DAV Class 8 Maths Solution Chapter 4 Direct and Inverse variation
DAV School Books Class 8 Maths Solution Chapter 4 direct and inverse variation all Question Answer. DAV Class 8 4th chapter direct and inverse variation full Chapter explanation by expert teacher.
DAV School Books Class 8 Maths Solution Chapter 4 direct and inverse variation:
Worksheet 1
1) Fill in the missing terms in the following tables, if x and y vary directly:
i)
X | 6 | 8 | 12 | ? |
Y | 18 | ? | ? | 63 |
ii)
X | 10 | 30 | 46 | |
Y | 5 | 10 |
iii)
X | 6 | 8 | 10 | |
Y | 15 | 20 | 40 |
Ans.
i)
X | 6 | 8 | 12 | ? |
y | 18 | ? | ? | 63 |
According to the question we know that x and y varies directly. Which means that x=ky where k is constant.
From the first set of x and y given we can see that
6= 18 × k
K = 1/3
So now we can fill the blank places of the box.
X | 6 | 8 | 12 | ? = 21 |
y | 18 | ? = 24 | ? = 36 | 63 |
ii)
X | 10 | 30 | 46 | |
Y | 5 | 10 |
According to the question we know that x and y varies directly. Which means that x=ky where k is constant.
From the first set of x and y given we can see that
10= 5 × k
K = 2
So now we can fill the blank places of the box.
X | 10 | ? = 5 | 30 | 46 |
Y | 5 | 10 | ? = 60 | ? = 92 |
iii)
X | 6 | 8 | 10 | ? |
Y | 15 | 20 | ? | 40 |
According to the question we know that x and y varies directly. Which means that x=ky where k is constant.
From the first set of x and y given we can see that
6= 15 × k
K = (2/5)
So now we can fill the blank places of the box.
X | 6 | 8 | 10 | ? = 16 |
Y | 15 | 20 | ? = 25 | 40 |
2) 5 bags of rice weigh 150kg. How many such bags of rice will weigh 900 kg?
Ans.
Let’s assume that m number of bags can contain 900 kg rice.
No. of bags of rice | 5 | m |
Weight of the bags | 150 | 900 |
This is a case of direct relation. As the number of rice bags increases, the total weight of the bags also increases.
Therefore,
(5/150) = (m/900)
m = 30
Therefore, 30 bags will contain 900 kg of rice.
3) The cost of 3 kg of sugar is Rs. 54. What will be the cost of 15 kg of sugar?
Ans.
Let’s assume that the cost of 15 kg of sugar is Rs. y.
Weight of sugar (in kg) | 3 | 15 |
Cost of sugar (in rupees) | 54 | y |
This is a case of direct relation. As the weight of the sugar increases, the total cost of the sugar also increases.
Therefore,
(3/54) = (15/y)
y = 270
Therefore, the cost of 15 kg of sugar is Rs. 270.
4) A motorboat covers 20 km in 4 hrs. What distance will it cover in 7 hrs. (speed remaining the same)?
Ans.
Let’s assume that the motorboat will cover y km in 7 hrs.
Distance covered by the speedboat (in km) | 20 | y |
Time taken to cover the distance (in hrs.) | 4 | 7 |
This is a case of direct relation. As the time increases the distance covered by the motorboat also increases
Therefore,
(20/4)=(y/7)
Or, y = 35
Therefore, the distance covered by the speedboat in 7 hours is 35 km.
5) A motorbike travels 210 km on 30 liters of petrol. How far would it travel on 7 liters of petrol?
Ans.
Let’s assume that the motorbike will travel y km on 7 liters of petrol.
Distance covered by the motorbike (in km) | 210 | y |
Petrol required to cover that distance (in liters) | 30 | 7 |
This is a case of direct relation. As the motorbike will cover more distance on more petrol.
Therefore,
(210/30) = (y/7)
Or, y = 49
Therefore, the distance covered by the motorbike on 7 liters of petrol is 49 km.
6) if 12 women can weave 15 meters of cloth in a day, how many meters of cloth can be woven by 20 women in a day?
Ans.
Let’s assume that y meters of cloth can be woven by 20 women in a day.
Length of the cloth woven in a day (in meter) | 15 | y |
Number of women required to weave that cloth | 12 | 20 |
This is a case of direct relation. As the length of the cloth woven in a day will increase with the increase in number of the women who are weaving it.
Therefore,
(15/12) = (y/20)
Or, y = 25
Therefore, 20 women will weave 25 meters of cloth in a day.
7) If the weight of 5 sheets of paper is 20 g. how many sheets of paper would weigh 2.5 kg?
Ans.
Let’s assume that y number of sheets weigh 2.5 kg.
Number of paper sheets | 5 | y |
Weight of those sheets (in kg) | (20/1000)
= 0.02 |
2.5 |
This is a case of direct relation. As the number of paper sheet increases the weight of the papers also increases.
Therefore,
(5/0.02) = (y/2.5)
Or, y = 625
Therefore, 625 sheets of paper will weigh 2.5 kg.
8) Reena types 600 words in 4 minutes. What time will she take to type 3,150 words?
Ans.
Let’s assume that Reena can type 3150 words in y minutes.
Time required to type (in minutes) | 4 | y |
Number of words typed | 600 | 3150 |
This is a case of direct relation as Reena will be able to type more words in more time.
Therefore,
(4/600) = (y/3150)
Or, y = 21
Therefore, Reena will take 21 minutes to type 3150 words.
9) Rohan takes 14 steps in covering a distance of 2.8m. What distance he would cover in 150 steps?
Ans.
Let’s assume that Rohan can cover y meters in 150 steps.
Distance covered by Rohan (in meters) | 2.8 | y |
Steps taken to cover that distance | 14 | 150 |
This is a case of direct relation as Rohan will be able to cover more distance if he takes more steps.
Therefore,
(2.8/14) = (y/150)
Or, y = 30
Therefore, Rohan will cover 30 meters in 150 steps.
10) 48 refined oil cans (of 5 liters) can be packed in 8 cartoons of the same. How many such cartoons will be required to pack 216 cans?
Ans.
Let’s assume that y number of cartoons will be required to pack 216 cans.
Number of cartoons | 8 | y |
Number of oil cans | 48 | 216 |
This is a case of direct relation as more cartoons can contain more number of oil cans.
Therefore,
(8/48) = (y/216)
Or, y = 36
Therefore, 36 cartoons will be required to contain 216 cans.
11) The total cost of 15 newspapers is Rs.22.50. Find the cost of 25 newspapers.
Ans.
Let’s assume that the cost of 25 newspaper is y Rupees.
Cost of newspapers (in Rupees) | 22.50 | y |
Number of newspapers | 15 | 25 |
This is a case of direct relation as the cost of newspaper will increase with the increase of the number of newspapers.
Therefore,
(22.50/15) = (y/25)
Or, y = 37.5
Therefore, the cost of 25 newspapers will be 37.5 Rupees.
12) A laborer gets Rs.675 for 9 days work. How many days should he work to get Rs. 900?
Ans.
Let’s assume that the laborer will have to work for y days to get Rs. 900.
Days of work | 9 | y |
Wage received (In Rupees) | 675 | 900 |
This is a case of direct relation as the wage of the laborer will increase with the increase of the days of work.
Therefore,
(9/675) = (y/900)
Or, y = 12
Therefore, the laborer will have to work for 12 days to get Rs. 900.
Worksheet 2
1) In the following tables, a and b vary inversely. Fill in the missing values.
i)
a | 7 | ? | 28 |
b | 8 | 4 | ? |
ii)
a | 2.5 | 4 | 0.5 |
b | 8 | ? | ? |
iii)
a | 10 | ? | 12 |
b | 6 | 15 | ? |
Ans.
i)
a | 7 | ? | 28 |
b | 8 | 4 | ? |
According to the question we know that a and b varies inversely. This means that
a=k/b , where k is constant.
From the first set of a and b given we can see that
7 = k / 8
Or, k = 56
So now we can fill the blank places of the box.
a | 7 | ? = 14 | 28 |
b | 8 | 4 | ? = 2 |
ii)
a | 2.5 | 4 | 0.5 |
b | 8 | ? | ? |
According to the question we know that a and b varies inversely. This means that
a=k/b , where k is constant.
From the first set of a and b given we can see that
2.5 = k / 8
Or, k = 20
So now we can fill the blank places of the box.
a | 2.5 | 4 | 0.5 |
b | 8 | ? = 5 | ? = 40 |
iii)
a | 10 | ? | 12 |
b | 6 | 15 | ? |
According to the question we know that a and b varies inversely. This means that
a=k/b , where k is constant.
