Direct and Inverse Proportions

Mohan prepares tea for himself and his sister. He uses 300 ml of water, 2 spoons of sugar, 1 spoon of tea leaves and 50 ml of milk. How much quantity of each item will he need, if he has to make tea for five persons?

So how do we find out the quantity of each item needed by Mohan? In a school 1 teacher is there for 15 students. How many teachers are required for 45 students? To answer such questions, we now study some concept of Proportions.

1.0Ratio and proportion

Ratio: The ratio of two quantities $a$ and $b$ in the same units, is the fraction $\frac{a}{b}$ and we write it $asa:b$. In the ratio $\mathrm{a}:\mathrm{b}$, we call ' a ' as the first term or the antecedent and ' b ' as the second term or the consequent. Proportion: The equality of two ratios is called proportion. If $\mathrm{a}:\mathrm{b}=\mathrm{c}:\mathrm{d}$, we write, $\mathrm{a}:\mathrm{b}::\mathrm{c}:\mathrm{d}$ and we say that $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}$ are in proportion. Here a and d are called extremes, while b and c are called mean terms. a:b :: c:d $\Rightarrow$ Product of means = Product of extremes $\Rightarrow (b\times c)=(a\times d)$.

• For finding the ratio of 2 quantities both quantities must be in same units.
• Q. Check whether 2, 3, 4 and 6 are in proportion or not. Explanation: If $2:3=4:6$ Product of extremes $=2\times 6=12$ Product of means $=3\times 4=12$ So, 2, 3, 4, 6 are in proportion.
• Q. What should be added to each of the four numbers: 6, 14, 18 and 38 to make them in proportion? Explanation: Let the number to be added $=x$ $6+x,14+x,18+x,38+x$ are in proportion. So, Product of extremes $=$ Product of means $(6+x)(38+x)=(14+x)(18+x)$ $\Rightarrow 228+6\mathrm{x}+38\mathrm{x}+{\mathrm{x}}^2=252+14\mathrm{x}+18\mathrm{x}+{\mathrm{x}}^2$ $\Rightarrow 44\mathrm{x}+228=252+32\mathrm{x}$ $\Rightarrow 44\mathrm{x}-32\mathrm{x}=252-228$ $\Rightarrow 12\mathrm{x}=24$ $\Rightarrow \mathrm{x}=2$
• Q. If $a:b=4:1$, then find $\frac{(a-3b)}{(2a-b)}$. Solution: $\frac{\mathrm{a}}{\mathrm{b}}=\frac{4}{1}$ $\frac{a-3b}{2a-b}=\frac{\frac{a}{b}-\frac{3b}{b}}{\frac{2a}{b}-\frac{b}{b}}$ (divide by $b$ in both numerator and denominator) $\Rightarrow \frac{\frac{\mathrm{a}}{\mathrm{b}}-3}{\frac{2\mathrm{a}}{\mathrm{b}}-1}=\frac{4-3}{2\times 4-1}=\frac{1}{7}$
• Q. A sum of money is to be distributed among $A,B,C,D$ in the ratio of $5:2:4:3$. If $C$ gets Rs. 1000 more than D, what is B's share? Solution: Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively. ATQ, $4\mathrm{x}-3\mathrm{x}=1000$ $\mathrm{x}=1000$ $\therefore$ B's share $=$ Rs. $2\mathrm{x}=$ Rs. $(2\times 1000)=$ Rs. 2000

2.0Variation

The word vary means change and the word rate means how a quantity (variable) changes with another quantity (variable). If the values of two quantities depend on each other in such a way that a change in one results corresponding change in the other, then the two quantities are said to be in variation.

Direct variation

Two quantities are said to vary directly if the increase (or decrease) in one quantity leads to a corresponding increase (or decrease) in the other quantity. We say that x is directly proportional to y , if $\mathrm{x}=\mathrm{ky}$ for some constant k and we write, $x\propto y\Rightarrow x=ky$, $\frac{x}{y}=k$ (constant) Rule : If two quantities $x$ and $y$ vary directly and $y_1,y_2$ are two values of $y$ corresponding to the values ${\mathrm{x}}_1$ and ${\mathrm{x}}_2$ of x , then $\frac{{\mathrm{x}}_1}{{\mathrm{y}}_1}=\frac{{\mathrm{x}}_2}{{\mathrm{y}}_2}$. Example of direct variation: (i) The cost of articles varies directly as the number of articles. (ii) The distance covered by a moving object varies directly as its speed. (Time should be same)

