Comparing Quantities

1.0Introduction

$1\mathrm{kg}=1000\mathrm{gms}$ $\therefore 3\mathrm{kg}=3\times 1000$ $=3000\mathrm{gms}$ $3000\mathrm{gms}>2700\mathrm{gms}$ $\therefore 3\mathrm{kg}>2700\mathrm{gms}$ So, weight of bananas is greater than the weight of apples. "We are already familiar with ratio and percentage. In a simple case the profit or loss equals the increase or decrease in the company's assets."

In this chapter we will deal with different topics like

• Percentage
• Profit and Loss
• Discount
• VAT/ST
• Simple Interest
• Compound Interest

2.0Percentage

The word percent is an abbreviation of the Latin phrase / word 'per centum' which means per hundred or hundredths. Thus, the word 'percentage' literally means 'per hundred or every hundred'. Therefore, whenever we calculate something as a part of 100 that part is numerically termed as 'percentage'. Symbol : In short 'percentage' is written as P.C. and symbolically it is denoted as $\%$.

Conversion of percentage into fraction

To express $\mathrm{a}\%$ as a fraction, divide ' a ' by 100 i.e., a $\%=\frac{\mathrm{a}}{100}$ Ex. We have, $35\%=\frac{35}{100}$ Thus, a fraction with its denominator 100 is equal to that percent as in the numerator. So, $8\%=\frac{8}{100},12\%=\frac{12}{100},60\%=\frac{60}{100}=\frac{3}{5}=3:5$.

Conversion of a fraction into percentage

To convert a fraction into a percent, we multiply the fraction by 100 and put the percent sign $\%$. Ex. Thus $\frac{4}{5}=\left(\frac{4}{5}\times 100\right)\%=80\%,\frac{9}{20}=\left(\frac{9}{20}\times 100\right)\%=45\%$.

Conversion of a percentage into decimal form

To convert a given percent in decimal form, we express it as a fraction with denominator as 100 and then the fraction is written in decimal form. Ex. $65\%=\frac{65}{100}=0.65,7.4\%=\frac{7.4}{100}=0.074$.

Conversion of a decimals into percentage

In order to convert a given decimal into a percent, we move the decimal point on the right side by two digits and put the percent sign $\%$. Ex. $0.122=12.2\%,0.275=27.5\%,0.037=3.7\%$.

To find the value of a certain percent of any quantity

To find the value of a certain percent of a given quantity, we first convert the percent into a fraction and then multiply the result with the given quantity. i.e., $\mathrm{P}\%$ of $\mathrm{x}=\frac{\mathrm{P}}{100}\times \mathrm{x}$

• Q. (i) Find $12\%$ of ₹ $1200$. Explanation : (i) $12\%$ of ₹ $1200=\frac{12}{100}\times ₹1200=₹144$ (ii) $45\%$ of $500=\frac{45}{100}\times 500=225$
• Q. If $60\%$ students completed homework, then what is the percentage of students who did not complete? Explanation : Total $=100\%$ $\%$ Students not completed homework $=(100-60)\%=40\%$
• Q. In a school out of 600 students, $12\frac{1}{2}\%$ were absent. Find the number of students absent. Explanation : Total number of students $=600$ Percentage of absent students $=12\frac{1}{2}\%$ i.e., $\frac{25}{2}\%$ Number of absent students $=12\frac{1}{2}\%$ of $600=\frac{25}{2}\times \frac{1}{100}\times 600=75$
• Q. Malvika gets 98 marks in her exams. This amounts to $56\%$ of the total marks. What are the maximum marks? Solution : Let the maximum marks be x . ATQ, $56\%$ of $\mathrm{x}=98$ $\Rightarrow \frac{56}{100}\times x=98$ $\Rightarrow x=98\times \frac{100}{56}$ $\Rightarrow \mathrm{x}=175$ Hence, the maximum marks are 175.
• Q. (i) $15\%$ of which number is 45 ? (ii) What percentage of 364 is 273 ? (iii) What percentage of 75 is 125 ? Explanation : (i) Suppose the number is x . ATQ, $15\%$ of x is 45 or $\frac{15}{100}\times x=45$ or $x=\frac{45\times 100}{15}=300$ $\therefore$ The required number is 300 . (ii) Percentage $=\frac{273}{364}\times 100=75\%$ Alternate Method: Let ' $y\%$ ' of 364 is 273. ATQ,y% of $364=273$ $\frac{\mathrm{y}}{100}\times 364=273$ $y=\frac{273\times 100}{364}=75$ $\therefore 75\%$ of 364 is 273. (iii) Percentage $=\frac{125}{75}\times 100=166.66\%$ Alternate Method : Let ' $\mathrm{z}\%$ ' of 75 is 125 . ATQ, z% of $75=125$ $\frac{\mathrm{z}}{100}\times 75=125$ $\mathrm{z}=\frac{125\times 100}{75}=166.66$ $\therefore 166.66\%$ of 75 is 125 .
• Q. A football team won 10 games from the total games they played. This was $40\%$ of the total. How many games were played in all? Solution : Let total games played be x. ATQ, $40\%$ of $\mathrm{x}=10$ or $\frac{40}{100}\times \mathrm{x}=10$ or $\mathrm{x}=\frac{10\times 100}{40}=25$ Thus, total games were 25 .