From the first set of a and b given we can see that
10 = k / 6
Or, k = 60
So now we can fill the blank places of the box.
a | 10 | ? = 4 | 12 |
b | 6 | 15 | ? = 5 |
2) The science teacher asked the student of Class-VII to make a project report on pollution. When 10 students work on it, the work gets finished in 3 days. How many students are required so that work finishes in 2 days?
Ans:
Let’s assume that y students are required to do the same job in just 2 days.
Number of students | 10 | y |
Days required to complete the whole project | 3 | 2 |
Clearly as the number of student increases the time required to complete the whole project decreases. So we can say that it is a case of inverse relation.
Therefore,
10 × 3 = y × 2
Or, y = 15
Therefore, total 15 students are required to finish the project.
3) Running at an average speed of 40 km/hr, a bus completes a journey in 4.5 hours. How much time will the return journey take if the speed is increased to 45 km/hr?
Ans:
Let’s assume that if the speed is increased to 45 km/hr then the return journey will take y hours.
Time required for the journey (hours) | 4.5 | y |
Speed of the bus (km/hr) | 40 | 45 |
Clearly as the speed of the bus increases the time required to complete the journey decreases. So we can say that it is a case of inverse relation.
Therefore,
4.5 × 40 = y × 45
Or, y = 4
Therefore, the total time required for the return journey is 4 hours with the increased speed of 45 km/hr.
4) Disha cycles to her school at an average speed of 12 km/hr. It takes her 20 minutes to reach the school. If she wants to reach her school in 15 minutes, what should be her average speed?
Ans:
Let’s assume that if she wants to reach her school in 15 minutes the average speed of her should be y km/hr.
Average speed (in km/hr) | 12 | y |
Time (in hours) | (20/60) = (1/3) | (15/60)
=(1/4) |
If she increases her speed then the time required to reach her school decreases. So we can say that it is a case of inverse relation.
Therefore,
12 × (1/3) = y × (1/4)
Or, y = 16
Therefore, to reach school in 15 minutes her average speed should be 16 km/hr.
5) 15 men can repair a road in 24 days, how long will it take 9 men to repair the same road?
Ans:
Let’s assume that it takes y days to complete road if 9 men work on it.
Number of days | 24 | y |
Number of men | 15 | 9 |
Clearly we can see that if the number of men increases the number of days required for completing the job decreases. So we can say that it is a case of inverse relation.
Therefore,
24×15 = y×9
Or, y = 40
Therefore, if 9 men work on repairing the road then it will take 40 days to complete the job.
6) If 30 goats can graze a field in 15 days, how many goats will graze the same field in 10 days?
Ans:
Let’s assume that y goats can graze the whole field in 10 days.
Number of goats | 30 | y |
Number of days | 15 | 10 |
Clearly we can see that if the number of goats increases the time required for grazing the whole field decreases. So we can say that it is a case of inverse relation.
Therefore,
30×15 = y×10
Or, y = 45
Therefore, it will take 45 goats to completely graze the field in 10 days.
7) A contractor with a work force of 420 men can complete a work of construction of a building in 9 months. Due to request by the owners he was asked to complete the job in 7 months. How many extra men he must employ to complete the job?
Ans:
Let’s assume that y men can complete the whole job in 7 months.
Number of men | 420 | y |
Number of days (in months) | 9 | 7 |
Clearly we can see that if the number of men increases the days required for completing the work decreases. So we can say that it is a case of inverse relation.
Therefore,
420×9 = y×7
Or, y = 540
Therefore, it will take 540 men to complete the whole construction of the building in 7 months.
So the contractor has to employ additional (540-420) = 120 men to complete the building in 7 months.
8) Uday can finish a book in 25 days if he reads 18 pages every day. How many days will he take to finish it, if he reads 15 pages every day?
Ans:
Let’s assume that it will take y days to finish the whole book if he reads 15 pages every day.
Number of days | 25 | y |
Number of pages read per day | 18 | 15 |
Clearly it will take more days to complete the book if the number of pages read every day decreases. So we can say that it is a case of inverse relation.
Therefore,
25×18 = y×15
Or, y = 30
Therefore, it will take 30 days to complete the book if he reads 15 pages every day.
9) A shopkeeper has enough money to buy 40 books, each costing Rs. 125. How many books he may buy if he gets a discount of Rs. 25 on each book?
Ans:
Let’s assume that he can buy y number of books if he gets a discount of 25 rupees on each book.
Number of books | 40 | y |
Cost of one book (in rupees) | 125 | 125-25=100 |
Clearly if the price of the book decreases then the number of books that can be brought with a fixed amount of money will definitely increase. So we can say that it is a case of inverse relation.
Therefore,
40×125 = y×100
Or, y = 50
Therefore, the shopkeeper will be able to buy 50 books with the amount of money he has.
10) 6 pumps working together empty a tank in 28 minutes. How long will it take to empty the tank if 4 such pumps are working together?
Ans:
Let’s assume that it will take y hours to empty the tank if only 4 pumps are working.
Time to empty the tank (in minutes) | 28 | y |
Number of pumps | 6 | 4 |
Clearly if the number of pumps decrease than the time required to empty the tank will increase. So we can say that it is a case of inverse relation.
Therefore,
28×6 = y×4
Or, y = 42
Therefore, the time required to completely empty the tank is 42 minutes.
11) A train moving at a speed of 75 km/hr covers a certain distance in 4.8 hours. What should be the speed of the train to cover the same distance in 3 hours?
Ans:
Let’s assume that the speed of the train will be y km/hr when it covers the same distance in 3 hours.
Speed of the train (in km/hr) | 75 | y |
Time taken to cover the distance
(in hours) |
4.8 | 3 |
Clearly if the speed of the train increases the time required to cover the same distance will decrease. So we can say that it is a case of inverse relation.
Therefore,
75×4.8 = y×3
Or, y = 120
Therefore, the speed of the train would be 120 km/hr to cover the distance in 3 hours.
12) A garrison of 120 men has provision for 30 days. At the end of 5 days, 5 more men just joined them. How many days can they sustain on the remaining provision?
Ans.
Let’s assume that the garrison can sustain y days on the remaining provision.
Number of days | (20-5)=25 | y |
Number of men | 120 | (120+5)=125 |
Clearly if the number of men increases the number of days they can sustain on a fixed amount of provision will decrease. So we can say that it is a case of inverse relation.
Therefore,
25×120 = y×125
Or, y = 24
Therefore, on the provision the garrison can survive for 24 days.
Worksheet 3
1) Ramit can finish his work in 25 days, working 8 hours a day. If he wants to finish the same work in 20 days, how many hours should he work in a day?
Ans:
Let’s assume that Ramit will have to work for y hours a day if he wants complete the work in 20 days.
Number of hours | 8 | y |
Number of days | 25 | 20 |
Clearly, the if the number of days available to complete the same work decreases then the hours required per day to complete the work will increase. So it is a inverse relation.
∴ 8 × 25 = y × 20
Or, y = 10
∴ Ramit has to work for 10 hours every day for 20 days to complete the work.
2) Udit can complete his work in 10 days. What amount of work will be completed in 8 days?
Ans:
Let’s assume that Udit can complete y amount of work in 8 days.
Amount of work | 1 | y |
Number of days | 10 | 8 |
More amount of work can be completed in more days and less amount of work can be done in less days. So, the relation between amount of work and number of days is direct.
∴ (1/10) = (y/8)
Or, y = 8/10
Or, y = 4/5
Therefore the amount of work done in 8 days is (4/5)th of the whole work.
3) 20 men can build a wall in 9 days. How long would it take 12 men to build the same wall?
Ans:
Let’s assume that 12 men can build the wall in y days.
Number of days | 9 | y |
Number of men | 20 | 12 |
When the number of men working on a task increases the time required to complete the task decreases. So, we can clearly say that this is a case of inverse relation.
∴ 9 × 20 = y × 12
Or, y = 15
Therefore, we can say that 12 men can complete the construction of the wall in 15 days.
4) Geetika weaves 20 baskets in 30 days. In how many days she will weave 120 baskets?
Ans:
Let’s assume that Geetika can weave 120 baskets in y days.
Number of days | 30 | y |
Number of baskets | 20 | 120 |
Clearly a person will take more days to weave more baskets. So it is clearly a case of direct relation.
∴(30/20) = (y/120)
Or, y = 180
Therefore, Geetika will take 180 days to weave 120 baskets.
5) A train 280 meters long is running at a speed of 42 km/hr. How much time will it take to pass a man standing on a platform?
Ans:
The length of the train is given in meters. Hence, we have to convert the speed of the train in meters/second.
∴ Speed = 42 km/hr
= (42×1000)/3600 m/s [∵ 1 km = 1000 m, 1 hr= 3600s]
= 35/3 m/s
To cross a man standing on the platform, the train has to cover a distance equal to its length, i.e. 280 m.
Let’s assume that the time required to cover 280 m is y seconds.
Time (in seconds) | 1 | y |
Distance (in meters) | 35/3 | 280 |
The train covers more distance in more time if the speed of the train is constant. So it is clearly a case of direct relation.
∴ 1/(35/3) = y/280
Or, 3/35 = y/280
Or, y = 24
So, the train will take 24 seconds to cross a man standing on the platform.
6) A train 350 meters long crosses an electric pole in 28 seconds. Find the speed of the train in km/hr?
Ans:
To cross an electric pole the train has to cover a distance equal to its length i.e. 350 meters.
So, the train covers 350 meters in 28 seconds.
∴ The speed of the train
= (distance covered/time taken)
= (350/28) m/s
= (350×3600)/(28×1000) km/hr
[∵ 1 km = 1000 m, 1 hr= 3600s]
=45 km/hr
∴ The speed of the train is 45 km/hr
7) A train 150 meters long is running at 72km/hr. It crosses a bridge in 13 seconds. Find the length of the bridge?
Ans:
The length of the train is in meters. Hence, we have to convert the speed of the train in meters/seconds
∴ Speed of the train = 72 km /hr
= (72×1000)/3600 m/s [∵ 1 km = 1000 m, 1 hr = 3600 s]
= 20 m/s
To completely cross a bridge a train has to cover the distance equal to the total length of train and bridge.
Total distance covered = time taken × speed of train
= (13 × 20) m
= 260 m
∴ The train covered 260 meters in 13 seconds. According to the question the length of the train is 150m so the length of the bridge is (260 – 150)=110m
Therefore the length of the bridge is 110 m.
8) How long will a train 120m long take to clear a platform 130 m long, if its speed is 50 km/hr?
Ans:
The length of the train is in meters. Hence, we have to convert the speed of the train in meters/seconds
∴ Speed of the train = 50 km / hr
= (50×1000)/3600 m/s [∵ 1 km = 1000 m, 1 hr = 3600 s]
= (125/9) m/s
To clear a platform a train has to cover a distance equal to the total length of train and platform which is in this case is (120+130)m = 250m
Let’s assume that the train takes y seconds to cross the platform.
Time (in seconds) | 1 | y |
Distance (in meters) | 125/9 | 250 |
The train covers more distance in more time if the speed of the train is constant. So it is clearly a case of direct relation.
∴ 1/ (125/9) = y/250
Or, 9/125 = y /250
Or, y = 18
So a train of 120m length, running at a speed of 50 km/hr, will take 18 seconds to cross a platform of 130m.
9) A train 210 m long took 12 seconds to pass a 90 m long tunnel. Find the speed of the train?
Ans:
To pass a 90 m long tunnel a train of 120 m length will have to cover (90+120)m = 210 m distance
To pass the tunnel train takes 12 seconds so the speed of the train is
= distance covered/ time taken
= (210/12) m/s
= 17.5 m/s
Therefore the speed of the train is 17.5 m/s
10) A train 270 m long is running at 80 km/hr. How much time will it take to cross a platform 130 m long?
Ans:
The length of the train is in meters. Hence, we have to convert the speed of the train in meters/seconds
∴ Speed of the train = 80 km /hr
= (80×1000)/3600 m/s [∵ 1 km = 1000 m, 1 hr = 3600 s]
= (200/9) m/s
To clear a platform a train has to cover a distance equal to the total length of train and platform which is in this case is (270+130)m = 400m
Let’s assume that the train takes y seconds to cross the platform.
Time (in seconds) | 1 | y |
Distance (in meters) | 200/9 | 400 |
The train covers more distance in more time if the speed of the train is constant. So it is clearly a case of direct relation.
∴ 1/ (200/9) = y/400
Or, 9/200 = y /400
Or, y = 18
Therefore, a train of 270m length, running at a speed of 80 km/hr, will take 18 seconds to cross a platform of 130m.
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