• Q. Given $x\propto y$ and when $x=15,y=3$, find (i) y when $\mathrm{x}=35$ (ii) x when $\mathrm{y}=4$ Explanation: If $\mathrm{x}\propto \mathrm{y}$ means $\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{k}$, where k is the constant of variation. $\mathrm{x}=\mathrm{ky}$ (i) When $\mathrm{x}=15$, and $\mathrm{y}=3$. This mean $15=\mathrm{k}\times 3$ or $\mathrm{k}=5$ Using this value of $k$, we have $x=5y$ So, when $\mathrm{x}=35,35=5\mathrm{y}$ or $\mathrm{y}=7$ (ii) When $\mathrm{y}=4,\mathrm{x}=\mathrm{ky}$ $x=5\times 4,x=20$
• Q. Find the cost of 8 books if 6 books cost Rs. 480. Solution: We can use unitary method $\because$ Cost of 6 books $=$ Rs. 480 $\therefore$ Cost of 1 book $=$ Rs. $\frac{480}{6}$ $\therefore$ Cost of 8 books $=$ Rs. $\frac{480\times 8}{6}=$ Rs. 640

Alternate method

We know the cost of books increases when the number of books increases. Let Rs. $x$ be the cost of 8 books.

$\therefore 6:8=480:x$ or $6\times x=480\times 8$ or $x=\frac{480\times 8}{6}=640$ Hence, cost of 8 books = Rs. 640

• Q. A car travels 432 km on 48 litres of petrol. How far would it travel on 20 litres of petrol? Explanation: Suppose the car travels $x$ km on 20 litres of petrol. Then, the above information can be put in the following tabular form:

We observe that the lesser the petrol consumed, the smaller the number of kilometres travelled. So, it is a case of direct variation. Ratio of petrol consumed = Ratio of distance travelled $\Rightarrow 48:20=432:x$ $\Rightarrow \frac{48}{20}=\frac{432}{x}$ $\Rightarrow 48\times x=20\times 432$ [By cross-multiplication] $\Rightarrow \mathrm{x}=\frac{20\times 432}{48}=180$ Hence, the car would travel 180 km on 20 litres of petrol.

• Q. If x and y vary directly, find the missing entries in the following tables.

(i)

(ii)

Solution: (i) It is given that $x$ and $y$ are in direct variation. Therefore, the ratio of the corresponding values of x and y remain constant. We have, $\frac{2.5}{5}=\frac{1}{2}$ So, x and y are in direct variation with the constant of variation equals to $\frac{1}{2}$. This means that x is half of y or y is twice of x . Thus, the required entries are $\frac{8}{2},\frac{24}{2}$ and $21\times 2$ i.e., 4,12 and 42 . (ii) We have, $\frac{9}{4.5}=\frac{15}{7.5}=2$ So, x and y are in direct variation such that x is twice of y . Thus, the missing entries are $3\times 2=6$ and $13.25\times 2=26.5$.

• Q. A train is moving at a uniform speed of $75\mathrm{km}/$ hour. (i) How far will it travel in $20$ minutes? (ii) Find the time required to cover a distance of $250\mathbf{km}$. Explanation: Let the distance travelled (in km ) in 20 minutes be x and time taken (in minutes) to cover 250 km be $y$.

Since the speed is uniform, therefore, the distance covered would be directly proportional to time. (i) We have $75:\mathrm{x}=60:20$ $\Rightarrow \frac{75}{\mathrm{x}}=\frac{60}{20}$ $\Rightarrow \frac{75\times 20}{60}=\mathrm{x}$ or $\mathrm{x}=25$ So, the train will cover a distance of 25 km in 20 minutes. (ii) Also, 75 : 250 :: $60:y$ $\Rightarrow \frac{75}{250}=\frac{60}{y}$ or $\mathrm{y}=\frac{250\times 60}{75}=200$ minutes or 3 hours 20 minutes. Therefore, 3 hours 20 minutes will be required to cover a distance of 250 kilometres. Alternately, when $x$ is known, then one can determine $y$ from the relation $:\frac{x}{20}=\frac{250}{y}$