Note:

• Increase (in %) $=\left(\frac{\text{ Increase }}{\text{ Original Value }}\times 100\right)\%$
• Decrease (in %) $=\left(\frac{\text{ Decrease }}{\text{ Original Value }}\times 100\right)\%$

• Q. The price of a scooter which was ₹ 34000 last year increased by $20\%$ this year. What is the price now? Solution : Let the original price be ₹ 100 . $20\%$ of $100=\frac{20}{100}\times 100=20$ $\therefore$ Increased price $=₹100+₹20=₹120$ If original price is ₹ 100 , increased price $=₹120$ If original price is ₹ 34000 , increased price $=₹\frac{120}{100}\times 34000=₹40800$.
• Q. The population of a town increases by $6\%$ every year. If the present population is $15900$, find its population a year ago. Solution : Let the population of the town be 100, a year ago. Then, Increases in population $=6\%$ of $100=\frac{6}{100}\times 100=6$ $\therefore$ Present population $=100+6=106$ If present population is 106 , population a year ago $=100$ If present population is 1 , population a year ago $=\frac{100}{106}$ If present population is 15900 , population a year ago $=\frac{100}{106}\times 15900=15000$ Hence, the population of the town a year ago was 15000 .

3.0Profit and Loss

Some important formulae

(a) Gain $\%=\left(\frac{\text{ Gain }}{\text{ C.P. }}\times 100\right)\%$ or Gain $=\frac{\text{ C.P. }\times \text{ Gain }\%}{100}$

(b) Loss $\%=\left(\frac{\text{ Loss }}{\text{ C.P. }}\times 100\right)\%$ or Loss $=\frac{\text{ C.P. }\times \text{ Loss }\%}{100}$

(c) To find C.P. when S.P. and gain% or loss% are given. (i) C.P. $=\frac{100}{100+\text{ Gain }\%}\times$ S.P. (ii) C.P. $=\frac{100}{100-\text{ Loss }\%}\times$ S.P.

(d) To find S.P. when C.P. and gain% or loss% are given. (i) S.P. $=\frac{100+\text{ Gain }\%}{100}\times$ C.P. (ii) S.P. $=\frac{100-\text{ Loss }\%}{100}\times$ C.P.

• Q. If the C.P. of $25$ chairs is equal to the S.P. of $30$ chairs, find the loss percent. Explanation : Let the C.P. of each chair be ₹ 1 . Then, C.P. of 30 chairs $=₹30$ It is given that, S.P. of 30 chairs $=$ C.P. of 25 chairs $\Rightarrow$ S.P. of 30 chairs $=₹25$ Clearly, S.P. < C.P. So, there is loss given by Loss = C.P. - S.P. $=₹(30-25)=₹5$ Now, Loss $\%=\left(\frac{\text{ Loss }}{\text{ C.P. }}\times 100\right)\%=\left(\frac{5}{30}\times 100\right)\%=16\frac{2}{3}\%$ Hence, Loss $\%=16\frac{2}{3}\%$

