SOLVING WORD PROBLEMS ON DIRECT AND INVERSE PROPORTION

Problem 1 :

A worker is paid $200 for 8 days. If he works for 20 days, how much will he get?

Solution :

Let x be the worker for 20 days work.

Number of days             Amount

         8                              200

        20                               x

When number of days is increasing, then the amount they will receive also work. It comes under the concept of direct proportion.

8 ⋅ x = 200 ⋅ 20

x = (200 ⋅ 20)/8

x = 500

So, for 20 days of work, they will earn $500.

Problem 2 :

A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km?

Solution:

Let x be the time taken to cover the distance.

Distance covered (km)        Time taken (min)

     51                                45

   221                                 x

If the distance to be covered is increasing then time taken will also increase. It comes under the concept of direct proportion.

51x = 45(221)

x = 45(221)/51

x = 195 minutes

x = (180 + 15) minutes

180 minutes = 3 hours

Soo, time taken is 3 hours and 15 minutes.

Problem 3 :

x and y are in inverse proportion. If y = 15 then x = 3, find the value of y when x = 9.

Solution:

When y varies inversely to x, y = k/x and k = xy

x and y are inversely and x = 3 and y = 15

xy = 3 × 15

xy = 45

Constant of variation (k) = 45

Now, if x = 9,then

In y = k/x, applying the value of k, we get

y = 45/9

y = 5

Problem 4 :

If xy = 10, then x and y vary ___ with each other.

Solution:

Given that xy = 10 and its known that 10 is a constant.

Hence, it can be concluded that x and y vary inversely.

Problem 5 :

Time taken to cover a distance by a car and speed of the car are said to be in ___ variation.

Solution:

If the speed of the car is increasing, then time taken to reach the distance will decrease. So, it comes under indirect variation.

Problem 6 :

55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days?

Solution:

Let the number of cows that can graze the field in 10 days be x. If number of days is reducing then the number of cows should increase. It is in inverse variation.

Hence, 88 cows can graze the same field in 10 days.

Problem 7 :

If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same would weigh 2 1/2 kilograms?

Solution:

Let x be the number of sheets.

1 kilo gram = 1000 grams

2.5 kilograms = 2500 grams

Number of sheets          weight(grams)

        12                            40

                                    x                           2500

Weight of one sheet = x g

12 ⋅ 2500 = x ⋅ 40

x = (12 ⋅ 2500)/40

x = 750 sheets

Problem 8 :

Observe the following table and find if x and y are in direct variation.

Solution:

Since the ratios are not constant. 

So, x and y are not direct variation.

Problem 9 :

A labourer gets $675 for nine days work. How many days should he work to get $900?

Solution:

Number of days of working       payment

                  9                               675

                  x                                900

To get more pay, the number of days of working should be improved. It comes under direct proportion.

9(900) = x(675)

x = 9(900)/675

x = 12

x = 12

So, he should work for 12 to get $900.

Problem 10 :

By walking for 30 minutes in the morning a person covers 2 kilometers. How much distance will he/she cover in 20 minutes by walking at the same ? 

Solution:

   Time taken (m)              Distance covered (km)

30                                  2 

20                                  x

If time taken is reducing then distance covered will also decrease. It comes under direct variation.

30 x = 2(20)

x = 40/30

= 1.33 km

Problem 11 :

If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Solution:

Each child gets = 5 sweets

24 children will get 24 × 5 = 120 sweets

Total number of sweets = 120

Number of sweets to be distributed = 120

Number of children they get = 24 - 4 ==> 20

each child will get sweets = 120/20 = 6 sweets

Problem 12 :

A car travels 60 km in 1 hr 30 min. How long it take to cover a distance of 100 km at the same speed?

Solution:

Distance covered(km)               Time taken(hours)

             60                                    1.5

         100                                   x

Since it is the same speed, if distance covered is increased then time taken will also increase. It comes under direct variation.

60x = 100(1.5)

x = 100(1.5) / 60

x = 2.5 hours

So, 100 km distance can be covered in 2.5 hours.

Problem 13 :

11 men can dig 6.75 m long trench in one day. How many men should be employed for digging 27 m long trench of the same type in one day?

Solution:

Number of men working               depth of trench

                 11                                         6.75

               x                                          27

Since length of trench is increasing then number of people working will also increase. It comes under direct proportion.

11 (27) = x (6.75)

x = 11(27) / 6.75

x = 44

So, number of men required is 44.

Problem 14 :

120 men had food provisions for 200 days. After 5 days, 30 men die due to an epidemic. How long will the remaining food last?

Solution:

Provision available for 120 men for 200 days .

Total provision can be consumed by 1 man = 120 × 200 = 24000 days.

Since , after 5 days 30 men died

Total provision consumed in 5 days= 120 × 5 = 600 men days

Total provision left = 24000 - 600 = 23400 men days

Persons left after 5days = 120 - 30 = 90 men

No of days it can last for remaining people = 23400/90 = 260 days

Thus, the remaining provision will last for 260 days.

Problem 15 :

A train 400 m long is running at a speed of 72 km/hr. How much time does it take to cross a telegraph post?

Solution:

Length of the train = 400 m

Speed of the train = 72 km/hr

So, time taken to cross a telegraph post 

So, it take 20 sec to cross a telegraph post.