NCERT Solutions for Class 8 Maths - Direct and Inverse Proportions - 1

Try These(Page No. 132)

Q1: Observe the following tables and find if x and y are directly proportional.

i.e. Each ratio is the same.
∴ x and y are directly proportional

i.e. All the ratios are not the same.
∴ x and y are not directly proportional.

i.e. All the ratios are not the same.
∴ x and y are not directly proportional.

Q2: Principal = Rs 1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) change in direct proportion with time period.

Solution. Case of Simple Interest [P = Rs 1000, r = 8% p.a.]

For simple interest, SI = P × r × t / 100.
Here P and r are fixed, so SI is proportional to t.
∵ In each case the ratio is the same.
∴ The simple interest changes in direct proportion with time period.
Case of Compound Interest [P = Rs 1000, r = 8% p.a.]

For compound interest the amount after n years is A = P(1 + r/100)ⁿ.
Thus CI = A - P = P{(1 + r/100)ⁿ - 1}.
When P, r and n are fixed, the bracket {(1 + r/100)ⁿ - 1} is a constant and CI is proportional to P but not to t when t (time) varies in the compound case shown in the table.
∵

  is not the same in each case.
∴ The compound interest does not change in direct proportion with time period.

Think, Discuss and Write 

Question: If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution. For simple interest,

Since rate and time period are constant, SI = P × (r × t /100), so SI changes directly with P.
Thus, the simple interest changes in direct proportion with the principal.
For compound interest,

When rate and time period are fixed, the amount A = P(1 + r/100)ⁿ and the compound interest CI = A - P = P{(1 + r/100)ⁿ - 1}.

Here {(1 + r/100)ⁿ - 1} is a constant, so CI = P × (constant).
i.e. CI changes proportionally with P.
Thus, the compound interest also changes directly in proportion with principal when rate and time are fixed.

Exercise 11.1

Q1: Following are the car parking charges near a railway station upto

Check if the parking charges are in direct proportion to the parking time.
Sol: We have,
Compute the charge per hour for each entry. If charges per hour are the same, they are directly proportional.
Since    (charges per hour are not constant),
∴ The parking charges are not in direct proportion with the parking time.

Q2: A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Sol: Let the red pigment be represented by x₁, x₂, x₃, ... and the base represented by y₁, y₂, y₃, ... .
Since the ratio base : red = 8 : 1, the base required = 8 × (parts of red).
∴ The quantities vary directly (base increases as red increases).

Now, for given red parts: if red = 4 → base = 8 × 4 = 32; red = 7 → base = 8 × 7 = 56; red = 12 → base = 8 × 12 = 96; red = 20 → base = 8 × 20 = 160.

Thus, the required parts of base are: 32, 56, 96 and 160.

Q3: In question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Sol:

Here, x₁ = 1 part red requires y₁ = 75 mL base. For y₂ = 1800 mL base let red be x₂.
For direct proportion, x₁/x₂ = y₁/y₂ or x₂ = x₁ × y₂/y₁.

Thus, x₂ = 1 × 1800/75 = 24 parts of red pigment.

Q4: A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution: 

Number of bottles filled	Number of hours
840	6
x	5

The number of bottles filled is directly proportional to time.
So x = 840 × 5 / 6.
               

Thus, the required number of bottles = 700.

Q5: A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Solution: Let the actual length of the bacteria = x cm

Number of times photograph enlarged	Length (in cm)
1	x
50,000	5

The enlargement is directly proportional to the length.
So 50,000 : 1 = 5 : x or x = 5 / 50,000.

Compute x: x = 5 ÷ 50,000 = 0.0001 cm.
For 20,000 times enlargement, enlarged length = actual × 20,000 = 0.0001 × 20,000 = 2 cm.
Again,

Q6: In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28m, how
long is the model ship?

Sol: Let the required length of the model of the ship = x cm
We have:

Length of the ship	Height of the mast
28	12
x	9

Heights and lengths are proportional. Convert actual mast height to cm: 12 m = 1200 cm, and actual ship length 28 m = 2800 cm.
Scale factor = model mast / actual mast = 9 / 1200.
So model length x = actual length × scale = 2800 × 9 / 1200.