Overhead charges

• Q. A man purchases two fans for ₹ 2160 . By selling one fan at a profit of $15\%$ and the other at a loss of $9\%$, he neither gains nor losses in the whole transaction. Find the cost price of each fan. Solution : Let the cost price of first fan be $₹x$ . Then, Cost price of second fan $=₹(2160-x)$ It is given that In the whole transaction, the man neither gains nor losses. $\therefore$ Gain on the sale of first fan $=$ Loss in the sale of second fan $\Rightarrow 15\%$ of $₹x=9\%$ of $₹(2160-x)$ $\Rightarrow \frac{15}{100}\times x=\frac{9}{100}\times (2160-x)\Rightarrow 15x=9(2160-x)$ $\Rightarrow 5\mathrm{x}=3(2160-\mathrm{x})\Rightarrow 5\mathrm{x}=6480-3\mathrm{x}$ $\Rightarrow 5\mathrm{x}+3\mathrm{x}=6480\Rightarrow 8\mathrm{x}=6480\Rightarrow \mathrm{x}=\frac{6480}{8}\Rightarrow \mathrm{x}=810$ $\therefore$ C.P. of first fan $=₹810$ C.P. of second fan $=₹(2160-x)=₹(2160-810)=₹1350$.
• Q. Neera sold a sewing machine to Meera at $16\%$ profit. Meera sold it to Madhu at a loss of $10\%$. If Madhu paid ₹ $1670.40$ for the machine, find the cost price of machine for Neera. Explanation : Cost price for Madhu $=$ Selling price for Meera $=₹1670.40$ Meera suffered a loss $=10\%$ means if C.P. is 100 , then S.P. is 90 . The S.P. for Meera is ₹ 90 then the C.P. is ₹ 100 $\therefore$ The S.P. for Meera is $₹1670.40$ then the C.P. is $\frac{100}{90}\times 1670.40=₹1856$. $\therefore$ The selling price of sewing machine for Neera is ₹ 1856 Neera earned a profit $=16\%$ means if C.P. is 100 , then S.P. is 116. $\because$ The S.P. for Neera is ₹ 116 then the C.P. is ₹ 100 . $\therefore$ The S.P. for Neera is ₹ 1856 then the C.P. is $\frac{100}{116}\times 1856=₹1600$ $\therefore$ The cost price of sewing machine for Neera is ₹ 1600 .
• Q. Mohan purchased an article for ₹ 125 and sold it to Sohan at $12\%$ profit. Sohan then sold it to Rohan at $10\%$ loss. Find the price paid by Rohan. Solution : For Mohan, cost price of the article is ₹ 125 . Profit $=12\%$ means $\{\begin{array}{c}\text{ If C.P. is }100\\ \text{ then S.P. is }112\end{array}$ $\because$ When the C.P. is ₹ 100 then the S.P. will be ₹ 112 $\therefore$ When the C.P. is $₹125$ then the S.P. will be $\frac{112}{100}\times 125=₹140$ $\therefore$ C.P. for Sohan $=₹140$ Loss $=10\%$ means $\{\begin{array}{l}\text{ If C.P. is }100\\ \text{ then S.P. is }90\end{array}$ $\because$ For Sohan when C.P. is ₹ 100 then the S.P. is ₹ 90 $\therefore$ For Sohan when C.P. is ₹ 140 then the S.P. is $₹\frac{90}{100}\times 140=₹126$ $\therefore$ For Rohan the cost price is ₹ 126 Hence, Rohan had to pay ₹ 126 for the article.
• Q. On selling 20 articles a shopkeeper gains the cost price of 4 articles. Find his gain percent. Solution : S.P. of 20 articles $=$ C.P. of 20 articles + C.P. of 4 articles $=$ C.P. of 24 articles. Let C.P. of one article be ₹ 1 $\therefore$ C.P. of 24 articles $=₹24$ S.P. of 20 articles $=$ C.P. of 24 articles $=₹24$ $\therefore$ S.P. of one article $=₹\frac{24}{20}=₹\frac{6}{5}$ $\therefore$ Gain $=\frac{6}{5}-1=\frac{6-5}{5}=₹\frac{1}{5}$ $\therefore$ Gain $\%=\frac{\text{ Total gain }}{\text{ C.P. }}\times 100=\frac{\frac{1}{5}}{1}\times 100=\frac{1}{5}\times 100=20\%$

4.0Discount

The deduction made on the marked price is called discount. Discount is generally given as a certain percent of the marked price. It is always calculated on the marked price or list price.

