Class 5

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.5

Question 1.
Write the like and unlike fractions for the following groups :
(a) \(\frac{3}{7}\), \(\frac{5}{7}\), \(\frac{1}{7}\), ……………….
(b) \(\frac{6}{9}\), \(\frac{4}{9}\), \(\frac{1}{9}\), ……………….
(c) \(\frac{9}{12}\), \(\frac{7}{11}\), \(\frac{7}{10}\), ……………….
(d) \(\frac{7}{10}\), \(\frac{6}{10}\), \(\frac{8}{10}\), ……………….
(e) \(\frac{5}{3}\), \(\frac{5}{7}\), \(\frac{5}{9}\), ……………….
Solution:
(a) Like fractions,
(b) Like fractions,
(c) Unlike fractions,
(d) Like fractions,
(e) Unlike fractions.

Question 2.
Write two like fractions for the following :
(a) \(\frac{1}{5}\), \(\frac{4}{5}\), \(\frac{3}{4}\), -, –
(b) \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{7}{9}\), -, –
(c) \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{9}{7}\), -, –
Solution:
(a) \(\frac{2}{5}\) and \(\frac{6}{5}\)
(b) \(\frac{1}{9}\) and \(\frac{5}{9}\)
(c) \(\frac{1}{7}\) and \(\frac{4}{7}\)

Question 3.
Write the unit fraction, whose denominator is as follows :
(a) 7
(b) 5
(c) 8
(d) 3
(e) 15.
Solution:
(a) \(\frac{1}{7}\),
(b) \(\frac{1}{5}\),
(c) \(\frac{1}{8}\),
(d) \(\frac{1}{3}\),
(e) \(\frac{7}{9}\)

Question 4.
Which of the following fractions are proper and improper fractions :
(a) \(\frac{7}{12}\)
(b) \(\frac{8}{3}\)
(c) \(\frac{12}{18}\)
(d) \(\frac{3}{5}\)
(e) \(\frac{7}{9}\)
Solution:
(a) Proper fraction,
(b) Improper fraction,
(c) Proper fraction,
(d) Proper fraction,
(e) Proper fraction.

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.4

Question 1.
Check whether the following fractions are in its lowest form or not :
(a) \(\frac{12}{14}\)
(b) \(\frac{21}{3}\)
(c) \(\frac{13}{17}\)
(d) \(\frac{25}{50}\)
(e) \(\frac{14}{21}\)
(f) \(\frac{8}{13}\)
(g) \(\frac{7}{15}\)
(h) \(\frac{14}{27}\)
(i) \(\frac{25}{35}\)
(j) \(\frac{18}{23}\)
Solution:
(a) To find whether the fraction \(\frac{12}{14}\) is in its lowest term or not, we will find the HCF of 12 and 14.
Now, HCF of 12 and 14 = 2
Because the HCF of numerator and denominator is not 1.
∴ The fraction is not in its lowest term. To change the fraction into its lowest term, we will divide the numerator (12) and denominator (14) by HCF (2)
Fraction = \(\frac{12}{14}\) = \(\frac{12 \div 2}{14 \div 2}\) = \(\frac{6}{7}\)
∴ The lowest form of \(\frac{12}{14}\) is \(\frac{6}{7}\).

(b) To find whether the fraction \(\frac{21}{35}\) in its lowest term or not, we will find the HCF of 21 and 35 Now, HCF of 21 and 35 = 7
Now, HCF of 21 and 35 = 7

Because the HCF of numerator and denominator is not 1
∴ The fraction is not in its lowest terms. To change the fraction into its lowest terms, we will divide the numerator (21) and denominator (35) by their HCF (7)
Fraction = \(\frac{21}{35}\) = \(\frac{21 \div 7}{35 \div 7}\) = \(\frac{3}{5}\)
∴ The lowest form of \(\frac{21}{35}\) is \(\frac{3}{5}\)

(c) To find whether the fraction \(\frac{13}{17}\) is in its lowest terms or not, we will find the HCF of 13 and 17 Now, HCF of 13 and 17 = 1
Therefore, the fraction \(\frac{13}{17}\) is in its lowest terms.