Compute: 2800 × 9 / 1200 = 2800 × (3/400) = 21 cm.
Thus, the required length of the model = 21 cm

Q7: Suppose 2 kg of sugar contains 9 * 10⁶ crystals. How many sugar crystals are there in
(i) 5 kg of sugar? 
(ii) 1.2 kg of sugar?
Sol: Let the required number of sugar crystals be x in 5 kg of sugar.
We have:

More sugar ⇒ more crystals, so direct variation between mass and number of crystals.
Given 2 kg → 9 × 10⁶ crystals. For 5 kg: x = 9 × 10⁶ × 5 / 2 = (45/2) × 10⁶ = 22.5 × 10⁶ = 2.25 × 10⁷ crystals.

Thus, the required number of sugar crystals = 2.25 × 10⁷.
(ii) Let the number sugar crystals in 1.2 kg of sugar be y.
∴ We have:

For direct variation, y = 9 × 10⁶ × 1.2 / 2 = 9 × 10⁶ × 0.6 = 5.4 × 10⁶ crystals.
∴ For a direct variation,
                 

Q8: Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Sol: Let the required distance covered in the map be x cm.
We have:

Direct proportion: x / 72 = 1 / 18, so x = 72 × 1 / 18 = 4 cm.
Since, it is a case of direct variation,

Thus, the required distance on the map is 4 cm.

Q9: A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time(i) the length of the shadow cast by another pole 10 m 50 cm high and (ii) the height of a pole which casts a shadow 5 m long.
Sol: (i) Let the required length of shadow be x cm.
We have:

Convert to cm: pole1 height = 5 m 60 cm = 560 cm, shadow1 = 3 m 20 cm = 320 cm, pole2 height = 10 m 50 cm = 1050 cm.
Since height and shadow are proportional, x / 1050 = 320 / 560.

Compute x = 1050 × 320 / 560 = 1050 × (32/56) = 1050 × (4/7) = 600 cm = 6 m.
Thus, the required length of the shadow = 600 cm or 6 m

(ii) Let the required height of the pole be y cm.
∴ We have:

Use proportion y / 500 = 560 / 320 (since shadow of 5 m = 500 cm).

Compute y = 500 × 560 / 320 = 500 × (56/32) = 500 × (7/4) = 875 cm = 8 m 75 cm.
Thus, the required height of the pole = 875 cm or 8 m 75 cm

Q10: A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Sol: Since the speed is constant, distance varies directly with time.
Let the required distance for 5 hours (= 300 minutes) be x km.
We have:

Distance (in km)	Time (in minutes)
14 x	25 300

So x = 14 × 300 / 25 = 14 × 12 = 168 km.
∴ The required distance = 168 km.

Try These(Page No. 139)

Ques: Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

(i) 

(ii)

(iii)

Sol:
(i) ∵ x₁ = 50 and y₁ = 5 ⇒ x₁y₁ = 50 * 5 = 250
       x₂ = 40 and y₂ = 6 ⇒ x₂y₂ = 40 * 6 = 240
       x₃ = 30 and y₃ = 7 ⇒ x₃y₃ = 30 * 7 = 210
       x₄ = 20 and y₄ = 8 ⇒ x₄y₄ = 20 * 8 = 160
Also 250 ≠ 240 ≠ 210 ≠ 160 or
x₁y₁ ≠ x₂y₂ ≠ x₃y₃ ≠ x₄y₄
∴ x and y are not in inverse proportion

and x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄

∴ x and y are in inverse proportion.

and x₁y₁ = x₂y₂ = x₃y₃ ≠ x₄y₄ ≠ x₅y₅ ≠ x₆y₆

∴ x and y are not in inverse variation.

Exercise 11.2

Q1: Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Sol:

REMEMBER
If an increase in one quantity brings about a corresponding decrease in the other and vice versa, then the two quantities vary inversely.
(i) If number of workers are increased then time to complete the job would decrease.
∴ It is a case of inverse variation.

(ii) For longer distance, more time would be required.
∴ It is not a case of inverse variation

(iii) For more area of land, more crops would be harvested.
∴ It is not a case of inverse variation.

