Adverse Effect on Biodiversity and Conservation Methods

The adverse effects on biodiversity and methods to mitigate them may differ from place to place because of different geographical regions. This note has information about the adverse effect on biodiversity and conservation methods.

Summary

The adverse effects on biodiversity and methods to mitigate them may differ from place to place because of different geographical regions. This note has information about the adverse effect on biodiversity and conservation methods.

Things to Remember

  • The adverse effects on biodiversity and methods to mitigate them may differ from place to place because of different geographical regions.
  • The deterioration of the natural environment causes the loss of mobility of birds and animals.
  • Habitat plays a significant role in retaining biodiversity.
  • When we protect ecosystem, it helps to protect the biodiversity.
  • Programmes for the conservation of biodiversity should be conducted at national, local and individual levels.

MCQs

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Subjective Questions

Q1:

The work done by Ram is 3 days can be done by Shyam in 4 days. Again, the work done by Shyam in 5 Days can be done by Hari in 6 days. In how many days the work done by Hari in 16 days can be done by Ram?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here, by chain rule<br>The work done by Ram in 3 days = the work done by Shyam in 4 days. <br>The work done by Shyam in 5 days = the work done by Hari in 6 days.<br>Let, the work done by Hari 16 days = the work done by Ram in x days<br><br>Now , 3 \(\times\) 5 \(\times\)16 = 4 \(\times\) 6 \(\times\) x<br><br>\(\therefore\) x= \(\frac{3 \times 5 \times 16}{4 \times 6}\) = 10. <br><br>Hence, the work done by Hari in 16 days can be done by Ram in 10 days. Ans.</p>

Q2:

The price of 3 ducks is equal to the price of 4 hens. The price of 4 peasants and the price of 7 ducks are equal. If the price of hens is Rs. 750 then what is the price of one peasant ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , solving the given problem by chain rule , <br>The price of 3 ducks = Price of 4 hens<br>Price of 2 hens= Rs. Rs.750<br><br>Let , Rs.x= price of 1 peasants<br>price of 4 peasants= Price of 7 ducks<br>Now , 3 \(\times\) 2 \(\times\) x \(\times\) 4 = 4 \(\times\) 750 \(\times\) 1 \(\times\) 7<br><br>or , x= \(\frac{4 \times 750\times 1 \times 7}{3 \times 2 \times 4}\) = 875. <br><br>The price of 1 peasant = Rs. 875.</p>

Q3:

12 oxen can eat as much as 24 sheep where 15 sheep can eat as much as food as 25 goats. Again the food for 17 goats is equal to the food for baby elephants. Similarly   , the food eaten by 8 baby elephants is equal to the food eatn by 12 horses.  How many horses can eat the food required for 153 oxen  ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , solving the given problem by chain rule ,<br>Food for x horses =food for 153 oxen<br>Food for 12 oxen = food for 24 sheep<br>Food for 15 sheep = food for 25 goats<br>Food for 17 goats= food for 3 baby elephants<br>Food for 8 baby ele</p>

Q4:

When A does \(\frac{1}{3}\) of a work , B does \(\frac{1}{4}\) of the same work. Similarly , when B does \(\frac{1}{5}\) of work , C does \(\frac{1}{2}\) of work . If A finishes the work in 20 hours then in how many days C finishes the work ?


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Let C take x hours to finish the work.<br>Here , solving the given problem by chain rule , <br>The time taken by A to do \(\frac{1}{3}\) work = the time taken by B to do \(\frac{1}{4}\) work. <br>The time taken by B to do \(\frac{1}{5}\) work = the time taken by C to do \(\frac{1}{2}\) work. <br>The time taken by C to do 1 work = x hours (suppose)<br>20 hours = time taken by A to do 1 work<br><br>Hence , \(\frac{1}{3}\) \(\times\) \(\frac{1}{5}\) \(\times\) 1 \(\times\) 20 = \(\frac{1}{4}\) \(\times\) \(\frac{1}{2}\) \(\times\) x \(\times\) 1<br><br>or , \(\frac{4}{3}\) = \(\frac{x}{8}\)<br><br>\(\therefore\) x = \(\frac{32}{3}\) = 10 \(\frac{2}{3}\). <br><br>\(\therefore\) C can complete 1 work in 10 \(\frac{2}{3}\) hours.</p>

Q5:

3 ducks eggs can be exchanged for 4 hen eggs. Similarly, with 4 eggs of the swan , 7 eggs of duck can be exchanged. If the cost of 2 hens eggs is Rs. 7.50 then what is the cost of a swan's egg?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Let, x eggs of Swan's can be exchanges with two eggs of the hen. <br>Here, solving the given problem by chain rule, <br>4 eggs of hen = 3 eggs of duck. <br>7 eggs of duck= 4 eggs of a swan. <br><br>Hence , x &times; 4 &times; 7 = 2 &times; 3 &times; 4<br><br>or , 28x = 24<br>x = \(\frac{24}{28}\) = \(\frac{6}{7}\)<br><br>Now, using unitary method<br>\(\frac{6}{7}\) eggs of swan = 2 eggs of hen<br><br>\(\therefore\) cost of \(\frac{6}{7}\) eggs of swan = Rs. 7.50<br><br>or , cost of 1 eggs of swan = Rs. 7.5 \(\times\) \(\frac{7}{6}\) = Rs. 8.75 Ans.</p>

Q6:

Two types of Rice costing Rs. 15 and Rs. 20 per kg are mixed in the ratio 2:3 . What is the cost of mixture of rice per kg ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Let , the cost of mixture after 2x kg of rice costing Rs. 15 and 3x kg of rice costing Rs. 20 be Rs. y per kg. <br><br>Now , Rs. 15 &times; 2x + Rs. 20 &times; 3x = Rs. y (2x + 3x)<br>or , (30 x + 60x) = 5xy<br>or , y= \(\frac{90x}{5x}\) = 18<br><br>\(\therefore\) The cost of mixture is Rs. 18 per kg.</p>

Q7:

When two types of oil costinf Rs. 80 and Rs. 90 per l are mixed in the ratio 9 : 8 , what if the cost of mixture of oil ?


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , cost of first type of oil = Rs. 80 per litre<br>Cost of second type of oil = rS. 90 per litre<br>Let , cost of mixture = Rs. x per litre. <br><br>\(\frac{quantity\;of\;second\;type\;of\;oil}{quantity\;of\;first\;type\;of\;oil}\) = \(\frac{9}{8}\)<br><br>By formula , <br>\(\frac{quantity \;of \;fist \;type \;of \;oil}{quantity \;of \;second \;type\; of\; oil}\) = \(\frac{cost of second type of oil- cost of mixture}{cost of mixture - cost of first type of oil }\)<br>or , \(\frac{9}{8}\) = \(\frac{Rs.90 -x }{x - Rs. 80}\)<br>or , 9x - Rs. 720 = Rs. 720 - 8x<br>or , 9x + 8x = Rs. 720 + Rs. 720<br>or , 17x = Rs. 1440<br>\(\therefore\) x = \(\frac{1440}{17}\) = Rs. 84.70</p>

Q8:

Rahul mixed three types of tea in the ratio 3: 4: 7. If the price of tea per kg is Rs. 80,  Rs. 100 And  Rs. 150 respectively, find the price of 35 kg of mixed tea.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Let the quantities of three types of tea in kg be 3x , 4x and 7x respectively , <br>Total weight of mixture = (3x + 4x + 7x) = 14x kg.<br>Total cost of mixture = Rs. (3x \(\times\) 80 \(\times\) + 4x \(\times\) 100 + 7x \(\times\) 150)<br><br> = Rs. (240x + 400x + 1050x)<br> = Rs. 1690x. <br><br>Here , cost of 14 x kg of mixture = Rs. 1690 x<br>cost of 1 kg of mixture = \(\frac{1690x}{14x}\)<br><br>Cost of 35 kf of mixture = Rs. \(\frac{1690x}{14x}\) \(\times\) 35 = Rs. 4225 <br><br><br></p>

Q9:

In a mixture of 56 l of milk and water , the ratio of milk and water is 4 : 3  , Find the quantity of milk and water in the mixture.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , the sum of the propertional parts = 56l<br>Quantity of milk = \(\frac{4}{7}\) \(\times\) 56 l = 4 \(\times\) 8 l= 32l <br>Quantity of water = \(\frac{3}{7}\) \(\times\) 56 l = 3 \(\times\) 8 l = 24l Ans.</p>

Q10:

There is Rs. 600 in a bag which consists of rupee , mohar and suki in the ratio od 3 : 4 : 12. Find the number of suki inside the bag.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Let , the number of rupee , Mohar and suki be 3x , 4x , and 12 x respectively. <br>Let us convert the mohor and suki into rupee. <br><br>Then , Mohar 4x = Rs. \(\frac{4x}{2}\) = Rs. 2x <br><br>and suki 12x = Rs. \(\frac{12x}{4}\) = Rs. 3x <br><br>Hence , Rs. 3x + mohar 4x + Suki 12 x = Rs. 3x + 2x + Rs. 3x = Rs. 8x<br><br>By question , <br>Rs. 8x = Rs. 600<br><br>\(\therefore\) x = \(\frac{600}{8}\) = 75. <br><br>Hence , the number of suki in bag = 12x = 12 \(\times\) 75 = 900. </p>

Q11:

A shopkeeper has three types of black pulse costing Rs. 60 , Rs. 71 and Rs. 90 per kg respectively. Out of them the first two types of pulse are mixed in the ratio of 2 : 3 . In what ratio  the third type of pulse should be mixed so that the cost f mixture will be Rs. 78 per kg  ?


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , the fisrt twot ypes of pulse are mixed in the ratio of 2 : 3 , so let 2x kg and 3x kg be quantity of first , and second type of pulse respectively . Also , let y kg of third type of pulse should be mixed in the mixture so that cost of mixture will be Rs. 78 per kg .<br><br>Now , The cost of 2x kg of pulse = Rs. 60 \(\times\) 2x = Rs. 120x<br>The cost of 3x kg of pulse = 3x \(\times\) 72 = Rs. 216 x<br>The cost of y kg of pulse = Rs. 90y<br>Total quantity of mixed pulse = (2x+ 3x + y) = (5x + y)<br><br>Total cost of mixed pulse = Rs. 78 (5x + y)<br>By question , 120x + Rs. 216x + Rs. 90y = Rs. 78 (5x + y) <br>or , 336x + 90y = 390x + 78y<br>or , 90y - 78y = 390x - 336x<br>or , 12y = 54x<br>or , \(\frac{12}{54}\) = \(\frac{x}{y}\)<br>or , \(\frac{x}{y}\) = \(\frac{2}{9}\) \(\therefore\) y = 4.5x<br><br>Now , in final mixture , the ratio of first , se</p>

Q12:

In 20 litres of milk , 69% is pure milk and the remaining is water . What amount of water should be added in the milk so that pure milk in the ratio mixture is 40% ?


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Quantity of pure milk = 20 \(\times\) \(\frac{60}{100}\) = 12l <br>Quantity of water= 20 - 12 = 8l<br>Let x l of water should be added in 20 l of milk so that pure milk in the mixture is 40&amp; or quantity of milk is 40% in (20 + x ) l of mixture. <br><br>\(\therefore\) 12 l = (20 = x) \(\frac{40}{100}\)<br>or , 1200l = 800 + 40x<br>or , 400l = 40x<br>\(\therefore\) x = \(\frac{400}{40}\) = 10l Ans.</p>

Q13:

The milk previoulsy sold for Rs. 22.50 per litre is to be sold for Rs. 18. What is the ratio of water and milk in the mixture ?


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Let , the quantity of pure milk = x l<br>Now , cost of x l of milk costing Rs. 22.50 per litre = 22.5x<br>Let , y l of water x l of milk be adde in so that cost of per litre of mixture is Rs. 18<br><br>Then , quantity of mixture = (x + y) l<br>Cost of per litre of mixture = Rs. 18<br><br>Now , cost of (x + y) l milk = Rs. 18 (x + y)<br>or , Rs. 22.5x = Rs. 18 (x + y)<br>or , 4.5x = 18y<br>or , \(\frac{4.5}{18}\) = \(\frac{y}{x}\)<br><br>\(\therefore\) \(\frac{y}{x}\) = \(\frac{1}{4}\) <br><br>Hence , the required ratio of water and milk is 1 : 4</p>

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Adverse Effect on Biodiversity and Conservation Methods

Adverse Effect on Biodiversity and Conservation Methods

Adverse effect on biodiversity and mitigating measures

The adverse effects on biodiversity and methods to mitigate them may differ from place to place because of different geographical regions. However, there are some similarities.

The adverse effects on biodiversity and its mitigating measures are as follows: -

  1. Degradation of ecosystem

    There are different types of living beings in the terrestrial and aquatic ecosystems. Cattle, insects, plants, trees, etc. grow in the terrestrial ecosystem while fish, some insects, small plants live in the aquatic ecosystem. The terrestrial ecosystem is degraded due to forest fire, destruction of forests, use of insecticides and pesticides, etc. Aquatic ecosystem is spoilt by falling stones and soil when construction work is done and by explosions. Thus, the living beings on both land and water are affected. Because of the degraded ecosystem, there will be the lack of habitat for the living beings. So, human beings should not degrade the ecosystem. We should not discharge harmful things like rubbish, insecticides, etc directly to water sources. When we protect ecosystem, it helps to protect the biodiversity.

  2. Loss of habitat

    Habitat plays a significant role in retaining biodiversity. Natural habitat is ruined because of human activities like construction work, extension of agricultural work, destruction of forests, etc. which results in bad effects in animal’s habitat, food cycle, climate and reproduction progress. The numerical growth of living beings and their life cycle are hampered. It is unwise to destroy the natural habitat while conducting development construction and human activities. The habitat which has been destroyed should be restored.

  3. Loss of mobility

    The deterioration of the natural environment causes the loss of mobility of birds and animals. The causes of deterioration of the environment are the destruction of the forest, expansion of agricultural land and urbanization. All these factors limit the space for the animals and birds for roaming. That’s why, the natural habitat for the several animals and birds have been diminished. In this situation, they have to stay within a limited area. So, people should not ruin the natural environment if they want to conserve animals and birds. We should try to make the dry and uncultivated land suitable for the habitat of animals and birds.

  4. Limitation of expansion of vegetation

    Land for cultivation is extended due to the population growth. Various types of development and construction work are continuing. Similarly, settlement areas are expanding and urbanization is taking place. Grazing land is declining in size. These activities harm to the forest, vegetation, herbs, etc. Because of these adverse effects, biodiversity is degrading. People should grow trees in the surrounding of their houses, either side of the road and bare fields. Pasture lands should not be damaged. Grass should be grown in the bare place for the cattle. Thus, the greenery can be increased.


National and Local Biodiversity Conservation Programmes

Programmes for the conservation of biodiversity should be conducted at national, local and individual levels. NGOs and civic societies must be encouraged. Conservation programmes can be effectively implemented with the joint effort of all.

Some of the major conservation programmes and activities are given below: -

  1. National park, wildlife reserve and conservation area

    National parks, wildlife reserves and conservation area have been established as protected areas in different parts of Nepal. There are total 10 national parks, 3 wildlife reserves, 4 conservation areas, 1 hunting reserve and 11 buffer zones. The wildlife reserves, conservation areas and national parks are helping to conserve the living beings in in-situ, that is, in their original places. The living beings get proper natural environment, food and habitat there. Chitwan National Park, Langtang National Park, Parsa Wildlife Reserve, Annapurna Conservation Area, Dhorpatan Hunting Reserve, etc. are some examples.

  2. Artificial conservation site

    The artificial environment (ex-situ conservation) should be created for the conservation of rare living beings as in-situ conservation is not possible for all types of living beings. In order to conserve the disappearing and rare animals, birds and vegetation, government has established botanical garden, parks and zoos. Attempts are being made to create suitable environment in the central zoo to conserve animals and birds. Similarly, botanical gardens are preserving different types of vegetation. These activities contribute to the conservation of biodiversity through protecting the genes of the living beings and assisting them to grow and reproduce.

  3. Awareness rising programme and conservation programme at local level

    Government and non-governmental organizations are involved in biodiversity conservation. They create awareness on it forest and Soil Conservation Ministry has been conducting natural environment and watershed conservation programme. Similarly, Ministry of Environment, Science and Technology formulates the policy on environmental conservation and implements it. It co-ordinates the programmes contributing to the conservation of ecosystem and biodiversity. Biodiversity is a common property. So, its conservation is the duty of all. Joint efforts can help to achieve the success. Conservation programme should be conducted at community levels. National Trust for Nature Conservation (NTNC) conducts programmes at local level whereas international organizations such as IUCN- The World Conservation Union, World Wildlife Fund (WWF) are also involved in conservation programmes. These programmes help in the conservation of biodiversity and environment as well.

Lesson

Biodiversity

Subject

EPH

Grade

Grade 10

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