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Maths 4 Mark Question

Compound Interest · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 21 questions.

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Question 14 Marks
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.
Answer
We know that:
$\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
$\therefore\ \text{P}=\frac{1200\times100}{5\times3}$
$=8,000$
Now,
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=80,000\Big(1+\frac{5}{100}\Big)^{3}$
$=8,000(1.05)^{3}$
$=9,261$
Now,
CI = A - P
= 9,261 - 8,000
= 1,261
Thus, the required compound interest is Rs. 1,261.
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Question 24 Marks
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs. 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Answer
Given:
P = Rs. 64,600
R = 5% p.a.
n = 1.5 years
When the interest is compounded half−yearly, we have:
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }64,000\Big(1+\frac{5}{200}\Big)^{3}$
$=\text{Rs. }64,000(1.025)^3$
$=\text{Rs. }68,921$
Also,
CI = A - P
= Rs. 68,921 - Rs. 64,000
= Rs. 4,921
Thus, the required interest is Rs. 4,921.
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Question 34 Marks
Amit borrowed Rs. 16000 at $17\frac{1}{2}\%$ per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually what dose he gain at the end of 2 years
Answer
Amount to be paid by Amit:
$\text{SI}=\frac{\text{PRT}}{100}$
$=\frac{16000\times17.5\times2}{100}$
$=\text{Rs. }5,600$
Amount gained by Amit:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }16,000\Big(1+\frac{17.5}{100}\Big)^2$
$=\text{Rs. }16,000(1.175)^2$
$=\text{Rs. }22,090$
we know that:
Cl = A - P
= Rs. 22,090 - Rs. 16,000
= Rs. 6090
Amit's gain in the whole transaction = Rs. 6,090 - Rs. 5,600
= Rs. 490
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Question 44 Marks
Find the amount and the compound interest on Rs. 8000 for $1\frac{1}{2}$years at 10% per annum, compounded half-yearly.
Answer
Given:
P = Rs. 8,000
R = 10% P.a.
n = 1.5 years
When compounded half−yearly, we have:
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }8,000\Big(1+\frac{10}{200}\Big)^{3}$
$=\text{Rs. }8,000(1.05)^3$
$=\text{Rs. }9,261$
Also,
CI = A - P
= Rs. 9,261 - Rs. 8,000
= Rs. 1,261
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Question 54 Marks
Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded quarterly.
Answer
Given:
P = Rs. 15,625
R = 16% p.a. $=\frac{16}{4}=4\%$ quarterly
n = 9 months = 3 quarters
We know that:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }15,625\Big(1+\frac{4}{100}\Big)^{3}$
$=\text{Rs. }15,625(1.04)^{3}$
$=\text{Rs. }17,576$
Also,
CI = A - P
= Rs. 17,576 - Rs. 15,625
= Rs. 1,951
Thus, the required compound interest is Rs. 1,951.
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Question 64 Marks
The difference in simple interest and compound interest on a certain sum of money at $6\frac{2}{3}\%$ per annum for 3 years is Rs. 46 Determine the sum.
Answer
Given:
$\text{CI}-\text{SI}=46$
$\text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Big]-\frac{\text{PRT}}{100}=46$
$\text{P}\Big[\Big(1+\frac{20}{300}\Big)^{3}-1\Big]-\frac{\text{P}\times20\times3}{3\times100}=46$
$\frac{4,096}{3,375}\text{ P}-\frac{\text{P}}{5}-\text{P}=46$
$\frac{(4,096-3,375-675)\text{P}}{3,375}=46$
$\text{P}=46\times\frac{3,375}{46}$
$=3,375$
Thus, the required sum is Rs. 3,375.
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Question 74 Marks
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Answer
  1. Population of the city in 2001 $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$=6760000\Big(1+\frac{4}{100}\Big)^{2}$

$=6760000(1.04)^{2}$

$=7311616$

Thus, Population of the city in 2001 is 7311616.
  1. Population of the city in 1997 $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{-2}$
$=6760000\Big(1+\frac{4}{100}\Big)^{-2}$

$=6760000(1.04)^{-2}$

$=6250000$

Thus, Population of the city in 1997 is 6250000.
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Question 84 Marks
Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs. 210 as compound interest, but paid Rs. 200 only as simple interest. Find the sum and the rate of interest.
Answer
Let the sum be Rs. P and the rate of interest be R%.
We know that Kamla paid Rs. 200 as simple interest.
$\therefore\ 200=\frac{\text{PR(2)}}{100}$
$\text{PR}=10,000...(1)$
Also, Kamla received Rs. 210 as compound interest.
$\therefore\ 210=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{2}-1$
$210(10,000)=\text{P}(\text{R}^{2}+200\text{R})$
$210\text{ R}=\text{ R}^{2}+200\text{R}$ [from (1)]
R = 10% p.a.
Putting the equation in (1), we get:
P = 1,000
Thus, the required sum is Rs. 1,000 and the rate of interest is 10%.
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Question 94 Marks
Find the compound interest on Rs. 1000 at the rate of 8% per annum for $1\frac{1}{2}$ years when interest is compounded half-yearly.
Answer
Given:
P = Rs. 1,000
R = 8%p.a.
n = 1.5 years
We know that:
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{2\text{n}}$
$=1,000\Big(1+\frac{8}{200}\Big)^{3}$
$=1,000(1.04)^{3}$
$=\text{Rs. }1,124.86$
Now,
CI = A - P
= Rs. 1,124.86 - Rs. 1,000
= Rs. 124.86.
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Question 104 Marks
The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs. 163.20.
Answer
Let the time period be n years.
Then, we have:
$\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$163.20=2,000\Big(1+\frac{4}{100}\Big)^{\text{n}}-2,000$
$2,163.20=2,000(1.04)^{\text{n}}$
$(1.04)^{\text{n}}=\frac{2,163.20}{2,000}$
$(1.04)^{\text{n}}=1.0816$
$(1.04)^{\text{n}}=(1.04)^{2}$
On comparing both the sides, we get:
n = 2
Thus, the required time is two years.
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Question 114 Marks
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?
Answer
Given:
$R_1=10 \%$
$R_2=-8 \%$
$R_3=12 \%$
$P=\text { Original count of bacteria }=13,125,000$
We know that:
$\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$\therefore$ Bacteria count after three hours:
$=13,125,000\Big(1+\frac{10}{100}\Big)\Big(1-\frac{8}{100}\Big)\Big(1+\frac{12}{100}\Big)$
= 13,125,000(1.10)(0.92)(1.12)
= 14,876,400
Thus, the bacteria count after three hours will be 14,876,400.
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Question 124 Marks
Jitendra set up a factory by investing Rs. 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, compute his total profit.
Answer
Profit at the end of the first year $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)$
$=2,500,000\Big(1+\frac{5}{100}\Big)$
$=2,500,000(1.05)$
$=2,625,000$
Profit at the end of the second year $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)$
$=2,625,000\Big(1+\frac{10}{100}\Big)$
$=2,625,000(1.10)$
$=2,887,500$
Total profit = Rs. 2,887,500 − Rs. 2,500,000
= Rs. 387,500.
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Question 134 Marks
At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs.410 in 2 years?
Answer
Let the rate percent be R.
We know that:
$\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$410=4,000\Big(1+\frac{\text{R}}{100}\Big)^{2}-4,000$
$4,410=4,000\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=\frac{4,410}{4,000}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=1.1025$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=(1.05)^{2}$
$1+\frac{\text{R}}{100}=0.05$
$\frac{\text{R}}{100}=0.05$
$\text{R}=5$
Thus, the required rate percent is 5.
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Question 144 Marks
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Answer
Here,
$P=\text { Initial population }=50,000$
$R_1=4 \%$
$R_2=5 \%$
$R_3=3 \%$
$n=\text { Number of years }=3$
$\therefore$ Population after three years:
$=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$=50,000\Big(1+\frac{4}{100}\Big)\Big(1+\frac{5}{100}\Big)\Big(1+\frac{3}{100}\Big)$
$=50,000(1.04)(1.05)(1.03)$
$=56,238$
Hence, the population after three years will be 56,238.
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Question 154 Marks
Find the amount that David would receive if he invests Rs. 8192 for 18 months at $12\frac{1}{2}\%$ per annum, the interest being compounded half-yearly.
Answer
Given:
P = Rs. 8,192
R = 12.5% p.a.
n = 1.5 years
When the interest is compounded half−yearly,
we have:
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }8,192\Big(1+\frac{12.5}{200}\Big)^{3}$
$=\text{Rs. }8,192(1.0625)^{3}$
$=\text{Rs. }9,826$
Thus, the required amount is Rs. 9,826.
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Question 164 Marks
Find the amount of Rs. 4096 for 18 months at $12\frac{1}{2}\%$ per annum, the interest being compounded semi-annually
Answer
Given:
P = Rs. 4,096
R = 12.5%p.a.
n = 18 months = 1.5 years
We have:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
When the interest is compounded semi-annually, we have:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{2\text{n}}$
$=\text{Rs. }4,096\Big(1+\frac{12.5}{200}\Big)^3$
$=\text{Rs. }4,096(1.0625)^3$
$=\text{Rs. }4,913$
Thus, the required amount is Rs. 4,913.
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Question 174 Marks
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year?
Answer
Number of workers = 6,400
At the end of the first year, 25% of the workers were retrenched.
$\therefore$ 25% of 6,400 = 1,600
Number of workers at the end of the first year = 6,400 − 1,600 = 4,800
At the end of the second year, 25% of those working were retrenched.
$\therefore$ 25% of 4,800 = 1,200
Number of workers at the end of the second year = 4,800 − 1,200 = 3,600
At the end of the third year, 25% of those working increased.
$\therefore$ 25% of 3,600 = 900
Number of workers at the end of the third year = 3,600 + 900 = 4,500
Thus, the number of workers during the fourth year was 4,500.
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Question 184 Marks
At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?
Answer
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$1102.50=1000\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$\frac{1102.50}{1000}=(1+0.01{\text{ R}})^{2}$
$(1+0.01{\text{ R}})^{2}=1.1025$
$(1+0.01{\text{ R}})^{2}=(1.05)^{2}$
On comparing both the sides, we get:
1 + 0.01R = 1.05
n = 5
Thus, the required rate percent is 5.
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Question 194 Marks
Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs. 12000 as simple interest.
Answer
$\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
According to the given values, we have:
$=\frac{12,000\times100}{5\times3}$
$=80,000$
The principal is to be compounded annually.
So,
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=80,000\Big(1+\frac{5}{100}\Big)^{3}$
$=80,000(1.05)^{3}$
$=92,610$
Now,
CI = A - P
= 92,610 - 80,000
= 12,610
Thus, the required compound interest is Rs. 12,610.
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Question 204 Marks
Simple interest on a sum of money for 2 years at $6\frac{1}{2}\%$ per annum is Rs. 5200. What will be the compound interest on the sum at the same rate for the same period?
Answer
$\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
$\therefore\ \text{P}=\frac{5,200\times100}{6.5\times2}$
$=40,000$
Now,
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=40,000\Big(1+\frac{6.5}{100}\Big)^{2}$
$=40,000(1.065)^{2}$
$=45,369$
Also,
CI = A - P
= 45,369 - 40,000
= 5,369
Thus, the required compound interest is Rs. 5,369.
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Question 214 Marks
Rekha deposited Rs. 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Answer
Given:
P = Rs. 16,000
R = 20% p.a.
n = 1 year
We know that:
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
When compounded quarterly, we have:
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{400}\Big)^{\text{4n}}$
$=\text{Rs. }16,000\Big(1+\frac{20}{400}\Big)^{4}$
$=\text{Rs. }16,000(1.05)^{4}$
$=\text{Rs. }19,448.10$
Also,
CI = A - P
= Rs. 19,448.1 - Rs. 16,000
= Rs. 3,448.10
Thus, the interest received by Rekha after one year is Rs. 3,448.10.
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