varoation

Contents

1. INTRODUCTION
2. DIRECT VARIATION
3. INVERSE VARIATION
4. JOINT VARIATION
5. VARIATION BY PARTS

Variation

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INTRODUCTION

Variation is an extension of the concept of ratio and proportion. When one quantity is changing with another quantity, then we say that the two quantities are varying with each other.

DIRECT VARIATION

When an increase (or decrease) in one quantity (say y) results in a proportionate increase (or decrease) in another quantity (say x), then we say that the two quantities are in direct variation and the relation between them is denoted by:

and is read as “x is directly proportional to y” or “x varies directly as y”.

In this case, if y increases to twice its original value, then x will also increase to twice its original value. Similarly, if y reduces to half of its original value, then x too will reduce to half of its original value. In general, if y changes to n times its value, then x will change to n times its value.

In equation form, direct variation is written as:

x = ky

where k is known as the constant of proportionality.

Whenever two quantities are in direct variation, their corresponding values are in the same ratio.

Some practical examples of direct variation:

• Number of men and the work done by them, are in direct proportion, if other things like the number of days and rate of doing the work remain constant.

• Distance travelled is directly proportional to the time taken to cover the distance, if speed is constant. In other words, if a car is travelling at a constant speed then the time taken by the car is directly proportional to the distance covered by the car.

The following graph shows the typical nature of the relationship between two quantities x and y, which are in direct proportion. The slope of the curve is the constant of proportionality. A closer look would reveal that as we move from the left to right of the graph, the values of x and y increase simultaneously, whereas if we move from the right to left, the values of x and y decrease simultaneously.

Example 1:

If 10 kg sugar costs Rs. 200, what is the cost of 12 kg sugar? Solution: Let the cost of the sugar be c and its quantity be q. Since the cost of sugar is directly proportional to its quantity, their relationship can be expressed as follows, c = k q Substituting the first set of values of c and q, we have, k = 20 Hence, the relation between c and q can now be expressed as, c = 20 q When q = 12 the cost is c = 20 12 = 240 Hence, the price of 12 kg sugar is Rs. 240. Alternatively, Cost of sugar is directly proportional to its quantity. Since quantity increases from 10 kg to 12 kg, i.e. by a factor of 1.2 times, the cost of sugar will also increase by the same factor i.e. 1.2 times of its original value. Hence, the cost of sugar would be = 200 1.2 = Rs. 240.

REMEMBER:

• When one quantity (say x) is directly varying with some power (say square) of another quantity (say y), then we do not say that x is directly proportional to y; instead, we say that x is directly proportional to the square of y. It is represented as

x y²

For example, area of an equilateral triangle is directly proportional to the square of its side and the area of a circle is directly proportional to the square of its radius.

Example 2:

According to Kepler’s third law of planetary motion, the square of the orbital period of a planet (time taken by the planet to complete one revolution) is directly proportional to the cube of the planet’s distance from the Sun. If the distances of Mercury and Earth from the Sun are in the ratio 2 : 5, then what is the orbital period of Mercury (in Earth years)? Solution: Now, for any planet, we have T2 D3, where T is the orbital period and D is the planet’s distance from the Sun. Thus, T2 = k D3 Let the orbital periods of Earth and Mercury be TE and TM respectively; also, let their corresponding distances from the Sun be DE and DM. Hence, Since the orbital period of Earth is one year (i.e. 1 Earth year), Hence, the orbital period of Mercury is 0.253 Earth years.

INVERSE VARIATION

When an increase in one quantity (say x) results in a decrease in another quantity (say y) or vice-versa, then the two quantities are said to be in inverse proportion or inverse variation and the relationship between them is expressed as,

which is read as “x is inversely proportional to y” or “x varies inversely as y”.

In this case, if y increases to twice its original value, then x will decrease to half of its original value. However, if y reduces to half of its original value, then x will increase to twice its original value. In general, if y changes to n times y, then x will change to 1/n times x.

In equation form, inverse variation is written as

where k is the constant of proportionality.

Whenever two quantities are in inverse variation, the product of their corresponding values will remain constant.

Some practical examples of inverse variation:

• Number of men and the days required to complete a work are inversely proportional, if other things like amount of work and rate of doing work remain constant.

