `:.` At the rate of 9% interest per annum the amount of Rs.x for 2 years `=Rs. (1+9/100)^(2)` `:.` The compound interet `=Rs. {x(1+9/100)^(2)-x}=Rs. X{(1+9/100)^(2)-1}` `=Rs. X{(1+9/100)^(2)-(10^(2))=Rs. X{(1+9/100+1)(1+9/100-1)}` `=Rs.x(2+9/100)xx9/100=Rs. x xx209/100xx+9/100` `=Rs. (1881x)/10000` Again at the rate of 9% interest per annum. the simple of interest of Rs. x for 2 years `=Rs. (x xx9xx2)/100` `=Rs. (9x)/50` As per question `(1881x)/10000-(9x)/50=129.60` `implies(1881x-1800x)/10000=129.60` `implies81x=1296000` `impliesx=1296000/81=16000` `:.`The required sum of money `=Rs. 16000` |
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| 7. |
Determine the difference between compound interest and simple interest on Rs. 10000 for 3 years at 5% per annum. |
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Answer» The amount of Rs. 10000 for 3 years at 5% compound interest per annum `=Rs. 10000xx(1+5/100)^(3)=Rs. 10000xx(105/100)^(3)` `=Rs.1000xx(105xx105xx105)/(100xx100xx100)=rs. 11576.25` `:.` The compound interest `=Rs. (11576.25-10000)=Rs. 1576.25` Again the simple interest for 3 years of Rs. 10000 at 5% per annum `=Rs. (10000xx5xx3)/100=Rs. 1500` Hence the required difference `=Rs. (1576.25-1500)=Rs. 76.25` |
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| 8. |
Fill in the blanks: a. Principal and compound interest are_________________proportional. b. Interest `=` Amount `-`_________. If principal `=Rs. P` rate of compound interest be r% per annum and time period be `n` years, then _____`=("prt")/100` |
| Answer» directly, principal, I (interest) | |
| 9. |
Divide Rs. 6305 into there parts in such a way that at the rate of compound interest of 5% per annum the amount of 1st part in 2 years, the amounts of 2nd part in 3 years and the amounts of 3rd part in 4 years are all equal. |
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Answer» Let the three parts be Rs.x, Rs.y and Rs.z respectively. `:.x+y+z=6305…………………….`1 At the rate of 5% compound interest per annum. The amountof Rs.x in 2 years `=Rs.(1+5/100)^(2)=Rs.x xx(21/20)^(2)` The amount of Rs.y in 3 years `=Rs.y(1+5/100)^(3)=Rs.yxx(21/20)^(3)` The amount of Rs.z is 4 years `=Rs.z(1+5/100)^(4)=Rs.xx(21/20)^(4)` As per question `x xx(21/20)^(2)=yxx(21/20)^(3)=zxx(21/20)^(4)=k`(let) `:.x xx(21/20)=kimpliesx=kxx(20/21)^(2)=(400k)/441` `yxx(21/20)^(3)=kimpliesy=kxx(20/21)^(3)=(8000k)/9261` `zxx(21/20)^(4)=kimpliesz=kxx(20/21)^(4)=(160000k)/194481` Then from 1 we get `=(400k)/441+(8000k)/9261+(160000k)/194481=6305` `implieskxx400/441(1+20/21+400/441)=6305` `implies(kxx400)/441xx1261/441=6305` `impliesk=(6305xx441xx441)/(400xx1261)=(441xx441)/80` `:.x=400/441xx(441xx441)/80=2205` `y=8000/9261xx(441xx441)/80=2100` `z=16000/194481xx(441xx441)/80=2000` `:.` The required three parts of Rs. 6305 ar eRs. 2205, Rs. 2100, Rs. 2000 |
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| 10. |
In the case of compound interestA. The principal remains the same in every yearB. The principal changes every yearC. Every year the principal may remain the same or may changeD. Noen of the above |
| Answer» The principal changes every year. | |
| 11. |
The simple interest of a sum of money at the rate of 4% per annum in 3 years is Rs. 303.60. Then find the compound interest of the same principal at the same rate of compound interest in the same period of time. |
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Answer» Let the principal be Rs. `x`. `:.` The simple interes of `Rs.x` at the of 4% simple interest per anuum in 3 years. `=Rs.(x xx4xx3)/100=Rs. (3x)/25` As per question `(3x)/25=303.60impliesx=(303.60xx25)/3` `impliesx=2530` Henc ethe principal `=Rs. 2530` Now the compound interest of Rs.x in 3 years at the rate of 4% compound interest per annum. `=Rs.2530xx(1+4/100)^(3)-Rs.2530` `impliesRs. 2530xx{(26/25)^(3)-1]` Rs.`=Rs. 2530xx(17576/15625-1)=Rs.2530xx(17576-15625)/15625` `=Rs.2530xx1951/15625` `=Rs.315.905=Rs.315.91` (approx) Hence the required compound interet `=Rs. 315.91` (approx) |