(iv) If speed is more, time to cover a fixed distance would be less.
∴ It is a case of inverse variation.

(v) For more population, less area per person would be required.
∴ It is a case of inverse variation.

Q2: In a television game show, the prize money of Rs 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Sol: If more the number of winners, less is the prize money per person. Prize per person = 1,00,000 / (number of winners).
∴ It is a case of inverse proportion.

Thus, the table is completed as under:

(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Sol: (i) More the number of spokes, less is the measure of the angle between consecutive spokes.
Angle between consecutive spokes = 360° / (number of spokes).
∴ It is a case of inverse variation.
Thus,

Thus, the table is completed as under

(ii) Let the required measure of angle be x°.
For 15 spokes, x = 360° / 15 = 24°.

(iii) Let required number of spokes be n.
Given angle = 40°, so n = 360° / 40° = 9.

∴ The required number of spokes = 9

Q4: 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?
Sol: Reduced number of children = 24 - 4 = 20
Total sweets = 24 × 5 = 120.
New share per child = 120 / 20 = 6 sweets.
Since, more the number of children, less is the quantity per child, it is inverse variation.
i.e. 24 × 5 = 20 × x
or  

∴ Each student will get 6 sweets. 

Q5: A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Sol: Number of animals added = 10
∴ Now, the total number of animals = 10 + 20 = 30
Total animal-days available = 20 × 6 = 120.
With 30 animals days = 120 / 30 = 4 days.
Thus, the food will now last for 4 days.

Q6: A contractor estimates that 3 persons could rewire jasminder's house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Sol: More persons ⇒ less time. Total person-days = 3 × 4 = 12.
With 4 persons, days = 12 / 4 = 3 days.
Thus, the required number of days = 3

Q7: A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, How many boxes would be filled?

Sol: Total bottles = 25 × 12 = 300.
With 20 bottles per box, number of boxes = 300 / 20 = 15.
Since more bottles per box ⇒ fewer boxes required, it is inverse variation.
Thus, the required number of boxes = 15

Q8: A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Sol: Let the number of machines required be x.
For a fixed total work, machines × days = constant. So 42 × 63 = x × 54.
Thus x = 42 × 63 / 54 = 42 × (7/6) = 49.
Therefore, the required number of machines = 49

Q9: A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Sol: For a fixed distance, time is inversely proportional to speed.
Time at 80 km/h = 2 × 60 / 80 = 2 × 3/4 = 1.5 hours = 1 hour 30 minutes.

Thus the required number of hours =  

Question 10. Two persons could fit new window in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Sol: (i) Two persons take 3 days ⇒ total person-days = 2 × 3 = 6. If only one person works, days = 6 / 1 = 6 days.
(ii) To finish in 1 day, persons required = total person-days / 1 = 6 persons.
∴ 1 person will complete the job in 6 days and 6 persons are required to complete the job in 1 day.

∴ 1 person will complete the job in 6 days.
(ii) We have

∴ 6 persons will be required to complete the job in 1 day.

Q11: A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school houres to be the same?
Sol: Total school minutes = 8 × 45 = 360 minutes. For 9 periods, duration per period x = 360 / 9 = 40 minutes.
Thus, the required duration per period = 40 minutes.

Do This(Page No. 143)

Q1: Take a sheet of paper. Fold it as shown in the figure. Count the number of parts and the area of a part in each case.
Tabulate your observations and discuss with your friends. Is it a case inverse proportion? Why?

Solution:

Here, more the number of parts, lesser is the area of each part.
Area of each part = (area of paper) / (number of parts).
∴ It is a case of "inverse proportion".

Q2: Take a few containers of different sizes with circular bases. Fill the same amount of water in each container. Note the diameter of each container and the respective height at which the water level stands. Tabulate your observations. Is it a case of inverse proportion?

Sol:

For the same volume of water, a container with a smaller base area gives a higher water level and vice versa.
Thus, as diameter decreases, the height of water increases; this is an inverse relation between diameter (through base area) and height of water.
∴ It is a case of inverse proportion.