Marked Price (M.P.) or List Price (L.P.)

Selling Price/Net Price

• Discount is given on the marked price only. S.P. = M.P. - Discount Discount $=$ Marked price (M.P.) - Selling price (S.P.) Discount $\%=\frac{\text{ Discount }}{\text{ M.P. }}\times 100$ or Discount $=\frac{\text{ M.P. }\times \text{ Discount }\%}{100}$ M.P. $=\frac{\text{ S.P. }\times 100}{100-\text{ Discount }\%}$ or S.P. $=\frac{\text{ M.P. }\times (100-\text{ Discount }\%)}{100}$ Discount $\%=\frac{\text{ M.P. }-\text{ S.P. }}{\text{ M.P. }}\times 100$ Successive discounts : Two successive discounts of $\mathrm{x}\%$ and $\mathrm{y}\%$ allowed on an item are equivalent to a single discount of $\left(x+y-\frac{xy}{100}\right)\%$
• This discount is always less than the sum of individual discounts.

• Q. A dealer buys an article for ₹ 380 . At what price must he mark it so that after allowing a discount of $5\%$, he still makes a profit of $25\%$ ? Explanation : We have, C.P. of the article $=₹380$, Gain $=25\%$ $\therefore$ S.P. of the article $=(\frac{100+\text{ Gain }\%}{100}\times$ C.P. $)$ $\Rightarrow$ S.P. of the article $=₹\left(\frac{100+25}{100}\times 380\right)=₹\left(\frac{125}{100}\times 380\right)=₹475$ Now, suppose the dealer marks ₹ 100 as the price of the article. He allows 5% discount on it. Discount $=\frac{\text{ M.P. }\times \text{ Discount }\%}{100}=\frac{100}{100}\times 5$ $\therefore$ Discount $=₹5$ $\therefore$ S.P. $=$ M.P. - Discount $=₹100-₹5=₹95$ Thus, If S.P. is ₹ 95 , then M.P. $=₹100$ If S.P. is ₹ 1 , then M.P. $=₹\frac{100}{95}$ If S.P. is ₹ 475 , then M.P. $=₹\frac{100}{95}\times 475=₹500$
• Q. Jasmine allows 4% discount on the marked price of her goods and still earns a profit of $20\%$. What is the cost price of a shirt, marked at ₹ 850 ? Solution : M.P. of shirt = ₹ 850 Rate of discount $=4\%$ $\therefore$ Discount allowed $=\frac{4}{100}\times 850=₹34$ Thus, S.P. of shirt = ₹ $850-₹34=₹816$ Now, profit% of jasmine = 20% Therefore, C.P. $=\frac{100\times \text{ S.P. }}{(100+\text{ profit }\%)}=₹\frac{100\times 816}{(100+20)}=₹\frac{100\times 816}{120}=₹680$ Thus, C.P. of shirt = ₹ 680.

5.0Sales Tax (S.T.)/ Value Added Tax (VAT)

Sales Tax

• Calculation of Sales Tax : Sales tax $=$ Selling price $\times \frac{\text{ Rate of sales tax }}{100}$ Rate of sales tax $=\frac{\text{ Sales tax }}{\text{ S.P. }}\times 100$ Price including sales tax $=$ Selling price + Sales tax.

