Mathematics Semester ii Solutions Solutions for Class 7 Math Chapter 2 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among class 7 students for Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Semester ii Solutions Book of class 7 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Semester ii Solutions Solutions. All Mathematics Semester ii Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 18:

Question O1:

The weights of Shankar and Shyla are 45 kg and 30 kg respectively. Express these weights as a ratio.

Answer:

3 : 2

[Ratio of their weights 45 : 30 = ]

Page No 18:

Question W1:

Write in the form of ratio.

The team of a school won 3 Kabaddi matches, lost 3 and drew one match in a tournament, find.

(1) the ratio between total number of matches played to the matches won.

(2) the ratio of total number of matches played to the matches lost.

(3) the ratio of matches won to those lost.

(4) the ratio of total number of matches played to the matches drawn.

(5) the ratio of matches won to the match drawn.

Answer:

Number of Kabaddi matches won = 3

Number of Kabaddi matches lost = 3

Number of Kabaddi matches drawn = 1

Total number of matches played = 3 + 3 + 1 = 7

(1) Ratio between total number of matches played to the matches won = 7: 3

(2) Ratio of the total number of matches played to the matches lost = 7: 3

(3) Ratio of the number of matches won to those lost = 3: 3 = 1:1

(4) Ratio of the total number of matches played to the matches drawn = 7: 1

 

(5) Ratio of the number of matches won to the matches drawn = 3: 1

Page No 18:

Question O2:

In the ratio 60 : 30, which is the consequent?

Answer:

In the ratio 60:30, the consequent is 30.

Page No 18:

Question W2:

Indicate antecedents and consequents in the following.

Answer:

(1) 15: 23

Antecedent = 15

Consequent = 23

(2) x: y

Antecedent = x

Consequent = y

(3) a: 3

Antecedent = a

Consequent = 3

(4)

Antecedent = 8

Consequent = 5

(5)

Antecedent = a

Consequent = b

Page No 18:

Question O3:

Express 30 : 50 in its simplest form.

Answer:

3 : 5 .

Page No 18:

Question W3:

Reduce these ratios to their simplest form.

Answer:

(1) 15: 20

∴ 15: 20 = 3: 4

(2) 45: 15

∴ 45: 15 = 3: 1

(3) 35: 56

∴ 35: 56 = 5: 8

(4) 60: 80

∴ 60: 80 = 3: 4

(5) 24: 18

∴ 24: 18 = 4: 3

(6) 5: 80

∴ 5: 80 = 1: 16

(7)

=1: 4

(8)

= 5: 9

(9)

=2: 1

(10)

=3: 5



Page No 19:

Question W4:

Express the following in the form of ratio and then simplify.

Answer:

(1) 75 paise, Re.1

75 paise : Re.1 = 75 paise: 100 paise = 75: 100 = = 3: 4

[H.C.F. of 75 and 100 is 25]

Thus, the required ratio is 3:4.

(2) 80 cm, 1 m

1 m = 100 cm

∴Required ratio = 80 cm: 100 cm = 80 : 100 = = 4: 5

[H.C.F. of 80 and 100 is 20]

(3) 1 km, 500 m

1 km = 1000 m

∴Required ratio = 1000 m: 500 m = 1000: 500 = = 2: 1

[H.C.F. of 500 and 1000 is 500]

(4) 1 kg, 700 g

1 kg = 1000 g

∴Required ratio = 1000 g: 700 g = 1000 : 700 = = 10: 7

[H.C.F. of 1000 and 700 is 100]

(5) 2 kg, 600 g

1 kg = 1000 g 2 kg = 2000 g

∴Required ratio = 2000 g: 600 g = 2000 : 600 = = 10: 3

[H.C.F. of 2000 and 600 is 200]

Page No 19:

Question W5:

Express the following ratios in fractions.

Answer:

(1)

(2)

(3)

(4)

(5)

Page No 19:

Question W6:

Express the following fractions in ratios.

Answer:

(1)

(2) = 3: 4

(3) = 5: 3

(4) = 7: 8

(5) = 7: 3

Page No 19:

Question W7:

Write one equal ratio for the following.

Answer:

(1) 1 : 3 =

∴ 1 : 3 = 2 : 6

(2) 3 : 5 =

∴ 3 : 5 = 6 : 10

(3) 1 : 2 =

∴ 1: 2 = 2: 4

(4) 7 : 2 =

∴ 7 : 2 = 14: 4

(5) 1.5 : 2 =

∴ 1.5 : 2 = 3 : 4

Page No 19:

Question W8.4:

Solve the following.

