SOLVING WORD PROBLEMS INVOLVING DIRECT AND INVERSE PROPORTION

Problem 1 :

The number of pencils sold varies directly as the cost. If 5 pencils cost $0.45, find the cost of 7 pencils.

Solution :

Cost of 5 pencils = 0.45

Cost of 7 pencils = ?

Let x be the number of pencils and y be the cost of pencils.

y ∝ x

y = kx -----(1)

0.45 = 5k

k = 0.45/5

k = 0.9

Applying the value of k in (1)

y = 0.9x

Applying x = 7, we get

y = 0.9(7)

y = 6.3

So, the cost of 7 pencils is $6.3.

Problem 2 :

On a scale drawing, 2 feet represents 30 yards. How many yards are represented by 3 feet?

Solution :

2 feet = 30 yards

Let x be the number of feet and y be the number of yards.

y = kx -----(1)

30 = 2k

k = 30/2

k = 15

Applying the value of k in (1)

y = 15x

When x = 3, y = 15 (3)

= 45 yards

So, 3 feet = 45 yards.

Problem 3 :

On a map, 180 miles are represented by 4 inches. How many miles are represented by 6 inches?

Solution :

4 inches = 180 miles

Let x be the number of inches and y be the number of miles.

y = kx -----(1)

180 = 4k

k = 180 / 4

k = 45

Applying the value of k in (1)

y = 45x

When x = 6, y = 45 (6)

= 270 inches

So, 6 feet = 270 inches

Problem 4 :

y varies directly as the square of x. If y is 25 when x is 3, find y when x is 2.

Solution :

y = kx² -----(1)

When x = 3, y = 25

25 = k(3)²

k = 25/9

Applying the value of k in (1), we get

y = (25/9)x² 

When x = 2

y = (25/9)(2)² 

y = 100/9

Problem 5 :

Laura has a mass of 60 kg and is sitting 265 cm from the fulcrum of a seesaw. Bill has a mass of 50 kg. How far from the fulcrum must he be to balance the seesaw? (Hint: The distance from the fulcrum varies inversely as the mass).

Solution :

Distance of fulcrum be F and mass be M.

F = k/M

k = FM

When M = 60 kg, then F = 265 cm

k = 60(265)

k = 15900

Applying the value of k in (1), we get

F = 15900/M

When M = 50 kg, then F = ?

F = 15900/50

F = 318 cm

So, the required length is 318 cm.

Problem 6 :

Time varies inversely as speed if the distance is constant. A trip takes 4 hours at 80 km/h. How long does it take at 64 km/h?

Solution :

Let T be the time and S be the speed. Let distance be the constant, that is k.

T = kS ----(1)

4 = k (80)

k = 4/80

k = 1/20

Applying the value of k in (1), we get

T = (1/20) S

T = (1/20) 64

T = 3.2 hours.

Problem 7 :

The number of hours required to do a job varies inversely as the number of people working. It takes 8 hours for 4 people to paint the inside of a house. How long would it take 5 people to do the job?

Solution :

Number T be the number of hours required and N be the number of people.

T = kN ----(1)

When N = 4, T = 8

8 = k (4)

k = 8/4

k = 2

Applying the value of k in (1), we get

T = 2N

When N = 5, then T = ?

T = 2(5)

T = 10 hours.