WRITE A DIRECT VARIATION EATION THAT RELATES THE TWO VARIABLES

The variables x and y vary directly. Write an equation that relates x and y. Then find y when x = 12.

Problem 1 :

x = 4, y = 8

Solution :

Equation of direct variation is y = kx

8 = k(4)

8/4 = k

2 = k

So, the equation that relates x and y is y = 2x.

To find y :

y = 2x

Substitute the value of x.

y = 2(12)

y = 24

So, the value of y is 24 when x = 12.

Problem 2 :  

x = -3, y = -5

Solution :

The equation is in the form y = kx

-5 = k(-3)

-5/-3 = k

5/3 = k

So, the equation that relates x and y is y = 5/3x.

To find y :

y = (5/3)x

Substitute the value of x.

y = 5/3(12)

y = 20

So, the value of y is 20 when x = 12.

Problem 3 :

x = 35, y = -7

Solution :

The equation is in the form y = kx

-7 = k(35)

-7/35 = k

-1/5 = k

So, the equation that relates x and y is y = -1/5x.

To find y :

y = -1/5x

Substitute the value of x.

y = -1/5(12)

y = -12/5

y = -2.4

So, the value of y is -2.4 when x = 12.

The variables x and y vary directly. Write an equation that relates x and y. Then find x when y = -4.

Problem 1 :

x = 5, y = -15

Solution :

The equation is in the form y = kx

-15 = k(5)

-15/5 = k

-3 = k

So, the equation that relates x and y is y = -3x.

To find x :

y = -3x

Substitute the value of y.

-4 = -3x

-4/(-3) = x

4/3 = x

1.33 = x

So, the value of x is 1.33 when y = -4

Problem 2 :

x = -6, y = 8

Solution :

The equation is in the form y = kx

8 = k(-6)

8/-6 = k

-4/3 = k

So, the equation that relates x and y is y = -4/3x.

To find x :

y = -4/3x

Substitute the value of y.

-4 = -4/3x

-4 × (-3/4) = x

3 = x

So, the value of x is 3 when y = -4

Problem 3 :

x = -18, y = -2

Solution :

The equation is in the form y = kx

-2 = k(-18)

-2/-18 = k

1/9 = k

So, the equation that relates x and y is y = 1/9x.

To find x :

y = 1/9x

Substitute the value of y.

-4 = 1/9x

(-4) × 9 = x

-36 = x

So, the value of x is -36 when y = -4.

Problem 4 :

Write a direct variation equation that relates x inches to y centimeters.

Solution :

x-inches and y-centimeters

y ∝ x

y = kx

When x = 1 inch, y = 2.54 cm

2.54 = k(1)

k = 2.54

y = 2.54 x

So, the required relationship is y = 2.54 x.

Problem 5 :

m varies as directly as cube of x, when m = 200, x = 2. Find the value of m when x = 0.4.

Solution :

m ∝ x³

m = kx³

When m = 200, x = 2

200 = k(2)³

200/8 = k

k = 25

Applying the value of k, we get

m = 25x³

When x = 0.4

m = 25(0.4)³

= 25(0.064)

m = 1.6

So, the required value of m is 1.6.

Problem 6 :

Lauren is paid $225 for 25 hours of work. Use direct proportion to calculate how much she is paid for 30 hours of work.

Solution :

Let t be the number of hours he is working and A be the amount paid.

A ∝ t

A = kt

When t = 25, A = 225

225 = k(25)

k = 225/25

k = 9

Applying the value of k, we get

A = 9t

Calculating the amount for 30 hours of work :

A = 9(30)

A = 270

Problem 7 :

A car covers a distance in 40 minutes with an average speed of 60 km per hour. The average speed to cover the same distance in 30 minutes is

(a) 80 km/h    (b) 45/2 km/h    (c) 70 km/h       (d) 45 km/h

Solution :

Average speed of the car = 60 Km/hour

Distance covered in 40 minutes = 60 × (40/60) Km = 60 × (2/3) = 40 km

Speed required to cover 40 km in 30 minutes

= 40/(30/60)

= (40 × 60)/30 Km/h

= 80Km/hour

Problem 8 :

Which of the following is in direct proportion?

(a) One side of a cuboid and its volume.

(b) Speed of a vehicle and the distance travelled in a fixed time interval.

(c) Change in weight and height among individuals.

(d) Number of pipes to fill a tank and the time required to fill the same tank.

Solution :

Option a :

When side length of the cuboid increases its volume will also increase, then it is direction proportion. So, option a is correct.