3 Mark Question - Maths STD 8 Questions - Vidyadip

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Three years ago, Beeru purchased a buffalo from Surjeet for Rs. 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually?

Answer

Price of a buffalo (P) = Rs. 11000
Rate of interest (r) = 8% p.a
Period (n) = 3 years
Price of buffalo at present
(A) $=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }11000\times\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }11000\times\Big(\frac{11}{10}\Big)^3$
$=\text{Rs. }11000\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}$
$=\text{Rs. }14641$

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Question 23 Marks

A sum of money amounts to Rs. 10240 in 2 years at $6\frac{2}{3}\%$ per annum, compounded annually. Find the sum.

Answer

Amount (A) = Rs. 10240
Rate (r) $=6\frac{2}{3}\%=\frac{20}{3}\%\text{ p.a}$
Period (n) = 2 years
Let sum - P, then
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow10240=\text{p}\Big(1+\frac{20}{3\times100}\Big)^2$
$\Rightarrow10240=\text{p}\Big(1+\frac{1}{15}\Big)^2$
$\Rightarrow10240=\text{p}\times\Big(\frac{16}{15}\Big)^2$
$\therefore\text{P }=\text{Rs. }10240\times\Big(\frac{15}{16}\Big)^2$
$=\text{Rs. }10240\times\frac{15}{16}\times\frac{15}{16}$
$=\text{Rs. }9000$
$\therefore\text{Sum}=-\text{Rs. }9000$

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Question 33 Marks

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population?

Answer

3 years ago, the population was $=50000$ Rate of increase successively $\left(r_1, r_2, r_3\right)=4 \%, 5 \%$ and $3 \%$ p.a..
Period (n) = 3 years
Present Population
$=\text{P}\Big(1+\frac{\text{r}_1}{100}\Big)\Big(1+\frac{\text{r}_2}{100}\Big)\Big(1+\frac{\text{r}_3}{100}\Big)$
$=50000\Big(1+\frac{5}{100}\Big)\Big(1+\frac{4}{100}\Big)\Big(1+\frac{3}{100}\Big)$
$=50000\times\frac{21}{20}\times\frac{26}{25}\times\frac{103}{100}$
$=56238$

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Question 43 Marks

A car is purchased for Rs. 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?

Answer

Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year
$\therefore(\text{R}_2)=20\%$ p.a. for second year
$\therefore$ Value after 2 years
$=\text{P}\Big(1-\frac{\text{R}_1}{100}\Big)\Big(1-\frac{\text{R}_2}{100}\Big)$
$=\text{Rs. }348000\Big(1-\frac{10}{100}\Big)\Big(1-\frac{20}{100}\Big)$
$=\text{Rs. }348000\times\frac{9}{10}\times\frac{4}{5}$
$=\text{Rs. }250560$

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Question 53 Marks

By using the formula, find the amount and compound interest on.
Rs. 62500 for 2 years 6 months at 12% per annum compounded annually.

Answer

Principal (P) = Rs. 62500 Rate (R) = 12% p.a Period (n) = 2 years 6 months $\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }62500\Big(1+\frac{12}{100}\Big)^2\Big(1+\frac{12\times1}{100\times2}\Big)$ $=\text{Rs. }62500\times\Big(\frac{28}{25}\Big)^2\times\frac{53}{50}$ $=\text{Rs. }62500\times\frac{28}{25}\times\frac{28}{25}\times\frac{53}{50}$ $=\text{Rs. }83104$ $\therefore$ C.I = A - P = Rs. 83104 - Rs. 62500= Rs. 20604.

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Question 63 Marks

Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per annum, compounded annually.

Answer

Principal (p) = Rs. 2500
Rate (r) = 10% p.a
Period (t) = 2 years
Interest for the first year $=\frac{\text{prt}}{100}=\frac{2500\times10\times1}{100}$
$=\text{Rs. }250$
Amount at the end of the first year = Rs. (2500 + 250 ) = Rs. 2750
Principal for the second year = Rs. 2750
Interest for the second year $=\text{Rs. }=\frac{2500\times10\times1}{100}$
$=\text{Rs. }2750$
Amount at the end of the second year = Rs. 2750 + 275 = Rs. 3025
And compound interest for the 2 years = Rs. 3025 - 2500 = Rs. 525

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Question 73 Marks

The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011?

