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11th Biology Chapter 1 Exercise Living World Solutions Maharashtra Board

June 25, 2026June 24, 2026 by Prasanna

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 1 Living World Textbook Exercise Questions and Answers.

Living World Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 1 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 1 Exercise Solutions

1. Choose correct option

Question A.
Which is not a property of living being?
a. Metabolism
b. Decay
c. Growth
d. Reproduction
Answer:
b. Decay

Question B.
A particular plant is strictly seasonal plant. Which one of the following is best suited if it is to be studied in the laboratory?
a. Herbarium
b. Museum
c. Botanical garden
d. Flower exhibition
Answer:
a. Herbarium

Question C.
A group of students found two cockroaches in the classroom. They had a debate whether they are alive or dead. Which life property will help them to do so?
a. Metabolism
b. Growth
c. Irritability
d. Reproduction
Answer:
c. Irritability

Question 2.
Distinguish between botanical gardens, zoological park and biodiversity park with reference to characteristics.
Answer:

No.	Botanical Gardens	Zoological Parks	Biodiversity Parks
1.	Plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner in a botanical garden.	Zoological parks are places where wild animals are kept in captivity.	It is an assemblage of species that form self-sustaining communities on degraded barren landscape.
2.	It is a type of ex situ conservation.	It is a type of ex situ conservation.	It is a type of in situ conservation.
3.	It is related to conservation of various	It is related to conservation of various fauna.	It is related to conservation of all biodiversity.

3. Answer the following questions

Question A.
Jijamata Udyan, the famous zoo in Mumbai has acclimatised humbolt penguins. Why should penguins be acclimatised when kept at a place away from their natural habitat?
Answer:

Zoological park (zoo) is a type of ex-situ conservation in which wild animals are kept in captivity.
Humboldt penguins are native to South America and the surrounding environment differs significantly at Jijamata Udyan (zoo) in Mumbai.
In order to ensure that these penguins survive longer and are healthy they need to be acclimatised (adjust) to their new environment slowly.
If they are not acclimatised or the facilities in the zoo where the penguins are kept are not optimal/ suitable, they may develop abnormal stress and exhibit unusual behaviours due to it.
These penguins may also be more prone to contracting certain diseases, since they are suited to living in a particular climatic condition.
The enclosure of these penguins consists of water pool, air handling units and a chiller system to maintain temperatures between 12 – 14°C, where the penguins were kept for around 8 to 10 days to get acclimatised to their new environment before allowing any visitors inside the zoo.

Hence, Humboldt penguins need to be acclimatised to their new surroundings, when kept at a place away from
their natural habitat.

Question B.
Riya found peculiar plant on her visit to Himachal Pradesh. What are the ways she can show it to her biology teacher and get information about it?
Answer:

Riya can press and mount the plant specimen on a herbarium sheet and preserve the dried plant material, until she returns back from her visit.
She can also write any available information regarding the collected specimen on the herbarium sheet, which can be useful for further studies with her biology teacher.
Various taxonomical aids can be useful to get information about this peculiar plant.
[Note: In order to conserve the local flora, Riya can collect photographs of plant and describe it’s structure to her teacher.]

Question C.
At Andaman, authorities do not allow tourists to collect shells from beaches. Why it must be so?
Answer:

Seashells are an important part of the coastal ecosystem and are crucial for the survival of various marine creatures.
They provide material for building nests of birds and also act as a substratum for attachment of algae, sea grass, sponges and various microbes.
Fishes use shells for hiding from predators, whereas hermit crabs use shells as temporary shelters.
Removal of seashells from seashores may also indirectly affect the rate of shoreline erosion.
Hence, in an attempt to protect the ecosystem, authorities in Andaman do not allow tourists to collect shells from beaches.

Question D.
Why do we have green house in botanical gardens?
Answer:

Greenhouse is a structure with suitable walls and a roof in which plants are grown under regulated climatic conditions.
Most botanical gardens exhibit ornamental plants which require stringent/ optimum climatic conditions for their growth and/or flowering.
The greenhouse associated with botanical gardens are also used to grow and propagate those plants that may not survive seasonal changes.

Hence, in order to provide optimum temperature for better growth and flowering and also to protect the plants from certain diseases, there are greenhouses in botanical gardens.

Question E.
What do you understand from terms like in situ and ex situ conservation?
Answer:

In situ conservation: It includes conservation of species in their natural habitats. Grazing, cultivation and collection of products from the forests is banned in such areas. Legally protected areas include national parks, wildlife sanctuaries and biosphere reserves.
Ex situ conservation: It includes conservation of species outside their natural habitats. Species are conserved in botanical gardens, culture collections and zoological parks.

4. Write short notes

Question A.
Role of human being in biodiversity conservation.
Answer:

Due to rapid increase in human population and industrialization, humans have over utilized natural resources; leading to degradation of the environment and hence only humans can help conserve the ecosystem.
Humans are capable of conserving and improving the quality of nature and thus, can play a major role in biodiversity conservation.
In order to conserve biodiversity and its environmental resources, humans must use the resources rationally and avoid excessive degradation of environment.
Human beings are stakeholders of the environment and need to come together to overcome pollution and improve the environment quality in order to conserve biodiversity. E.g. Ban or limit on use of harmful products (plastic, chemicals, etc.) that are toxic to various birds, animals, etc.
Human beings also play a role in conservation of biodiversity by establishment of various sites for in situ (national parks, wildlife sanctuaries and biosphere reserves) and ex situ (botanical gardens, culture collections and zoological parks) conservation.

Question B.
Importance of botanical garden.
Answer:
The importance of botanical gardens is as follows:

It is a place where there is an assemblage of living plants maintained for botanical teaching and research purpose.
Botanical gardens are important for their records of local flora.
Botanical gardens provide facilities for the collection of living plant materials for botanical studies.
Botanical gardens also supply seeds and material for botanical investigations.
The development of botanical gardens in any country is associated with its history of civilization, culture, heritage, science, art, literature and various other social and religious expressions.
Botanical gardens besides possessing an outdoor garden may contain herbaria, research laboratory, greenhouses and library.
Botanical gardens are not only important for botanical studies, but also to develop tourism in the country.

Question 5.
How can you, as an individual, prevent the loss of Biodiversity?
Answer:
As individuals, we can prevent loss of biodiversity in the following ways:

Increasing awareness about environmental issues. Making posters that provide more information about biodiversity conservation, to raise public awareness.
Increased support and/ or active participation in government policies and actions laid down for conservation of biodiversity.
Protect various plant and animal species in our surrounding.
Set up bird and bat houses wherever possible.
Prevent felling of trees especially native plants or trees in a particular area.
Reduce, recycle and reuse resources. Especially, reduce pollution and use of plastic bags and other materials that are potential threats for the environment.
Use environment friendly products, segregate and dispose garbage correctly.
Convince people about the importance of trees and the need to participate in tree plantation campaign.
Obey the rules that fall under Biodiversity Act.
[Students can use the given points as reference and mention additional preventive measures on their own.]

Practical / Project :

Question 1.
Make herbarium under the guidance of your teacher.
Answer:
Students are expected to perform the given activity by themselves under the guidance of their teacher.

Question 2.
Find out information about any one sacred grove (devrai) in Maharashtra.
Answer:
Sacred groves in Maharashtra are located in districts like Ahmednagar, Bhandara, Chandrapur, Jalgaon, Kolhapur, Nashik, Pune, Raigad, Ratnagiri, Sangli, Satara, Sindhudurg, Thane, Yavatmal.
[Source: Data as per C.P.R. Environment Education Centre, Chennai.]
e. g. Sacred grove of Parinche valley, Pune district of Maharashtra:

The Parinche valley region is comprised of the inaccessible rear part of the Purandhar fort and its surrounding valley region and is situated about 63 km to the southeast of Pune city and 18 km from Saswad town. The total area of the valley region is about 132 sq. km. Parinche is the biggest village and a nodal place in the valley. The majority (12) of the documented groves are located in the Kaldari and Pangare zones. The size of the sacred groves has however reduced due to various human related activities that have taken place in recent years.

The biggest sacred grove in the Parinche valley belongs to Buvasaheb of Tonapewadi and spreads over an area of 4.80 hectares. The forest types are unique to the groves. Presence of key species in the sacred groves varies from region to region. Two key tree species, i.e. Terminalia bellerica and Ficus spp., are present in these sacred groves which have almost disappeared from the surrounding areas. Large buttressed trees are another important feature of well-preserved sacred groves. The presence of these tree species indicates the vegetation of the past and also the type of potential vegetation that can be regenerated in these regions.

[Source: Waghchaure, C. K., Tetali, P., Gunale, V. R., Antia, N. H., & Birdi, T. J. (2006). Sacred Groves of Parinche Valley of Pune District of Maharashtra, India and their Importance. Anthropology & Medicine, 13(1), 55-76]
[Students can refer the given answer and search for more information about other sacred groves on their own.]

11th Biology Digest Chapter 1 Living World Intext Questions and Answers

Can you recall? (Textbook Page No. 1)

Question 1.
Whether all organism are similar? Justify your answer.
Answer:
No, all organisms are not similar.

Organisms on the earth exhibit great diversity.
Organisms are grouped as microbes, plants (autotrophs), animals (heterotrophs) and decomposers.
Different microbes and decomposers have various shapes and sizes.
Plants can be further classified on their shape, size, structure, mode of reproduction, etc. Plants also differ greatly based on the locations in which they are found, e.g. Snowy, desert, forest, aquatic, etc.
Even animals show a high degree of variation. They are classified as unicellular, multicellular, invertebrates, vertebrates, etc. Also, based on the environment in which they live, they are classified as terrestrial, aerial, aquatic and amphibians.

