## Monday, August 8, 2022

#### Direct And Inverse Proportions Solution of TS & AP Board Class 8 Mathematics

###### Exercise 10.1

Question 1.

The cost of 5 meters of a particular quality of cloth is Rs. 210. Find the cost of (i) 2 (ii) 4 (iii) 10 (iv) 13 meters of cloth of the same quality.

Given: Cost of 5 m of cloth = Rs. 210

To Find: Cost of 2m, 4m, 10m and 13m of cloth.

Formula:

(i) Substituting x1 = 5, y1 = 210 and x2 = 2 in the formula,

Gives y2 = 84

∴The cost of 2m cloth is Rs. 84

(ii) Substituting x1 = 5, y1 = 210 and x2 = 4 in the formula,

Gives y2 = 168

∴The cost of 4m cloth is Rs. 168

(iii) Substituting x1 = 5, y1 = 210 and x2 = 10 in the formula,

Gives y2 = 420

∴The cost of 10m cloth is Rs. 420

(iv) Substituting x1 = 5, y1 = 210 and x2 = 13 in the formula,

Gives y2 = 546

∴The cost of 2m cloth is Rs. 546

Question 2.

Fill the table.

Given: Cost of 1 Apple = Rs. 8

To Find: Cost of 4, 7, 12, 20 apples.

Formula:

(i) Substituting x1 = 1 y1 = 8 and x2 = 4 in the formula,

Gives y2 = 32

∴The cost of 4 apples is Rs. 32

(ii) Substituting x1 = 1 y1 = 8 and x2 = 7 in the formula,

Gives y2 = 56

∴The cost of 7 apples is Rs. 56

(iii) Substituting x1 = 1 y1 = 8 and x2 = 12 in the formula,

Gives y2 = 96

∴The cost of 12 apples is Rs. 96

(iv) Substituting x1 = 1 y1 = 8 and x2 = 20 in the formula,

Gives y2 = 160

∴The cost of 20 apples is Rs. 160

Question 3.

48 bags of paddy costs Rs 16, 800 then find the cost of 36 bags of paddy.

Given: Cost of 48 bags of paddy = Rs. 16,800

To find: Cost of 36 bags of paddy

Formula:

Substituting x1 = 48, y1 = 16800, x2 = 36 in the formula,

Gives y2 =

= 12600

∴The cost of 36 bags of paddy is Rs. 12,600

Question 4.

The monthly average expenditure of a family with 4 members is Rs. 2,800. Find the monthly average expenditure of a family with only 3 members.

Given: Monthly average expenditure of 4 members = Rs. 2800

To find: Monthly average expenditure of 3 members

Monthly average expenditure of 1 member = Rs. 2800/4 = Rs. 700

Monthly average expenditure of 3 members = 3 × Rs. 700 = Rs. 2100

Question 5.

In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is 9 cm what is the length of the model ship?

Given: Height of the mast in a 28m long ship = 12m

To find: Height of the mast in a 9cm long ship

Formula:

Substituting x1 = 28, y1 = 12, x2 = 9 in the formula,

Gives y2 =  cm

= 3.85 cm

∴The height of the mast in a 9 cm long ship is 3.85 cm

Conclusion: Always maintain consistency in the units of x1 and y1, x2 and y2.

Question 6.

A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (i) the length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts a shadow 5m long.

Given: (i) Pole Height = 5.6m, Shadow Length = 3.2m
(ii) Pole Height = 10.5 m

(iii) Pole Height

(i)

The length of the shadow cast by a pole 10.5m long is 6m.

(ii)

The height of the pole that casts a 5m long shadow is 8.75m.

Question 7.

A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Answer: Given: Distance traveled in 25 minutes = 14 km
To Find: Distance traveled in 5 hours

The more distance will be traveled in more time, so the distance and time are in direct proportion.

Therefore,

Distance traveled in 1 minute = 14/25 km

Distance traveled in 5 hours (5 × 60 min = 300 minutes) = 14/25 × 300 km

Distance traveled in 5 hours = 168 km

Question 8.