Cost of 100g of coffee powder is Rs. 24 and 1kg of tea powder costs Rs.80. Find the ratio of the cost of coffee powder to the tea?

Answer:

Cost of 100 g of coffee powder = Rs. 24

∴ cost of 1 kg or 1000 g of coffee powder = Rs. 24 × 10 = Rs. 240

Cost of 1 kg of tea powder = Rs. 80

∴Required ratio=

Page No 19:

Question W8.3:

Solve the following.

The ratio of two numbers is 3 : 5. Sum of two numbers is 40. Find the numbers.

Answer:

Sum of two numbers = 40

Ratio between the two numbers = 3 : 5

Sum of the parts = 3 + 5 = 8

∴First number =

Second number = 40 − 15 = 25

Page No 19:

Question W8.2:

Solve the following.

Divide Rs.75 between A and B in the ratio 3 : 2. Find the share of B.

Answer:

Total amount of money = Rs.75

Ratio of the share between A and B = 3: 2

Sum of the parts = 3 + 2 = 5

∴ Share of B =

Page No 19:

Question W8.1:

Solve the following.

The population of Mallapur village is 4,500, of which 1,350 are literates. Find

(i) the ratio of total population to the number of illiterates in the simplest form.

(ii) the ratio of total population to the number of literates in the simplest form.

(iii) the ratio of literates to the illiterates in its simplest form.

Answer:

Population of Mallpur village = 4500

Number of illiterate people = 1350

Number of literate people = 4500 − 1350 = 3150

(i)

Ratio of total population to the number of illiterates

= 4500 : 3150

(ii)

Ratio of total population and number of literates

= 4500: 1350

(iii)

Ratio of number of literates to number of illiterates

= 1350: 3150

= 3: 7

Page No 19:

Question W8.5:

Solve the following.

The ratio of copper to the zinc in an alloy is 5 : 3. The weight of copper in the alloy is 30g. Find the weight of zinc.

Answer:

Ratio of copper to zinc = 5: 3

Weight of copper = 30 g

Here,

Therefore, the weight of zinc is 18 g.

Page No 19:

Question W8.6:

Solve the following.

The ratio of length to the breadth of a play ground is 5 : 2. If the breadth is 40m. What is its length?

Answer:

Ratio of length to the breadth = 5: 2

Breadth = 40 m

∴ Length = 100 m

Page No 19:

Question W8.7:

Solve the following.

The ratio of income to the savings of a family is 7 : 1. Find the savings, if the income is Rs.1,400.

Answer:

Ratio of income to the saving = 7: 1

Income = Rs.1400

∴ Savings = Rs.200

Page No 19:

Question W8.8:

Solve the following.

The ratio of adults to the youths travelling in a bus is 4 : 7. There are 20 adults. What is the total number of persons travelling in the bus?

Answer:

Ratio of adults to the youths = 4: 7

Number of adults = 20

∴ Number of youths = 35

∴ Total number of persons travelling in the bus = 20 + 35 = 55



Page No 25:

Question O1:

Mention the rule of proportion.

Answer:

The rule of the proportion states that if , then .

Page No 25:

Question W1:

Fill in the blanks with suitable signs = or .

Answer:

(1)

(2)

(3)

(4)

(5)

Here, .

3 × 6 = 2 × 9 = 18

Page No 25:

Question O2:

Give one equivalent proportion of 2 : 3

Answer:

One of the equivalent proportions of 2:3 is given below.

Page No 25:

Question O3:

If what is the value of ‘x’?

Answer:

x = 6



Page No 26:

Question W2:

Fill in the blanks with suitable numbers.

Answer:

(1) Let the missing term be x.

(2) Let the missing term be x.

Then, 8 : 5 = 24: x

(3) Let the missing term be x.

(4) Let the missing term be x.

(5) Let the missing term be x.

Then, 4: x = 36: 27

Page No 26:

Question W3:

Identify which of the following sets are in proportion.

Answer:

(1) 12, 16, 6, 8

Here, 12 × 8 = 96 and 16 × 6 = 96.

12 × 8 16 × 6

Therefore, the given numbers are in proportion.