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| 12. |
A. Write true of false: a. If Principal `=Rs.p`, rate of compound interest per annum `=r%` and time `=n` years, then the amount after 2 years is `Rs. p(1+r/100)^(2)` b. If principal `=Rs. p` and rate of compound interest per annum for first, 2nd and 3rd year be `r_(1)%, r_(2)%` and `r_(3)%` respectively, then total amount for 3 years.`=Rs. p(1+(r_(1))/100)^(3)+Rs. p(1+(r_(2))/100)^(3)+Rs. p(1+(r_(3))/100)^(3)` |
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Answer» a. True since simple interest after 1 year `=Rs. (pxxrxx1)/100=Rs. (pr)/100` `:.` amount after 1 year `=Rs. P+Rs. (pr)/100=Rs. P(1+r/100)` Again simple interest of new principal `Rs. P(1+r/100)` for 2nd year `Rs. (p(1+r/100)xxrxx1)/100=Rs. (pr(1+r/100))/100` `:.` Amount after 2 year `=Rs. p(1+r/100)+Rs. (pr(1+r/100))/100` `=Rs. p(1+r/100)(1+r/100)=Rs. p(1+r/100)^(2)` Hence amount after 2 years `=Rs. p(1+r/100)^(2)` False: Since the total amount for 3 years `=Rs. p(1+(r_(1))/100)(1+(r_(2))/100)(1+(r_(3))/100)` |
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| 13. |
Rahamanchacha deposited some money in a Co-operative Bank at the rate of 9% compound interest and he received amount Rs. 29702.50 after 2 years. Find how much money Rahamanchacha had deposited in Co-operative Bank. |
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Answer» Let Rahamanchacha had deposited Rs. X in the bank. `:.` The amount of Rs.x in 2 years at the rate of compound of 9% per annum. `=Rs.x xx(1+9/100)^(2)=Rs. x xx(109/100)^(2)` `=Rs.x xx(109xx109)/(100xx100)` As per question `x xx(109/100)^(2)=29702.50` `impliesx xx(109xx109)/(100xx100)=29702.50` `impliesx=(29702.50xx100xx100)/(109xx109)=25000` Hence Rahamanchacha had deposited Rs. 25000 in the Co-operative Bank. |
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| 14. |
In the case of compound interest the rate of compound interest per annum for every years isA. equalB. unequalC. both equal or unequalD. none of the above |
| Answer» Answer: Unequal | |
| 15. |
Calculate the principal which amonts Rs. 9826 after 18 months at the rate of compound interest 2.5% per annum when interest is compunded at the interval of 6 months. |
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Answer» Let the principal be Rs.x. The rate of compound inteest per annum is `2.5%=5/2%` Time `=18` months `=18/12` years `=3/2` years Since interest is compounded at the interval of 6 months, the period of time is 2. `:.` The amount of Rs. `x`in `3/2` years `=Rs.x xx(1+(5/(2xx2))/100)^(3/2xx2)=Rs.x xx(1+1/80)^(3)` `impliesRs. x xx((80+1)/80)^(3)=Rs.x xx(81/80)^(3)` As per question `x xx(81/80)^(3)=9826` `=x xx(81xx81xx81)/(80xx80xx80)=9826` `impliesx=(9826xx80xx80xx80)/(81xx81xx81)` `impliesx=9466.54` (approx) Hence the required principal is Rs. 9466.54 (approx) |
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| 16. |
Find the principal which becomes Rs. 2648 after getting 10% compund interest per annum for 3 years. |
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Answer» Let the principal `=Rs. x` `:.` The amount of Rs. `x` in 3 years at the rate of compound interest of 10% per annum `=Rs.x xx(1+10/100)^(3)` `=Rs. x xx(11/10)^(3)` `impliesRs.x xx1331/1000` `:.` The compound interet `=Rs. (x xx1331/1000-x)` `=Rs. (1331x-1000x)/1000=Rs. (331x)/1000` As per question `(331x)/1000=2648impliesx=(2648xx1000)/331=8000` Hence the required principal `=Rs. 8000 |
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| 17. |