VAT

• Q. David purchased a pair of shoes for ₹ 441 including value added tax. If the marked price of the shoes is ₹ 420 , find the rate of value added tax. Explanation : Let the rate of value added tax be $\mathrm{x}\%$, then, Value added tax $=x\%$ of ₹ 420 $=₹\left(\frac{x}{100}\times 420\right)=₹\frac{21}{5}x$ $\therefore$ Selling price of shoes $=₹\left(420+\frac{21\mathrm{x}}{5}\right)$ But selling price of shoes is ₹ 441 $\therefore 420+\frac{21x}{5}=441\Rightarrow \frac{21x}{5}=21\Rightarrow x=5$ Hence, the rate of value added tax is $5\%$
• Q. Samir bought a shirt for ₹ 336, including 12% VAT and a neck-tie for ₹ 110, including 10% VAT. Find the printed price (without VAT) of shirt and neck-tie together. Solution : Let the printed price of the shirt be ₹ $x$ and that of neck-tie be ₹ $y$. Then, VAT on shirt $=12\%$ of $₹x=₹\frac{12x}{100}=₹\frac{3x}{25}$ VAT on neck-tie $=10\%$ of $₹y=₹\frac{10y}{100}=₹\frac{y}{10}$ $\therefore$ Selling price of shirt $=₹\left(x+\frac{3x}{25}\right)=₹\frac{28x}{25}$ and Selling price of neck-tie $=₹\left(y+\frac{y}{10}\right)=₹\frac{11y}{10}$ But, selling prices of shirt and neck-tie are ₹ 336 and ₹ 110 respectively. $\therefore \frac{28\mathrm{x}}{25}=336$ and $\frac{11\mathrm{y}}{10}=110\Rightarrow \mathrm{x}=\frac{336\times 25}{28}$ and $\mathrm{y}=\frac{110\times 10}{11}\Rightarrow \mathrm{x}=₹300$ and $\mathrm{y}=₹100$ Hence, the total printed price of the shirt and neck-tie $=₹(300+100)=₹400$
• Q. The cost of an article at a shop was ₹ 850 . The sales tax was charged to be $5\%$. Find the total bill amount. Explanation : Cost of an article $=₹850$ Sales tax charged $=5\%$ $\therefore$ Sales tax $=\frac{5}{100}\times 850=₹42.5$ Total bill amount $=850+42.5=₹892.5$
• Q. Shelly buys an article for ₹ $10,000$ and pays $6\%$ tax. She sells the same article for ₹ $13,000$ and charges 8% tax. Find the VAT paid by Shelly. Solution : Cost of the article $=₹10,000$ Tax paid by Shelly $=6\%$ of ₹ 10,000 $=\frac{6}{100}\times 10,000=₹600$ Selling price of the article $=₹13,000$ Tax charged by Shelly at $8\%=8\%$ of 13,000 $=\frac{8}{100}\times 13,000=₹1040Therefore,VAT=taxrecoveredonsale-Taxpaidonpurchase$=₹ 1040 \text { - ₹ } 600=₹ 440$

6.0Interest

Now, let's understand about interest. Interest is the amount of money a lender or financial institution receives for lending out money. When you lend money to borrowers, you get interest and when you borrow money from lenders, you pay interest. So, Interest is the extra money that the borrower pays for using the lender's money. There are two types of interest. (i) Simple interest (ii) Compound interest

Simple Interest (S.I.)

The interest charged on the initial principal is called simple interest. If P is the principal, R is the rate of interest per annum and T is the time in years, then the simple interest is given by S.I. $=$ $\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$ where $\mathrm{P}=$ Principal, $\mathrm{R}=$ Rate, $\mathrm{T}=$ Time, S.I. $=$ Simple interest

• Q. Find the principal at the rate of $10\%$, the simple interest on it for $3$ years is ₹ $150$. Explanation : Given : Rate $=10\%$, Time $=3$ years, Interest $=₹150$ To find : Principal (P) From the formula Simple interest $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$ or $150=\mathrm{P}\times \frac{10}{100}\times 3$ or $\mathrm{P}=\frac{150\times 100}{3\times 10}=500$ Hence, the required principal is ₹ 500 .
• Q. 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 ₹ $360$ more. Find the sum. Solution : Let sum $=P$ and original rate $=R$ Then ATQ, $\left(\frac{\mathrm{P}\times (\mathrm{R}+2)\times 3}{100}\right)-\left(\frac{\mathrm{P}\times \mathrm{R}\times 3}{100}\right)=360$ $\Rightarrow \frac{3\mathrm{P}\times (\mathrm{R}+2)-3\mathrm{PR}}{100}=360$ $\Rightarrow 3\mathrm{PR}+6\mathrm{P}-3\mathrm{PR}=36000$ $\Rightarrow 6\mathrm{P}=36,000$ $\Rightarrow P=₹6000$ Hence, the sum is ₹ 6000 .