(2) 7, 27, 18, 3

Here, 7 × 3 = 21 and 27 × 18 = 486.

7 × 3 27 × 18

Therefore, the given numbers are not in proportion.

(3) 3, 4, 26, 6

Here, 3 × 6 = 18 and 4 × 26 = 104.

3 × 6 4 × 26

Therefore, the given numbers are not in proportion.

(4) 33, 44, 66, 88

Here, 33 × 88 = 2904 and 44 × 66 = 2904.

33 × 88 = 44 × 66

Therefore, the given numbers are in proportion.

(5)

Here, and .

Therefore, the given numbers are not in proportion.

Page No 26:

Question W4:

Find the value of ‘a’ in the following proportion.

Answer:

(1) 20 : 30 = 100 : a

(2) a : 6 = 55: 11

(3) 18 : a = 27 : 3

(4) 16: 18 = a : 90

(5) a : 3 = 5 : 6

(6) 32 : a = a : 8

Page No 26:

Question W5.1:

Solve the following.

If the first, third and fourth terms of a proportion are 32, 112 and 217. Find the second term.

Answer:

Let the second term be x.

Then, 32 : x : : 112 : 217

So, the second term is 62.

Page No 26:

Question W5.2:

Solve the following.

If the first, second and fourth terms of a proportion are 51, 81 and 108. Find the third term.

Answer:

Let the third term be x.

Then, 51 : 81 : : x : 108

So, the third term is 68.

Page No 26:

Question W5.3:

Solve the following.

Find the mean proportion between 9 and 49.

Answer:

Let the mean proportion be x.

Then, 9 : x : : x : 49

So, the mean proportion is 21.

Page No 26:

Question W5.4:

Solve the following.

Find the third term of 48 and 36.

Answer:

Let x be the third term of 48 and 36.

Then, 48 : 36 : : 36: x

Thus, the third term of 48 and 36 is 27.

Page No 26:

Question W5.8:

Solve the following.

Antony and Budenmia invests Rs. 24,000 and Rs. 36,000 in a partnership (Small scale industry). At the end of the year, they get Rs.10,000 in the form of profit. What is the amount of profit each gets?

Answer:

Antony’s investment = Rs. 24000

Budenmia’s investment = Rs. 36000

Profit = Rs. 10000

Ratio of their investments = 24000 : 36000 = 2 : 3

So, the profit must be divided in the ratio 2 : 3.

Sum of the parts = 2 + 3 = 5

∴Profit of Antony =

Profit of Budenmia = Rs. (10000 − 4000) = Rs. 6000

Page No 26:

Question W5.7:

Solve the following.

A person works 5 days and earns Rs. 180. Find the amount he earns in 28 days.

Answer:

Let x be the amount the person earned in 28 days.

Then, 5 : 180 : : 28 : x

So, the amount the person earned in 28 days is Rs. 1008.

Page No 26:

Question W5.6:

Solve the following.

The cost of 30m of cloth is Rs. 450. Find the cost of 10m of cloth.

Answer:

Let x be the cost of 10 m cloth.

Then, 30 : 450 : : 10 : x

So, the cost of 10 m cloth is Rs. 150.

Page No 26:

Question W5.5:

Solve the following.

The length of the shadow of an electric pole is 20m. At the same time, the length of the shadow of a tree of height 6m is 8m. Find the height of the electric pole.

Answer:

Let x be the length of the electric pole.

Then, x : 20 : : 6 : 8

So, the length of the electric pole is 15 m.



Page No 30:

Question O1:

In the given quantities, if one quantity increases and the other quantity decreases, what kind of a proportion is it?

Answer:

Inverse proportion

Page No 30:

Question W1.1:

Fill in the blanks:

In direct proportion, if one quantity increases another quantity also ...........

Answer:

In direct proportion, if one quantity increase another quantity also increases.

Page No 30:

Question W1.2:

Fill in the blanks:

In inverse proportion, if one quantity increases the other quantity ...........

Answer:

In inverse proportion, if one quantity increases the other quantity decreases.

Page No 30:

Question W1.3:

Fill in the blanks:

a : b and c : d are in inverse proportion, then a : b = ...........

Answer:

a : b and c : d are in inverse proportion, then a: b = d: c.

Page No 30:

Question W1.4:

Fill in the blanks:

If the number of people increases, the food decreases. This is an example for ........... proportion.

Answer:

If the number of people increases, the food decreases. This is an example for inverse proportion.

Page No 30:

Question W2.4:

Solve the following problems:

60 men can complete the job in 54 days, how many days will 20 workers take for completing the same job?

Answer:

Let x be the number of days in which 20 men can complete the job.

Ratio of number of days = 54: x

Ratio of number of men = 60: 20

The quantities involved in the situation are in inverse proportion.

∴ 54: x = 20: 60

Thus, 20 men can complete the job in 162 days.

Page No 30:

Question W2.3:

Solve the following problems:

If one can have 4 pens instead of 2 balls. How many pens can one have instead of 10 balls?

Answer:

Let x be the number of pens one can have instead of 10 balls.

The quantities involved in the situation are in direct proportion.

∴ 4 : 2 = x : 10

Thus, one can have 20 pens instead of 10 balls.

Page No 30:

Question W2.2:

Solve the following problems:

The cost of 5 kites is Rs.25 Find the cost of 25 kites.

Answer:

Let x be the cost of 25 kites.

The quantities involved in the situation are in direct proportion.

∴ 5 : 25 = 25 : x

Thus, the cost of 25 kites is Rs. 125.

Page No 30:

Question W2.1:

Solve the following problems:

If 3 articles can be purchased for Rs. 15. How many articles can be purchased for Rs. 75?

Answer:

Let x articles can be purchased for Rs. 75.

The quantities involved in the situation are in direct proportion.

∴ 3: x = 15: 75

Thus, 15 articles can be purchased for Rs. 75.

Page No 30:

Question W1.5:

Fill in the blanks:

If 10 : 12 and 24 : x is in inverse proportion, then 10 : 12 = ...........

Answer:

If 10 : 12 and 24 : x is in inverse proportion, then 10 : 12 = x : 24.

Page No 30:

Question O2:

The cost of 5 litres of milk is Rs. 55 and 7 litres of milk is Rs. 77. What kind of a proportion does this statement indicate?

Answer:

Direct proportion

Reason: When the quantity of milk increases, the cost of milk also increases.

Page No 30:

Question O3:

a : b and c : d is in inverse proportion. Express this in direct proportion.

Answer:

If a : b and c : d is in inverse proportion, then .

Page No 30:

Question W2.5:

Solve the following problems:

Food grains in a hostel lasts for 30 days for 10 persons. How long does it last for 25 persons?

Answer:

Let x be the number of days for which the grains will last for 25 people.

Ratio of days = 30: x

Ratio of persons = 10: 25

The quantities involved in the situation are in inverse proportion.

x: 30 = 10: 25

Page No 30:

Question W2.6:

Solve the following problems:

A Scooter covers a certain distance at a speed 40km/hr in 6 hours. How long does it take to travel the same distance at the speed of 60 km/hr?

Answer:

Let the motorcycle take x hours to cover the particular distance at 60 km/h.

Ratio of speeds = 40: 60

Ratio of time durations = 6: x

The quantities involved in the situation are in inverse proportion.

x: 6 = 40: 60

Thus, the motorcycle will take 4 hours to cover the distance at 60 km/h.

Page No 30:

Question W2.7:

Solve the following problems:

10 workers can complete a job in 24 days. How many days less will 12 workers take to complete the same job?

Answer:

Let 12 workers take x days to complete the given job.

Ratio of number of days = 24 : x

Ratio of number of workers = 10 : 12

The quantities involved in the situation are in inverse proportion.

∴ 24 : x = 12 : 10

In 20 days, 12 workers will complete the given job, which is 4 days less than that was taken by 10 workers.

Page No 30:

Question W2.8:

Solve the following problems:

6 workers can complete a job in 10 days. How many days will 5 workers take to complete the same job?

Answer:

Let 5 workers take x days to complete the job.

Ratio of number of days = 10 : x

Ratio of number of workers = 6 : 5

The quantities involved in the situation are in inverse proportion.

∴ 6 : 5 = x : 10

Therefore, 5 workers will take 12 days to complete the job.



Page No 34:

Question W1:

Solve the following.

A person buys 5 sheep at Rs.3,250. Find the cost of a sheep.

 

Answer:

Cost of 5 sheep = Rs. 3250

 

Page No 34:

Question W2:

Solve the following.

If 25 labourers harvest 20 acres of maize crop in a day, then how many acres do 35 labourers harvest during the same period?