If interest is compounded at the interval of 3 months, the amount of the principal Rs. A. at the rate of compound interest of `b`% per annum after c years isA. `Rs. A(1-b/400)^(4c)`B. `Rs.a(1+b/100)^(4c)`C. `Rs.a(1+b/100)^©`D. `Rs.a(1+b/100)^(c)` |
| Answer» Rs.`a(1+b/100)^(4c)` | |
| 18. |
A sum of money was invested in a monetary fund for 2 years at the rate of compound interest of 20% per annum. If interest had been compounded at the interval of 6 months , then the compound interest would be Rs. 482 more of the same principal at the same rate of interest. Find the sum of money invested. |
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Answer» Let the sum of the money invested be Rs. x Then for the first case the amount of Rs.x after 2 years `=Rs.x xx (1+20/100)^(2)` `=Rs.x xx(1+1/5)^(2)` `5^(2)` `=Rs.x xx(6/5)^(2)=Rs.(36x)/25` Also for the second case, the amount of Rs. x after 2 years, the amount `=Rs.x xx(1+(20/2)/100)^(2xx2)` `=Rs. x xx(1+1/10)^(4)=Rs.x xx(11/10)^(4)` As per question `(14641x)/10000-(36x)/25=482` `implies(14641x-14400x)/10000=482` `implies(241x)/1000-482` `impliesx=(482xx10000)/241` `impliesx=20000` Hence Rs. 20000 was invested. |
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| 19. |
Calculate the compound interest and amount on Rs. 1600 for `1 1/2` years at the rate of 10% compound interest per annum compounded at the interval of 6 months. |
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Answer» Here principal `=Rs. 1600` rate of interest `=10%` time `=1 1/2` years `=3/2` years, period `=12/6=2` Hence the required amount `=Rs. 1600(1+(10/2)/100)^(3/2xx2)` `=Rs. 1600(1+1/20)^(3)=Rs. 1600(21/20)^(3)` `=Rs. (1600xx21xx21xx21)/(20xx20xx20)=Rs. 1852.20` `:.` Required compound interest `=Rs. (1852.20-1600)=Rs. 252.20` Hence the required compound interest `=Rs. 252.20` and the required amount `=Rs. 1852.20` |
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| 20. |
Calculate the compound interest on Rs. 6250 for 9 months at the rate of 10% compound interest per annum compounded at the interval of 3 months. |
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Answer» Here the principal is R. 6250 The rate of compound interest is 10%. Time `=9` months` =9/12` year `=3/4` year. Since the interest is compounded at the interval of 3 months, the period is 4. `:.` The amount of Rs. 6250 in 9 months at the rate of 10% per annum `=Rs. 6250xx(1+(10/4)/100)^(3/4xx4)` `=Rs. 6250xx(1+10/(4xx100)^(3)` `=Rs. 6250xx((40+1)/40)^(3)=Rs. 6250xx(41/40)^(3)` `=Rs. 6250xx(41xx41xx41)/(40xx40xx40)Rs. 3760.57` (approx) `:.` The compound interest `=Rs. (6730.57-6250)=Rs. 480.57` Hence the compound interest of Rs. 6250 in 9 months is Rs. 480.57 (approx). |
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| 21. |
Calculate in how many years will Rs. 300000 amount to Rs. 399300 at the rate of 10% compound interest per annum. |
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Answer» here principal `=Rs. 300000` Rate of compound interest `=10%` per annum let the required time be `t` years. `:.` The amount of Rs. 300000 after `t` years `=Rs. 300000xx(1+10/100)^(t)` `=Rs. 300000xx(1+1/10)^(t)` `=Rs. 300000xx(11/10)^(t)` As per questinon `300000xx(11/10)^(t)=399300` or `(11/10)^(t)=399300/30000` `=1331/1000=(11/10)^(3)` `:.(11/10)^(t)=(11/10)^(3)impliest=3` Hence the required time `=3` years. |
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| 22. |
Calculate the compound interest on Rs. 6000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months. |
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Answer» Here the principal is Rs. 6000 and the rate of compound inteest is 8% per annum. Since the interest is compounded at the interval of 6 months, the period of time is 2 and the time is 1 year. `:.` The required amount `=Rs. 6000xx(1+(8/2)/100)^(2xx1)` `=Rs. 6000xx(1+1/25)^(2)` `=Rs. 6000xx(26/25)^(2)` `=Rs. 6000xx26/25xx26/25=Rs. 6489.60` `:.` The compound interest `=Rs.(6489.40-6000)=Rs. 489.40` Hence the compound interest `=Rs. 489.40` |