Inverse variation

Two quantities are said to vary inversely if the increase (or decrease) in one quantity causes a corresponding decrease (or increase) in the other quantity. If $x_1$ and $x_2$ are two different values of $x^2$ and $y_1$ and $y_2$ be the corresponding values of $y$ then $x_1{y}_1=x_2{y}_2$ $\Rightarrow \frac{x_1}{x_2}=\frac{y_2}{y_1}$ This result is useful in solving inverse variation problems $\frac{{\mathrm{x}}_1}{{\mathrm{x}}_2}=\frac{{\mathrm{y}}_2}{{\mathrm{y}}_1}$ or ${\mathrm{x}}_1:{\mathrm{x}}_2={\mathrm{y}}_2:{\mathrm{y}}_1$ We say that x is inversely proportional to y , if $\mathrm{xy}=\mathrm{k}$ for some constant k and we write, $\mathrm{x}\propto \frac{1}{\mathrm{y}}$ Example of Inverse Variation: (i) The time taken to finish a piece of work varies inversely as the number of men at work. (ii) The speed varies inversely as the time taken to cover a distance.

• When two quantities vary inversely their product is constant.
• Q. 120 men had food provision for 200 days. After 5 days, $30$ men died due to an epidemic. How long will the remaining food last? Explanation: Since 30 men died after 5 days. Therefore, the remaining food is sufficient for 120 men for 195 days. Suppose the remaining food lasts for x days for the remaining 90 men. Thus, we have the following table:

We note that more men will consume the food in less number of days and less number of men will consume the food in more number of days. So, it is a case of inverse variation. $\therefore$ Ratio of number of men = Inverse ratio of number of days $\Rightarrow$ 120:90 = x: 195 $\Rightarrow \frac{120}{90}=\frac{\mathrm{x}}{195}\Rightarrow \frac{120\times 195}{90}=260$ Hence, the remaining men will consume the food in 260 days.

• Q. Ruchi has enough money to send 14 parcels each weighing 500 grams. Find the weight of each parcel if she can send 20 parcels for the same money. Also, find the constant of variation. Solution: Suppose the weight of each parcel is x grams. | Number of parcels | 14 | 20 | |--------------------------|------|------| | Weight (in gm) | 500 | x |

Obviously, it is a case of inverse variation. $\therefore 20\times \mathrm{x}=14\times 500$ or $\mathrm{x}=\frac{14\times 500}{20}=350$ $\therefore$ Weight of each parcel is 350 grams. Constant of variation $=x\times y=14\times 500=7000$

• Men and days are inversely proportional to each other.
• Q. 10 men dig a trench in 6 hours. How many men will dig the same trench in 12 hours? (Assume that all men work at the same rate.) Explanation: In 6 hours the trench is dug by 10 men. In 1 hour the trench is dug by $10\times 6$ men. In 12 hours the trench is dug $\frac{10\times 6}{12}$ or 5 men. $\therefore 5$ men will dig the trench.

Alternate method

Let x men are required to dug the trench in 12 hours.

Ratio of hours = Inverse ratio of men or $6:12=x:10$. or $12\times x=6\times 10$ or $\mathrm{x}=\frac{10\times 6}{12}=5$ $\therefore 5$ men will dig the trench.

Time and work

If $A$ can do a piece of work in ' $n$ ' days, then A's 1 day work $=\frac{1}{n}$ If A's 1 day work $=\frac{1}{n}$, then $A$ can finish the work in ' $n$ ' days. If $A$ is thrice as good a workman as $B$, then : Ratio of work done by $A$ and $B=3:1$. Ratio of time taken by $A$ and $B$ to finish a work $=1:3$.

• If ratio of work done by $A$ and $B=m:n$ Then, the ratio of time taken by A and B to finish a work $=\mathrm{n}:\mathrm{m}$
• Q. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long would both $A$ and $B$, working together take to do the same job? Explanation: A's 1 hour work $=\frac{1}{8}$, B's 1 hour work $=\frac{1}{10}$ $(A+B)$ 's 1 hour work $=\left(\frac{1}{8}+\frac{1}{10}\right)=\frac{9}{40}$ Both A and B will finish the work in $\frac{40}{9}=4\frac{4}{9}\mathrm{hrs}$.
• Q. 35 Men could do a certain piece of work in 40 days. If after 4 days, $7$ of the men suddenly left, in how many days will the remaining work be completed? Solution: 35 men needed 40 days to complete the work. Since 4 days work is done, the remaining work would have needed 35 men and 36 days to complete. Now there are 35-7=28 men. Let the number of days needed be x . This is a case of inverse variation. So, $35\times 36=28\times x$ or $\mathrm{x}=45$ The remaining work will be completed in 45 days.
• Q. Manoj can complete $\frac{1}{4}$ of a work in 5 days and Pinto can complete $\frac{1}{5}$ of the same work in 6 days. How long would both of them take to complete that work, if they work together? Explanation: Manoj does $\frac{1}{4}$ of work in 5 days Manoj completes the full work in $=5\times 4$ or 20 days $\therefore$ Work done by Manoj in 1 day $=\frac{1}{20}$ Pinto does $\frac{1}{5}$ of work in 6 days Pinto completes the full work in $=6\times 5$ or 30 days. $\therefore$ Work done by Pinto in 1 day $=\frac{1}{30}$ From (i) and (ii), we get Work done by both of them in 1 day $=\frac{1}{20}+\frac{1}{30}=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12}$ $\therefore$ Both of them complete the work in 12 days. Time, Speed and Distance Distance $=$ Speed $\times$ Time Speed $=\frac{\text{ Distance }}{\text{ Time }}$, Time $=\frac{\text{ Distance }}{\text{ Speed }}$

• To convert speed from $\mathrm{km}/\mathrm{hr}$ to $\mathrm{m}/\sec$ multiply the given number by $\frac{5}{18}$. $>$ To convert speed from $\mathrm{m}/\sec$ to $\mathrm{km}/\mathrm{hr}$ multiply the given number by $\frac{18}{5}$.

• Q. A train 100 m long is running at a speed of $30\mathrm{km}/\mathrm{hr}$. Find the time taken by it to cross a man standing near the railway line.
  
  Explanation: Speed of the train $=\left(30\times \frac{5}{18}\right)\mathrm{m}/\sec =\left(\frac{25}{3}\right)\mathrm{m}/\sec$. Distance moved in passing the standing man $=100\mathrm{m}$. Required time taken $=\frac{100}{\left(\frac{25}{3}\right)}=\left(100\times \frac{3}{25}\right)\sec =12\sec$
• If a train crosses a pole, tree or any person etc., then distance covered by it is equal to its own length.
• If a train crosses any bridge, other train tunnel etc., then distance covered by it is equal to the sum of length of both.
• Q. Two trains 50 m and 110 m long are going at $34\mathrm{km}/\mathrm{hr}$ and $30\mathrm{km}/\mathrm{hr}$ respectively in opposite directions. How long would it take them to cross each other?
  
  Explanation: When two trains go in opposite directions, we add their speeds to get relative speed per hour. The distance covered is the sum of their lengths. Now in this case Relative speed $=34+30=64\mathrm{km}/\mathrm{hr}=\frac{64\times 5}{18}\mathrm{m}/\mathrm{s}=\frac{160}{9}\mathrm{m}/\mathrm{s}$ Distance covered $=50+110=160\mathrm{m}$ $\therefore$ Time taken to pass each other $=\frac{160}{\left(\frac{160}{9}\right)}=\frac{160\times 9}{160}=9$ seconds
• Q. Two goods trains 150 m and 160 m long are going in the same direction at $73\mathrm{km}/\mathrm{hr}$ and $42\mathrm{km}/\mathrm{hr}$ respectively. How long would it take to pass each other if the faster train is behind the slower train?
  
  Solution: Here, the relative speed will be taken as the difference of their speeds and the distance covered will be the sum of their lengths. Now in this case Relative speed $=73-42=31\mathrm{km}/\mathrm{hr}=\frac{31\times 5}{18}\mathrm{m}/\mathrm{s}=\frac{155}{18}\mathrm{m}/\mathrm{s}$ Distance covered $=150+160=310\mathrm{m}$ $\therefore$ Time taken to pass each other $=\frac{\text{ Distance }}{\text{ Speed }}=\frac{310}{\left(\frac{155}{18}\right)}=\frac{310}{155}\times 18=36$ seconds

• If 2 trains moving in same direction with speed $u\mathrm{km}/\mathrm{h}$ and $\mathrm{v}\mathrm{km}/\mathrm{h}$, then their relative speed is $(u-v)km/h[u>v]$. If 2 trains moving in opposite directions with speed $u\mathrm{km}/\mathrm{h}$ and $\mathrm{vkm}/\mathrm{h}$, then their relative speed is $(u+v)km/h$.