If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16  kilograms?

Given: Weight of 12 sheets of paper = 40 grams

To find: Number of papers weighing

Formula:

Substituting x1 = 12, y1 = 40, y2 =  in the formula,

Gives x2 =

= 5000

∴Number of papers weighing 16 kg are 5000.

Question 9.

A train moves at a constant speed of 75 km/hr.

(i) How far will it travel in 20 minutes?

(ii) Find the time required to cover a distance of 250 km.

Given: Speed of train = 75kmph

To Find: Distance Travelled in 20 minutes, Time required to cover 250km.

Formulation: Speed of the train is 75kmph, i.e, it requires 1 hour (60 minutes) for the train to cover 75km.

Formula:

(i) Substituting x1 = 60 y1 = 75 and x2 = 20 in the formula,

Gives y2 = 25 km

∴The train will move 25km in 20 minutes

(ii) Substituting x1 = 60 y1 = 75 and y2 = 250 in the formula,

Gives x2 = 200 minutes or 3 hours 20 minutes

∴The time required to cover a distance of 250km is 3 hours 20 minutes.

Question 10.

The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the microchip?

Given: Length of the microchip design = 18cm, Scale Ratio = 40:1

To find: Actual Length of microchip

Formula:

Substituting x1 = 1, y1 = 40 y2 = 18 in the formula,

Gives x2 =

= 0.45 cm

∴The actual length of the microchip is 0.45 cm

Question 11.

The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, find the ratio of the number of doctors to the number of lawyers.

Given:

Average age of doctors and lawyers = 40 Average age of doctors = 35

Average age of lawyers = 50

To find: Ration of number of doctors to the number of lawyers

Let a = Number of doctors

b = Number of lawyers

x = total age of doctors

y = total age of lawyers

According to the problem,

Ã° x = 35a and y = 50b

Total age of doctors and lawyers = x+y

According to the problem,

Ã°

Ã°

Ã°

Ã°

Ã°

Therefore, the ratio of ages of doctors and lawyers is 2:1

###### Exercise 10.2

Question 1.

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

x1 = 50, y1 = 5  x1y1 = 250

x2 = 40, y2 = 6  x2y2 = 240

x3 = 30, y3 = 7  x3y3 = 210

x4 = 20, y4 = 8  x4y4 = 160

x1y1≠x2y2≠x3y3≠x4y4

Therefore, x and y are not in inverse proportions.

Question 2.

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

x1 = 100, y1 = 60  x1y1 = 6000

x2 = 200, y2 = 30  x2y2 = 6000

x3 = 300, y3 = 20  x3y3 = 6000

x4 = 400, y4 = 15  x4y4 = 6000

x1y1 = x2y2 = x3y3 = x4y4

Therefore, x and y are in inverse proportions.

Question 3.

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

x1 = 90, y1 = 10  x1y1 = 900

x2 = 60, y2 = 15  x2y2 = 900

x3 = 45, y3 = 20  x3y3 = 900

x4 = 30, y4 = 25  x4y4 = 750

x5 = 20, y5 = 30  x5y5 = 600

x6 = 5, y6 = 25  x6y6 = 175

x1y1 =x2y2 =x3y3≠x4y4≠x5y5≠x6y6

(x1,y1), (x2,y2), (x3, y3) are in inverse proportions.

Question 4.

A school wants to spend Rs 6000 to purchase books. Using this data, fill the following table.

Formula: xnyn = 6000 where n = 1,2,3….

x1 = 40, y1 = 150  x1y1 = 6000

x2 = 50, y2 = ?  x2y2 = 6000  y2 =

x3 = ?, y3 = 100  x3y3 = 6000  x3 =

x4 = 75, y4 = ?  x4y4 = 6000  y4 =

x5 = ?, y5 = 75  x5y5 = 6000  x5 =

The completed table would be

Question 5.