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| 23. |
At the rate of compound interest of 8% per annum in how many years will the principles Rs. 40000 amount Rs. 46656? |
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Answer» Here the principal is Rs. 40000 The rate of compound interest `=8%` Let the required time be `t` years. `:.` The amount of Rs. 40000 in `t` years `=rs. 40000xx(1+8/100)^(t)` `=Rs. 40000xx(1+2/25)^(t)` `=Rs. 40000xx(27/25)^(t)` As per questio `40000xx(27/25)^(t)=46656` or `(27/25)^(t)=46656/40000` or `(27/25)^(t)=729/625=(27/25)^(2)impliest=2` Hence the required time `=2` years. |
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| 24. |
Calculate the amount on Rs. 5000 at the rate of 8% compound interest per annum for 3 years. |
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Answer» The required amount `=Rs. 5000 (1+8/100)^(3)=Rs. 5000(1+2/25)^(3)` `=Rs. 5000xx(27/25)^(3)=Rs. 6298.56` |
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| 25. |
If the rate of compound interest for the first and second year are 5% and 6% respectively, then find the compound interest on Rs. 5000 for 2 years. |
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Answer» The principal at the beginning `=Rs. 5000` The rate fo the compound interest `=5%` period of time `=1` year `:.` Amount after first year `=Rs. 5000(1+5/100)` ltbr. `=Rs. 5000(1+1/20)=Rs. 5000xx21/20=Rs. 5250` `:.` Principal at the end of the second year `=Rs. 5250` The rate of compound interest `=6%` Period of time `=1` year `:.` The amount after second year `=Rs. 5250 (1+6/100)` `=Rs. 5250xx106/100=Rs. 5565` `:.` Compound interest `=Rs.(5565-5000)` `=Rs. 565` `:.` The required compound interest of Rs. 5000 for 2 years is Rs. 565. |
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| 26. |
Gautambabu has taken a loan of Rs. 2000 for 2 years at the rate of compound interest fo 6% per annum. Find the compound interest he has to pay after 2 years. |
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Answer» Gautambabu has taken a loan of Rs. 2000. `:.` Here principal `(P)=Rs. 2000` Rate of compound interest per annum `=6%` Period of time `=2` years. `:.` After 2 years the amount `=Rs. 2000xx(1+6/100)^(2)` `=Rs. 2000xx(106/100)^(2)` `=Rs. (2000xx106xx106)/(100xx100)` `=Rs. 2247.2` `:.` The compound interest `=rs. (2247.2-2000)=Rs. 247.2` Hence Gautambabu has to pay compound interest Rs. 247.2 after 2 years. |
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| 27. |
A person took a loan of R. 25220 from a bank under this condition that the would repay the loan b three equal instalments. If the rate of compound interest of the bank be 5% per annum, then find the amount of money in each instalment. |
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Answer» Let each of the instalment be Rs. 100 and the principal of the first instalment be Rs. `x`. `:.x xx(1+5/100)=100impliesx xx((100+5)/100)=100` `impliesx xx105/100=100` `impliesx=(100xx100)/105=2000/21` Similarly the principal of the second instalment `=Rs. (100xx100)/105xx100/105=Rs.40000/441` And the principal of the third instanlment. `=Rs. (100xx100)/105xx100/105xx100/105=Rs. 800000/9261` `:.` The total principal `=Rs.(2000/21+40000/441+800000/9261)` `=Rs. 2000/21(1+20/21+400/441)` `=Rs.2000/21((441+420+400)/441)` `=Rs. 2000/21xx1261/441=Rs. 2522000/9261` `:.` If the principal be `Rs. 2522000/9261`, then the instalments is Rs. 100 `:.` If the principal be Rs. 1, then the instalment is `Rs.(100xx9261)/2522000` If the principal be Rs. 25220 then the instalment is `Rs.(100xx9261xx25220)/2522000=Rs.9261` Hence the amount of each instalment is Rs. 9261. |
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| 28. |
Weightof Sovanbabu is 80 kg. In order to reduce his weight, he started regular morning walk. He decided to reduce his weight every year by 10%. Calculaate the weight of Sovanbabu after 3 years. |
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Answer» The present weight of Sovanbabu`=80` kg `:.` The weight of Sovanbabu will be after 3 years `=80xx(1-10/100)^(3)` kg `=80xx(9/10)^(3)` kg `=80xx(9xx9xx9)/(10xx10xx10)kg=58.32kg` Hence after 3 years the required weight of Savanbabu will be 58.32 k. |
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| 29. |
As a result of Sarba Siksha Abhiyan the students leaving the school before completion, are re-admitted, so the students in a year is increased by 5% in comparison to its previous year. If the number of such re-admitted students in a district be 3528 in the present year. Find the number of students re -admitted 2 years before in this manner. |
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Answer» Let the number of re-admitted students be x. `:.` At the present year the number of students re-adimitted `=x xx(1+5/100)^(2)=x xx(1+1/20)^(2)` `=x xx(21/20)^(2)=(441x)/400` As per question `(441x)/400=3528` or `x=(3528xx400)/441` or `x=3200` Hence the required number of sudents re-admitted before 2 years `=3200` |
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| 30. |
A sum ofmoney lent at compound interest for 2 years at 20% per annum would fetch Rs482 more, if the interest was payable half-yearly than if it was payableannually. The sum is:(a) Rs10,000 (b) Rs 20,000 (c) Rs 40,000 (d) Rs 50,000 |
| Answer» As we know,`CI=P(1+R/200)^(2n) - P``= (1+10/100)^4 - x``=14641x-x``=14641x/10000-x = 4641/10000x``A = P(1+R/100)-P = x(1+20/100)^2-x``= x*36/25-x``=11/25x``H-A = 482``4641x/10000 - 11/25x = 482``4641x/10000-4400x/10000=482``x=482*10000/241`hence, the sum is eqal to`x= Rs. 20,000` | |
| 31. |
For the families having no electricity in their house, a Panchayat samity of village Bakultala accepted a plan to offer electric connections, 1200 families in these village have no electric connection in their houses. In comparison to its previous years, it is possible to arrange electricity every year for 75% of the families having no electricity. Find the number of families without electricity after 2 years. |
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Answer» At present the no. of families having no electric connections `=1200` Every year the electric connectiion increases `=75%` `:.` Every year the no of families having no electric connections decreases `=100%-75%=25%` `:.` After 2 years the number of families having no electric connections will be `1200xx(1-25/100)^(2)=1200xx(1-1/4)^(2)` `=1200xx(3/4)^(2)` `=1200xx(3xx3)/(4xx4)=675` Hence the number of families without electricity after 2 years will be 675. |
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| 32. |
Ay what rate of compound interest per annum Rs. 1000 will amount Rs. 13310 in 3 year? |
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Answer» Let the rate be r% per annum As per question `Rs. 10000(1+r/100)^(3)=Rs. 13310` or `(1+r/100)^(3)=13310/10000=1331/1000` or `(1+r/100)^(3)=(11/10)^(3)` `implies 1+r/100=11/10impliesr/100=11/10-1` r/`100=1/10` or `r=10` Hence the required rate `=10%` |
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| 33. |
In how much time rs. 6400 will amount Rs. 6561 at the rate of 5% compound interest when interest is compounded quaterly? |
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Answer» Let the required tiem `=n` years As per question `Rs. 6400(1+(5/4)/100)^(4n)=Rs. 6561` Or `(1+1/80)^(4n)=6561/6400` or `(81/80)^(4n)=6561/6400` or `(81/80)^(4n)=(81/80)^(2)` `implies4n=2impliesn=2/4=2/4xx12` months `=6` months Hence the required time `=6` months. |
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| 34. |
This year the number of blood donners in a hospital is 24000. If the number of blood donners increases by 5% in every 6 months, then in how much time will the number of blood donners be 27783? |
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Answer» This year the number of blood donners is 24000. The number of blood donners increases by 5% in every 6 months, i.e. 10% in every year. Let the required time be `t` years to reach the number of blood donners form 24000 to 27783. As per question `24000xx(1+(10/2)/100)^(2t)=27783` `implies24000xx(1+1/20)^(2t)=27783` `implies((20+1)/20)^(2t)=27783/24000` `implies(21/20)^(2t)=27783/24000=9261/8000=(21/20)^(3)` `implies(21/20)^(2t)=(21/20)^(3)implies2t=3impliest=3/2=1 1/2` Hence after `1 1/2` years the number of blood donners will be 27783. |
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| 35. |
The present populatiion of a town is 16000. If the rate of increase of population be 5% per annum, then what will be the population of the town after 2 years? |
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Answer» The present population `=16000` Rate of increase of population `=5%` per annum. Period of time `=2` years. `:.` The population of the town after 2 years `=16000xx(1+5/100)^(2)` `=16000xx(1+1/20)^(2)=16000xx(21/20)^(2)` `=16000xx(21xx21)/(20xx20)` `=17640` Hence the population of the two after 2 years `=17640` |
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| 36. |
The number of bicycles produced in the year 2012 and in the year 2015 are 80000 and 92610 respectively. Then find the rate of increase of production in percentage per annum. |
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Answer» The number of bicycles produced in 2012 is 80000. The number of bicycles produced in 2015 is 92610. Let the rate of increase of production in every year be r% and the time is 3 years. As per question `800xx(1+r/100)^(3)=92610` `(1+r/100)^(3)=92610/80000` `implies(1+r/100)^(3)=(21/20)^(3)` `implies1=r/100=21/20` `impliesr/100=21/20-1` `impliesr/100=(21-20)/20` `impliesr/100=1/20impliesr=5` Hence,the rate of increase of production of bi-cycles is 5% per annum. |
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| 37. |
Fill in the blanks: i. If the increase of any object occurs at the certain rate with respect to certain period of time. Then it is called………..increase. ii. If the decrease of any object occurs at a certain rate with respect to certain period of time, then it is called uniform......... iii. Compound interest `=............-` principal.iv. If the rate of compound interest decreases, then the quantity of compound interest also.......... ltbr. v. For the 1st year the simple and ocmpound interest are the............ |
| Answer» Uniform, decrease, Amount, decreases, Same | |
| 38. |
At the end of 2nd and 3rd yea, the compound interest of a sum of money is Rs. 880 and Rs. 968 respectively. What will be the rate of interest per annum? |
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Answer» Here the interest of Rs. 880 in 1 year `=(968-880)=Rs. 88`. Let the rate of interest be r% per annum. `:.880(1=+r/100)-880=88` or `880(1+r/100-1)=88` or `880xxr/100=88` or `r=(88xx100)/880=10` Hence the rate of interest is 10% per annum. |
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| 39. |
The value of a machien being decreased in `y` years at the rate of `r%` per year, becomes Rs. V. Find the value of the machine before `n` years. |
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Answer» Let the value of the machine before `n` years was Rs.`x` As per questin `x(1-r/100)^(n)=V` `impliesx=V((1-r/100)^(n))=V(1-r/100)^(-n)` Hence before `n` years the value of the machine was Rs. `V(1-r/100)^(-n)` |
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