First half of Exercise 4(a)…

Question 26: The compound interest on a sum of money for $2$ years is $Rs. \ 1050$ and the simple interest on the same sum for the same period and at the same rate is $Rs. \ 1000$. Find: (i) the rate of interest (ii) the sum.

Compound Interest:

Principal $= P$, Rate of Interest $= r\%$ per annum, $T = 2 \ years$

$A = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$

Compound Interest $1050 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2 - P$

$\Rightarrow 1050 = P \Big[ (1+$ $\frac{r}{100})^2$ $- 1 \Big]$

$\Rightarrow 1050 = P (2+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ $)$ … … … … … (i)

Simple Interest:

Principal $= P$, Rate of Interest $: R=r\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$1000 =$ $\frac{P \times r \times 2}{100}$ $\Rightarrow 50000 = Pr$

$\Rightarrow P =$ $\frac{50000}{r}$ … … … … … (ii)

Now solving (i) and (ii) we get

$1050 =$ $\frac{50000}{r}$ $(2+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ \$latex )

$\Rightarrow 1050 = 1000 + 5r$

$\Rightarrow r = 10\%$

$\Rightarrow P =$ $\frac{50000}{10}$ $= 5000 \ Rs.$

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Question 27: Find the rate percent per annum if $Rs. \ 2000$ amounts to $Rs. \ 2662$ in $1\frac{1}{2}$ years, interest being compounded half-yearly.

Principal $= 2000 \ Rs.$, Rate of Interest $: R=r\%,$ semi-annually, $T= 1.5 \ years$ year  $Amount = 2662 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$26620 = 2000 \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.331 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.10 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 20\%$ per annum

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Question 28: Find the rate at which a sum of money will double itself in $3$ years if the interest is compounded annually.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 3$ year  $Amount = 2x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$2x = x \Big( 1+$ $\frac{r}{100 }$ $\Big)^{3}$

$2 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^3$

$1.2599 = 1 +$ $\frac{r}{100}$ $\Rightarrow r = 25.99\%$ per annum

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Question 29: Find the rate at which a sum of money will become four amount times the original in $2$ years if the interest is compounded half yearly.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 2$ year  $Amount = 4x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$4x = x \Big( 1+$ $\frac{r}{100 \times 2 }$ $\Big)^{2 \times 2}$

$4 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^4$

$1.4142 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 82.84\%$ per annum

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Question 30: A sum compounded annually becomes $\frac{25}{16}$ times of itself in $2$ years. Determine the rate of interest.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 2$ year  $Amount = \frac{25}{16}x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$\frac{25}{16}$ $x = x \Big( 1+$ $\frac{r}{100}$ $\Big)^{2}$

$\frac{25}{16}$ $= \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^2$

$\frac{5}{4}$ $= 1 +$ $\frac{r}{100}$ $\Rightarrow r = 25\%$ per annum

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Question 31: Rishi invested $Rs. \ 30000$ in a finance company and received $Rs. \ 39930$ after $1\frac{1}{2}$ years. Find the rate of interest per annum compounded semi-annually.

Principal $= 30000 \ Rs.$, Rate of Interest $: R=r\%,$ semi-annually, $T= 1.5 \ years$ year  $Amount = 39930 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$39930 = 30000 \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.331 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.10 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 20\%$ per annum

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Question 32: In how much time would $Rs. \ 5000$ amounts to $Rs. \ 6655$ at $10\%$ per annum compound interest?

Principal $= 5000 \ Rs.$, Rate of Interest $: R=10\%,$ annually, $T= n \ years$ year  $Amount = 6655 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$6655 = 5000 \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$1.331 = \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$1.1^3 = \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$\Rightarrow n = 3 \ years$

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Question 33: In what time will $Rs. \ 4400$ become $Rs. \ 4576$ at $8\%$ per annum interest compounded half-yearly?

Principal $= 4400 \ Rs.$, Rate of Interest $: R=8\%,$ annually, $T= n \ years$ year  $Amount = 4576 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$4576 = 4400 \Big(1+$ $\frac{8}{100 \times 2}$ $\Big)^{n \times 2}$

$1.04 = \Big( 1+$ $\frac{4}{100}$ $\Big)^{2n}$

$1.04 = ( 1.04)^{2n}$

$\Rightarrow 2n = 1 \ year \Rightarrow n =$ $\frac{1}{2}$ $\ year$

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Question 34: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 1320$ and for the third year is $Rs. \ 1452$. Calculate the rate of interest and the original sum of money.

$1^{st} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_1$

$A_1 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^1$ … … … … (i)

$2^{nd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_2$

$A_2 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ … … … … (ii)

$3^{rd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_3$

$A_3 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ … … … … (iii)

Given: $A_2 - A_1 = 1320$  and $A_3 - A_2 = 1452$

Therefore

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 1320$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ $) = 1320$ … … … … (iv)

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 1452$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)^2($ $\frac{r}{100}$ $) = 1452$ … … … … (v)

Dividing (v) by (iv) we get

$\frac{1452}{1320}$ $= 1 +$ $\frac{r}{100}$

$\Rightarrow r = 1.1 - 1 = 10\%$

Substituting in (iv) $P =$ $\frac{1320 \times 100}{10 \times 1.1}$ $= 12000 \ Rs.$

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Question 35: On what sum of money will the difference between the compound interest and simple interest for $2$ years be equal to $Rs. \ 25$, if the rate of interest charged for both is $5\%$ per annum?

Compound Interest:

Principal $= P$, Rate of Interest $= 5\%$ per annum, $T = 2 \ years$

$A = P \Big( 1+$ $\frac{5}{100}$ $\Big)^2$

$A = P \Big( 1+$ $\frac{5}{100}$ $\Big)^2 = 1.1025 P$

Compound Interest $= 1.1025 P - P = 0.1025P$

Simple Interest:

Principal $= P$, Rate of Interest $: R=5\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$=$ $\frac{P \times 5 \times 2}{100}$ $\Rightarrow = 0.1P$

Given: $25 = 0.1025 P - 0.1P \Rightarrow P = 10000 \ Rs.$

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Question 36: Mr Kumar borrowed $Rs. \ 15000$ for two years. The rate of interest for the two successive years are $8\%$ and $10\%$ respectively. If he repays $Rs. \ 6200$ at the end of the first year, find the outstanding amount at the end of second year.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 15000$, Rate of Interest $= 8\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{15000 \times 8 \times 1}{100}$ $= 1200 \ Rs.$

Therefore, the Amount at the end of $1^{st} year =15000 + 1200 = 16200 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 16200 - 6200 = 10000$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{10000 \times 10 \times 1}{100}$ $=1000 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 10000 + 1000 = 11000 \ Rs.$

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Depreciation Problems

Question 37: The value of a machine depreciates at the rate of $10\%$ per annum. what will be its value $2$ years hence if the present value is $Rs. \ 100000$? Also, find the total depreciation during this period.

Present Value $V_0 = 100000 \ Rs.$  Rate of Depreciation $= 10\%$ No of Years $= 2$

$V_2 = V_0 \Big( 1-$ $\frac{R}{100}$ $\Big)^n$

$V_2 = 100000 \Big( 1-$ $\frac{10}{100}$ $\Big)^2$

$V_2= 81000 \ Rs.$

Total depreciation $= 100000 - 81000 = 19000 \ Rs.$

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Question 38: Pritam bought a plot of land for $Rs. \ 640000$. Its value is increasing by $5\%$ of its previous value after every six months. What will be the value of the plot after $2$ years?

Present Value $= 640000 \ Rs.$, Rate of Interest $= 5\%$ half yearly , $T = 2 \ years$

$V_2 = P \Big( 1+$ $\frac{r}{100k}$ $\Big)^{nk}$

$V_2 = 640000 \Big( 1+$ $\frac{10}{100 \times 2}$ $\Big)^{2 \times 2}$

$V_2 = 777924 \ Rs.$

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Question 39: The value of a machine depreciates at the rate of $10\%$ per annum. It was purchased $3$ years ago. If its present value is $Rs. \ 43740$, find its purchase price.

Present Value $V_0 = 43740 \ Rs.$  Rate of Depreciation $= 10\%$ No of Years $= 3$ Value 3 years back  $V_{-3} = x$

$V_0 = V_{-3} \Big( 1-$ $\frac{R}{100}$ $\Big)^3$

$43740 = V_{-3} \Big( 1-$ $\frac{10}{100}$ $\Big)^3$

$V_{-3} = 60000 \ Rs.$

Total depreciation $= 60000 - 43740 = 16260 \ Rs.$

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Question 40: The cost of a T.V. set was quoted $Rs. \ 17000$ at the beginning of 1999. In the beginning of 2000 the price was hiked by $5\%$. Because of decrease in demand the cost was reduced by $4\%$ in the beginning of 2001. What was the cost of the T.V. set in 2001?

Cost of TV at the beginning of 1999 $= 17000 \ Rs.$

Cost of TV at the beginning of 2000 $= 17000 \times 1.05 = 17850 \ Rs.$

Cost of TV at the beginning of 2001 $= 17850 \times 0.96 = 17136 \ Rs.$

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Question 41: Ashish staffed the business with an initial investment of $Rs. \ 500000$. In the first he incurred a loss of $4\%$. However, during the second year he earned a profit of $5\%$ which in third year rose to $10\%$. Calculate the net profit for the entire period of $3$ years.

Initial Investment at the beginning of 1st Year $= 500000 \ Rs.$

Capital at the beginning of 2nd Year $= 500000 \times 0.96 = 480000 \ Rs.$

Capital at the beginning of 3rd Year $= 480000 \times 1.05 = 504000 \ Rs.$

Capital at the beginning of 4th Year $= 504000 \times 1.10 = 554400 \ Rs.$

Net profit $= 554400 - 500000 = 54400 \ Rs.$

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Population Questions

Question 42: The present population of a town is $28000$. If it increases at the rate of $5\%$ per annum, what will be its population after $2$ years?

Population $P = 28000$, Rate $R = 5\%$ per annum, $n = 2$ years

Population after $n$ years $= P \Big(1+$ $\frac{R}{100}$ $\Big)^{n}$

Population after $2$ years $= 28000 \Big(1+$ $\frac{5}{100}$ $\Big)^{2} = 30870$

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Question 43: The present population of a town is $25000$. It grows at first $4\%, 5\%$ and $8\%$ during year, second year and third year respectively. Find lts population after $3$ years.

Current Population $P = 25000$, Rate $R_1 = 4\%, R_2=5%, R_3 = 8\%$ per annum, $n = 3$ years

Population after $3^{rd} \ Year = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Population after $3^{rd} \ Year = 25000 \Big(1+$ $\frac{4}{100}$ $\Big).\Big(1+$ $\frac{5}{100}$ $\Big).\Big(1+$ $\frac{8}{100}$ $\Big)$

$= 25000 \times 1.04 \times 1.05 \times 1.08 = 29484$

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Question 44: There is a continuous growth in population of a village at the rate of $5\%$ per annum. If its present population is $9261$, what was it $3$ years ago?

Population 3 years ago $= P_{-3}$ Current Population $P_0 = 9261$, Rate $R = 5\%$ per annum, $n = 3$ years

Therefore $P_{0} = P_{-3} \Big(1+$ $\frac{R}{100}$ $\Big)^{n}$

$9261 =P_{-3} \Big(1+$ $\frac{5}{100}$ $\Big)^{3}$

$\Rightarrow P_{-3} = 8000$

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Question 45: In a factory the production of scooters rose to $46305$ from $40000$ in $3$ years. Find the annual rate of growth of the production of scooters.

$P = 40000 \ A = 46305 \ n = 3 \ years, Rate = R\%$

$46305 = 40000 \Big(1+$ $\frac{R}{100}$ $\Big)^{3}$

$1.05 = 1 +$ $\frac{R}{100}$

$\Rightarrow R = 5\%$

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Question 46: The population of a town increases at the rate of $50$ per thousand. Its population after $2$ years will be $22050$. Find its present population.

Rate of Increase $= 5% \ \ A$ (population in 2 years) $= 22050, \ P$ (Initial population) $= P$

$22050 = P \Big(1+$ $\frac{5}{100}$ $\Big)^{2}$

$\Rightarrow P = 20000$

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Question 47: The count of bacteria in a culture grows by $10\%$ in the first hour, decreases by $8\%$ in the second hour and again increase by $12\%$ in third hour. If the count of bacteria in the sample $13125000$, what will be the count of bacteria after $3$ hour.

Current bacteria Population $P = 13125000$, Rate $R_1 = 10\%, R_2=-8%, R_3 = 12\%$ per annum, $n = 3$ hours

Bacteria Population after $3^{rd} \ hours = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Bacteria Population after $3^{rd} \ hours = 13125000 \Big(1+$ $\frac{10}{100}$ $\Big).\Big(1+$ $\frac{-8}{100}$ $\Big).\Big(1+$ $\frac{12}{100}$ $\Big)$

$= 13125000 \times 1.10 \times 0.92 \times 1.12 = 14876400$

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Question 48: $6400$ workers were employed to construct a river bridge in four years. At the end First year, $25\%$ workers were retrenched. At 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 $25\%$ the end of the third year. How many workers were working during the fourth year?

Current worker Population $P = 6400$, Rate $R_1 = -25\%, R_2=-25%, R_3 = 25\%$ per annum, $n = 3$ years

Worker Population after $3^{rd} \ years = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Worker Population after $3^{rd} \ years = 6400 \Big(1+$ $\frac{-25}{100}$ $\Big).\Big(1+$ $\frac{-25}{100}$ $\Big).\Big(1+$ $\frac{25}{100}$ $\Big)$

$= 64000 \times 0.75 \times 0.75 \times 1.25 = 4500$

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Question 49: A man started a factory with an initial investment of $Rs. \ 100000$. In the first year, he incurred a loss of $5\%$. However, during the second year; he earned a profit of $10\%$ which in the third year rose to $12\%$. Calculate the profit for the entire period of three years.

Initial Investment $P = 100000$, Rate $R_1 = -5\%, R_2=10%, R_3 = 12\%$ per annum, $n = 3$ years

Worker Population after $3^{rd} \ years = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Worker Population after $3^{rd} \ years = 100000 \Big(1+$ $\frac{-5}{100}$ $\Big).\Big(1+$ $\frac{10}{100}$ $\Big).\Big(1+$ $\frac{12}{100}$ $\Big)$

$= 100000 \times 0.95 \times 1.10 \times 1.12 = 117040 \ Rs.$

Therefore net profit $= 117040 - 100000 = 17040 \ Rs.$

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Question 50: 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.

$P (1999) = 6760000 \ , Rate = 4\%$
$P_{2001} = 6760000 \Big(1+$ $\frac{4}{100}$ $\Big)^{2} = 7311616$
$6760000= P_{1997} \Big(1+$ $\frac{4}{100}$ $\Big)^{2}$
$\Rightarrow P_{1997} = 6250000$
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