Take a squared paper and arrange 48 squares in different number of rows as shown below:

What do you observe? As R increases, C decreases

(i) Is R1 : R2 = C2 : C1?

(ii) Is R3 : R4 = C4 : C3?

(iii) Is R and C inversely proportional to each other?

(iv) Do this activity with 36 squares.

(i) Is R1 : R2 = C2 : C1?

Yes

(ii) Is R3 : R4 = C4 : C3?

Yes

(iii) Is R and C inversely proportional to each other?

Yes

(iv) Do this activity with 36 squares.

Yes

###### Exercise 10.3

Question 1.

Siri has enough money to buy 5 kg of potatoes at the price of Rs 8 per kg. How much can she buy for the same amount if the price is increased to Rs 10 per kg?

Given: Price of potatoes = Rs. 8 per kg, Siri can buy = 5kg

To find: Amount of potatoes if Price is Rs. 10 per kg

Formula:

Substituting x1 = 8, y1 = 5 x2 = 10 in the formula,

Gives y2 =

∴Amount of potatoes Siri can buy is 4 kg

Question 2.

A camp has food stock for 500 people for 70 days. If 200 more people join the camp, how long will the stock last?

Given: Number of people living off the food stock for 70 days = 500

To find: Number of days the stock will last if 700 people live off.

Formula:

Substituting x1 = 500, y1 = 70 x2 = 700 in the formula,

Gives y2 =

∴If 200 more people join, the food will last for 50 days.

Question 3.

36 men can do a piece of work in 12 days. In how many days 9 men can do the same work?

Given: Number of men required to do a work in 12 days = 36

To find: Number of days required for 9 men to do the same work

Formula:

Substituting x1 = 36, y1 = 12 x2 = 9 in the formula,

Gives y2 =

∴Number of days required for 9 men to do the work is 48 days.

Question 4.

A cyclist covers a distance of 28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.

Given: Distance covered in 2 hours = 28km

To find: Time taken to cover 56km in same speed.

Formula:

Substituting x1 = 28, y1 = 2 x2 = 56 in the formula,

Gives y2 = 4 hours

∴It takes 4 hours for the cyclist to cover 56km.

Question 5.

A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 meters).

Given: Speed in 10 hours = 16 nautical miles per hour

To find: Increase in speed required if the ship has to cover the same distance in 8 hours.

Formula:

Substituting x1 = 10, y1 = 16 x2 = 8 in the formula,

Gives y2 = 20 nautical miles per hour

Change in speed = Final Speed – Initial Speed

20-16 = 4 nautical miles per hour.

∴ The speed must be increased by 4 nautical miles per hour for the ship to cover the same distance in 8 hours.

Question 6.

5 pumps are required to fill a tank in 1hours. How many pumps of the same type are used to fill the tank in half an hour.

Given: Number of pumps required to fill a tank in 1.5 hours = 5

To find: Number of pumps require to fill the tank in 0.5 hours

Formula:

Substituting x1 = 1.5, y1 = 5 x2 = 0.5 in the formula,

Gives y2 = 15 pumps

∴ 15 pumps are required to fill the tank in 0.5 hours.

Question 7.

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?

Given: Number of workers that can build a wall in 48 hours = 15

To find: Number of workers that can build a wall in 30 hours

Formula:

Substituting x1 = 48, y1 = 15 x2 = 30 in the formula,

Gives y2 = 24

∴ 24 workers can build the wall in 30 hours.

Question 8.

A School has 8 periods a day each of 45 minutes’ duration. How long would each period become, if the school has 6 periods a day? (assuming the number of school hours to be the same)

Given: Number of periods per day = 8

Duration of each period = 45 minutes

To find: Duration of each period if there are 6 periods per day

Formula:

Substituting x1 = 8, y1 = 45 x2 = 6 in the formula,

Gives y2 = 60

∴ Duration of each period will be 60 minutes if there are 6 periods.

Question 9.

If z varies directly as x and inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.

