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Question 1: A mixture of paint is prepared by mixing 1 part of green pigments with 6 parts of the base. In the following table, find the parts of base needed to be added.
Parts of green pigment  1  4  5  6 
Parts of base  6  x_{1}  x_{2}  x_{3} 
Solution: Here, as the base increased, the required number of green pigments will also increase.
∴ The quantity vary directly:
Thus, the required unknown quantities are: x_{1} = 24, x_{2} = 30, and x_{3} = 36.
Question 2: A machine fills 540 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Numbers of bottles filled  Number of hours 
540  6 
x  5 
Let the required number of bottles to be filled in 5 hours = x
Since, more number of bottles, more number of hours would be required.
∴ The given quantities very directly.
= 5 * 90 = 450
Thus, the required number of bottles = 450.
Question 3: Jagmeet has a road map with a scale of 1 cm = 20 km. He drives on a road for 72 km. What would be his distance covered in the map?
Solution: Let the distance covered on the map be ‘x’ km.
∴ We have:
Actual distance covered on the road (in km)  Distance covered (represented) on the map ( in cm) 
20  1 
72  x 
It is a case of direct variation.
Thus, the required distance on the map is represented as 3.4 cm.
Question 4: In a PG House, the food provision for 20 persons is for 10 days. How long would the food provision last if there were 5 more persons in that PG house?.
Solution: Number of persons added = 5
∴ Present number of persons = 20 + 5 = 25
Since, for more persons, the food will last less number of days.
∴ It is a case of inverse variation.
∴25 * x = 20 * 10
∴ The food will now last for 8 days.
Question 5: A contractor estimates that 5 persons complete a task in 4 days. If he uses 4 persons instead of 5, how long should they take to complete the task?
Solution: More is the number of persons, less is the time to complete the task.
∴ It is a case of inverse variation,
Now, we have:
Number of presons  Number of days to complete the task 
5  4 
4  x 
5 * 4 = 4 * x
Thus, the required number of days = 5.
Question 6: A school has 9 periods a day each of 50 minutes duration. How many period will there be, if the duration of every period is reduced by 5 minutes?
Solution: Presents duration each period = (50 – 5) minutes = 45 minutes.
Let the present number of period be ‘x’.
Since, more the number of periods, less is the duration of a period.
∴ It is a case of inverse variation.
We have:
∴ 9 * 50 = x * 45
Thus, the required number of periods = 10.
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