Speaking in general terms, two variables, x and y, are said to vary directly if when one of them
increases, the other also increases proportionally and if one of them decreases, the other also decreases proportionally.
The word ‘proportionally’ is important here. If one of them doubles, the other should also double; if one of them becomes one-third, the other should also become one-third.
E.g. Speed and Distance covered in same time, say 1 hr, behave in the above manner. Consider I cover a certain distance at a certain speed. If the speed doubles, the distance covered also doubles and if the speed becomes one-third, the distance covered also becomes one-third.
If two variables, x and y vary directly, then the relation is depicted as x α y.
As the variable x assumes different values, say x1, x2, x3, …, the variable y will also change and let’s say it assumes y1, y2, y3, … values correspondingly.
E.g. The variables could be ‘the number of men working’ and ‘the number of chairs made’. Now from common sense we know that the number of chairs build and the number of men working are directly proportional (number of men increasing means the number of chairs build also increases proportionally). Next the number of men working could vary i.e. the number of men working could be 5 or 8 or 10 or 25. Accordingly the number of chairs build would also keep changing. Thus, if number of men is denoted as x1, x2, x3, x4, in the different scenarios, the number of chairs build in the respective scenarios would be denoted as y1, y2, y3, y4. Thus, each instance of a certain number of men working and the corresponding number of chairs made will be a pair like (x1, y1), (x2, y2), (x3, y3), (x4, y4).
In the case of direct relation, x α y, we would have the relation 1 = 2 = 3 = 4 =
1 2 3 4 ... x x x x y y y y . Alternately, x k=
y or x = ky, where k is a constant i.e. the pair (x, y) could be any of (x1, y1), (x2, y2), (x3, y3), (x4, y4) but for any pair, the constant k, will be the same.
In questions we would typically have only two instances of the variables, x and y, assuming different values e.g. in the question ‘if 20 people can build 30 chairs, find the number of chairs that 50 people can build?’, there are two instances of the data – instance 1: 20 people, 30 chairs; instance 2: 50 people, x chairs.
One can use any of the following approaches: Approach 1: Use ratios being equal i.e. 1= 2
1 2
x x
y y or 12 = 12
x y
x y (both are the same, use any relation that one
happens to write then). Thus, in this case 20 50= ⇒
30 ? x = 75. Or one can write the ratios as = 20 30
50 x , which again gives x = 75.
Approach 2: Use x = ky, where k is a constant.
One set of corresponding values (x, y) will be given – 20 people, 30 chairs. Use these values to find the value of the constant k. i.e. 30 = 20k i.e. k = 1.5
Then use the relation again with, with the already found value of k, to find the unknown in the other set of data i.e. x = 50 × 1.5 i.e. x = 75.
36 | Direct & Inverse Variation
E.g. 1: y varies directly as x and when x = 6, y = 24. What is the value of y, when x = 5?
Approach 1: Using 1 = 2 1 2 x x y y We have = ⇒ 2 6 5 24 y y2 = 5 × 4 = 20. Approach 2: Using x = ky
From the set of given values of (x, y), we have 6 = k × 24 i.e. = 1 4
k
Using this value of k in the other set of data (5, y), we have 5=1× ⇒
4 y y = 5 × 4 = 20.
E.g. 2: The value of a diamond varies directly as the square of its weight. If a diamond weighing
1.5 gms has a value of Rs. 9,000, what will be the value of a diamond weighing 2.5 gms? Approach 1: Using 1 = 2
1 2
x x
y y
Denoting value by v and weight by w, in this case we have 12 = 22
1 2
v v
w w
Plugging values, we get = 2 ⇒ = × = ×
2 2 2 9000 6.25 9000 25 9000 1.5 2.5 2.25 9 v v =25,000. Approach 2: Using x = ky
Since the value of the diamond is directly proportional to square of weight, we will have v = k × w2
From the set of given values of (v, w), we have 9000 = k × 1.52 i.e. = = ×
× 9000 9000 100
2.25 9 25
k =
4,000.
Using this value of k in the other set of data (v2, 2.5), we have v2 = 4000 × 2.52 = 4000 × 6.25 = 25,000.
E.g. 3: The volume of a sphere is directly proportional to the cube of its radius. If the volumes of
two spheres are in the ratio 8 : 1, find the ratio of the radii of the spheres.
Representing volume of the sphere by ‘v’ and radius by ‘r’, based on the given proportionality relation we would have = =
3 3 1 1 1 1 3 2 2 2 2 i.e. v r v r
v r v r . (We could also have written the relation as
follows 1 = 2
3 3
1 2
v v
r r , which when rearranged to result in v1/v2, given in the question, would have resulted in the same expression)
We are given that 1 =
2 8 1 v v . Thus, = ⇒ = 3 1 1 2 2 8 2 1 1 r r r r .
Direct & Inverse Variation | 37
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Inverse Variation
Speaking in general terms, two variables, x and y, are said to vary inversely if when one of them increases, the other decreases proportionally and if one of them decreases, the other increases proportionally.
The word ‘proportionally’ is important here. If one of them doubles, the other should halve; if one of them becomes one-third, the other should become thrice.
E.g. Speed and Time taken to cover a certain distance behave in the above manner. Consider I take a certain time to go from home to office at a certain speed. If the speed doubles, the time taken will halve and if the speed becomes one-third, the time taken would become thrice.
If two variables, x and y vary inversely, then the relation is depicted as x α 1
y.