(d) To find whether the fraction \(\frac{25}{50}\) is in its lowest terms or not, we will find the HCF of 25 and 50.
Now, HCF of 25 and 50 = 25

Because, the HCF of numerator and denominator is not 1, therefore, the fraction is not in its lowest terms. To change the fraction into its lowest terms, we will divide the numerator (25) and denominator (50) by their HCF (25)
Fraction = \(\frac{25}{50}\) = \(\frac{25 \div 25}{50 \div 25}\) = \(\frac{1}{2}\)
∴The lowest form of \(\frac{25}{50}\) is \(\frac{1}{2}\)

(e) In \(\frac{14}{21}\), HCF of numerator (14) and denominator (21) = 7

The fraction \(\frac{14}{21}\) is not in its lowest terms because HCF of numerator (14) and denominator (21) is not 1. To change, the fraction, into its lowest terms we will divide the numerator (14) and denominator (21) by their HCF (7)
\(\frac{14}{21}\) = \(\frac{14 \div 7}{21 \div 7}\) = \(\frac{2}{3}\)
∴ The lowest form of fraction \(\frac{14}{21}\) is \(\frac{2}{3}\).

(f) To find whether the fraction \(\frac{8}{13}\) is in its lowest terms or not, we will find the HCF of numerator (8) and denominator (13)
Now, HCF of 8 and 13 = 1

Therefore, the fraction \(\frac{8}{13}\) is in its lowest form.

(g) To find whether the fraction \(\frac{7}{15}\) is in its lowest terms or not, we will find the HCF of numerator 7 and denominator 15.
Now, HCF of 7 and 15 = 1

Therefore, the fraction is in its lowest form.

(h) To find whether the fraction \(\frac{14}{27}\) is in its lowest terms or not, we will find the HCF of numerator (14) and denominator (27).
Now, HCF of 14 and 27 = 1

Therefore, the fraction \(\frac{14}{27}\) is in its lowest terms.

(i) To find whether the fraction \(\frac{25}{35}\) is in its lowest term or not, we will find the HCF of numerator (25) and denominator (35).
Now, the HCF of 25 and 35 = 5

Because, the HCF of numerator and denominator is not 1, therefore, the fraction \(\frac{25}{35}\) is not m its lowest terms.
To change the fraction into its lowest terms, we will divide the numerator and denominator by their HCF (5).
\(\frac{25}{35}\) = \(\frac{25 \div 5}{35 \div 5}\) = \(\frac{5}{7}\)
Therefore, the lowest form of \(\frac{25}{35}\) is \(\frac{5}{7}\).

(j) To find whether the fraction \(\frac{18}{23}\) is in its lowest terms or not, we will find the HCF of numerator (18) and denominator (23).
Now, HCF of 18 and 23 = 1

Therefore, the fraction \(\frac{18}{23}\) is in its lowest terms.

Question 2.
Write the lowest form of following fractions :
(a) \(\frac{4}{8}\)
(b) \(\frac{12}{18}\)
(c) \(\frac{15}{20}\)
(d) \(\frac{35}{45}\)
(e) \(\frac{24}{36}\)
(f) \(\frac{8}{12}\)
(g) \(\frac{18}{21}\)
(h) \(\frac{25}{45}\)
(i) \(\frac{6}{12}\)
(j) \(\frac{9}{27}\)
Solution:
(a) HCF of 4 and 8 = 4
Therefore, the lowest form of \(\frac{4}{8}\)
= \(\frac{4 \div 4}{8 \div 4}\) = \(\frac{1}{2}\)

(b) HCF of 12 and 18 = 6
Therefore, the lowest form of \(\frac{12}{18}\)
= \(\frac{12 \div 6}{18 \div 6}\) = \(\frac{2}{3}\)

(c) HCF of 15 and 20 = 5
Therefore, the lowest form of \(\frac{15}{20}\)
= \(\frac{15 \div 5}{20 \div 5}\) = \(\frac{3}{4}\)

(d) HCF of 35 and 45 = 5
Therefore, the lowest form of \(\frac{35}{45}\)
= \(\frac{35 \div 5}{45 \div 5}\) = \(\frac{7}{9}\)

(e) HCF of 24 and 36 = 12
Therefore, the lowest form of \(\frac{24}{36}\)
= \(\frac{24 \div 12}{36 \div 12}\) = \(\frac{2}{3}\)