 

Answer:

Let x acres of maize crop be harvested by 25 labourers.

Ratio of labourers = 25: 35

Ratio of acres = 20 : x

This situation is a direct proportion.

∴25 : 35 = 20 : x

Thus, 25 acres of maize crop can be harvested by 25 labourers during the same period.

 

Page No 34:

Question W3:

Solve the following.

If 8 oxen can plough 45 acres of land in 15 days, then how many acres of land can 12 oxen plough in 9 days?

 

Answer:

Let 12 oxen plough x acres of land in 9 days.

 

If the number of acres of land increase, then the required number of oxen increases.

This is direct proportion.

If the number of acres increase, then the number of days will also increase.

This is also a direct proportion.

The detailed proportions are:

Oxen

Days

Acres

8

15

45

12

9

x

Thus, 12 oxen can plough of land in 9 days.

 



Page No 35:

Question W4:

Solve the following.

5 men can build a wall 20ft long in 7 days. Find the length of the wall built by 13 men in 21 days?

 

Answer:

Let x be the length of wall made by 13 men in 21 days.

Length of wall increases, when the required number of men increase.

This is direct proportion.

The detailed proportions are:

Men

Days

Length

5

7

20

13

21

x

Thus, length of the wall made by 13 men in 21 days is 156 ft.

 

Page No 35:

Question W5:

Solve the following.

In 10 hours 8 pumps can empty 74,400 gallons of water. How many gallons of water can be emptied in 6 hours by 12 pumps?

 

Answer:

Let x gallons of water can be emptied in 6 hours by 12 pumps.

Number of gallons of water that is going to be emptied increases when the number of hours increase.

This is direct proportion.

Number of gallons of water that is going to be emptied increases when the number of pumps increase.

This is also a direct proportion.

The detailed proportions are:

No. of pumps

No. of hours

No. of gallons

8

10

74400

12

6

x

 

Thus, 66960 gallons of water can be emptied in 6 hours by 12 pumps.

 

Page No 35:

Question W6:

Solve the following.

If a certain quantity of rice can serve for 8 men for 20 days at a cost of Rs. 480, find the cost of rice required for 12 men for 15 days.

 

Answer:

Let x be the cost of rice required for 12 men for 15 days.

Cost of rice increases when the number of days increase.

This is direct proportion.

Cost of rice increases when the number of men increase.

This is direct proportion.

The detailed proportions are:

Men

Days

Cost

8

20

480

12

15

x

 

Thus, the cost of rice required for 12 men for 15 days is Rs.540.

 

Page No 35:

Question W7:

Solve the following.

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

 

Answer:

Let 15 men construct 45 km road in x days.

Number of days increase when the length of road increases.

This is direct proportion.

Number of days decrease when the number of men increases.

This is inverse proportion.

The detailed proportions are:

Men

Km (Road)

Days

10

75

5

15

45

x

 

Thus, 15 men can construct 45 km road in 2 days.

 

Page No 35:

Question W8:

Solve the following.

24 men can complete a job, working 7 hours a day in 27 days. In how many days can 14 men, complete the same job working 9 hours a day.

 

Answer:

Let x be the number of days in which 14 men complete the job, working 9 hours a day.

Number of days decrease when the working hours increase.

This is inverse proportion

Number of days decrease when the number of men increases.

This is inverse proportion

The detailed proportions are:

Men

Hours

Days

24

7

27

14

9

x

 

Thus, the number of days in which 14 men will complete the job, working 9 hours a day, is 36.

 



Page No 40:

Question W1.4:

Find the speed.

The distance travelled is 200m in 20seconds.

Answer:

Distance = 200 m, time = 20 sec

Let v be the speed, then .

Page No 40:

Question W1.3:

Find the speed.

The distance travelled is 240km at 4 hours.

Answer:

Distance = 240 km, time = 4 hours

Let v be the speed, then .

Page No 40:

Question W1.2:

Find the speed.

The distance travelled is 95km at hours.

Answer:

Distance = 95 km, time = = 2.5 hours

Let v be the speed, then .

Page No 40:

Question W1.1:

Find the speed.

The distance travelled is 12 km in 2 hours.

Answer:

Distance = 12 km, time = 2 hours

Let v be the speed, then .

Page No 40:

Question O1:

What is the constant obtained when distance is divided by time known as?

Answer:

Speed

Page No 40:

Question O2:

What is the time taken by a scooter travelling at 50 km/hr to cover a distance of 200 km.

Answer:

4 hours

Page No 40:

Question W1.5:

Find the speed.

The distance travelled is 500m at 10seconds.

Answer:

Distance = 500 m, time = 10 sec

Let v be the speed, then .

Page No 40:

Question W2.2:

Calculate the distance travelled in each of the following.

40 minutes at 48 km/hr.

Answer:

Speed = 48 km/h, time = 40 minutes =

Let d be the distance covered.

Then,

Page No 40:

Question W2.3:

Calculate the distance travelled in each of the following.

12 seconds at 36 km/hr.

Answer:

Speed = 36 km/h, time = 12 sec =

Let d be the distance covered.

Then,

Page No 40:

Question O3:

What is the distance covered by a vehicle travelling at 40 km/hr in hours?

Answer:

100 km

Page No 40:

Question W2.1:

Calculate the distance travelled in each of the following.

4 hrs at 56 km/hr.

Answer:

Speed = 56 km/h, time = 4 hours

Let d be the distance covered.

Then, .



Page No 41:

Question W2.4:

Calculate the distance travelled in each of the following.

A vehicle moves 10m, in 5 seconds. Find the distance moved by it in hours.

Answer:

When distance = 10 m, time taken = 5 sec

Let v be the speed, then .

Let d be the distance covered in .

Page No 41:

Question W3.4:

To find the time to travel a distance.

Find the time, taken to travel a distance of 50 km if the speed is 5 m/sec.

Answer:

Distance = 50 km = 50,000 m, speed = 5 m/ sec

Let t be the time taken.

Page No 41:

Question W3.3:

To find the time to travel a distance.

Find the time, if the distance travelled is 50km and the speed is 50km/hr.

Answer:

Distance = 50 km, speed = 50 km/ hr

Let t be the time taken.

Then,

Page No 41:

Question W3.2:

To find the time to travel a distance.

What is the time taken to travel a distance of 90 km at a speed of 36 km/hr.

Answer:

Distance = 90 km, speed = 36 km/ hr

Let t be the time taken.

Then,

Page No 41:

Question W3.1:

To find the time to travel a distance.

What is the time taken to travel a distance of 120 km at a speed of 50 km/hr.

Answer:

Distance = 120 km, speed = 50 km/ hr

Let t be the time taken.

Page No 41:

Question W2.5:

Calculate the distance travelled in each of the following.

A Vehicle moving with a speed of 48 km/hr takes 8 hours to reach the place Find the distance travelled by the vehicle.

Answer:

Speed = 48 km/h, time = 8 hours

Let d be the distance covered.

Then,

Page No 41:

Question W3.5:

To find the time to travel a distance.

Find the time taken to travel a distance of 150 km and the speed is 30m/sec.

Answer:

Distance = 150 km = 150000 m, speed = 30 m/ sec

Let t be the time taken.

Page No 41:

Question W4:

Fill in the blanks.

Sl. No.

Distance

km

Time

hours

Speed

km.hr

1.

120

3

2.

360

30

3.

4

50

4.

400

4

5.

400

64

Answer:

1.

2.

3.

4.

5.

Thus, the complete table is:

Sl. No.

Distance (km)

Time (hours)

Speed (km/h)

1.

120

3

40

2.

360

12

30

3.

200

4

50

4.

400

4

100

5.

400

6 hrs 15 min

64

Page No 41:

Question W5.1:

Solve the following.

Convert 72km/hr into m/sec.

Answer:

Page No 41:

Question W5.2:

Solve the following.

Convert 25m/sec into km/hr.

Answer:

Page No 41:

Question W5.3:

Solve the following.

A student covers a distance of 2km in the first hour, 2.5km and 1.5km in the second and the third hour respectively. Find the average speed.

Answer:

Total distance covered = (2 + 2.5 + 1.5) km = 6 km

Total time taken = 3 hrs

∴Average speed,

Page No 41:

Question W5.4:

Solve the following.

A train 350m long, passes a person standing on the platform in 28 seconds. Find the speed.

Answer:

Length of the train = 350 m

Time taken = 28 sec

∴Speed of the train,

Page No 41:

Question W5.5:

Solve the following.

Which is greater 60 km/hr or 30 m/sec?

Answer:

∴ 30 m/sec > 60 km/ hr



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