• Time required to cover a distance is inversely proportional to the speed, if the distance is constant.

REMEMBER:

• When one quantity (say x) is inversely varying with some power (say square) of another quantity (say y), then we say that x is inversely proportional to the square of y and it is represented as

The following graph shows the typical nature of the relation between two quantities x and y, which are inversely proportional.

Example 3:

If 50 men are required to construct a bridge in 50 days, what is the number of men required to construct the same bridge in 10 days? Solution: Let the number of men be m and the number of days be d. Since these quantities are inversely proportional, their relationship can be expressed as follows, Substituting the first set of values of m and d we have, k = 2500 Hence, the relation between m and d can now be expressed as, When d = 10 the number of men required are, Hence, the number of men required is 250. Alternatively, Number of men and the days required are inversely proportional. Since the number of days becomes 1/5 times (10/50 = 1/5), the number of men required will increase to 5 times, i.e. 5 50 or 250 men.

Example 4:

The speed of a car varies inversely with the time taken by the car to reach its destination. If the driver wants to decrease his time by 20%, then by what percent should he increase his speed? Solution: Let S be the speed of the car, and T be the time it takes to reach its destination. Since the time is decreased by 20%, his new time will be 0.8T. Hence, Hence, he should increase his speed by 25%.

JOINT VARIATION

When change (increase or decrease) in two or more quantities (say y and z) results in a corresponding change in another quantity (say x), then we say that x is in joint variation with y and z. We denote the joint variation between x, y and z as:

• 

when x is directly proportional to both y and z,

when x is directly proportional to y and inversely proportional to z,

when x is inversely proportional to both y and z.

Some practical examples of joint variation:

• Number of men is directly proportional to the work done and inversely proportional to the days required, if rate of doing work remains constant.

• Time required is directly proportional to the distance and inversely proportional to the speed.

• Area of a rectangular plot is directly proportional to both its length and breadth.

Example 5:

If 20 men can complete a piece of work in 10 days, how many men are required to complete a work twice that of the previous one, in 5 days? Solution: Number of men is directly proportional to the work and inversely proportional to the days required. Since the work becomes twice and the days are halved, the number of men required will become 2 1/0.5 or 4 times the original work, i.e. 4 20 = 80 men.

Example 6:

Ohm’s law can be combined with Joule’s law to show that Power varies directly as the square of electric current, when the resistance is constant; and it varies directly as resistance, when the current is kept constant. In an experiment involving two separate circuits, it was found that the ratio of power across circuit 1 to circuit 2 was 3 : 5 and the resistances in circuit 1 and circuit 2 were in the ratio 5 : 12. Find the ratio of currents across the two circuits (circuit 1 : circuit 2). Solution: Let power be denoted by P, electric current by I and resistance by R. Then, P I2R or P = k1 I2R I2 = (1/k1) (P/R) Thus, Hence, the currents across the two circuits are in the ratio 6 : 5.

Example 7:

The speed of a train varies directly as the fourth power of the amount of coal fed to it and inversely as the square of number of compartments attached to it. If the quantity of coal is doubled and the number of compartments is tripled, then what is the ratio of the original speed to the new speed of the train? Solution: Let the Speed, amount of coal and number of compartments be denoted by s, c and n respectively. Then, Since the amount of coal was doubled and the number of compartments tripled, hence Hence, the ratio of the original speed to the new speed of the train was 9 : 16.

VARIATION BY PARTS

When one quantity is changing with a change in another quantity but the change in both the quantities is not proportional, it is a case of variation by parts.

For example:

• When x is partially constant but partially varying directly with y, it is expressed as:

• When x is partially constant but partially varying inversely with y, it is expressed as:

• When x is partially varying directly with y and partially varying directly with z, it is expressed as:

• When x is partially varying inversely with y and partially varying inversely with z, it is expressed as:

• When x is partially varying directly with y and partially varying inversely with z, it is expressed as:

Some real life examples of variation by parts:

• Cost of production is partially constant (due to fixed cost) and partially varying directly with the number of units produced.

• Total expense of a dinner party is partially constant (due to fixed cost) and partially varying directly with the number of guests.

• Perimeter of a rectangular plot is partially varying directly with its length and partially varying directly with its breadth (or you could also say that the perimeter varies directly with the sum of its length and breadth).

The following graph shows the typical nature of the relation between two quantities x and y, which are directly related, but not varying proportionately. Notice that the graph is not passing through the origin.

Example 8:

If the cost of producing 100 items is Rs. 2,500 and that of producing 200 items is Rs. 4,500, what will be the cost of producing 300 items? Solution: Cost of production = fixed cost + variable cost, where Variable cost = number of items produced per unit cost 2500 = fixed cost + 100 per unit cost … (i) 4500 = fixed cost + 200 per unit cost … (ii) On subtracting equation (i) from equation (ii), we get, 2000 = 100 per unit cost Per unit cost = 2000/100 = Rs. 20 On substituting the value of per unit cost in equation (i) we get, Fixed cost = 2500 100 20 = 2500 2000 = Rs. 500 Hence, the cost of producing 300 items = 500 + 300 20 = 500 + 6000 = Rs. 6,500

Example 9:

The amount that a tourism bus service charges its clients includes both fixed and variable costs. For each batch of tourists, the variable cost per person is directly proportional to the number of tourist spots in the area and inversely proportional to the number of tourists present in that batch. A batch of 15 tourists, visit both Rajasthan and Goa, where the number of tourist spots are 13 and 18 respectively. In Rajasthan, the total amount that each tourist paid was Rs. 1500; whereas in Goa, the amount paid by each tourist was Rs. 2000. If the same batch wants to visit Kashmir, where the number of tourist spots is 25, then how much would each tourist have to pay? Solution: Charges per tourist = Fixed charges + Variable charges For Goa, we have: For Rajasthan, we have: Subtracting (ii) from (i), we have, k = 1500 Fixed charges = 2000 – (1500/15) 18 = 2000 – 100 18 = 200 Hence, for Kashmir, we have: Thus, each tourist would have to pay Rs. 2700 if the batch were to visit Kashmir.

Example 10:

The speed of railway engine is 42 km per hour when no compartment is attached, and the reduction in speed is directly proportional to square root of the number of compartments attached. If the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached, then the maximum number of compartments that can be carried by the engine is [CAT 1999] (1) 49(2) 48(3) 46 (4) 47 Solution: Let S denote the speed of the train when no compartment is attached to the engine, and n denote the number of compartments attached. Thus, S = 42 km/hr and When 9 compartments are attached, the speed of the train reduces from 42 km/hr to 24 km/hr; i.e. a decrease of 18 km/hr. Hence, 18 = k k = 18/3 = 6 Hence, the equation becomes: Now, maximum number of compartments can be carried by the engine when the speed is reduced to its minimum. Hence, we have, 42 = 6 = 7 or n = 49 When 49 compartments are added to the engine, the speed of the train is reduced to 0 km/hr (i.e. the train is not moving). Hence, the maximum number of compartments that can be added, such that the train keeps moving, is 48. Hence, option 2.

Example 11:

Total expenses of boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders? [CAT 1999] (1) Rs. 550(2) Rs. 560 (3) Rs. 540(4) Rs. 570 Solution: Total Expenses of the Boarding House = Fixed Expenses + Variable Expenses Also, Variable Expenses Number of boarders Variable Expenses = kn, where n is the number of boarders and k is a constant Total Expenses of the Boarding House = Fixed Expenses + kn Now, when there are 25 boarders, the expense per boarder is Rs. 700. Hence, the total expenses of the boarding house would have been Rs. 700 25 700 25 = Fixed Expenses + 25k… (i) Similarly, when there are 50 boarders, we get: 600 50 = Fixed Expenses + 50k… (ii) Subtracting equation (i) from (ii), we get, 25k = (600 50) – (700 25) = (600 50) – (350 50) = 50 250 = 12500 k = 500 and Fixed Expenses = Rs. 5000 Hence, we have: Total Expenses of the Boarding House = 5000 + 500n Thus, when there are 100 boarders, Total Expenses of the Boarding House = 5000 + 500(100) = 5000 + 50000 = Rs. 55000 The average expense per boarder will be 55000/100 = Rs. 550 Hence, option 1.