(f) HCF of 8 and 12 = 4
Therefore, the lowest form of \(\frac{8}{12}\)
= \(\frac{8 \div 4}{12 \div 4}\) = \(\frac{2}{3}\)

(g) HCF of 18 and 21 = 3
Therefore, the lowest form of \(\frac{18}{21}\)
= \(\frac{18 \div 3}{21 \div 3}\) = \(\frac{6}{7}\)

(h) HCF of 25 and 45 = 5
Therefore, the lowest form of \(\frac{25}{45}\)
= \(\frac{25 \div 5}{45 \div 5}\) = \(\frac{5}{9}\)

(i) HCF of 6 and 12 = 6
Therefore, the lowest form of \(\frac{6}{12}\)
= \(\frac{6 \div 6}{12 \div 6}\) = \(\frac{1}{2}\)

(j) HCF of 9 and 27 = 9
Therefore, the lowest form of \(\frac{9}{27}\)
= \(\frac{9 \div 9}{27 \div 9}\) = \(\frac{1}{3}\)

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.3

Question 1.
Check whether the following fractions are equivalent or not :
(a) \(\frac{3}{7}\) and \(\frac{6}{14}\)
(b) \(\frac{11}{14}\) and \(\frac{77}{98}\)
(c) \(\frac{6}{9}\) and \(\frac{24}{36}\)
(d) \(\frac{5}{8}\) and \(\frac{10}{24}\)
(e) \(\frac{7}{12}\) and \(\frac{14}{21}\)
(f) \(\frac{8}{9}\) and \(\frac{40}{54}\)
Solution:
(a) Numerator of first fraction × Denominator of second fraction
= 3 × 14 = 42
Denominator of first fraction × Numerator of second fraction
= 7 × 6 = 42
Both the products are equal. So these fractions are equivalent.

(b) Numerator of first fraction × Deno-minator of second fraction
= 11 × 98 = 1078
Denominator of first fraction × Num-erator of second fraction
= 14 × 77 = 1078
Both the products are equal, so these fractions are equivalent.

(c) Numerator of first fraction × Denominator of second fraction
= 6 × 36 = 216
Denominator of first fraction × Numerator of second fraction
= 9 × 24 = 216
Both the products are equal, so these fractions are equivalent.

(d) Numerator of first fraction × Denominator of second fraction
= 5 × 24 = 120
Denominator of first fraction × Numerator of second fraction
= 8 × 10 = 80
Both the products are not equal, so these fractions are not equivalent.

(e) Numerator of first fraction × Denominator of second fraction
= 7 × 21 = 147
Denominator of first fraction × Numerator of second fraction
= 12 × 14= 168
Both the products are not equal, so these fractions are not equivalent.

(f) Numerator of first fraction × Denominator of second fraction.
= 8 × 54 = 432
Denominator of first fraction × Numerator of second fraction
= 9 × 40 = 360
Both the products are not equal, so these fractions are not equivalent.

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.2

Question 1.
Match the following:

Solution:

Question 2.
Match the following:

Solution:

Question 3.
Fill in the blanks:

(a) \(\frac{1}{3}\) part of 9 guavas = _____ guavas
(b) \(\frac{1}{6}\) part of 12 toffees = ___ toffees
(c) \(\frac{1}{6}\) part of 18 ice-creams = ____ ice-creams
(d) \(\frac{1}{4}\) part of 16 pencils = ___ pencils
(e) \(\frac{1}{10}\) part of ₹ 20 = ₹ ___
(f) \(\frac{1}{10}\) part of 100 pencils = ___ pencils
(g) \(\frac{1}{10}\) part of 100 cm = ___ cm
(h) \(\frac{1}{8}\) part of 32 laddoos = ___ laddoos.