Compound Interest

The interest charged every year on the amount of last year is called compound interest. When the interest is not paid in the specified period but is added to the principal for the calculation of future interest, such an interest is known as compound interest. Compound interest $=$ Amount of the last time period - principal of the first time period C.I. $=\mathrm{A}-\mathrm{P}$. Where, $A=P{\left(1+\frac{R}{100}\right)}^n$ Where $\mathrm{P}=$ Principal, $\mathrm{R}=$ Rate of interest, $\mathrm{n}=$ Time. C.I. at the end of a certain specified period is equal to the difference between the amount at the end of the period and the original principal. Hence, C. I. $=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}-1\right]$

Type-I

• C.I. is equal to S.I. for the period of one year and it is always greater than S.I. for more than 1 year, when interest is calculated annually.

• Q. Find the compound interest on ₹ $10,000$ for 2 years at rate of $8\%$ per annum. Explanation : $P=10,000,R=8\%,n=2$ years $A=P{\left(1+\frac{R}{100}\right)}^n=10000{\left(1+\frac{8}{100}\right)}^2=10000\times \frac{108}{100}\times \frac{108}{100}=11664$ C.I. $=\mathrm{A}-\mathrm{P}=₹11664-₹10000=₹1664$

Type-II

• Q. Find the compound interest on ₹ 1500 for $2\frac{1}{2}$ years at $10\%$ per annum, when the interest is payable annually. Solution : $P=₹1500,n=2\frac{1}{2}$ years, $R=10\%$ $\because$ Interest is paid annually $A=P{\left(1+\frac{R}{100}\right)}^n$ $=1500\times {\left(1+\frac{10}{100}\right)}^2\left(1+\frac{10}{200}\right)$ $=1500\times \frac{110}{100}\times \frac{110}{100}\times \frac{105}{100}=₹1905.75$ [ $\because$ Rate of interest is $10\%$ annually, hence $5\%$ rate of interest will be charged in half year.] Compound interest $=$ Amount - Principal $=1905.75-1500=₹405.75$ Alternate Method : $A=P{\left(1+\frac{R}{100}\right)}^n$ $A=1500{\left(1+\frac{10}{100}\right)}^{2\frac{1}{2}}$ Now, we will find amount for 2 years using C.I. and then use this as principal to get S.I. for $\frac{1}{2}$ year more. $A=1500{\left(1+\frac{10}{100}\right)}^2=1500\times \frac{11}{10}\times \frac{11}{10}$ = ₹ 1815 S.I. $=\frac{1815\times 10\times 1/2}{100}=₹90.75$ Total interest $=₹(1815-1500)+₹90.75$ = ₹ 315 + ₹ 90.75 = ₹ 405.75 Type-III : Computation of compound interest when the interest is compounded half yearly.
• Q. Kapil deposited ₹ 1600 in a bank on $1^{\text{st }}$ January 2005. Find the amount in his bank account on $1^{\text{st }}$ January 2006, if the bank pays interest at $8\%$ per annum and the interest is calculated every year on $30^{\text{th }}$ June and $31^{\text{st }}$ December. Solution: $\mathrm{P}=$ ₹ $1600,\mathrm{n}=1$ year, Rate $=8\%$ Interest is calculated every year on $30^{\text{th }}$ June and $31^{\text{st }}$ December, the interest is payable half yearly. Time $=1$ year $=2$ half years Rate $=8\%$ per annum $=4\%$ half yearly $$\begin{gathered} \text{ Amount }=1600\times {\left(1+\frac{4}{100}\right)}^{2\times 1} \\ =1600\times \frac{104}{100}\times \frac{104}{100}=₹1730.56 \end{gathered}$$ Amount in Kapil's account on $1^{\text{st }}$ Jan. $2006=₹1730.56$

Note:

• Compound interest can also be calculated by adding the interest for each year.
• When the interest is compounded half yearly, the rate of interest became half i.e., $\left(\frac{R}{2}\right)\%$ and the time doubles i.e., (2n)

• Q. How much would a sum of ₹ 16000 amounts to in 2 years time at $10\%$ per annum compound interest, interest being payable half-yearly? Solution : Here, $\mathrm{P}=₹16000,\mathrm{R}=10\%$ per annum and $\mathrm{n}=2$ years. Interest being payable half yearly, Time $=2$ years $=4$ half years, Rate $=10\%$ per annum $=5\%$ half yearly $\therefore$ Amount after 2 years $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{200}\right)}^{2\mathrm{n}}$ $=₹16000\times {\left(1+\frac{10}{200}\right)}^{2\times 2}$ $=₹16000\times {\left(1+\frac{1}{20}\right)}^4$ $=₹16000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}=₹19448.10$ Hence, a sum of ₹ 16000 amounts to ₹ 19448.10 in 2 years.
• Q. Find the principal, if the compound interest at the rate of $5\%$ p.a. for 3 years is ₹ 1261 . Explanation : Suppose the principal is ₹ 100 . Rate $=5\%$ per annum, Time ( $n$ ) $=3$ years $\therefore$ Amount $=₹100{\left(1+\frac{5}{100}\right)}^3=₹100{\left(1+\frac{1}{20}\right)}^3=₹100{\left(\frac{21}{20}\right)}^3$ $=₹100\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}=₹\frac{9261}{80}$ $\therefore$ Compound interest $=₹\frac{9261}{80}-₹100=₹\frac{9261-8000}{80}=₹\frac{1261}{80}$ If the C.I. is ₹ $\frac{1261}{80}$, then the sum $=₹100$ If the C.I. is ₹ 1 , then the sum $=₹100\times \frac{80}{1261}$ If the C.I. is $₹1261$ , then the sum $=₹100\times \frac{80}{1261}\times 1261=₹8000$

7.0Growth and depreciation

Growth Rate

The growth rate for a certain period is the ratio of the change in that period and the magnitude just before this period, which is expressed as percent. $P=$ Final value after $n$ years. Then rate of growth $=\frac{\text{ Final value }-\text{ Initial value }}{\text{ Initial value }}\times 100\%$ or $R=\frac{P-P_0}{P_0}\times 100\%$

Results on population

Let the present population of town be ${\mathrm{P}}_0$ and let there be an increase or decrease of $\mathrm{R}\%$ per annum, then Population after $n$ years $=P_0{\left(1\pm \frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ Increase : $\mathrm{P}={\mathrm{P}}_0\times {\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ Decrease : $\mathrm{P}={\mathrm{P}}_0\times {\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ ${\mathrm{P}}_0=$ Original value, $\mathrm{R}=$ Rate of increase or decrease, $\mathrm{n}=$ Period, $\mathrm{P}=$ Final value. $(+)\to$ increase,$(-)\to$ decrease

Depreciation

The relative decrease in the value of a machine over a period of time is called its depreciation. If ${\mathrm{P}}_0$ is the value of an article at a certain time and $\mathrm{R}\%$ per annum is the rate of depreciation, then the value $P_n$ at the end of $n$ years is given by $P_n=P_0{\left(1-\frac{R}{100}\right)}^n$

• Q. The value of a flat worth ₹ 500000 is depreciating at the rate of $10\%$ per annum. In how many years will its value be reduced to ₹ 364500 ? Explanation : We have, Present value = ₹ 500000, Depreciated value $=₹364500$ Rate of depreciation $=10\%$ per annum. Let the depreciation period be of $n$ years. Then, $364500=500000{\left(1-\frac{10}{100}\right)}^n$ $\Rightarrow \frac{3645}{5000}={\left(\frac{9}{10}\right)}^n$ $\Rightarrow \frac{729}{1000}={\left(\frac{9}{10}\right)}^n$ $\Rightarrow {\left(\frac{9}{10}\right)}^3={\left(\frac{9}{10}\right)}^{\mathrm{n}}$ $\Rightarrow \mathrm{n}=3$ Hence, in 3 years the value of the flat will be reduced to $₹364500$ .