Given: z is directly proportional to x, z  x

z is inversely proportional to y, z

Therefore z

To find: Percentage increase in z if there is 12% increase in x and 20% decrease in y

(Assuming the proportionality constant to be 1)

Change in x  100x+12x

Change in y  100y-20y

The new z value becomes

Change in z value = New z value – Initial z value

Percentage increase in z =

40%

∴ Percentage increase in z is 40%

Question 10.

If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.

Given: Number of days required for x+1 men to do the work = x+1

To find: Number of days required for x+2 men to do the work

Formula:

Substituting x1 = x+1, y1 = x+1 x2 = x+2 in the formula,

) (

------------

0 +1

Gives y2 =  days

∴ Number of days required for x+2 men to complete the work is  days

Question 11.

Given a rectangle with a fixed perimeter of 24 meters if we increase the length by 1m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.

Given: Perimeter of the rectangle P = 2(l+b) = 24m

To find: Variation of width and area with length if perimeter is fixed.

Area = lxb

(Note: All lengths are in cm, all areas are in cm2)

l = 3, P = 24 2(l+b) = 24 = 2(3+b)  b = 9, A = 27

l = 4, P = 24 2(l+b) = 24 = 2(4+b)  b = 8, A = 32

l = 5, P = 24 2(l+b) = 24 = 2(5+b)  b = 7, A = 35

l = 6, P = 24 2(l+b) = 24 = 2(6+b)  b = 6, A = 36

l = 7, P = 24 2(l+b) = 24 = 2(7+b)  b = 5, A = 35

l = 8, P = 24 2(l+b) = 24 = 2(8+b)  b = 4, A = 32

l = 9, P = 24 2(l+b) = 24 = 2(9+b)  b = 3, A = 27

Observation: Length and breadth are inversely proportional to each other if perimeter is kept constant. Area first increases with length then decreases.

###### Exercise 10.4

Question 1.

Rice costing Rs 480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?

Given: Cost of Rice for 8 members for 20 days = Rs. 480

To find: Cost of Rice for 12 members for 15 days.

Cost of Rice for 8 members for 1 day = Rs.  = Rs.24

Cost of Rice for 1 member for 1 day = Rs. = Rs. 3

Cost of Rice for 12 members for 1 day = Rs. 3x12 = Rs. 36

Cost of Rice for 12 members for 15 days = Rs. 36x15 = Rs.540

∴ Cost of Rice for 12 members for 15 days is Rs. 540

Question 2.

10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?

Answer: 1 men can lay a road in 5 days = 75/10 = 7.5 km

Now, 15 men lay a road of 45 km

1 men can lay a road of = 45/15 = 3 km

According to the question,

Question 3.

24 men working at 8 hours per day can do a piece 20 men of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?

Given: Number of men working 8 hours a day for 15 days = 24

24 men working at 8 hours per day can do the work in 15 days

Amount of work = 24 x 8 x 15 = 2880 man-hour

Find the number of days needed for 20 men working 9 hours a day:

Time number of man-hour = 20 x 9 = 180

Number of days needed = 2880 ÷ 180 = 16 days

Hence, 20 men working on 9 hours can complete the work in 16 days.

Question 4.

175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?

Given: Number of days required for 175 men to dig a canal 3150m long = 36

To find: Number of men required to dig a canal of 3900m in 24 days.

Formula:  = Constant where W = Work, WF = Workforce, T = Time

Gives WF2 = 325

∴ Number of men required to dig a canal of 3900m in 24 days is 325

Question 5.

If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?

Given: Number of days required for 14 typists to type the manuscript 6 hours a day = 12 days

To find: Number of days required for 4 typists to type working 7 hours a day.

14 typists work 72 hours to complete the manuscript of a work.

14 typists complete  of the manuscript in 1 hour.

1 typist completes  of the manuscript in 1 hour.

4 typists complete 4× of the work in 1 hour.

4 typists complete 6× of the work in 6 hours per day

It will take  = 42 days for 4 typists to complete the whole manuscript.

∴ Number of days required for 4 typists to type working 7 hours a day is 42 days.

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