Solution:

(a) \(\frac{1}{3}\) × 9 = 3,
(b) \(\frac{1}{6}\) × 12 = 2,
(c) \(\frac{1}{6}\) × 18 = 3,
(d) \(\frac{1}{4}\) × 16 = 4,
(e) \(\frac{1}{10}\) × 20 = 2,
(f) \(\frac{1}{10}\) × 100 = 10,
(g) \(\frac{1}{10}\) × 100 = 10,
(h) \(\frac{1}{8}\) × 32 = 2,

Question 4.
Neha’s uncle brought a chocolate which looks as following diagram:

(a) Neha gave half part of her chocolate to her sister Nidhi. How many pieces of chocolate will she give to Nidhi?
(b) Neha gave one-eighth part of this chocolate to her grand mother. How many pieces will she give to her?
(c) Neha gave one-fourth part of this chocolate to her mother. How many pieces will she give to her?
(d) After giving pieces to all, rest of pieces she ate. How many pieces will she get for herself?
Solution:
(a) Total number of pieces of chocolate = 16
Number of pieces given to Nidhi = \(\frac{1}{2}\) × 16 = 8
(b) Number of pieces given to her grand
mother = \(\frac{1}{8}\) × 16 = 2
(c) Number of pieces given to her mother = 16 × \(\frac{1}{4}\) = 4
(d) Number of pieces she will get for herself = 16 – (8 + 2 + 4)
= 16 – 14 = 2 .

Question 5.
Arjun studies in 5th class. He spends his day according to the following schedule.

• One fourth of a day in the school.
• One third of a day for sleeping.
• \(\frac{1}{12}\)th of a day for watching T.V.
• \(\frac{1}{12}\)th of a day for playing.
• \(\frac{1}{8}\)th of a day for school homework. 8
• \(\frac{1}{8}\)th of a day he spends with grand parents.

After the following :

(a) How much time Arjun spends in School ?
(b) For how many hours Arjun takes sleep ?
(c) For how many hours Arjun watches T.V. ?
(d) For how many hours Arjun plays ?
(e) For how many hours Aijun does his homework.
(f) How many hours Arjun spends with his grand parents ?
Note. Teacher will tell the students that there are 24 hours in a day.
Solution:
(a) Time spent by Aijun in the school = One fourth of a day.
= \(\frac{1}{4}\) × 24 hours = 6 hours.
(b) Number of hours for which Aijun takes sleep = one third of a day
= \(\frac{1}{3}\) × 24 hours = 8 hours.
(c) Number of hours for which Aijun watches T.V. = One-twelfth of a day
= \(\frac{1}{12}\) × 24 hours = 2 hours.
(d) Number of hours for which Arjun , plays = One twelfth of a day
= \(\frac{1}{12}\) × 24 hours = 2 hours.
(e) Number of hours for which Aijun does his homework = One eighth of a day
= \(\frac{1}{8}\) × 24 hours = 3 hours.
(f) Number of hours Aijun spends with his grandparents = one eighth of a day
= \(\frac{1}{8}\) × 24 hours = 3 hours.

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 6 Measurement Intext Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 6 Measurement Intext Questions

Page No. 130

Question 1.
How many pieces of 2 m can be cut from a 30 m long rope? How many times will you cut the rope?
Solution:
Length of rope = 30 m
Length of each piece = 2 m
No. of pieces that can be cut = 30 ÷ 2 = 15
No. of times we will cut the rope = 15 – 1 = 14

Question 2.
Observe the following table and fill ups :

Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 6 Measurement MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 6 Measurement MCQ Questions

Multiple Choice Questions

Tick (✓) the right answer :

Question 1.
Convert 8 m into centimetres,
(a) 80 cm
(b) 800 cm
(c) 8000 cm
(d) 8 cm.
ans:
(b) 800 cm

Question 2.
Convert 16 kl into litres.
(a) 160 l
(b) 1600 l
(c) 16000 l
(d) 160000 l
Ans:
(c) 16000 l

Question 3.
Convert 10 da.g.into grams.
(a) 100 g
(b) 1000 g
(c) 10 g
(d) 10000 g.
Ans:
(c) 10 g

Question 4.
How many kg are there in 1000 g?
(a) 100 kg
(b) 10 kg
(c) 20 kg
(d) 1 kg.
Ans:
(d) 1 kg.