Question 2.
What is the difference between living and non-living things?
Answer:

	Living Things	Non-living Things
a.	Living things show growth from within.	Non-living things show growth by accumulation of materials on their surface.
b.	They reproduce asexually or sexually, except mules, sterile worker bees, infertile males.	They do not reproduce.
c.	They perform metabolism in order to obtain energy.	No metabolic changes occur in non-living things.
d.	They show irritability and respond to changes in their surroundings.	They do not show irritability.
e.	They undergo ageing and eventually die.	Non-living things do not have a finite life span.

Question 3.
Enlist the characters of living organisms.
Answer:
The basic principles of life are as follows:

Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
Death: As the body loses its capacity to perform metabolism, an organism dies.
Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Can you tell? (Textbook Page No. 1)

Question 1.
Whether all organisms prepare their own food?
Answer:
No, all organisms do not prepare their own food. Organisms that prepare their own food are known as autotrophs (e.g. Green plants, certain microbes). These organisms prepare their own food in the presence of sunlight, water and carbon dioxide.

Question 2.
Which feature can be considered as all-inclusive characteristic of life? Why?
Answer:
Metabolism can be considered as an all-inclusive (defining) feature of life since it is exhibited by all living organisms and does not take place in non-living things.

Another all-inclusive characteristic of life is responsiveness or irritability. This is a unique property of living beings since all living beings are conscious of their surroundings.

Question 3.
How can we study large number of organisms at a glance?
Answer:
Systematic study of organisms with the help of taxonomical aids can be used to study a large number of organisms at a glance.

Can we call? (Textbook Page No. 1)

Question 1.
Reproduction as inclusive character of life?
Answer:
No, we cannot call reproduction as an inclusive character of life. Certain organisms like mules and worker bees do not reproduce and are still living. Thus, reproduction cannot be considered as an all inclusive defining characteristic of living organisms.

Think about it (Textbook Page No. 1)

Question 1.
Can metabolic reactions demonstrated in a test tube (called ‘in vitro’ tests) be called living?
Answer:

The sum total of all the chemical reactions occurring in the body is known as metabolism and no non¬living object exhibits metabolism.
However, metabolic reactions can be demonstrated outside the body in a test tube (cell-free medium).
Thus, isolated metabolic reaction (s) outside the body of an organism, performed in a test tube is neither living nor non-living.
Metabolic reactions occurring in vitro are living reactions but not living things.

Question 2.
Now a days patients are declared ‘brain dead’ and are on life support. They do not show any sign of self-consciousness. Are they living or non-living?
Answer:
The brain controls all life processes. Hence, when a patient is declared as ‘brain dead’, he does not carry out any of the inclusive defining characters of living things (e.g. metabolism, consciousness, etc.) and is completely dependent on machines. Since, such patients do not show any sign of self-consciousness, these patients cannot exactly be called as living.

Internet my friend (Textbook Page No. 2)

Question 1.
Collect information about Prof. Almeida, Prof. V. N. Naik, Dr. A. V. Sathe, Dr. P. G. Patwardhan with reference to their taxonomic work and biodiversity conservation.
Answer:
i. Prof. Almeida:
Prof. (Dr.) Marselin R. Almeida was a renowned Plant Taxonomist and Medicinal Plant Consultant of India. He was a curator at the Blatter Herbarium (Mumbai). He discovered four new species of pteridophytes from Bombay presidency. His work includes – Pteridophytes of Maharashtra and Flora of Mahabaleshwar. He has contributed to the Flora of Maharashtra, Sawantwadi and its adjoining areas along with Dr. S. M. Almeida.

ii. Prof. V. N. Naik:
Prof. V. N. Naik is a renowned ‘Angiosperms Taxonomist’ of India. He completed the Flora of Marathwada. He has produced 15 Ph.D., 110 research articles and 6 books. His book on ‘Taxonomy of Angiosperms’ (Tata McGraw-Hill Education, 1984) is widely used throughout the world. He is currently a faculty of Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.
[Source: http://www. bamu. ac. in/dept-of-botany/Achievements, aspxj]

iii. Dr. A. V. Sathe:
Collection and taxonomic studies of mushrooms in Maharashtra started around 1974. Prof. A.V. Sathe and his team were amongst the first to begin these studies. They recorded 75 species distributed in 43 genera. These species were collected from Maharashtra, Karnataka and Kerala. The collection of these species was documented in the form of a Monograph on Agaricales.
[Source: Borkar P., Doshi A., Navathe D. (2015) Mushroom diversity of Konkan region of Maharashtra, India. Journal of Threatened Taxa. 7(10): 7625-7640]

iv. Dr. P. G. Patwardhan:
Dr. Patwardhan and his associates at the M.A.C.S. Research Institute, Pune-renamed as Agharkar Research Institute (ARI), Pune have performed detailed studies on lichens. His school is in possession of over 600 species of crustose lichens, obtained after intensive collection programmes. These specimens have been deposited in the Ajarekar Mycological Herbarium in the Department of Mycology and Plant Pathology at the M.A.C.S. Research Institute, Pune.
[Source: http://lib.unipune.ac.in:8080/xmlui/bitstreamfhandle/l23456789/7451/07_introduction.pdf? sequence=7&is Allowedly]
[Students are expected to find more information on their own.]

Can you tell? (Textbook Page No. 3)

Question 1.
What are the essentials of a good herbarium?
Answer: The essentials of a good herbarium are as follows:

It is essential to identify and label the collected specimen correctly.
Specimens should be stored in a dry place.
The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.
It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet. Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 2.
Why does the loss of biodiversity matter?
Answer:

The loss of biodiversity is an moral and ethical issue.
Biodiversity helps to maintain stability in an ecosystem.
Humans share the environment with various other organisms and harm to these species can result in loss of biodiversity.
The loss of even one variety of organisms can affect the entire ecosystem.
Hence, due to all these reasons, loss of biodiversity matters.

Question 3.
Why should we visit botanical gardens, museums and zoo?
Answer:

Botanical gardens, museums and zoos are taxonomical aids which can be used to study biodiversity.
Botanical gardens have a wide range of plant species that are protected and preserved which can be observed and studied.
Museums help gain information about various plants and animals that are preserved and may even be extinct. They act as reference hubs for biodiversity studies.
Zoos provide information about various animals. They also harbour certain endangered animals and help us understand the role of biodiversity conservation. They can also be visited to study the food habits and behaviour of animals.
Hence, we should visit botanical gardens, museums and zoos.

Find out (Textbook Page No. 4)

Question 1.
Human being is at key position in maintaining biodiversity of earth. Find out more information about the following.
i. Laws to protect and conserve biodiversity in India.
Answer:
a. Forest (Conservation) Act, 1980
b. Biological Diversity Act, 2002
c. Wildlife (Protection) Act, 1972
d. Environment Protection Act, 1986
[Students can find out more laws to protect and conserve Biodiversity in India ]

ii. Environmental effects of ambitious projects like connecting rivers or connecting cities by constructing roads.
Answer:
Connecting rivers or connecting cities by constructing roads have the following environmental effects:
a. They form barriers to animals.
b. Construction of roads requires cutting down of trees and results in large scale deforestation.
c. They occupy large land resources resulting in loss of habitat of various species.
d. It can alter the water flow pattern and damage many ecosystems.
e. Increase in air, water, soil and noise pollution can disturb various animals and birds, thus affecting their behavioural pattern.

iii. Did bauxite mining in Western Ghats affect critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp. ?
Answer:
a. The Western Ghats, is one of the global biodiversity hotspots and retains more than 30% of all plant, aquatic, reptile, amphibian and mammal species found in India.

b. Recently, this ecologically sensitive region has been subjected to various developmental activities that have adversely affected the flora and fauna of the region.

c. Bauxite mining is one such activity which has had significant negative impact on the local environment. To access bauxite ore deposits, the above-ground vegetation needs to be completely removed, causing large scale deforestation. The vegetation in the adjoining area is also affected due to dumping.

d. The major threats of this activity include vegetation loss, forest fragmentation and biodiversity loss.
e. Since most mines fall in Eco-Sensitive Zones (ESZ), it has seriously affected the flora and fauna of the Western Ghats.

f. Black panthers have frequently been spotted at various locations in the Western Ghats and mining in these areas can seriously affect their health and numbers.

g. Certain species of Ceropegia and Eriocaulon that are endemic in the area have been reported to be critically endangered.

[Source: Chandore A. (2015) Endemic and threatened flowering plants of Western Ghats with special reference to Konkan region of Maharashtra. Journal of Basic Sciences. 2 (21-25)]
Hence it is most likely that bauxite mining in Western Ghats has adversely affected the critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp.

Internet my friend (Textbook Page No. 4)

Question 1.
i. Collect information about botanical gardens, zoological parks and biodiversity hotspots in India.
Answer:
a. Botanical gardens in India:

No.	Botanical Gardens of India	Location
1.	Acharya Jagadish Chandra Bose Indian Botanic Garden	Kolkata
2.	Lloyd Botanical Garden	Darjeeling
3.	National Botanical Research Institute	Lucknow
4.	Botanical Garden of the Forest Research Institute	Dehradun
5.	The State Botanical Garden	Odisha
6.	Botanical Garden	Saharanpur
7.	Government Botanical Garden	Ootacamund

b. Zoological Parks in India:

No.	Zoological parks	Location	Type of animals
1.	Rajiv Gandhi Zoological Park	Pune [Katraj]	Reptiles, mammals, birds. They have a snake park.
2.	Jijamata Udyan	Mumbai	Endangered species of animals and birds.
3.	Nehru Zoological Park	Hyderabad	3500 species of birds, animals and reptiles.
4.	Indira Gandhi Zoological Park	Vishakhapatanam	Primates, carnivores, small mammals, reptiles and birds.
5.	Padmja Naidu Himalayan Zoological Park	Darjeeling	Endangered animals like snow leopards, red pandas, gorals (mountain goat), Siberian tigers and a variety of endangered bird species.
6.	Allen Forest Zoo	Kanpur	Hyena, Bear, Rhinoceros, Hippopotamus, Langoor, Musk deer. Ostrich, Emu, Crane etc.
7.	Lucknow Zoo	Lucknow	Royal Bengal Tiger, White Tiger, Gibbon, Black Bear, Asiatic Elephant, Great pied, Horn Bill etc.
8.	Alipore Zoological Gardens	Kolkata	Royal Bengal Tiger, African Lion, Hippopotamus, Great Indian One-homed Rhinoceros.
9.	The Madras Crocodile Bank Trust	Chennai	Crocodiles and many species of turtles, snakes and lizards.
10.	Parassinikkadavu Snake Park	Kannur	Spectacled Cobra, King Cobra, Russell’s Viper, Krait and Pit Viper.

c. Biodiversity hotspots in India:

No.	Biodiversity Hotspots
1.	The Eastern Himalayas (Arunachal Pradesh, Bhutan, Eastern Nepal)
2.	Indo – Burma (Purvanchal Hills, Arakan Yoma, Eastern Bangladesh)
3.	The Western Ghats and Srilanka

ii. Collect information of endemic flora and fauna of India.
Answer:
a. Endemic flora:
Albizia sikharamensis (Mimosaceae), Argvreia arakuensis (Convolvulaceae), Arundinella setosa (Poaceae), Acacia diadenia (Mimosaceae), Citrus assamensis (Rutaceae), Magnolia bailloni (Magnoliaceae), etc.
[Source: http://www. bsienvis. nic. in/Database/E_3942. aspx]

b. Endemic fauna:
Bare Bellied Hedgehog (Paraechinus nudiventris), Andaman Shrew (Crocidura andamanensis), Aruanchal Macaque (Macaca munzala), Car Nicobar Rat (Rattus palmarum), Peter’s Tube-nosed Bat (Harpiola grisea) etc.
[Source: http://faunaofindia.nic.in/PDFVolumes/spb/056/index.pdf]
[Students are expected to use the given sources and find more information on their own.]

11th Std Biology Questions And Answers:

Problem Set 46 Class 5 Maths Chapter 11 Problems on Measurement Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Problems on Measurement Class 5 Problem Set 46 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 11 Problems on Measurement

Question 1.
Add :

(1) ₹ 9, 50 paise + ₹ 14, 60 paise
Solution:

₹	Paise
1
9 + 14	5 0 6 0
2 4	1 0

50 paise + 60 paise
= 110 paise
= 1 ₹ 10 paise
∴ ₹ 24, 10 paise

(2) 6 cm 5 mm + 7 cm 9 mm
Solution:

cm	mm
1
6 + 7	5 9
1 4	4

5 mm + 9 mm
= 14 mm 14 mm
= 1 cm 4 mm
∴ 14 cm 4 mm

(3) 22 m 50 cm + 25 m 75 cm
Solution:

m	cm
1
2 2 + 2 5	5 0 7 5
4 8	2 5

50 cm + 75 cm
= 125 cm
= 1 m 25 cm
∴ 48 m 25 cm

(4) 15 km 740 m + 13 km 950 m
Solution:

km	m
1
1 5 + 13	7 4 0 9 5 0
2 9	6 9 0

740 m + 950 m
= 1690 m 1690 m
= 1km 690 m
∴ 29 km 690 m

(5) 25 kg 650 g + 29 kg 770 g
Solution:

kg	gm
1
2 5 + 29	6 5 0 7 7 0
5 5	4 2 0

650 gm + 770 gm
= 1420 gm
= 1 kg 420 gm
∴ 55 kg 420 gm

(6) 19l 840ml + 25l 250ml
Solution:

l	ml
1 1
1 9 + 2 5	8 4 0 2 5 0
4 5	0 9 0

840 ml + 250 ml
= 1090 ml
= 11 + 90 ml
∴ 45 l 90 ml

Question 2.
Subtract :

(1) ₹ 19, 50 paise – ₹ 12, 60 paise
Solution:

₹	Paise
1 8	1 5 0
~~1 9~~ – 1 2	~~5 0~~ 6 0
6	9 0

We cannot subtract 60 paise from 50 paise. So convert 1 ₹ into 100 paise.
₹ 6, 90 paise

∴ ₹ 6, 90 paise

(2) 24 cm 2 mm – 3 cm 8 mm
Solution:

cm	mm
2 3	1 2
~~2 4~~ – 3	2 8
2 0	4

We cannot subtract 8 mm from 2 mm. So, convert 1 cm = 10 mm

∴ 20 cm 4 mm

(3) 20 m 30 cm – 17 m 60 cm
Solution:

m	cm
1 9	1 3 0
~~2 0~~ – 1 7	~~3 0~~ 6 0
2	. 7 0

We cannot subtract 60 cm from 30 cm. So, convert 1 m = 100 cm

∴ 2 m 70 cm

(4) 40 km 255 m – 17 km 960 m
Solution:

km	m
3 9	12 2 5
~~4 0~~ -1 7	2 2 5 9 6 0
2 2	2 6 5

We cannot subtract 960 m from 225 m. So, convert 1 km = 1000 m

∴ 22 km 265 m

(5) 35 kg 150 g – 26 kg 470 g
Solution:

kg	gm
3 4	1 1 5 0
~~3 5~~ – 2 6	~~1 5 0~~ 4 7 0
8	6 8 0

We cannot subtract 470 gm from 150 gm. So, convert I kg= 1000gm

∴ 8 kg 680 gm

(6) 46 l 200 ml – 38 l 750 ml
Solution:

l	ml
4 5	1 2 0 0
~~4 6~~ – 3 8	~~2 0 0~~ 7 5 0
7	4 5 0

We cannot subtract 750 ml from 200 ml. So, convert 1 l = 1000 ml

∴ 7 l 450 ml

Word problems

Study the following examples.

Example (1) If a shopkeeper has 150 kg 500 g of rice and sells 75 kg 750 g, how much rice will be left?
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 1

74 kg 750 g of rice is left.

Example (2) A can of milk has 20 l 450 ml of milk. Another can has 18 l 800 ml. How much milk is there in the two cans altogether?
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 2

The total quantity of milk is 39l 250ml.

Example (3) At a speed of 90 km per hour, what distance will a train cover in two and a half hours?

The speed of the train is 90 kmph. That is, it travels 90 km in one hour. It travels 90 more km in the second hour.
In the next half an hour, 90 ÷ 2 = 45 km
The total distance travelled is 90 + 90 + 45 = 225 km.

Example (4) If one dress requires 3 m 25 cm of cloth, how much do 4 dresses need?

Manju’s method :
3 m 25 cm for the 1st dress
+ 3 m 25 cm for the 2nd dress
+ 3 m 25 cm for the 3rd dress
3 m 25 cm for the 4th dress
_________
12 m 100 cm
1 m is 100 cm, therefore 12 + 1 = 13 m

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 3

Example (5)
If a wire that is 9 m 50 cm long is cut into pieces of 5 cm each, how many pieces will be made?
9 m 50 cm = (900 + 50) cm
To find out how many pieces of 5 cm can be made from a wire 950 cm long, let us use division.
190 pieces will be made.
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 4

Example (6) A play started at 30 minutes past 6 in the evening and finished two and three quarter hours later. What time did the play get over?
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 5

The play got over at 15 minutes past 9 at night.

Note : The units for length, mass and capacity are written in decimal form. This makes it easy to carry out addition and subtraction of length, mass and capacity.

Units of measuring time are not in decimal form. It is a little more difficult to carry out additions and subtractions of those quantities.

Problems on Measurement Problem Set 46 Additional Important Questions and Answers

Add the following:

(1) 12 km 880 m + 7 km 620 m
Solution:

km	m
1
1 2 + 7	8 8 O 6 2 0
2 0	5 0 0

880m + 620 m = 1500 m
= 1km 500 m
∴ 20 km 500 m

(2) ₹ 62, 45 paise + ₹ 37, 55 paise
Solution:

₹	Paise
1
6 2 + 3 7	4 5 5 5
1 0 0	0 0

45 paise + 55 paise
100 paise = 1 ₹
∴ 100 rupees

Subtract the following:

(1) 15 m 15 cm – 4 m 65 cm
Solution:

kg	gm
1 4	1 1 5
~~1 5~~ – 4	~~1 5~~ 6 5
1 0	5 0

We cannot subtract 65 cm from 15 cm. So, convert l m = 100 cm
∴ 10 m 50 cm

(2) 29 kg 880 gm – 8 kg 900 gm
Solution:

kg	gm
2 8	1 8 8 0
~~2 9~~ – 8	~~8 8 0~~ 9 0 0
2 0	9 8 0

We cannot subtract 900 gin from 880 gm. So, convert 1 kg = 1000 gm
∴ 20 kg 980 gm

Class 5 Maths Solution Maharashtra Board

Problem Set 15 Class 5 Maths Chapter 4 Multiplication and Division Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Multiplication and Division Class 5 Problem Set 15 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 4 Multiplication and Division

Question 1.
Solve the following and write the quotient and remainder.
(1) 1284 ÷ 32
Solution :
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 1
Quotient = 40
Remainder = 4

(2) 5586 ÷ 87
Solution :
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 2
Quotient = 64
Remainder =18

(3) 1207 ÷ 27
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 3
Quotient = 44
Remainder =19

(4) 8543 ÷ 41
Solution :
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 4
Quotient = 208
Remainder =15

(5) 2304 ÷ 43
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 5
Quotient = 53
Remainder = 25

(6) 56,741 ÷ 26
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 6
Quotient =2182
Remainder = 9

Question 2.
How many hours will it take to travel 336 km at a speed of 48 km per hour?
Solution:
Time = Distance ÷ Speed
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 9
Answer:
It will take 7 hours.

Question 3.
Girija needed 35 cartons to pack 1400 books. There are an equal number of books in every carton. How many books did she pack into each carton?
Solution:
No. of cartons x No. of books in each carton = Total no. of books 35 x No. of books in each carton = 1400 No. of books in each carton = 1400 35
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 10
Answer:
She packs 40 books in each carton.

Question 4.
The contribution for a picnic was 65 rupees each. Altogether, 2925 rupees were collected. How many had paid for the picnic?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 11
Answer:
45 persons paid for the picnic.

Question 5.
Which number, on being multiplied by 56, gives a product of 9688?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 12
Answer:
173

Question 6.
If 48 sheets are required for making one notebook, how many notebooks at the most will 5880 sheets make and how many sheets will be left over?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 13
Answer:
122 notebooks can be made and 24 sheets left over.

Question 7.
What will the quotient be when the smallest five-digit number is divided by the smallest four-digit number?
Solution:
Smallest five-digit number is 10,000 and smallest four-digit number is 1,000.
So, 10000 ÷ 1000 = 10
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 14
Answer:
Quotient = 10

Mixed examples

A farmer brought 140 trays of chilli seedlings. Each tray had 24 seedlings. He planted all the seedlings in his field, putting 32 in a row. How many rows of chillies did he plant?

Let us find out the total number of seedlings when there were 24 seedlings in each of the 140 trays. We shall multiply 140 and 24.
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 17
Total number of seedlings 3,360.
To find out how many rows were planted with 32 seedlings in each row, we shall divide 3,360 by 32.
The quotient is 105.
Therefore, the number of rows is 105.
Carry out the multiplication of 105 × 32 and verify your answer.
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 18

Multiplication and Division Problem Set 15 Additional Important Questions and Answers

Solve the following and write the quotient and remainder.

(1) 9148 ÷ 37
Solution
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 7
Quotient = 247
Remainder = 9

(2) 1175 ÷ 15
Solution :
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 8
Quotient =78
Remainder = 5

Solve the following word problems:

(1) If 45 kg of sugar cost 1305 rupees, what is the rate of sugar per kg?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 15
Answer:
The rate per kg of sugar is 29 rupees.

(2) 17 people spent ₹ 83,475. How much did each person spend and what is the amount left?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 16
Answer:
Each person spent ₹ 4,910 and the amount left is ₹ 5

Class 5 Maths Solution Maharashtra Board

Problem Set 6 Class 5 Maths Chapter 2 Number Work Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Number Work Class 5 Problem Set 6 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Write the proper symbol, ‘<’ or ‘>’ in the box.
(1) 5,705 [ < ] 15,705
(2) 22,74,705 [ ] 12,74,705
(3) 35,33,302 [ ] 35,32,302
(4) 99,999 [ ] 9,99,999
(5) 4,80,009 [ ] 4,90,008
(6) 35,80,177 [ ] 35,88,172
Answer:
(1) <
(2) >
(3) >
(4) <
(5) <
(6) <

Question 2.
Solve the problems given below.

(1) The Swayamsiddha Savings Group made 3,45,000 papads while the Swabhimani Group made 2,95,000. Which group made more papads?
Answer:
Here, 3,45,000 > 2,95,000
Hence, the Swayamsiddha saving group made more papads.

(2) Children of the Primary School in Ahmadnagar District collected 2,00,000 seeds while those in Pune District collected 3,25,000. Which children collected more seeds?
Answer:
Here, 3,25,000 > 2,00,000
Hence, Pune District children collected more seeds.

(3) The number of people who took part in the Republic Day flag hoisting ceremony was 2,01,306 in Pandharpur taluka and 1,97,208 in Malshiras taluka. In which taluka did a larger number of people participate?
Answer:
Here, 2,01,306 > 1,97,208
Hence, people of Pandharpur taluka participated in larger number

(4) At an exhibition, the Annapoorna Savings Group sold goods worth 5,12,345. The Nirman Group sold goods worth 4,12,900. This figure was 4,33,000 for the Srujan Group and 5,11,937 for the Savitribai Phule group.

Which group had the largest sales?

Which group had the smallest?

Write the sales figures in ascending order.
Answer:
Among the numbers 5,12,345; 4,12,900; 4,33,000; 5,11,937

5,12,345 is largest and 4,12,900 is smallest. Hence, Annapoorna group had the largest sale and Nirman Group had the smallest sales.

Sales in ascending order

4,12,900 < 4,33,000 < 5,11,937 < 5,12,345

Introducing crores

99,99,999 is the biggest seven-digit number. On adding the number 1 to it, we get the smallest eight-digit number, 1,00,00,000. We read this number as ‘one crore’. The new place created to write this number is called the ‘crores’ place.

From the following examples, you can learn to read eight-digit numbers.

Number – Reading

8,45,12,706 – Eight crore forty-five lakh twelve thousand seven hundred and six
5,61,63,589 – Five crore sixty-one lakh sixty-three thousand five hundred and eighty-nine
6,09,04,034 – Six crore nine lakh four thousand and thirty-four

Something more

On the left of the crores place are the places for ten crores, abja, and ten abja in that order. The place value of each of these is ten times the value of the one on its right. According to the Census of the year 2011, the population of our country is 1,21,01,93,422. We read this as ‘one Abuja twenty-one crore one lakh ninety-three thousand four hundred and twenty-two.
Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 5

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Write the proper symbol, ‘<‘ or ‘>’ in the box.
(1) 68,34,170 [ ] 8,43,170
(2) 5,04,132 [ ] 5,04,123
(3) 1,01,001 [ ] 1,00,101
(4) 14,55,432 [ ] 4,54,532
Answer:
(1) >
(2) >
(3) >
(4) >

Question 2.
Write the numbers in words.

(1) 15,97,21,409
Answer:
Fifteen crore, ninety-seven lakh, twenty-one thousand, four hundred and nine

(2) 99,99,99,999
Answer:
Ninety-nine crore, ninety-nine lakh, ninety- nine thousand, nine hundred and ninety nine.

(3) 7,54,21,607
Answer:
Seven crore, fifty-four lakh, twenty-one thousand, six hundred and seven.

(4) 5,16,36,854
Answer:
Five crore, sixteen lakh, thirty-six thousand, eight hundred and fifty four.

Question 3.
Write in figures.

(1) One crore, fifteen lakh, fifty-nine thousand, seven hundred and four
Answer:
1,15,59,704

(2) Sixty-five crore, seventy lakh, fifty thousand and thirty nine
Answer:
65,70,50,039

(3) Four crore, fifty-nine lakh, fourty-three thousand, five hundred and thirty four
Answer:
4,59,43,534

(4) Eighteen crore, seventy-six lakh, fifty-four thousand and one
Answer:
18,76,54,001

Question 4.
Fill in the blanks in the table below:
Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 1
Answer:

Question 5.
Write the following numbers in words.
(1) 17,301
(2) 45,019
(3) 40,018
(4) 28,740
Answer:
(1) Seventeen thousand, three hundred and one.
(2) Forty-five thousand and nineteen.
(3) Forty thousand and eighteen
(4) Twenty-eight thousand seven hundred and forty

Question 6.
How many rupees do they make?
(1) 8 notes of rupees 2,000, 3 notes of rupees 100,11 notes of rupees 10.
Answer:
16,410

(2) 9 notes of rupees 2,000, 18 notes of rupees 100,18 notes of rupees 50,18 notes of rupees 10.
Answer:
20,880

(3) Write the smallest and the biggest five-digit numbers that can be made using the digits only once.
(a) 6, 8, 0,1, 9
(b) 3, 5,1,2, 8
Answer:
Smallest number : (i) 10,689 (ii) 12358
Biggest number : (i) 98,610 (ii) 85321

(4) Write the smallest and the biggest number from the following numbers.
(a) 35,798
(b) 39,785
(c) 39,587
(d) 35,789
Answer:
Smallest number : 35,789
Biggest number : 39,785

(5) Write the number from the given number which is neither biggest nor smallest.
(a) 45, 798
(b) 45, 789
(c) 45, 897
Answer:
45,798.

(6) Write the biggest and the smallest three-digit numbers that can be made using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 only once.
Answer:
Biggest three-digit number : 987
Smallest three-digit number : 102

Question 7.
Read the numbers and write them in words.
(1) 2,65,048
(2) 1,80,794
(3) 1,06,709
(4) 8,80,006
Answer:
(1) Two lakh sixty-five thousand and forty- eight,
(2) One lakh eighty thousand seven hundred and ninety-four.
(3) One lakh six thousand seven hundred and nine.
(4) Eight lakh eighty thousand and six.

Question 8.
Read the numbers and write them in figures.
(1) Two lakh five thousand three hundred and six.
(2) Six lakh and six
(3) Nine lakh forty thousand and thirty seven.
(4) Five lakh ninety-nine thousand and fifteen.
Answer:
(1) 2,05,306
(2) 6,00,006
(3) 9,40,037
(4) 5,99,015

Question 9.
Write six, six-digit numbers using the digits 0,.3,5,7,9,1 only once with 9 lakh fifty-seven thousand in all numbers.
Answer:
(1) 9,57,301
(2) 9,57,310
(3) 9,57,103
(4) 9,57,130
(5) 9,57,013
(6) 9,57,031

Question 9.
(A) Match the columns:

(A)	(B)
(1) Nine lakh nine thousand nine	(a) 9,09,090
(2) Nine lakh june thousand nine hundred nine	(b) 9,90,090
(3) Nine lakh nine thousand ninety	(c) 9,09,009
(4) Nine lakh ninety thousand ninety	(d) 9,09,909

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

(B) Match the columns:

(A)	(B)
(1) Thirty-three lakh, three thousand and three	(a) 33,30,300
(2) Thirty-three lakh, thirty thousand, three hundred	(b) 33,03,003
(3) Thirty lakh, three thousand and thirty.	(c) 30,30,003
(4) Thirty lakh, thirty thousand and three	(d) 30,03,030

Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

Question 10.
Read the numbers and write them in words.
(1) 34,87,569
(2) 70,85,039
(3) 48,07,102
(4) 67,40,960
(5) 88,00,080
(6) 40,40,004
Answer:
(1) Thirty-four lakh, eighty-seven thousand, five hundred and sixty-nine.
(2) Seventy lakh, eight-five thousand and thirty-nine.
(3) Forty-eight lakh, seven thousand, one hundred and two.
(4) Sixty-seven lakh, forty thousand, nine hundred and sixty.
(5) Eighty-eight lakh and eighty.
(6) Forty lakh, forty thousand and four.

Question 11.
Read the numbers and write them in figures.
(1) Fifty-nine lakh, seven thousand, seventeen.
(2) Twenty-two lakh, ten thousand, five hundred.
(3) Fifty-two lakh, twenty-five thousand, four hundred and fifteen.
(4) Thirty lakh, thirty thousand and thirty.
Answer:
(1) 59,07,017
(2) 22,10,500
(3) 52,25,415
(4) 30,30,030

Question 12.
Write the place value of the underlined digit.
(1) 68,03,512
(2) 3,42,157
(3) 84,52,170
(4) 79,345
(5) 38,14,093
(6) 8,10,618
(7) 35,10,387
Answer:
(1) 8,00,000
(2) 40,000
(3) 2,000
(4) 5
(5) 90
(6) 600
(7) 30,00,000

Question 13.
Write the numbers in their expanded form.
(1) 78,15,692
(2)50,95,182
(3)6,40,078
(4) 9,58,802
Answer:
(1) 70,00,000 + 8,00,000 + 10,000 + 5,000 + 600 + 90 + 2
(2) 50,00,000 + 90,000 + 5,000 + 100 + 80 + 2
(3) 6,00,000 + 40,000 + 70 + 8
(4) 9,00,000 + 50,000 + 8,000 + 800 + 2

Question 14.
Write the place name and place value of each digit in the following numbers.
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555
Answer:
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555

Question 15.
The expanded form of the number is given. Write the number.
(1) 70,000 + 6,000 + 500 + 40 + 8
(2) 8,00,000 +-30,000 + 5,000 + 400 + 3
(3) 60,00,000 + 2,00,000 + 70 + 4
(4) 20,00,000 + 5,00,000 + 900 + 5
Answer:
(1) 76,548
(2) 8,35,403
(3) 62,00,074
(4) 25,00,905

Question 16.
Considering the number 50,43,176.
Fill in the blanks.
(1) The digit in the ten thousand place is ……………………………………….. .
(2) Place value of 1 is ……………………………………….. .
(3) The digit in the lakhs place is ……………………………………….. .
(4) Place value of 5 is ……………………………………….. .
(5) The digit 7 is in ……………………………………….. place.
Answer:
(1) 4
(2) 100
(3) 0
(4) 50,00,000
(5) tens

Question 17.
Write the proper symbols ‘<‘ or ‘>’ in the box.
(1) 12,625 [ ] 21,526
(2) 23,564 [ ] 23,546
(3) 36,60,660 [ ] 36,60,606
(4) 89,14,507 [ ] 89,15,407
Answer:
(1) <
(2) >
(3) >
(d) <

Question 18.
Solve the problems given below.
(1) Population of city A is 8,57,238 and that of city B is 8,75,461. Population of which city is more?
Answer:
city B

(2) Yearly income of Rajnikant is? 3,48,600 and that of Shashikant is? 3,46,500. Whose income is less?
Answer:
Shashikant

Question 19.
Profit of the four companies A, B, C, D is as follows.
Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 3
Now, answer the following questions.
(1) Which company made maximum profit?
(2) Which company made minimum profit?
(3) Write the profit of the companies in the descending order.
Answer:
(1) B
(2) C
(3) profit of company B > D > A > C

Question 20.
In a certain election, candidates : Tavade, Patel, Chauhan, and Shinde got the votes as follows.
Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 4
Now, answer the following questions.
(1) Who got the highest number of votes?
(2) Who got the least number of votes?
(3) Write the number of votes obtained in the ascending order.
Answer:
(1) Patel
(2) Shinde
(3) 34,67,008 < 37,51,386 < 43,51,239 < 48,00,173

Question 21.
Compare the following using >, < or = signs.
(1) 3,97,48,632 [ ] 3,97,58,632
(2) 1,50,15,178 [ ] 1,50,15,780
(3) 3,74,98,561 [ ] 96,42,748
(4) 30,49,75,831 [ ] 30,49,00,831
Answer:
(1) <
(2) <
(3) >
(4) >

Question 22.
Circle the correct answer:
(1) Mark periods 617231801 according to the Indian Number system.
(a) 61,72,31,801
(b) 16,172,31
(c) 617,231,801
Answer:
(a) 61,72,31,801

(2) Mark periods 90289164 according to the international Number system.
(a) 9,0289,164
(b) 902891,64
(c) 90,289,164
Answer:
(c) 90,289,164

(3) 1,00,00,000 is read as ……………………………….. .
(a) ten crore
(b) one crore
(c) hundred thousand
Answer:
(b) one crore

Class 5 Maths Solution Maharashtra Board

Problem Set 23 Class 5 Maths Chapter 5 Fractions Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Fractions Class 5 Problem Set 23 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

Question 1.
What is \(\frac{1}{3}\) of each of the collections given below?

(1) 15 pencils
(2) 21 balloons
(3) 9 children
(4) 18 books
Answer:
(1) 15 pencils → \(\frac{1}{3}\) of 15 = 5, 15 ÷ 3 = 5 pencils.
(2) 21 baloons → \(\frac{1}{3}\) of 21 = 7,21 ÷ 3 = 7 baloons.
(3) 9 children → \(\frac{1}{3}\) of 9 = 3, 9 ÷ 3 = 3 chi1dren.
(4) 18 books → \(\frac{1}{3}\) of 18 = 6, 18 ÷ 3 = 6 books.

Question 2.
What is \(\frac{1}{5}\) of each of the following?
(1) 20 rupees
(2) 30 km
(3) 15 litres
(4) 25 cm
Answer:
(1) 20 rupees → \(\frac{1}{5}\) of 20 = 4, 20 ÷ 5 = 4 rupees.
(2) 30 km → \(\frac{1}{5}\) of 30 = 6, 30 ÷ 5 = 6km.
(3) 15 litres → \(\frac{1}{5}\) of 15 = 3, 15 ÷ 5 = 3 litres.
(4) 25 cm → \(\frac{1}{5}\) of 25 = 5, 25 ÷ 5 = 5cm.

Question 3.
Find the part of each of the following numbers equal to the given fraction.

(1) \(\frac{2}{3}\) of 30
Solution:
\(\frac{2}{3}\) x 30 So, we take \(\frac{1}{3}\) of 30, twice
\(\frac{1}{3}\) x 30 = 10, twice of 10 is 2 x 10 = 20
It means that \(\frac{2}{3}\) x 30 = 20

(2) \(\frac{7}{11}\) of 22
Solution:
\(\frac{7}{11}\) x 22 So, we take of 22, 7 times
\(\frac{1}{11}\) x 22 = 2, seven times of 2 is 2 x 7 = 14

(3) \(\frac{3}{8}\) of 64
Solution:
\(\frac{3}{8}\) x 64 So, we take \(\frac{1}{8}\) of 64, thrice
\(\frac{1}{8}\) x 64 = 8, 3 times 8 is 3 x 8 = 24

(4) \(\frac{5}{13}\) of 65
Solution:
\(\frac{5}{13}\) x 65 So, we take \(\frac{1}{13}\) of 65, 5 times
\(\frac{1}{13}\) x 65 = 55 times of 5 is 5 x 5 = 25

Mixed fractions

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 1
Half of each of the three circles is coloured. That is, 3 parts, each equal to \(\frac{1}{2}\) of the circle, are coloured.

The coloured part is \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\), that is, \(\frac{3}{2}\) or 1 + \(\frac{1}{2}\).

1 + \(\frac{1}{2}\) is written as 1 \(\frac{1}{2}\). 1 \(\frac{1}{2}\) is read as ‘one and one upon two’.

In the fraction 1 \(\frac{1}{2}\), 1 is the integer part and \(\frac{1}{2}\) is the fraction part. Hence, such fractions are called mixed fractions or mixed numbers. 2 \(\frac{1}{4}\), 3 \(\frac{2}{5}\), 7 \(\frac{4}{9}\) are all mixed fractions.

Fractions in which the numerator is greater than the denominator are called improper fractions.

\(\frac{3}{2}\), \(\frac{5}{3}\) are improper fractions. We can convert improper fractions into mixed fractions.

For example, Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 2

Activities
1. Colour the Hats.
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 3
In the picture alongside :
Colour \(\frac{1}{3}\) of the hats red.
Colour \(\frac{3}{5}\) of the hats blue.
How many hats have you coloured red?
How many hats have you coloured blue?
How many are still not coloured?

2. Make a Magic Spinner.
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 4
Take a white cardboard disc. As shown in the figure, divide it into six equal parts.

Colour the parts red, orange, yellow, green, blue and violet.

Make a small hole at the centre of the disc and fix a pointed stick in the hole.

Your magic spinner is ready.

What fraction of the disc is each of the coloured parts?
Give the disc a strong tug to make it turn fast. What colour does it appear to be now?

The Clever Poet

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 5

There was a king who had a great love for literature. A certain poet knew that if the king read a good poem it made him very happy. Then the king would give the poet an award. Once, the poet composed a good poem. He thought if he showed it to the king, he would win a prize. So, he went to the king’s palace. But, it was not easy to meet the king. You had to pass a number of gates and guards. The first guard asked the poet why he wanted to meet the king. So, the poet told him the reason. Seeing the chance of getting a share of the award, the guard demanded, ‘You must

give me \(\frac{1}{10}\) of your prize. Only then will I let you go in.’ The poet could do nothing but agree. The second guard stopped him and said, ‘I will let you go in only if you promise me \(\frac{2}{5}\) of your prize.’ The third guard, too, was a greedy man. He said, ‘I will not let you go, unless you promise me \(\frac{1}{4}\) of your prize.’ The king’s palace was just a little distance away. Now, the poet told the guard, ‘Why only \(\frac{1}{4}\), I shall give you half the prize!’ The guard was pleased and let him in.

The king liked the poem. He asked the poet, ‘What is the prize you want?’ ‘I shall be happy if Your Majesty awards me 100 lashes of the whip.’ The king was surprised. ‘Are you out of your mind!’ he exclaimed. ‘I have never met anyone so crazy as to ask for a whipping !’

‘Your Majesty, if you wish to know the reason, the three palace guards must be called here.’ When the guards came, the poet explained, ‘Your Majesty, all of them have a share in the 100 lashes that you have awarded to me. Each of them has fixed his own share of the prize I get. The first guard

must get \(\frac{1}{10}\) of the award, that is, [ ] lashes. The second must get \(\frac{2}{5}\), which is [ ], and the third must get half the award, that is, [ ] lashes !’ The king could now see how greedy the guards were and how clever the poet was. He saw to it that each guard got the punishment he deserved. He gave the poet a prize for his poem. He also gave him an extra 100 gold coins for exposing the greed of the guards.

What was the clever idea of the poet which the king appreciated so much?

Fractions Problem Set 23 Additional Important Questions and Answers

Question 1.
What is \(\frac{1}{3}\) of each of the collections given below?

(1) 24 marbles →
(2) 6 erasers →
Answer:
(1) 24 marbles → \(\frac{1}{3}\) of 24 = 8, 24 ÷ 3 = 8 marbles.
(1) 6 erasers → \(\frac{1}{3}\) of 6 = 2, 6 ÷ 3 = 2 erasers.

Question 2.
What is \(\frac{1}{5}\) of each of the following?

(1) 35 gm →
(2) 40m →
Answer:
(1) 35 gm → \(\frac{1}{5}\) of 35 = 7, 35 ÷ 5 = 7 gm.
(2) 40m → \(\frac{1}{5}\) of 40 = 8, 40 ÷ 5 = 8m.

Question 3.
Find the part of each of the following numbers equal to the given fraction:

(1) \(\frac{7}{9}\) of 45
Solution:
\(\frac{7}{9}\) x 45 So, we take \(\frac{1}{9}\) of 45, 7 times
\(\frac{1}{9}\) x 45 = 5, 7 times of 5 is 7 x 5 = 35

(2) \(\frac{3}{7}\) of 28
Solution:
\(\frac{3}{7}\) x 28 So, we take \(\frac{1}{7}\) of 28, thrice
\(\frac{1}{7}\) x 28 = 4, 3 times of 4 is 4 x 3 = 12

Question 4.
Find the proper number in the box:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 6

Answer:
(1) 3
(2) 36
(3) 3
(4) 7
(5) 8, 18
(6) 12, 6
(7) 9, 16, 20, 24
(8) 15, 20, 35, 36, 55

Question 5.
Find an equivalent fraction with denominator 3, for each of the following fractions.

Answer:

Question 6.
Find an equivalent fraction with numerator 30 for each of the following fractions.

Answer:

Question 7.
Find two equivalent fractions for each of the following fraction.
\(\text { (1) } \frac{5}{7}\)
\(\text { (2) } \frac{8}{9}\)
\(\text { (3) } \frac{7}{13}\)
Answer:
(1)
(2)
(3)

Question 8.
Match the columns (A) and (B) for having equivalent fractions:

	(A)	(B)
(1)	\(\frac{3}{4}\)	(a) \(\frac{15}{27}\)
(2)	\(\frac{5}{9}\)	(b) \(\frac{2}{3}\)
(3)	\(\frac{7}{11}\)	(c) \(\frac{27}{36}\)
(4)	\(\frac{8}{12}\)	(d) \(\frac{28}{44}\)

Answer:
(1) ↔ (c)
(2) ↔ (a)
(3) ↔ (d)
(4) ↔ (b)

Question 9.
Convert the given fractions into like fractions:
\(\text { (1) } \frac{1}{10}, \frac{2}{3}\)
\(\text { (2) } \frac{3}{7}, \frac{4}{5}\)
\(\text { (3) } \frac{1}{3}, \frac{3}{5}\)
\(\text { (3) } \frac{1}{4}, \frac{2}{5}\)
Answer:
\(\text { (1) } \frac{3}{30}, \frac{20}{30}\)
\(\text { (2) } \frac{15}{35}, \frac{28}{35}\)
\(\text { (3) } \frac{5}{15}, \frac{9}{15}\)
\(\text { (3) } \frac{5}{20}, \frac{8}{20}\)

Question 10.
Write the proper symbol from <, > or = in the box:

Answer:
(1) >
(2) >
(3) >
(4) >
(5) >

Question 11.
Add the following:
\(\text { (1) } \frac{1}{6}+\frac{2}{6}\)
\(\text { (2) } \frac{1}{4}+\frac{3}{4}\)
\(\text { (3) } \frac{5}{13}+\frac{2}{13}+\frac{3}{13}\)
\(\text { (4) } \frac{2}{9}+\frac{3}{7}\)
\(\text { (5) } \frac{3}{11}+\frac{2}{3}\)
\(\text { (6) } \frac{1}{10}+\frac{4}{5}\)
Answer:
\(\text { (1) } \frac{3}{6}\)
\(\text { (2) } \frac{4}{4}\)
\(\text { (3) } \frac{10}{13}\)
\(\text { (4) } \frac{41}{63}\)
\(\text { (5) } \frac{31}{33}\)
\(\text { (6) } \frac{9}{10}\)

Question 12.
Subtract the following:
\(\text { (1) } \frac{5}{6}-\frac{1}{6}\)
\(\text { (2) } \frac{3}{5}-\frac{2}{5}\)
\(\text { (3) } \frac{7}{16}-\frac{3}{16}-\frac{1}{16}\)
\(\text { (4) } \frac{5}{6}-\frac{7}{12}\)
\(\text { (5) } \frac{13}{16}-\frac{5}{8}\)
\(\text { (6) } \frac{4}{9}-\frac{3}{10}\)
Answer:
\(\text { (1) } \frac{4}{6}\)
\(\text { (2) } \frac{1}{5}\)
\(\text { (3) } \frac{3}{16}\)
\(\text { (4) } \frac{3}{12}\)
\(\text { (5) } \frac{3}{13}\)
\(\text { (6) } \frac{13}{90}\)

Question 13.
What is \(\frac{1}{4}\) of each of the collections given below:
(1) 20 marbles
(2) 12 pens
(3) 24 notebooks
(4) 8 ladoos
Answer:
(1) 5 marbles
(2) 3 pens
(3) 6 notebooks
(4) 2 ladoos

Question 14.
What is \(\frac{1}{6}\) of each of the following:
(1) 18 bananas
(2) 12 gms
(3) 30 metres
(4) 24 ₹
Answer:
(1) 3 bananas
(2) 2 gms
(3) 5 metres
(4) 4 ₹

Question 15.
Find the part of each of the following numbers equal to the given fraction.
(1) \(\frac{2}{5}\) of 25
(2) \(\frac{3}{7}\) of 21
(3) \(\frac{4}{9}\) of 36
(4) \(\frac{4}{17}\) of 34
Answer:
(1) 10
(2) 9
(3) 16
(4) 8

Question 16.
Printed price of. the book was 80. Vikram purchased the book by paying of the printed price of the book. How much he paid for the book?
Answer:
64 ₹

Class 5 Maths Solution Maharashtra Board

Problem Set 36 Class 5 Maths Chapter 9 Decimal Fractions Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Decimal Fractions Class 5 Problem Set 36 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 36 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Write the following mixed fractions in decimal form and read them aloud.

\(\text { (1) } 3 \frac{9}{10}\)
Answer:
3.9, Three-point nine.

\(\text { (2) } 1 \frac{4}{10}\)
Answer:
1.4, One point four.

\(\text { (3) } 5 \frac{3}{10}\)
Answer:
5.3, Five-point three.

\(\text { (4) } \frac{8}{10}\)
Answer:
0.8, Zero points eight.

\(\text { (5) } \frac{7}{10}\)
Answer:
0.5, Zero points five.

Hundredths

If \(\frac{1}{10}\) is divided into 10 equal parts, each part becomes \(\frac{1}{100}\) or one hundredth. Therefore, note that 1 tenth =10 hundredths, or 0.1=0.10. By multiplying \(\frac{1}{100}\) by 10 we get \(\frac{10}{100}\) = \(\frac{1}{10}\). Therefore, it is possible to create a hundredths place next to the tenths place. After creating a hundredths place we can write \(\frac{14}{100}\) as 0.14.

\(\frac{14}{100}=\frac{10+4}{100}=\frac{10}{100}+\frac{4}{100}=\frac{1}{10}+\frac{4}{100}\) meaning that when writing \(\frac{14}{100}\) in decimal form, 1 is written in the tenths place and 4 is written in the hundredths place. This fraction is written as 0.14 and is read as ‘zero point one four’. Similarly, 6 \(\frac{57}{100}\) is written as 6.57 and 50 \(\frac{71}{100}\) is written as 50.71.

While writing \(\frac{3}{100}\) in decimal form, we must remember that there is no number in the tenths place and so, we put 0 in that place, which means that \(\frac{3}{100}\) is written as 0.03.

Study how the decimal fractions in the table below are written and read.
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 36 1

Decimal Fractions Problem Set 36 Additional Important Questions and Answers

\(\text { (1) } 4 \frac{6}{10}\)
Answer:
4.6, Four point six. 7

\(\text { (2) } 4 \frac{6}{10}\)
Answer:
2.7, Two point seven.

\(\text { (3) } 4 \frac{6}{10}\)
Answer:
6.2, Six points two.

\(\text { (4) } 4 \frac{6}{10}\)
Answer:
21.1, Twenty-one point one.

\(\text { (5) } 4 \frac{6}{10}\)
Answer:
17.5, Seventeen points five.

Class 5 Maths Solution Maharashtra Board

Problem Set 4 Class 5 Maths Chapter 2 Number Work Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Number Work Class 5 Problem Set 4 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 4 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Read the numbers and write them in words.
(1) 25,79,899
(2) 30,70,506
(3) 45,71,504
(4) 21,09,900
(5) 43,07,854
(6) 50,00,000
(7) 60,00,010
(8) 70,00,100
(9) 80,01,000
(10) 90,10,000
(11) 91,00,000
(12) 99,99,999
Answer:
(1) Twenty-five lakh, seventy-nine thousand, eight hundred and ninety-nine.
(2) Thirty lakh, seventy thousand, five hundred and six.
(3) Forty-five lakh, seventy-one thousand, five hundred and four.
(4) Twenty-one lakh, nine thousand, nine hundred.
(5) Forty-three lakh, seven thousand, eight hundred and fifty-four.
(6) Fifty lakh.
(7) Sixty lakh and ten.
(8) Seventy lakh and one hundred.
(9) Eighty lakh and one thousand
(10) Ninety lakh and ten thousand
(11) Ninety-one lakh
(12) Ninety-nine lakh, ninety-nine thousand, nine hundred and ninety-nine.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.
Pune : ₹ 94,29,408
Nashik : ₹ 61,07,187
Nagpur : ₹ 46,53,570
Ahmadnagar : ₹ 45,43,159
Aurangabad : ₹ 37,01,282
Yavatmal : ₹ 27,72,348
Sindhudurg : ₹ 58,49,651
Answer:
Rupees ninety-four lakh, twenty-nine thousand, four hundred and eight.
Rupees sixty-one lakh, seven thousand, one hundred and eighty-seven
Rupees forty-six lakh, fifty-three thousand, five hundred and seventy.
Rupees forty-five lakh, forty-three thousand one hundred and fifty-nine.
Rupees thirty-seven lakh, one thousand, two hundred and eighty-two.
Rupees twenty-seven lakh, seventy two thousand, three hundred and forty-eight.
Rupees fifty-eight lakh, forty-nine thousand, six hundred and fifty-one.

The expanded form of a number and the place value of digits

Teacher : Look at the place value of each of the digits in the number 27,65, 043.
Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 4 1
Hamid : When we write the place values of the digits as an addition, we get the expanded form of the number. So, the expanded form of the number 27,65,043 is 20,00,000 + 7,00,000 + 60,000 + 5,000 + 0 + 40 + 3.

Teacher : Now tell me the expanded form of 95,04,506.

Soni : 90,00,000 + 5,00,000 + 0 + 4,000 + 500 + 0 + 6.

Teacher : Good! It can also be written as 90,00,000 + 5,00,000 + 4,000 + 500 + 6. Now write the number from the expanded form that I give you. 4,00,000 + 90,000 + 200

Asha : Here, we have 4 in the lakhs place, 9 in the ten thousands place and 2 in the hundreds place. There are no digits in the ten thousands place and in the tens and units places. Hence, we write 0 in those places. Therefore, the number is 4,90,200.

Teacher : Tell me the place value of the underlined digit in the number 59,30,478.
Soni : The underlined digit is 5. The digit is in the ten lakhs place. Hence, its place value is 50,00,000 or fifty lakhs.

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Read the numbers and write them in words:

(1) 80,91,001
Answer:
Eighty lakh, ninety-one thousand and one.

(2) 50,50,505
Answer:
Fifty lakh, fifty thousand, five hundred and five.

(3) 68,06,086
Answer:
Sixty-eight lakh, six thousand and eighty- six.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.

(1) Thane : 75,14,365
Answer:
Rupees seventy-five lakh, fourteen thousand, three hundred and sixty-five.

(2) Jalgaon : 39,42,180
Answer:
Rupees thirty-nine lakh, forty-two thousand, one hundred and eighty.

(3) Kalyan : 37,40,509
Answer:
Rupees thirty-seven lakh, forty thousand, five hundred and nine.

(4) Kolhapur: 16,05,430
Answer:
Rupees sixteen lakh, five thousand, four hundred and thirty.

Class 5 Maths Solution Maharashtra Board

Problem Set 11 Class 5 Maths Chapter 3 Addition and Subtraction Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Addition and Subtraction Class 5 Problem Set 11 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Subtract the following:

(1) 8,57,513 – 4,82,256
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 1
Answer:
3,75,257

(2) 13,17,519 – 10,07,423
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 2
Answer:
3,10,096

(3) 68,34,501 – 23,57,823
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 3
Answer:
44,76,678

(4) 45,43,827 – 12,05,938
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 8
Answer:
33,37,889

(5) 70,12,345 – 28,64,547
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 7
Answer:
41,47,798

(6) 38,01,213 – 37,54,648
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 6
Answer:
46,565

Study the following word problem.

In 2001, the population of a city was 21,43,567. In 2011, it was 28,09,878. By how much did the population grow?
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 9

The population grew by 6,66,311.

Addition and Subtraction Problem Set 11 Additional Important Questions and Answers

Subtract the following:

(1) 53,14,018 – 43,14,019
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 5
Answer:
9,99,999

(2) 67,05,136 – 34,56,789
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 4
Answer:
32,48,347

Class 5 Maths Solution Maharashtra Board

Problem Set 35 Class 5 Maths Chapter 8 Multiples and Factors Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Multiples and Factors Class 5 Problem Set 35 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 8 Multiples and Factors

Determine whether the pairs of numbers given below are co-prime numbers.
(1) 22, 24
Answer:
Common factors of 22 and 24 are 1 and 2. (Not only 1 common factor) So, 22, 24 are not co-prime numbers.

(2) 14, 21
Answer:
Common factors of 14 and 21 are 1 and 7. So, this pair is not co-prime numbers.

(3) 10, 33
Answer:
Common factors of 10 and 33 is only 1. So, 10 and 33 are co-prime numbers.

(4) 11, 30
Answer:
Common factors of 11 and 30 is only 1. So, 11 and 30 are co-prime numbers.

(5) 5, 7
Answer:
Common factor of 5 and 7 is only 1. So, 5 and 7 are co-prime numbers.

(6) 15, 16
Answer:
Common factors of 15 and 16 is only 1. So, 15 and 16 are co-prime numbers.

(7) 50, 52
Answer:
Common factors of 50 and 52 are 1 and 2. So, 50 and 52 are not co-prime numbers.

(8) 17, 18
Answer:
Common factors of 17 and 18 is only 1. So, 17 and 18 are co-prime numbers.

Activity 1 :

Write numbers from 1 to 60.
Draw a blue circle around multiples of 2.
Draw a red circle around multiples of 4.
Do all numbers with a blue circle also have a red circle around them?
Do all the numbers with a red circle have a blue circle around them?
Are all multiples of 2 also multiples of 4?
Are all multiples of 4 also multiples of 2?

Activity 2 :

Write numbers from 1 to 60.
Draw a triangle around multiples of 2.
Draw a circle around multiples of 3.
Now find numbers divisible by 6. Can you find a property that they share?

Eratosthenes’ method of finding prime numbers
Eratosthenes was a mathematician who lived in Greece about 250 BC. He discovered a method to find prime numbers. It is called Eratosthenes’ Sieve. Let us see how to find prime numbers between 1 and 100 with this method.

Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35 1

1 is neither a prime nor a composite number. Put a square [ ] around it
2 is a prime number, so put a circle around it.
Next, strike out all the multiples of 2. This tells us that of these 100 numbers more than half of numbers are not prime numbers.
The first number after 2 not yet struck off is 3. So, 3 is a prime number.
Draw a circle around 3. Strike out all the multiples of 3.
The next number after 3 not struck off yet is 5. So, 5 is a prime number.
Draw a circle around 5. Put a line through all the multiples of 5.
The next number after 5 without a line through it is 7. So, 7 is a prime number.
Draw a circle around 7. Put a line through all the multiples of 7.

In this way, every number between 1 and 100 will have either a circle or a line through it. The circled numbers are prime numbers. The numbers with a line through them are composite numbers.

One more method to find prime numbers

Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35 2

See how numbers from 1 to 36 have been arranged in six columns in the table alongside.

Continue in the same way and write numbers up to 102 in these six columns.

You will see that, in the columns for 2, 3, 4, and 6, all the numbers are composite numbers except for the prime numbers 2 and 3. This means that all the remaining prime numbers will be in the columns for 1 and 5. Now isn’t it easier to find them? So, go ahead, find the prime numbers!

Something more

Prime numbers with a difference of two are called twin prime numbers. Some twin prime number pairs are 3 and 5, 5 and 7, 29 and 31 and 71 and 73. 5347421 and 5347423 are also a pair of twin prime numbers.
There are eight pairs of twin prime numbers between 1 and 100. Find them.
Euclid the mathematician lived in Greece about 300 BC. He proved that if prime numbers, 2, 3, 5, 7, ……., are written in serial order, the list will never end, meaning that the number of prime numbers is infinite.

Multiples and Factors Problem Set 35 Additional Important Questions and Answers

Determine whether the pairs of numbers given below are co-prime numbers.

(1) (12,18)
Answer:
Common factors of 12 and 18 are 1, 2, 3, 6. Hence 12 and 18 are not co-prime numbers.

(2) (26, 39)
Answer:
Common factors of 26 and 39 are 1 and 13. Hence, 26 and 39 are not co-prime numbers.

(3) (23, 29)
Answer:
Common factor of 23 and 29 is only 1. Hence, 23 and 29 are co-prime numbers.

(4) (28, 32)
Answer:
Common factors of 28 and 32 are 1, 2, 4 (not only 1). Hence, 28, 32 are not co-prime numbers.

Class 5 Maths Solution Maharashtra Board

Problem Set 45 Class 5 Maths Chapter 10 Measuring Time Question Answer Maharashtra Board

April 21, 2025April 20, 2025 by Prasanna

Measuring Time Class 5 Problem Set 45 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 10 Measuring Time

Question 1.
Add the following :
(1) 2 hours 30 minutes + 4 hours 55 minutes
Solution:

Hrs.	Min.
1
2 + 4	3 0 5 5
7	2 5

85 minutes = 1 hr 25 min

(2) 3 hours 50 minutes + 4 hours 20 minutes
Solution:

Hrs.	Min.
3 + 4	5 0 2 0
7	7 0
8	1 0

70 minutes = 1 hr 10 min.

(3) 3 hours 45 minutes + 1 hour 35 minutes
Solution:

Hrs.	Min.
3 + 1	4 5 3 5
4	8 0
5	2 0

80 minutes = 1 hr 20 min

(4) 4 hours 15 minutes + 2 hours 50 minutes
Solution:

	Min.
4 + 2	1 5 5 0
6	6 5
7	0 5

65 minutes = 1 hr 05 min

Question 2.
Subtract the following :
(1) 3 hours 10 minutes – 2 hours 40 minutes

Hrs.	Min.
2	60 + 10
3 – 2	1 0 4 0
0	3 0

(2) 5 hours 20 minutes – 2 hours 35 minutes
Solution:

Hrs.	Min.
4	60 + 20
5 – 2	20 3 5
2	4 5

(3) 4 hours 25 minutes – 1 hour 55 minutes

Hrs.	Min.
3	60 + 25
4 – 1	25 5 5
2	3 0

(4) 6 hours 15 minutes – 2 hours 45 minutes

Hrs.	Min.
5	60 + 15
6 – 2	15 4 5
3	3 0

Question 3.
A government office opens at 7 in the morning and closes at 3 in the afternoon. How long is this office open?
Solution:

Hrs.	Min.
1 5 – 7	0 0 Closing time 0 0 Opening time
8	0 0

∴ Office remain open for 8 hours

Question 4.
A movie starts at 45 minutes past 3 in the afternoon and finishes two and a half hours later. At what time does the movie end?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 1
∴ Movie ends at 6:15 in the evening

Question 5.
Sakharam was ploughing the field from 8 in the morning. At 12:30 in the afternoon, he stopped and started for home. He reached home at 1:30. How long was he ploughing the field? How long did it take him to reach home from the field?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 2
∴ He ploughed for 4:30 hrs. He took 1 hour to reach home.

Question 6.
Rambhau started the water pump at ten-thirty at night and put it off the same night at a quarter to twelve. How long was the water pump on?
Solution:
Quater to 12 is 11:45 pm
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 3
∴ pump was on for 1 hour 15 minutes

Question 7.
Geeta taught in the classroom for 2 hours and 25 minutes in the morning and 1 hour and 45 minutes in the afternoon. How long was she teaching in all?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 4
∴ Total teaching of Geeta was for 4 hrs 10 min.

Question 8.
If a bank is open for business from 10 in the morning to 4:30 in the evening, how long is it open?
Solution:
Here, in 24 hours clock, 4:30 in the evening = 16:30
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 5
∴ Bank opens for 6 hrs 30 min.

Question 9.
If a shop is open from 9:30 am to 10 pm, how long is it open?
Solution:
Here, 10 pm in 24 hours clock is 22:00
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 6
∴ Shop opens for 12 hours 30 minutes

Question 10.
If the Maharashtra Express leaving from Kolhapur at 15:30 arrives at Gondia the next day at 20:15, how long is the journey from Kolhapur to Gondia?
Solution:
15:30 to next 15:30 is 24 hours
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 7
24 hours + 4 hr. 45 min. = 28 hr. 45 min.
∴ jurney from Koihapur to Gondiya is 28 hours and 45 minutes.

Measuring Time Problem Set 45 Additional Important Questions and Answers

Question 1.
Add the following.

(1) 5 hours 25 minutes + 2 hours 35 minutes
Solution:

Hrs.	Min.
5 + 2	2 5 3 5
7	6 0
8	0 0

60 minutes = 1 hr

(2) 6 hours 55 minutes + 2 hours 15 minutes
Solution:

Hrs.	Min.
6 + 2	5 5 1 5
8	7 0
9	1 0

70 minutes = 1 hr. 10 min

Question 2.
Subtract the following.

(1) 7 hours 30 minutes – 4 hours 50 minutes
Solution:

Hrs.	Min.
6	60 + 30
7 – 4	~~3 0~~ 5 0
2	4 0

(2) 2 hours 35 minutes – 1 hour 40 minutes
Solution:

Hrs.	Min.
1	60 + 35
2 – 1	~~3 5~~ 4 0
0	5 5

Question 3.
Solve the following:

(1) Supriya left for a picnic at 7:15 am. She came back at 6:45 pm. How long was she out for the picnic?
Here 6:45 pm = 18:45 (In 24 hours clock)
Solution:

Hrs.	Min.
1 8 – 7	4 5 1 5
1 1	3 0

∴ Total time of picnic is 11:30 hrs.

(2) In Dave’s school, the tree planting ceremony started at 10:00 in the morning and got over at 13:45. How long did the ceremony go on?
Solution:

Hrs.	Min.
1 3 – 1 0	4 5 1 5
3	4 5

∴ Ceremony of planting tree go on for 3 hrs 45 min

Question 4.
Write the time shown in each clock in the box given below it.
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 8
Answer:
35 minutes past 3

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 9
Answer:
Five minutes to 5

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 10
Answer:
Quarter to 2

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 11
Answer:
Half past eight

Question 5.
Draw the hands of the clock to show the time given in the box.
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 12
Answer:
Quater past 6

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 13
Answer:
50 minutes past 1

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 14
Answer:
Half past 3

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 45 15
Answer:
5 minutes to 5

Question 6.
The time below is given by the 12 hour clock. Write the same by the 24 hour clock.
(1) 45 minutes past 8 in the morning.
(2) 30 minutes past 2 in the evening.
(3) 50 minutes past 7 in the evening.
(4) 15 minutes past 11 in the evening.
(5) 25 minutes past after midnight.
(6) 25 minutes past 12 in afternoon.
Answer:
(1) 8:45
(2) 114:30
(3) 19:50
(4) 23:15
(5) 00:25
(6) 12:25

Question 7.
Match the following.

‘A’	‘B’
(1) 7:20 am	(a) 13:20
(2) 1:20 pm	(b) 22:10
(3) 6:10 pm	(c) 7:20
(4) 10:10 pm	(d) 6:10
(5) 6:10 am	(e) 18:10

Answer:
(1 – c),
(2 – a),
(3 – e),
(4 – b),
(5 – d)

Question 8.
Add the following.
(1) 3 hours 40 minutes + 2 hours 55 minutes
(2) 5 hours 25 minutes + 4 hours 35 minutes
(3) 6 hours 45 minutes + 1 hour 30 minutes
(4) 7 hours 50 minutes + 2 hours 30 minutes
(5) 9 hours 10 minutes + 3 hours 20 minutes
(6) 15 hours 45 minutes + 20 hours 15 minutes
Answer:
(1) 6 hrs 35 min
(2) 10 hrs
(3) 8 hrs 15 min
(4) 10 hrs 20 min
(5) 12 hrs 30 min
(6) 36 hours

Question 9.
Subtract the following.
(1) 4 hours 20 minutes – 1 hour 30 minutes
(2) 3 hours 25 minutes – 1 hour 45 minutes
(3) 5 hours 10 minutes – 2 hours 40 minutes
(4) 2 hours 15 minutes – 50 minutes
(5) 9 hours 10 minutes – 6 hours 10 minutes
(6) 17 hours 30 minutes – 5 hours 25 minutes
Answer:
(1) 2 hours 50 minutes
(2) 1 hour 40 minutes
(3) 2 hours 30 minutes
(4) 1 hour 25 minutes
(5) 3 hours
(6) 12 hours 05 minutes

Question 10.
Solve the following word problems.
(1) A play started at 9:50 at night and finished at 11:45 the same night. What was the duration of the play?
(2) Ramu went out at 10:45 in the morning and came back home at 7 pm. How long he was out of the home?
(3) Local train started from the Virar station at 8:35 am and reached at Churchgate at 10:30 am. Find the journey time taken by the train.
(4) Anita started her homework at 5:45 pm and completed the work at 7:30 pm. How much time is taken by Anita for the homework?
(5) One day test started at 9:15 am and the test ends at 4-.10 in the evening. How much time was taken for this test?
(6) Seema travelled for 2 hours and 20 minutes by train and 1 hour 30 niinutes by bus. What the total time of her journey?
(7) A train that starts from Mumbai at 17:50 reaches Nira at 2:10. How long does the Mumbai – Nira journey take?
Answer:
(1) 1 hour 55 minutes
(2) 8 hrs 15 min
(3) 1 hr 55 min
(4) 1 hr 45 min
(5) 6 hrs 55 min
(6) 3 hrs 50 min
(7) 8 hrs 20 min

Class 5 Maths Solution Maharashtra Board