Answer

Population of a city in 2013 = 120000 Increase in next year = 6% and decrease in the following year = 5% Population in 2015 $=\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)^\text{1}\Big(1+\frac{\text{R}_2}{100}\Big)^\text{1}$ $=120000\Big(1+\frac{6}{100}\Big)\Big(1-\frac{5}{100}\Big)$ $=120000\times\frac{53}{50}\times\frac{19}{20}$$=120840$

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Question 83 Marks

In how many years will Rs. 6250 amount to Rs. 7290 at 8% per annum, compounded annually?

Answer

Principal (P) = Rs. 6250Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then
$\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$\frac{7290}{6250}=\Big(1+\frac{8}{100}\Big)^\text{n}$
$\Rightarrow\frac{729}{625}=\Big(\frac{27}{25}\Big)^\text{n}$
$\Rightarrow\Big(\frac{27}{25}\Big)^\text{n}=\Big(\frac{27}{25}\Big)^\text{n}$
Comparing, we get n = 2
$\therefore$ Period = 2 years.

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Question 93 Marks

At what rate per cent per annum will Rs. 640 amount to Rs. 774.40 in 2 years when compounded annually?

Answer

Principal (P) = Rs. 640
Amount (A) = Rs. 774.40
Period (n) = 2 years
Let r be the rate percent annum
We know that,
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{774.40}{640.00}=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow\frac{121}{100}=\Big(1+\frac{\text{r}}{100}\Big)^2$ (Dividing by 64)
$\Rightarrow\Big(\frac{11}{10}\Big)^2=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow1+\frac{\text{r}}{100}=\frac{11}{10}$
$\Rightarrow1+\frac{1}{100}\times100=10$
$\therefore$ Rate percent is 10% p.a

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Question 103 Marks

A scooter is bought at Rs. 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years?

Answer

Value of scooter (P) = Rs. 56000Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
$\therefore$ Value of scooter after 3 years
$=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }56000\Big(1-\frac{10}{100}\Big)^\text{3}$
$=\text{Rs. }56000\times\frac{9}{10}\times\frac{9}{10}\times\frac{9}{10}$
$=\text{Rs. }40824$

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Question 113 Marks

The simple interest on a sum of money for 2 years at 8% per annum is Rs. 2400. What will be the compound interest on that sum at the same rate and for the same period?

Answer

Simple interest (S.I) = Rs. 2400
Rate (R) = 8% p.a
Period (T) = 2 years
$\therefore\text{Sum}(\text{P})=\frac{\text{S.I}\times100}{\text{R}\times\text{T}}$
$=\frac{2400\times100}{8\times2}=\text{Rs. }15000$
$\therefore$ Amount on compound interest,
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^4=15000\Big(1+\frac{8}{100}\Big)^2$
$=\text{Rs. }15000\times\Big(\frac{27}{25}\Big)^2$
$=\text{Rs. }15000\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }17496$
$\therefore$ C.I = A - P = Rs. 17496 - 15000
= Rs. 2496

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Question 123 Marks

In how many years will Rs. 1800 amount to Rs. 2178 at 10% per annum when compounded annually?

Answer

Principal (P) = Rs. 1800Amount (A) = Rs. 2178
Rate (r) = 10% p.a.
Let n be the number of years.
We know that
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}=\frac{\text{A}}{\text{P}}$
$\Rightarrow\Big(1+\frac{10}{100}\Big)^\text{n}=\frac{1}{1800}$
$\Rightarrow\Big(\frac{11}{10}\Big)^\text{n}=\frac{121}{100}=\Big(\frac{11}{10}\Big)$(Dividing by 18)
$\therefore$ While comparing, we get n = 2
$\therefore$ Period = 2 years.

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Question 133 Marks

By using the formula, find the amount and compound interest on.
Rs. 31250 for 3 years at 8% per annum compounded annually.

Answer

Principal (P) = Rs. 31250 Rate (R) = 8% p.a Period (n) = 3 years $\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }31250\Big(1+\frac{8}{100}\Big)^3$ $=\text{Rs. }31250\times\Big(\frac{27}{25}\Big)^3$ $=\text{Rs. }31250\times\frac{27}{25}\times\frac{27}{25}\times\frac{27}{25}$ $=\text{Rs. }39366$ C.I = A - P = Rs. 39366 - Rs. 31250= Rs. 8116

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Question 143 Marks

By using the formula, find the amount and compound interest on.
Rs. 6000 for 2 years at 9% per annum compounded annually.

Answer

Principal (P) = Rs. 6000
Rate (R) = 9% p.a
Period (n) = 2 years
Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }6000\Big(1+\frac{9}{100}\Big)^2$
$=\text{Rs. }6000\times\frac{109}{100}\times\frac{109}{100}$
$=\text{Rs. }\frac{71286}{10}=\text{ Rs. }7128.60$
C.I = A - P = Rs. 7128.60 - Rs. 6000
= Rs. 1128.60

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Question 153 Marks

The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.

Answer

Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (2 n) = hours
$\therefore$ Bacteria after 2 hours $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=500000\Big(1+\frac{2}{100}\Big)^2$
$=500000\times\Big(\frac{51}{50}\Big)^2$
$=500000\times\frac{51}{50}\times\frac{51}{50}$
$=520200$

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Question 163 Marks

Find the amount of Rs 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.

Answer

Principal (P) = Rs. 8000
Period (n) = 2 years
Rate $(R_1)$ = 9% for the first years
$R_2$= 10% the second years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)^1$
$=\Big(1+\frac{\text{R}_2}{100}\Big)^1$
$=8000\Big(1+\frac{9}{100}\Big)\Big(1+\frac{10}{100}\Big)$
$=\text{Rs. }8000\times\frac{109}{100}\times\frac{110}{100}$
$=\text{Rs. }11253$

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Question 173 Marks

By using the formula, find the amount and compound interest on.
Rs. 10240 for 3 years at $12\frac{1}{2}\%$ per annum compounded annually.

Answer

Principal (P) = Rs. 10240 Rate (R) $12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a}$ Period (n) = 3 years $\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }10240\Big(1+\frac{25}{2\times100}\Big)^3$ $=\text{Rs. }10240\times\Big(\frac{9}{8}\Big)^3$ $=\text{Rs. }10240\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}$ $=\text{Rs. }14580$ $\therefore$ C.I = A - P = Rs. 14580 - Rs. 10240= Rs. 4340.

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Question 183 Marks

The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.

Answer

Growth of bacteria in a culture $(R_1)$ = 10% in first hourDecrease in next hour $(R_2)$ = 10%
Increase in the third hour $(R_3)$ = 10%
Bacteria in the beginning = 20000
Bacteria after 3 hours
$=\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)\Big(1+\frac{\text{R}_2}{100}\Big)\Big(1+\frac{\text{R}_3}{100}\Big)$
$20000\Big(1+\frac{10}{100}\Big)\Big(1-\frac{10}{100}\Big)\Big(1+\frac{10}{100}\Big)$
$20000\times\frac{11}{10}\times\frac{9}{10}\times\frac{11}{10}$
$=21780$

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Question 193 Marks

What sum of money will amount to Rs. 21296 in 3 years at 10% per annum, compounded annually?

Answer

Amount (A) = Rs. 21296
Rate (r) = 10% p.a
Period (n) = 3 years
Let P be the sum, then
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow21296=\text{P}\Big(1+\frac{10}{100}\Big)^3$
$\Rightarrow21296=\text{P}\Big(\frac{11}{10}\Big)^3$
$\text{P}=\text{Rs. }21296\times\Big(\frac{10}{11}\Big)^3$
$=21296\times\frac{10}{11}\times\frac{10}{11}\times\frac{10}{11}$
$=\text{Rs. }16000$
$\therefore\text{Sum}=\text{Rs. }16000$

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Question 203 Marks

Find the amount and the compound interest on Rs. 15625 for 3 years at 12% per annum, compounded annually.

Answer

Principal (P) = Rs. 15625
Rate (R) = 12% p.a
Period (n) = years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }15625\Big(1+\frac{12}{100}\Big)^3$
$=\text{Rs. }15625\times\Big(\frac{28}{25}\Big)^3$
$=\text{Rs. }15625\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25}$
$=\text{Rs. }21952$
C.I = A - P = Rs. 21952 - 15625 = Rs. 6327

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Question 213 Marks

Anand obtained a loan of Rs. 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount wil he have to pay after 3 years to clear the debt?

Answer

Principal (P) = Rs. 1,25,000
Rate of interest (r) = 8% p.a
Period (n) = 3 years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }125000\times\Big(1+\frac{8}{100}\Big)^3$
$=\text{Rs. }125000\times\Big(\frac{27}{25}\Big)^3$
$=\text{Rs. }125000\times\frac{27}{25}\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }157464$

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Question 223 Marks

The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years?

Answer

Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years
$\therefore$ Population after 3 years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=125000\Big(1+\frac{2}{100}\Big)^3$
$=125000\times\Big(\frac{51}{50}\Big)^3$
$=125000\times\frac{51}{50}\times\frac{51}{50}\times\frac{51}{50}$
$=132651$

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Question 233 Marks

At what rate percent per annum will Rs. 4000 amount to Rs. 4410 in 2 years when compounded annually?

Answer

Amount (A) = Rs. 4000
Principal (P) = Rs. 4410
Period (n) = 2 years
Let r be the rate percent per annum
We know that,
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{4410}{4000}=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow\Big(\frac{21}{20}\Big)^2=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow1+\frac{\text{r}}{100}=\frac{21}{20}$
$\Rightarrow\frac{\text{r}}{100}=\frac{21}{20}-1=\frac{1}{20}$
$\Rightarrow\text{r}=\frac{1}{20}\times100=5$
$\therefore\text{Rate}=5\%\text{ p.a}$

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Question 243 Marks

By using the formula, find the amount and compound interest on.
Rs. 9000 for 2 years 4 months at 10% per annum compounded annually.

Answer

Principal (P) = Rs. 9000 Rate (R) = 10% p.a Period (n) = 2 years 4 months $=2\frac{1}{3}\text{ years}$ $\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }9000\Big(1+\frac{10}{100}\Big)^2\Big(1+\frac{10\times1}{100\times3}\Big)$ $=\text{Rs. }9000\times\frac{11}{10}\times\frac{11}{10}\times\frac{31}{30}$ $=\text{Rs. }11253$ $\therefore$ C.I = A - P = Rs. 11253 - Rs. 9000= Rs. 2253

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Question 253 Marks

Shubhalaxmi took a loan of Rs. 18000 from surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and $12\frac{1}{2}\%$ per annum during the second year, how much will she have to pay after 2 years?

Answer

Amount of loan taken (p) = Rs. 18000
Rate $(R_1)$ = 12% p.a during first year
$R_2$ = $12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a}$ during second year
Period (n) = 2 years
$\therefore$ Total amount (A) $=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{p}\Big(1+\frac{\text{R}_1}{100}\Big)^1\Big(1+\frac{\text{R}_2}{100}\Big)^1$
$=\text{Rs. }18000\times\Big(1+\frac{12}{100}\Big)\Big(1+\frac{25}{2\times100}\Big)$
$=\text{Rs. }18000\times\frac{28}{25}\times\frac{9}{8}$
$=\text{Rs. }22680$

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Question 263 Marks

Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Answer

Amount borrowed from Bank (p) = Rs. 24000
Rate (R) = 10% p.a
Period (n) = 2 years 3 mounths
$=2\frac{1}{4}\text{years}$
$\therefore$ Amount after the period (A) $=\text{p}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }24000\Big(1+\frac{10}{100}\Big)^2\Big(14+\frac{10\times1}{100\times4}\Big)$
$=\text{Rs. }24000\times\Big(\frac{11}{10}\Big)^2\times\Big(\frac{41}{40}\Big)$
$=\text{Rs. }29766\times\frac{11}{10}\times\frac{11}{10}\times\frac{41}{40}$
$=\text{Rs. }29766$

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Question 273 Marks

By using the formula, find the amount and compound interest on.
Rs. 10000 for 2 years at 11% per annum compounded annually.

Answer

Principal (P) = Rs. 10000 Rate (R) = 11% p.a Period (n) = 2 years $\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=10000\Big(1+\frac{11}{100}\Big)^2$$=\text{Rs. }10000\times\Big(\frac{111}{100}\Big)^2$
$=\text{Rs. }10000\times\frac{111}{100}\times\frac{111}{100}$ $=\text{Rs. }12321$ C.I = A - P = Rs. 12321 - Rs. 10000 = Rs. 2321.

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Question 283 Marks

The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased?

Answer

Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
Present value (A) = Rs. 291600
Value of machine 3 years ago
$(\text{P})=\text{A}\div\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }291600\div\Big(1-\frac{10}{100}\Big)^\text{3}$
$=\text{Rs. }291600\div\Big(\frac{9}{10}\Big)^3$
$=\text{Rs. }291600\times\frac{10}{9}\times\frac{10}{9}\times\frac{10}{9}$
$=\text{Rs. }400000$

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Question 293 Marks

A machine is purchased for Rs. 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?

Answer

Value of machine (P) = Rs. 625000 Rate of depreciation (R) = 8% p.a. Period (n) = 2 years$\therefore$ Value after 2 years $=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }625000\Big(1-\frac{8}{100}\Big)^\text{2}$
$=\text{Rs. }625000\Big(\frac{23}{25}\Big)^\text{2}$
$=\text{Rs. }625000\times\frac{23}{25}\times\frac{23}{25}$
$=\text{Rs. }529000$

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