Question 5.
Decimal formation of 3 l 175 ml.
(a) 31.75 l
(b) 317.5 l
(c) 3.175 l
(d) 0.3175 l
ans:
(c) 3.175 l

Question 6.
3.5 km = …………… m
(a) 350 m
(b) 3500 m
(c) 35 m
(d) 0.350 m.
Ans:
(b) 3500 m

Question 7.
Which unit is used by a shopkeeper to weigh vegetables ?
(a) litre and kl
(b) metre and km
(c) gram and kg
(d) none.
Ans:
(c) gram and kg

Question 8.
Which unit is used to measure liquids ?
(a) litre
(b) kilometre
(c) metre
(d) none.
Ans:
(a) litre

Question 9.
Kanwal bought 6 kg potatoes, 3 kg 500 g onions and 500 g tomatoes from the market How many kg of vegetables had he bought ?
(a) 10 kg
(b) 6 kg
(c) 3 kg
(d) 11 kg.
ans:
(a) 10 kg

Question 10.
Harpreet has bought 10 m cloth, he uses 6 m 50 cm cloth for her suit. How much cloth is left ?
(a) 2 m 50 cm
(b) 4 m
(c) 4 m 50 cm
(d) 3 m 50 cm.
ans:
(d) 3 m 50 cm

Question 11.
How many metre are in one milimetre ?
(a) \(\frac{1}{100}\)
(b) \(\frac{1}{1000}\)
(c) \(\frac{1}{10}\)
(d) 100.
Ans:
(b) \(\frac{1}{1000}\)

Question 12.
How many centimetres are in one hectometres ?
(a) 1000
(b) 10,000
(c) 100
(d) \(\frac{1}{1000}\)
Ans:
(b) 10,000

Question 13.
How many hectogram are in one kilogram ?
(a) 100
(b) \(\frac{1}{100}\)
(c) 10
(d) \(\frac{1}{10}\)
Ans.
(c) 10

Question 14.
How many decalitres are in one kilolitre ?
(a) 1000
(b) 500
(c) 200
(d) 100
Ans.
(d) 100

Question 15.
How many mililitres are in one decilitre ?
(a) 10
(b) 100000
(c) 100
(d) 1000
Ans:
(c) 100

Question 16.
How many days are in a leap year ?
(a) 364
(b) 366
(c) 365
(d) 363
Ans:
(b) 366

Question 17.
How many days are there in the month of February in a leap year ?
(a) 28
(b) 30
(c) 29
(d) 31.
Ans:
(c) 29

Question 18.
Write 3:10 pm according to 24 hour clock ?
(a) 23:10
(b) 25:10
(c) 15:10
(d) 13:10.
Ans:
(c) 15:10

Question 19.
Write 22:25 according to 12 hour clock.
(a) 10:25 PM
(b) 12:25 AM
(c) 12:25 PM
(d) 9: 25 PM.
Ans:
(a) 10:25 PM

Question 20.
How many seconds make one hour ?
(a) 60
(b) 3600
(c) 360
(d) 300.
Ans:
(b) 3600

Question 21.
The distance of your village dispensary from your school is 2 km, village community hall 955 m and distance of guruduara is 1500 m. Which of these is at a maximum distance from your school ?
(a) Dispensary
(b) Community centre
(c) Guruduara
(d) All are at equal distances.
Ans:
(a) Dispensary

Question 22.
See Carefully and tell :


(a) less than 500 ml
(b) In-between 500 ml and 1 litre
(c) In between 1 l and 2 l
(d) More than 2 l
Ans:
(c) In between 1 l and 2 l

Question 23.
How many metres are there in 3.5 kilometres ?
Ans:
3.5 kilometres = 3.5 × 1000 metre
= 3500 metre.

Question 24.
How many seconds are there in a day ?
Ans:
1 day = 24 hours = 24 × 60 minutes
= 24 × 60 × 60 seconds = 86400 seconds.

Question 25.
The map of a village which is at a distance from a city is given. Simran is moving in the village on his cycle.

Find the distance covered by Simran :
(a) From D to A (passing through B)
(b) From A to D (passing through B and C)
Ans:
(a) Distance covered by Simran from D to A (passing through B) = DB + BA.
= 1335 m + 1580 m
= 2915 m

(b) Distance covered by Simran from A to D (passing through B and C)
= AB + BC + CD
= 1580m + 1200m + 1315 m
= 4095m

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 6 Measurement Ex 6.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 6 Measurement Ex 6.7

1. Find the difference :

Question 1.
8 hours 30 min and 2 hours 10 min
Solution:

Question 2.
10 hours 30 min 20 sec and 8 hours 20 min 15 sec
Solution:

Question 3.
11 years 5 months and 6 years
Solution:

Question 4.
7 years 2 months and 3 years
Solution:

Note, ∵ 1 year = 12 months
∴ 12 months + 2 months = 14 months

2. Find the Time :

Question 1.
4 hours before 5:30 pm
Solution:
4 hours before 5:30 pm
4 hours = 30 minutes + 3 hours + 30 minutes
30 minutres before 5:30 pm = 5.00 pm
3 hours before 5:00 pm = 2:00 pm
30 minutes before 2:00 pm = 1:30 pm

Question 2.
2 hours after 11:00 am
Solution:
2 hours after 11:00 am
1 hour after 11:00 am = 12:00 noon
1 hour after 12:00 noon =1:00 pm
Second method :
2 hours after 11:00 am

i.e. 12:00 +1:00
= 1:00 pm

Question 3.
6 hours before 4:30 am
Solution:
6 hours before 4:30 am
6 hours = 30 minutes + 4 hours + 1 hour + 30 minutes
30 minutes before 4:30 am = 4:00 am
4 hours before 4:00 am = 12:00 mid night
1 hour before 12:00 mid night = 11:00 pm

Question 4.
1 hour 45 min after 8:30 am
Solution:
1 hour 45 min after 8:30 am
1 hour 45 minutes = 30 minutes + 1 hour + 15 minutes
30 minutes after 8:30 am = 9:00 am
1 hour after 9:00 am = 10: 00 am
15 minutes after 10:00 am = 10:15 am

3. Find the Time Gap :

Question 1.
From 3:00 am to 10:00 am
Solution:
Time gap from 3:00 am to 10:00 am = 7 hour

Question 2.
From 6:00 am to 1:30 pm
Solution:
Time gap from 6:00 am to 1:30 pm
Time gap between 6:00 am to 12:00 noon = 6 hours
Time gap from 12:00 noon to 1:00 pm = 1 hour
Time gap from 1:00 pm to 1:30 pm = 30 minutes
Hence, time gap from 6:00 am to 1:30 pm = 7 hours 30 minutes

Second Method :
1:30 pm = 1:30 + 12:00 = 13:30 o’clock
Time gap from 6:00 am to 1:30 pm

Question 3.
From 5:00 am to 10:45 pm
Solution:
Time gap from 5:00 pm to 10:45 pm
Time gap from 5:00 pm to 10:00 pm = 5 hours
Time gap from 10:00 pm to 10:45 pm = 45 minutes
Hence, time gap from 5:00 to 10:45 pm = 5 hours 45 minutes

Question 4.
From 9:00 am to 2:30 am (next morning)
Solution:
Time gap from 9:00 pm to 2:30 am (next morning)
Time gap from 9:00 pm to 12:00 mid night = 3 hours
Time gap from 12:00 mid night to 2:00 am = 2 hours
Time gap from 2:00 am to 2:30 am = 30 minutes
Hence, time gap from 9:00 pm to 2:30 am (next morning) = 5 hours 30 minutes

Question 4.
A bank opens at 9:30 am and closes at 5:00 pm. How many working hours are there ?
Solution:
Time gap from 9:30 am to 10:00 am
= 30 minutes Time gap from 10:00 am to 12:00 noon = 2 hours
Time gap from 12:00 noon to 5:00 pm = 5 hours
Hence, time gap from 9:00 am to 5:00 pm = 7 hours 30 minutes
Therefore, working hours of the bank = 7 hours 30 minutes

Question 5.
A bus starts from Chandigarh at 7:30 am and reaches Shimla at 10:50 am. How much time is taken by the bus to reach Shimla ?
Solution:
Time gap from 7:30 am to 8:00 am = 30 minutes
Time gap from 8:00 am to 10:00 am = 2 hours
Time gap from 10:00 am to 10:50 am = 50 minutes
Time gap from 7:30 am to 10:50 am
= 30 minutes + 2 hours + 50 minutes
= 2 hours + 80 minutes
= 2 hours + 60 minutes + 20 minutes
= 2 hours + 1 hour + 20 minutes
= 3 hours 20 minutes
Therefore, time taken by the bus to reach Shimla = 3 hours 20 minutes

Second Method :
The time at which the bus reaches Shimla = 10:50 am
The time at which the bus starts from Chandigarh = 7:30 am
= 3:20 hours
The time taken by bus to reach Shimla = 3 hours 20 min.

Question 6.
A boy goes to school at 7:30 am and returns back from school at 2:45 pm. How much time does he spend in the school ?
Solution:
The time when the boy goes to school = 7:30 am
The time when the boy returns back = 2.45 am
The time gap from 7:30 am to 8:00 am = 30 minutes
The time gap from 8:00 am to 12:00 noon = 4 hours
The time gap from 12:00 noon to 2:00 pm = 2 hours
The time gap from 2:00 pm to 2:45 pm = 45 minutes
Total time
= 30 minutes + 6 hours + 45 minutes
= 6 hours + 75 minutes = 6 hours + 60 minutes + 15 minutes
= 6 hours + 1 hour + 15 minutes
= 7 hours 15 minutes
Therefore, time spent in the school = 7 hours 15 minutes

Second Method :
2:45 pm = 2:45 + 12:00 = 14:45 o’clock
The time when the boy returns back = 14:45 o’clock
The time when the boy goes to school = 7:30 o’clock
The time spent in the school by boy

= 7 hours 15 minutes

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.1

Question 1.
Out of the following group of stars:

(a) Write the fraction of coloured stars. ____
Solution:
\(\frac{4}{9}\)

(b) Write fraction of stars without colour. ____
Solution:
\(\frac{5}{9}\)

Question 2.
In the following diagram

(a) Write fraction of coloured ice creams ____
Solution:
\(\frac{2}{5}\)

(b) Write fraction of ice creams without colour
Solution:
\(\frac{3}{5}\)

Question 3.
In the following diagram:

(a) Write fraction of coloured balls. ____
Solution:
\(\frac{6}{11}\)

(b) Write fraction of balls without colour.
Solution:
\(\frac{5}{11}\)

Question 4.
There are 12 balls in each of the following box. Colour the balls according to given fraction in the box and write number of coloured balls in blank box :

Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 6 Measurement Ex 6.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 6 Measurement Ex 6.6

1. Addition :

Question 1.
2 hours 10 min and 1 hour 20 min.
Solution:

Question 2.
4 hours 35 min and 3 hours 40 min.
Solution:

= 7 hours + 75 min
= 7 hours + 60 min + 15 min
= 7 hours + 1 hours + 15 min
= 8 hours 15 min

2. Add the following :

Question 1.
1 hour 10 min 20 sec and 3 hours 20 min
Solution:

Question 2.
2 hours 50 min 30 sec and 1 hour 10 min 30 sec
Solution:

3 hours + 60 min + 60 sec
= 3 hours + 1 hour + 1 min + 0 sec
= 4 hours + 1 min + 0 sec
= 4 hours 1 min

3. Add :

Question 1.
7 months and 2 years 3 months
Solution:

Question 2.
4 years 5 months and 1 year 8 months.
Solution:

= 5 years + 13 months
= 5 years + 12 months + 1 month
= 5 years + 1 year + 1 month
= 6 years 1 month

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions InText Questions and Answers.

PSEB 5th Class Maths Solutions Chapter 4 Fractions InText Questions

Try These : (Textbook Page No.86)

Question 1.
Write the fraction of coloured stars.

Solution:
\(\frac{1}{2}\)

Solution:
\(\frac{3}{4}\)

Solution:
\(\frac{5}{8}\)

Question 2.
Colour the diagram according to given fraction :

(a) \(\frac{2}{3}\)



Solution:

Question 3.
In fraction \(\frac{2}{3}\), numerator is and denominator is .
Solution:
In fraction \(\frac{2}{3}\), numerator is and denominator is .

Question 4.
In fraction \(\frac{1}{2}\), numerator is and denominator is .
Solution:
In fraction \(\frac{1}{2}\), numerator is and denominator is .

Question 5.
Write the fraction with numerator 4 and denominator 5:
Solution:
Write the fraction with numerator 4 and denominator 5: