Prasanna

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 44 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 44

Question 1.
The time below is given by the 12 hour clock. Write the same by the 24 hour clock.
(1) 30 minutes past 10 in the morning –
(2) 10 minutes past 8 in the morning –
(3) 20 minutes past 1 in the afternoon –
(4) 40 minutes past 5 in the evening –
Answer:
(1) [10:30]
(2) [8:10]
(3) [13:20]
(4) [17:40]

Question 2.
Match the following.
12 hour clock – 24 hour clock
(1) 9:10 am – 23:10
(2) 2:10 pm – 7:25
(3) 5:25 pm – 14:10
(4) 11:10 pm – 9:10
(5) 7:25 am – 17:25
Answer:
(1) – d
(2) – c
(3) – e
(4) – a
(5) – b

Examples of time measurement

Example (1) If Abdul started working on the computer at 11 in the morning and finished his work at 3:30 in the afternoon, how long did he work?

Method 1 :
From 11 in the morning to 12 noon, it is 1 hour. From 12 noon to 3:30 in the afternoon, it is 3 hours and 30 minutes. Therefore, the total time is 4 hours and 30 minutes.

Method 2 :
According to the 24 hour clock, 11’o’clock in the morning is 11:00 and 3:30 in the afternoon is 15:30.

Hr	Min
15 – 11	30 00
4	30

Abdul worked for a total of 4 hours and thirty minutes, or four and a half hours.

Example (2) Add : 4 hours 30 min + 2 hours 45 min

Example (3) Subtract : 5 hr 30 min – 2 hr 45 min

45 minutes cannot be subtracted from 30 minutes. Therefore, we borrow 1 hour and convert it into 60 minutes for the subtraction.

Example (4) Amruta travelled by bus for 3 hours 40 minutes and by motorcycle for 1 hour 45 minutes. How long did she spend travelling?

(60 + 25) minutes are 85 minutes, that is, 1 hour and 25 minutes.
Let us add this 1 hour to 4 hours.

Therefore, Amruta travelled for a total of 5 hours and 25 minutes.

Measuring Time Problem Set 44 Additional Important Questions and Answers

The time is given by the 12-hour clock. Write the same by the 24-hour clock.

(1) 15 minutes past 9 in the evening –
(2) 12 midnight –
Answer:
[21:15]
[0o:00]

The time below is given by the 24-hour clock. Write the same by the 12-hour clock.

(1) 20:20 =
(2) 9:30 =
(3) 23:00 =
(4) 4:00 =
(5) 12:00 =
(6) 00:00 =
Answer:
[8:20 pm]
[9:30 am]
[11 pm]
[4 am]
[12 noon] or [12:00]
[12 midnight]

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43

Question 1.
Write the time shown in each clock in the box given below it.
(1)
Answer:
25 minutes past 2

(2)
Answer:
50 minutes past 7

(3)
Answer:
5 minutes past 8

(4)
Answer:
40 minutes past 4

Question 2.
Draw the hands of the clock to show the time given in the box.
(1)
Four-thirty
Answer:

(2)
Quarter past nine
Answer:

(3)
Quarter to five
Answer:

(4)
20 minutes past 11
Answer:

Question 3.
If a bus that leaves Nashik at 5 o’clock in the morning reaches Pune that same day at ten-thirty in the morning, how long does the journey take?
Solution:

Hrs.	Min.
10 – 5	30 00
5	30

∴ Bus took 5 hrs 30 min

Question 4.
A play that was to start at nine fifteen at night was delayed by half an hour because of a power outage. What time did the play start?
Solution:

Hrs.	Min.
9 +	15 30
9	45

∴ Play started 9:45 at night

Question 5.
If a train leaves Mumbai at ten-fifteen at night and reaches Nagpur at one forty the next afternoon, how long does the journey take?
Solution:
Ten fifteen at night to 12 mid night is

Hrs.	Min.
1 2 -10	0 0 1 5
1	4 5

12 mid night to next 1:40 afternoon = 13 hours 40 minutes
Total times

Hrs.	Min.
1 + 13	45 40
15	25

∴ Total time of the journey is 15 hrs. 25 min:

Learning about seconds

This clock is showing 5 minutes past 3. We know this because of the position of the hour and minute hands. There is another hand in the picture called the second hand. This hand moves swiftly. The second is a very small unit used to measure time less than a minute.

The face of a clock is a circle divided into 60 equal parts. When the second-hand moves one part, it takes one second. When it completes one round of the clock face, it moves across all 60 parts. This takes 60 seconds. In the same time, the minute hand moves one place, which means that one minute is over.

It means that, 1 minute is equal to 60 seconds.

1 minute = 60 seconds

The clock in the picture above shows 5 minutes and 50 seconds past 3.

20 minutes and 10 seconds past 7
15 minutes and 40 seconds past 10

Seconds are used on various occasions such as measuring temperature with a thermometer, measuring heartbeats or timing a race.

Ante meridiem and post meridiem

Shripati was sitting at home at night, tired. There were guests at home. They asked, “You must have worked very hard in the fields today. How long were you working?”

Shripati said, “I was in the field from six o’clock to eight o’clock.” Someone asked, “You’re this tired even though you were in the field for only two hours?”

Shripati said, “No, no, I was in the field from 6 in the morning till 8 at night! Now tell me how many hours I spent in the field.”

The guests had not understood what Shripati said at first. To avoid such mistakes, it has been internationally agreed that as the clock strikes 12 midnight, one day ends and the next day begins. From that moment on, the clock shows the time for the next day. When one hour passes after 12 midnight, it is 1’o’clock. After that, it is 2, 3, 4, …, 12 o’clock in serial order. After 12 noon, again it is 1, 2, 3, …, 12 o’clock in serial order. The time before 12 noon is stated as ante meridiem or am. The time after 12 noon is stated as post meridiem or pm.

This method of measuring time is called the 12 hour clock.

Shripati was in the field from 6 am to 8 pm or for 14 hours.

The 24 hour clock
The 24 hour clock is used to avoid this division of the day into ante meridiem and post meridiem. This method is used in timetables for trains, planes, buses and long distance boat journeys. In this method, instead of starting again from 1, 2, 3 after 12 noon, we continue with 13, 14, 15,…,24. In a 24 hour digital watch, time is shown only in the form of numbers. It does not have hands. In such a clock, 20 minutes past 6 in the morning is shown as ‘6:20’ and 20 minutes past 6 in the evening is shown as ‘18:20’.

23:59 means 59 minutes after 23 and one minute later, 24 hours will be complete. The digital clock will show this time as 00:00 at midnight and the day will change. At that time, a 12 hour clock shows 12 midnight.

Study the following table to see how different times of the day are shown in the 12 hour and 24 hour clocks.

The timetables of some trains going from Badnera to Nagpur are given below. Observe the use of the 24 hour clock in the timetable.

Measuring Time Problem Set 43 Additional Important Questions and Answers

Write the time shown in each clock in the box given below it.
(1)
Answer:
15 minutes past 6

(2)
Answer:
30 minutes past 9

Draw the hands of the clock to sho the time given in the box.
(1) 5 minutes to four
Answer:

(2) 35 minutes past to 2
Answer:

(3) Sujata left home 6:30 and returned at 11. How much time did she spend away from home?
Solution:

Hrs.	Min*
11 – 6	00 30
4	30

∴ 4 hrs. 30min. spent away from home.

(4) A speech that started at 4:20 in the afternoon ended at 5:45. How long was the speech?
Solution:

Hrs.	Min.
5 + 4	4.5 20
1	25

∴ Speech was for 1 hr. 25 min.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40

Write the following fractions as decimal fractions.

(1) Two and a half
Answer:
2 \(\frac{1}{2}\) = 2.5 = 2.50

(2) Two and a quarter
Answer:
2 \(\frac{1}{4}\) = 2.25

(3) Two and three quarters
Answer:
2 \(\frac{3}{4}\) = 2.75

(4) Ten and a half
Answer:
10 \(\frac{1}{2}\) = 10.5 = 10.50

(5) Fourteen and three quarters
Answer:
14 \(\frac{3}{4}\) = 14.75

(6) Sixteen and a quarter
Answer:
16 \(\frac{1}{4}\) = 16.25

(7) Twenty-eight and a half
Answer:
28 \(\frac{1}{2}\) = 28.50 = 28.5

Adding decimal fractions

Sir : If the cost of one pencil is two and a half rupees and the cost of a pen is four and half rupees, what is the total cost?

Sumit : Two and a half rupees means two rupees and one half rupee. Similarly, four and a half rupees means four rupees and one-half rupee. 4 rupees and 2 rupees make 6 rupees and two half rupees make one rupee, so both objects together cost 6 + 1 = 7 rupees.

Sir : Correct ! Now, see how this is done using decimals.
The sum of the 0’s in the hundredths place is 0.
0.5 + 0.5 is the same as
\(\frac{5}{10}+\frac{5}{10}=\frac{5+5}{10}=\frac{10}{10}=\frac{1}{1}=1\)

This 1 is carried over to the units place. There is nothing in the tenths place, so we put a zero there. In the units place, 2 + 4 = 6 plus the carried over 1 makes 7.

So 2.50 rupees and 4.50 rupees add up to 7 rupees.

We use the decimal system to write whole numbers. We extend the same method to write fractions; therefore, we can add in the same way as we add whole numbers.

I will now show some more additions. Watch carefully.

Sumit : There is no carried over number in the first sum, but there are carried over numbers in the second and third sums.

Rekha : While adding whole numbers, we add units first. Similarly, here, tenths are added first. In the second example, the sum of the tenths place is 13. 13 tenths are 10 tenths + 3 tenths = 1 unit + 3 tenths.

Sumit : That is why, in the sum, 3 stayed in the tenths place and 1 was carried over to the units place. 6 + 5 plus 1 carried over makes 12.

Sir : Your observations are absolutely correct. We write digits one below the other according to their place values while adding whole numbers. We do the same thing here. Remember that while writing down an addition problem and the total, the decimal points should always be written one below the other.

Study the following additions. (Note that: 10 tenths = 1 unit. 10 hundredths = 1 tenth)

Example (1) Add : 7.09 + 54.93
First, add the digits in the 100ths place. 9 + 3 = 12.

The 1 from the sum 12 in the hundredths place is carried over to the tenths place and 2 is written in the hundredths place. Adding 1 + 9 gives 10 tenths or 1 unit. This 1 is carried over to the units place. 0 is left in the tenths place. Then, the addition is completed in the usual way.

Example (2) Add : 45.83 + 167.4
4 5 . 8 3 We arrange the numbers so that the places and
+
1 6 7. 4 decimal points come one below the other.

\(\frac{4}{10}=\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\) Therefore, to make the denominators of the fractions equal, 167.4 is written as 167.40 and then the fractions are added.

As usual, the digits with the smallest place values are added first and then those with bigger place values are added serially.

Example (3) 10.46 Rupees

Example (4) 48.80 m

Example (5) 7.5 cm

Decimal Fractions Problem Set 40 Additional Important Questions and Answers

(1) Thirty and a quarter
Answer:
30 \(\frac{1}{4}\) = 30.25

(2) Thirty and a half
Answer:
30 \(\frac{1}{2}\) = 30.50 = 30.5

(3) Thirty and three quarters
Answer:
30 \(\frac{3}{4}\) = 30.75

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 42 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 42

Question 1.
Subtract the following :

(1) 25.74 – 13.42
Solution:

(2) 206.35 – 168.22
Solution:

(3) 63.4 – 31.8
Solution:

(4) 63.43 – 31.8
Solution:

(5) 63.4 – 31.83
Solution:

(6) 8.23 – 5.45
Solution:

(7) 18.23 – 9.45
Solution:

(8) 78.03 – 41.65
Solution:

Question 2.
Vrinda was 1.48 m tall. After a year, her height became 1.53 m. How many centimeters did her height increase in a year?
Answer:

∴ 5 cm height has increased in a year.

Something more

Decimals used for measurement

We need to measure distance, mass (weight) and volume every day. We use suitable units for these measurements. Kilometre, metre and centimeter for distance; litre, millilitre for volume and kilogram and gram for mass are the units that are used most of the time.

All these units are decimal units. In this method, gram, metre and litre are taken as the basic units for mass, distance and volume respectively. Units larger than these increase 10 times at every step and smaller units become \(\frac{1}{10}\) of the previous unit at each step.

Look at the table of these units given below.

The origin of the terms kilo, hecto… milli is in the Greek or Latin language. Their English equivalents are given in brackets along with the terms.

Decimal Fractions Problem Set 42 Additional Important Questions and Answers

Subtract the following:

(1) 304.17 – 95.28
Solution:

(2) 72.84 – 36.96
Solution:

(3) 9.17 – 5.88
Solution:

(4) 100 – 49.99
Solution:

(5) Atul has 56.25 and Anup has 65. Whose amount is more? How much?
Solution:

∴ Anup’s amount is more by ₹ 8.75

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 41 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 41

Question 1.
Convert the following into decimal fractions and add them.

(1) ‘One and a half metre’ and ‘two and a half metres’
Solution:

(2) ‘Five and three quarter rupees’ and ‘seven and a quarter rupees’
Solution:

(3) ‘Six and a half metres’ and ‘three and three quarter metres’.
Solution:

Question 2.
(1) 23.4 + 87.9

(2) 35.74 + 816.6
Solution:

(3) 6.95 + 74.88
Solution:

(4) 41.03 + 9.98
Solution:

Question 3.
(1) 51.4 cm + 68.5 cm
Solution:

(2) 94.7 m + 1738.45 m
Solution:

(3) 5158.75 + `841.25
Solution:

Subtraction of decimal fractions

Study the subtraction of decimal fractions given below.

8 hundredths cannot be subtracted from 1 hundredth, so 1 tenth (or 10 hundredths) from 4 tenths are borrowed. The borrowed 10 hundredths and the original one hundredth make 11 hundredths. 11 hundredths minus 8 hundredths are 3 hundredths. They are written in the hundredths place under the line. The rest of the subtraction is carried out using the same method.

Decimal Fractions Problem Set 41 Additional Important Questions and Answers

Convert the following into decimal fractions and add them.

(1) ‘Fourteen and a half rupees’ and ‘Fifteen and a half rupees’.
Solution:

(2) ‘Three quarters’ and ‘a half’
Solution:

Add the following:

(1) 37.84 + 12.16
Solution:

(2) 328.69 + 84.84
Solution:

Solve the following:

(1) 304.86 m + 70.94 m
Solution:

(2) 79.56 cm + 19.65 cm
Solution:

(3) ₹ 64.79 + ₹  49.5
Solution:

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

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 36

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.

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.

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

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

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.

• 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

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.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 34 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 34

Question 1.
Write all the prime numbers between 1 and 20.
Answer:
2, 3, 5, 7, 11, 13, 17, 19.

Question 2.
Write all the composite numbers between 21 and 50.
Answer:
Composite numbers between 21 and 50 are 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49.

Question 3.
Circle the prime numbers in the list given below. 22, 37, 43, 48, 53, 60, 91, 57, 59, 77, 79, 97, 100
Answer:

Question 4.
Which of the prime numbers are even numbers?
Answer:
Only even prime number is 2. (the Rest of the even numbers are composites.)

Co-prime numbers

Dada : Tell me all the factors of 12 and 18.

Anju : I’ll tell the factors of 12: 1, 2, 3, 4, 6, 12.

Manju : I’ll give the factors of 18: 1, 2, 3, 6, 9, 18.

Dada : Now find the common factors of 12 and 18.

Anju : Common factors ?

Dada : 1, 2, 3 and 6 are in both groups, which means that they are common factors. Now tell me the factors of 10 and 21.

Sanju : Factors of 10 : 1, 2, 5, 10.

Manju : Factors of 21: 1, 3, 7, 21.

Dada : Which of the factors in these two groups are common?

Sanju : 1 is the only common factor.

Dada : Numbers which have only 1 as a common factor are called co-prime numbers, so 10 and 21 are co-prime numbers. The common factors of 12 and 18 are 1, 2, 3 and 6; which means that the common factors are more than one. Therefore, 12 and 18 are not co-prime numbers. Now tell me whether 8 and 10 are co-prime numbers.

Manju : The factors of 8 are 1, 2, 4 and 8 and the factors of 10 are 1, 2, 5 and 10. These numbers have two factors, 1 and 2, in common, so 8 and 10 are not co-prime numbers.

Multiples and Factors Problem Set 32 Additional Important Questions and Answers

Question 1.
21 to 50
Answer:
23, 29, 31, 37, 41, 43, 47

Question 2.
Which of the number is neither prime nor composite?
Answer:
1

Question 13.
Between nearest which two prime numbers the prime number 43 lies?
Answer:
43 lies between prime numbers 41 and 47.

Question 14.
Which of the prime numbers are odd numbers?
Answer:
All prime numbers are odd except 2.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39

Question 1.
Write how many rupees and how many paise.

(1) ₹ 58.43
Answer:
58 rupees 43 paise.

(2) ₹ 9.30
Answer:
9 rupees 30 paise.

(3) ₹ 2.30
Answer:
2 rupees 30 paise.

(4) ₹ 2.3
Answer:
2 rupees 30 paise.

Question 2.
Write how many rupees in decimal form.

(1) 6 rupees 25 paise
Answer:
₹ 6.25

(2) 15 rupees 70 paise
Answer:
₹ 15.70

(3) 8 rupees 5 paise
Answer:
₹ 8:05

(4) 22 rupees 4 paise
Answer:
₹ 22.04

(5) 720 paise
Answer:
₹ 7.20

Question 3.
Write how many metres and how many centimetres.

(1) 58.75 m
Answer:
58 m 75 cm

(2) 9.30 m
Answer:
9 m 30 cm

(3) 0.30 m
Answer:
30 cm

(4) 0.3 m
Answer:
30 cm

(5) 1.62 m
Answer:
1 m 62 cm

(6) 91.4 m
Answer:
91 cm 40 cm

(7) 7.02 m
Answer:
7 m 2 cm

(8) 0.09 m
Answer:
9 cm

Question 4.
Write how many metres in decimal form.

(1) 1 m 50 cm
Answer:
1.5 m

(2) 50 m 40 cm
Answer:
50.40 m

(3) 50 m 4 cm
Answer:
50.04 m

(4) 734 cm
Answer:
7.34 m

(5) 10 cm
Answer:
0.1 m

(6) 2 cm
Answer:
0.02 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 6.9 cm
Answer:
6 cm 9 mm

(2) 20.4 cm
Answer:
20 cm 4 mm

(3) 0.8 cm
Answer:
8 mm

(4) 0.5 cm
Answer:
5 mm

Question 6.
Write how many centimetres in decimal form.
(1) 7 cm 1 mm
Answer:
7.1 cm

(2) 16 mm
Answer:
1.6 cm

(3) 144 mm
Answer:
14.4 cm

(4) 8 mm
Answer:
0.8 cm

Writing half, quarter, three-quarters and one and a quarter in decimal form

‘Half’ is usually written as \(\frac{1}{2}\). To convert this fraction into decimal form, the denominator of \(\frac{1}{2}\) must be converted into an equivalent fraction with denominator 10.

\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) so the decimal form of \(\frac{1}{2}\) will be \(\frac{5}{10}\) or 0.5 Just as \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) = 0.5, note that \(\frac{1}{2}=\frac{1 \times 50}{2 \times 50}=\frac{50}{100}\) = 0.50

Therefore, ‘half’ is written as ‘0.5’ or 0.50’. ‘Quarter’ and ‘three quarters’ are written in fractions as \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively. Let us convert them into decimal fractions. 10 is not divisible by 4. Therefore, the denominators of \(\frac{1}{4}\) and \(\frac{3}{4}\) cannot be made into fractions with multiples of 10. However, 4 × 25 = 100, so the denominator can be 100.

A quarter \(=\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}=0.25\)
and Three quarters \(=\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}=0.75\)
One and a quarter = 1 \(\frac{1}{4}\) = 1.25
One and a half = 1 \(\frac{1}{2}\) = 1.50 = 1.5
One and three quarters = 1 \(\frac{3}{4}\) = 1.75
Seventeen and a half = 17 \(\frac{1}{2}\) = 17.50 = 17.5

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

Question 1.
Write how many rupees and how many paise.

(1) ₹ 147.5
Answer:
1 hundred and 47 rupees 50 paise.

(2) ₹ 40.4
Answer:
40 rupees and 40 paise.

Question 2.
Write how many rupees in decimal form.

(1) 105 paise
Answer:
₹ 1.05

(2) 6 rupees 6 paise
Answer:
₹ 6.06

(3) 20 rupees 20 paise
Answer:
₹ 20.2

Question 3.
Write how many metres and how many centimetres.

(1) 1.1m =
Answer:
1 m 10 cm

(2) 120 cm =
Answer:
1 m 20 cm

(3) 24.8 m =
Answer:
24 m 80 cm

(4) 0.5 m =
Answer:
50 cm

Question 4.
Write how many metres in decimal form.

(1) 110 cm =
Answer:
1.1m

(2) 60 cm =
Answer:
0.6 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 0.1 cm =
Answer:
1 mm

(2) 10.5 cm =
Answer:
10 cm 5 mm

Question 6.
Write how form. many centimetres in decimal
(1) 1 mm =
Answer:
0.1 cm

(2) 100 mm =
Answer:
10.0 cm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 33 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 33

Question 1.
(1) Write five three-digit numbers that are multiples of 2.
Answer:
100, 102, 104, 106, 108.

(2) Write five three-digit numbers that are multiples of 5.
Answer:
100, 105, 110, 115, 120.

(3) Write five three-digit numbers that are multiples of 10.
Answer:
100, 110, 120, 130, 140.

Question 2.
Write 5 numbers that are multiples of 2 as well as of 3.
Answer:
2 as well as of 3 means 2 and 3 that is multiples of 6.
They are 6, 12, 18, 24, 30.

Question 3.
A ribbon is 3 metres long. Can we cut it into 50 cm pieces and have nothing left over? Write the reason why or why not.
Answer:
3 metres = 300 cm.
We can cut it into 50 cm pieces.
Since 300 is exactly divisible 50.
That is 300 is multiples of 50.
300 ÷ 50 = 6
We will get 6 pieces, nothing is left over.

Question 4.
A ribbon is 3 metres long. I need 8 pieces of ribbon each 40 cm long. How many centimetres shorter is the ribbon than the length I need?
Answer:
1 piece of 40 cm, so for 8 pieces ribbon needed is 40 x 8 = 320 cm.
But ribbon is 3 metre = 300 cm long.
So ribbon is shorter by 320 – 300 = 20 cm.

Question 5.
If the number given in the table is divisible by the given divisor, put ✓ in the box. If it is not divisible by the divisor, put ✗ in the box.

Answer:

Prime and composite numbers

Some numbers are given in the tables below. Write all of their factors.

Dada : What do you notice on studying the table?

Ajay : The number 1 is a factor of every number. Some numbers have only 1 and the number itself as factors. For example, the only factors of 3 are 1 and 3. Similarly, the factors of 2 are only 1 and 2 and the factors of 19 are only 1 and 19. Some numbers have more than two factors.

Dada : Numbers like 2, 3, 19 which have only two factors are called prime numbers.

A number which has only two factors, 1 and the number itself, is called a prime number.

Ajay : What do we call numbers like 4, 6 and 16 which have more than two factors?

Dada : Numbers like 4, 6 and 16 are called composite numbers.

A number which has more than two factors is called a composite number.

Dada : Think carefully and tell me whether 1 is a prime or composite number.

Ajay : The number 1 has only one factor, 1 itself, so I can’t answer your question.

Dada : You’re right. 1 is considered neither a prime number nor a composite number.

1 is a number which is neither prime nor composite.

Multiples and Factors Problem Set 33 Additional Important Questions and Answers

Question 1.
Write five three-digit numbers that are multiples of 3.
Answer:
102, 105, 108, 111, 114.

Question 2.
Write five two-digit numbers that are multiples of 7.
Answer:
14, 21, 28, 35, 42

Question 3.
Write five three-digit numbers that are multiples of 4.
Answer:
112, 116, 120, 124, 128.

Question 4.
Write 5 numbers that are multiples of 3 as well as 5.
Answer:
3 as well as 5 means 3 and 5. i.e. multiples of 15.
They are 15, 30, 45, 60, 75.

Question 5.
A string is 4 metres long. Can we cut it into 50 cm pieces and have nothing left over?
Answer:
4 metres = 400 cm.
We can cut it into 50 cm pieces.
Since 400 is exactly divisible by 50.
That is 400 is multiple of 50 400 + 50 = 8
We will get 8 pieces. Nothing is left over.

Question 6.
A paper Is 2 metres long. I need 8 pieces of paper each 30 cm long. How many centimetres shorter is the paper than the length I need?
Answer:
A piece of 30 cm, so for 8 pieces paper needed is 30 x 8 = 240 cm.
But paper is 2 metre = 2 x 100 = 200 cm long.
So paper is shorter by 240 – 200 = 40 cm

Question 7.
If the number given in the table is divisible by the given divisor, put P in the box. If it is not divisible by the divisor, put in the box.
Answer: