Introduction to direct variation in math
This topic has more influence in our day today life. It represents the change of a variable with respect to the other or others. If there is an increase in two related variables, it is called the direct variation.
Example:
A person wages can be decided based on the number of hours they work. If they work for more number of hours, the wages will be more. If w represents the total wages, k the wage for an hour and h representing the number of hours, then it can be related as follows: `w/h` = k
Therefore the total wages is given by: w = kh.
Now let see few problems on this topic direct variation in math.
Examples on Direct Variation in Math
Ex 1: If w varies directly as h and given w = 125 and h = 25, find:
(i) the relation connecting w and h,
(ii) the value of w when h = 35,
(iii) the value of h when w = 250.
Sol: (i) Given: w ∞ h.
This implies, w = kh, where k is the constant of proportionality.
Therefore, when w = 125 and h = 25, we get:
w = kh
125 = k(25)
Therefore, k = `125/25` = 5.
Hence the equation is written as w = 5h.
(ii) When h = 35, w = 5(35)
= 175
Therefore, w = 175.
(iii) When w = 250, 250 = 5h
h = 250/5
h = 50.
Therefore, h = 50.
Ex 2: If s, the speed, varies directly as d, the distance and given s = 60 mph and d = 120 miles, find:
(i) the relation connecting s and d,
(ii) the value of s when d = 150,
(iii) the value of d when s = 45.
Sol: (i) Given: s ∞ d.
This implies, s = kd, where k is the constant of proportionality.
Therefore, when s = 60 and d = 120, we get:
s = kd
60 = k(120)
Therefore, k = 60/120 = 1/2.
Hence the equation is written as s = d/2.
(ii) When d = 150, s = ½(150)
= 75
Therefore, d = 75 mph.
(iii) When s = 45, 45 = ½ (d)
d = 2 * 45
d = 90.
Therefore, d= 90 miles.
More Example on Direct Variation in Math
Ex 3: The fare (F) of a taxi varies directly as the distance (D) travelled. If the distance is 50 km, the cost is $35. What will be the cost if the distance traveled is 125km?
Sol: Given: F ∞ D.
This implies that, F = kD
Therefore, when F = 35, D = 50, we have 35 = k(50)
K = 0.7.
Therefore, the equation can be written as F = 0.7D
Now, when D = 125km, F = 0.7× 125
= 87.5 dollars.
I hope the discussion on the direct variation would have been clear by now on going through those above examples.
This topic has more influence in our day today life. It represents the change of a variable with respect to the other or others. If there is an increase in two related variables, it is called the direct variation.
Example:
A person wages can be decided based on the number of hours they work. If they work for more number of hours, the wages will be more. If w represents the total wages, k the wage for an hour and h representing the number of hours, then it can be related as follows: `w/h` = k
Therefore the total wages is given by: w = kh.
Now let see few problems on this topic direct variation in math.
Examples on Direct Variation in Math
Ex 1: If w varies directly as h and given w = 125 and h = 25, find:
(i) the relation connecting w and h,
(ii) the value of w when h = 35,
(iii) the value of h when w = 250.
Sol: (i) Given: w ∞ h.
This implies, w = kh, where k is the constant of proportionality.
Therefore, when w = 125 and h = 25, we get:
w = kh
125 = k(25)
Therefore, k = `125/25` = 5.
Hence the equation is written as w = 5h.
(ii) When h = 35, w = 5(35)
= 175
Therefore, w = 175.
(iii) When w = 250, 250 = 5h
h = 250/5
h = 50.
Therefore, h = 50.
Ex 2: If s, the speed, varies directly as d, the distance and given s = 60 mph and d = 120 miles, find:
(i) the relation connecting s and d,
(ii) the value of s when d = 150,
(iii) the value of d when s = 45.
Sol: (i) Given: s ∞ d.
This implies, s = kd, where k is the constant of proportionality.
Therefore, when s = 60 and d = 120, we get:
s = kd
60 = k(120)
Therefore, k = 60/120 = 1/2.
Hence the equation is written as s = d/2.
(ii) When d = 150, s = ½(150)
= 75
Therefore, d = 75 mph.
(iii) When s = 45, 45 = ½ (d)
d = 2 * 45
d = 90.
Therefore, d= 90 miles.
More Example on Direct Variation in Math
Ex 3: The fare (F) of a taxi varies directly as the distance (D) travelled. If the distance is 50 km, the cost is $35. What will be the cost if the distance traveled is 125km?
Sol: Given: F ∞ D.
This implies that, F = kD
Therefore, when F = 35, D = 50, we have 35 = k(50)
K = 0.7.
Therefore, the equation can be written as F = 0.7D
Now, when D = 125km, F = 0.7× 125
= 87.5 dollars.
I hope the discussion on the direct variation would have been clear by now on going through those above examples.