As the variable x assumes different values, say x1, x2, the variable y will also change and let’s say it correspondingly assumes values y1, y2. By now you would be clear by the meaning of x1, x2, y1 and y2. In the case of inverse relation, we would have the relation x1 × y1 = x2 × y2.
Alternately, we could use the relation, x = k
y or x × y = k, where k is a constant i.e. the pair (x, y) could be
any of (x1, y1), (x2, y2), (x3, y3), (x4, y4), …… but for any pair, the constant k, will be the same.
In questions we would typically have only two instances of the variables, x and y, assuming different values e.g. in the question ‘if a wall can be build in 10 days by 18 people, how many people will be needed to build the wall in 15 days?’, the variables are ‘number of people’ and ‘days taken’ and they vary inversely. Further there are two instances of this set of data – instance 1: 18 people, 10 days; instance 2: x people, 15 days.
One can use any of the following approaches: Approach 1: Use x1 × y1 = x2 × y2
Thus, in this case 18 × 10 = 15 × x i.e. x = 12. Approach 2: Use x = k
y, where k is a constant.
One set of corresponding values (x, y) will be given – 18 people, 10 chairs. Use these values to find the value of the constant k.
= 18
10
k i.e. k = 180
Then use the relation again with, with the already found value of k, to find the unknown in the other set of data.
= 180 15
x i.e. x = 12.
E.g. 4: If y varies inversely as x, when x = 2, y = 3. What is the value of x if y = 1?
Approach 1: Use x1 × y1 = x2 × y2
Using this relation we will have 2 × 3 = x × 1 i.e. x = 6. Approach 2: Use x = k
y or x × y = k, where k is a constant.
From the set of given values of (x, y), we have 2 × 3 = k i.e. k = 6.
38 | Direct & Inverse Variation
E.g. 5: The number of four wheelers sold in a year vary inversely with the number of two
wheelers sold in that year. If 4000 four wheelers were sold in one particular year, 21000 two wheelers were sold that year. How many two wheelers were sold in a year, when 6000 four wheelers were sold?
Approach 1: Use x1 × y1 = x2 × y2
Using this relation we will have 4,000 × 21,000 = 6,000 × y i.e. y = 14,000. Approach 2: Use x = k
y or x × y = k, where k is a constant.
From the set of given values of (4000, 21000), we have 4000 × 21000 = k i.e. k = 84,000,000. Using this value of k in the other set of data (6000, y), we have 6000 × y = 84,000,000 i.e. y = 14,000.
E.g. 6: The cost per kg of rice varies inversely with the square of the quantity of rice produced
in a year. When 7 million tons of rice was produced, its cost was Rs. 36/kg. How much was the production in the year when the cost of rice was Rs. 49/kg?
Denoting price by p and quantity by q, the relation in this case will be p1 × q12 = p2 × q22. Plugging the given values, 36 × 72 = 49 × q22 i.e. q2 = 6.
Joint Variation
Sometimes more than two variables could be involved in the questions, say x, y and z and the proportionality relation is such that x is directly proportional to y and x is inversely proportional to z.
Since x is common to the two proportionality relation, we should form two separate relations
=
x
y some constant and x × z = another constant. And then the two relation can be clubbed into
one relation like x z ky× = . From this last relation, for difference instances of the three variables, (x1, y1, z1) and (x2, y2, z2), we will have
× × = 2 2 1 1 1 2 x z x z y y .
E.g. 7: The value of a circular coin varies directly with the square of it’s radius and varies
inversely with it’s thickness. What is the ratio of values of two coins with the ratio of their radii being 2 : 3 and the ratio of their thickness being 1 : 4.
If the value, radius and thickness is denoted by v, r and t respectively, using the proportionality relation we have 2
v
r = a constant and v × t = a constant.
Clubbing the two relations together we have ×2
v t
r = a constant.
Now we have two coins, the ratio of who’s values have to be found i.e. we need to find v1/v2. And we are given that 1 =
2 2 3 r r and 12 = 1 4 t t .
From the relation tying up the three variables, we have 1× 1 = 2× 2
2 2 1 2 v t v t r r . On re-arranging we have = × 2 2 1 1 t v r
. Plugging the given data, we find, = × =
2
1 2 4 16
Direct & Inverse Variation | 39
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Exercise
1. y varies directly as the square of x. When x = 4, y = 2. What is the value of y when x = 3
2? a. 9 32 b. 3 4 c. 16 3 d. 128 9
2. y varies inversely as the cube root of x. If y = 7, when x = 8, what is the value of x when y = 56?
a. 1/2 b. 1/8 c. 1/3 d. 1/64
3. The value of a bar of gold weighing 400 gm, varies inversely with the square root of the fraction of impurities in it. If the value of a bar containing 25 gms of impurities is Rs. 45000, how many gms of pure gold is there in a bar costing Rs. 90000?
a. 12.5 b. 387.5 c. 375 d. 393.75
4. The value of a silver coin varies directly as the square of its diameter, when thickness is constant and varies directly as its thickness when diameter remains constant. Two silver coins have the diameters in the ratio 4 : 3. Find the ratio of the thickness if the value of the first coin is four times the value of the second coin.
a. 3 : 4 b. 4 : 3 c. 16 : 3 d. 9 : 4
5. The value of a diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2 : 3, what is the loss percentage in the value?
40 | Assignment: Ratio Proportion Variation