Samacheer Class 6 Maths Measurements Book Back Answers & Solutions

Q: Fill in the blanks. (i) 250 ml + \frac{1}{2} ml = _____ l. (ii) 150 kg 200 g + 55 kg 750 g = ____ kg ____ g. (iii) 20 l – 1 l 500 ml = ____ l ___ ml (iv) 450 ml × 5 = ____ l ____ ml. (v) 50 Kg ÷ 100 g = ______

(i) \frac{3}{4} l (ii) 205 kg 950 g (iii) 18 l 500 ml (iv) 2l 250 ml (v) 500

Q: Convert into higher units: (i) 13000 mm (km, m, cm)

13000 mm = \frac{13000}{10} cm = 1300 cm = \frac{13000}{1000} m = 13000 mm = 13 m = \frac{13000}{1000000} km = 13000 mm = 0.013 km (ii) 8257 ml (kl, l) 8257 ml = \frac{8257}{1000} l = 8.257 l = 8257 ml = \frac{8257}{1000000} kl = 0.008257 kl

Q.1Fill in the blanks. (i) 250 ml + \(\frac{1}{2}\) ml = _____ l. (ii) 150 kg 200 g + 55 kg 750 g = ____ kg ____ g. (iii) 20 l – 1 l 500 ml = ____ l ___ ml (iv) 450 ml × 5 = ____ l ____ ml. (v) 50 Kg ÷ 100 g = ______v

Solution

(i) \(\frac{3}{4}\) l
(ii) 205 kg 950 g
(iii) 18 l 500 ml
(iv) 2l 250 ml
(v) 500
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(i) \(\frac{3}{4}\) l
(ii) 205 kg 950 g
(iii) 18 l 500 ml
(iv) 2l 250 ml
(v) 500
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.2True or False. (i) Pugazhenthi ate 100 g of nuts which is equal to 0.1 kg. (ii) Meena bought 250 ml of buttermilk which is equal to 2.5 l. (iii) Karkuzhali’s bag 1 kg 250 g and Poongkodi’s bag 2 kg 750 g. The total weight of their bags 4 kg. (iv) Vanmathi bought 4 books each weighing 500 g. Total weight of 4 books is 2 kg. (v) Gayathri bought 1 kg of birthday cake. She shared 450 g with her friends. The weight of cake remaining is 650 g.v

Solution

(i) True
(ii) False
(iii) True
(iv) True
(v) False
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(i) True
(ii) False
(iii) True
(iv) True
(v) False
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.3Convert into indicated units: (i) 10 l and 50 ml into ml (ii) 4 km and 300 m into m (iii) 300 mg into gv

Solution

(i) 10 l and 50 ml
= 10 × 1000 ml + 50 ml
= (10000 + 50)ml
= 10050 ml
(ii) 4 km and 300 m
= 4 × 1000 + 300 m
= (4000 + 300) m
= 4300 m
(iii) 300 mg
= \(\frac{300}{1000}\)g
= 0.3 g
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(i) 10 l and 50 ml
= 10 × 1000 ml + 50 ml
= (10000 + 50)ml
= 10050 ml
(ii) 4 km and 300 m
= 4 × 1000 + 300 m
= (4000 + 300) m
= 4300 m
(iii) 300 mg
= \(\frac{300}{1000}\)g
= 0.3 g
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.4Convert into higher units: (i) 13000 mm (km, m, cm)v

Solution

13000 mm
= \(\frac{13000}{10}\) cm
= 1300 cm
= \(\frac{13000}{1000}\) m = 13000 mm
= 13 m
= \(\frac{13000}{1000000}\) km = 13000 mm
= 0.013 km
(ii) 8257 ml (kl, l)
8257 ml
= \(\frac{8257}{1000}\) l
= 8.257 l
= 8257 ml
= \(\frac{8257}{1000000}\) kl
= 0.008257 kl
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

13000 mm
= \(\frac{13000}{10}\) cm
= 1300 cm
= \(\frac{13000}{1000}\) m = 13000 mm
= 13 m
= \(\frac{13000}{1000000}\) km = 13000 mm
= 0.013 km
(ii) 8257 ml (kl, l)
8257 ml
= \(\frac{8257}{1000}\) l
= 8.257 l
= 8257 ml
= \(\frac{8257}{1000000}\) kl
= 0.008257 kl
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.5Convert into lower units: (i) 15 km (m, cm, mm) (ii) 12 kg (g, mg)v

Solution

(i) 15 km = 15 × 1000 m = 15000 m
15 km = 15 × 100000 cm
= 1500000 cm
15 km = 15 × 1000000 mm
= 15000000 mm
(ii) 12 kg (g, mg)
12 kg = 12 × 1000 g
= 12000 g
12 kg = 12 × 1000000 mg
= 12000000 mg

Answer:

(i) 15 km = 15 × 1000 m = 15000 m
15 km = 15 × 100000 cm
= 1500000 cm
15 km = 15 × 1000000 mm
= 15000000 mm
(ii) 12 kg (g, mg)
12 kg = 12 × 1000 g
= 12000 g
12 kg = 12 × 1000000 mg
= 12000000 mg

Q.6Compare and put > or < or = in the following: (i) 800 g + 150 g ____ 1 kg (ii) 600 ml + 400 ml ____ 1 l (iii) 6 m 25 cm ____ 600 cm + 25 cm (iv) 88 cm ____ 8 m 8 cm (v) 55 g ____ 550 mgv

Solution

(i) 800 g + 150g < 3kg
(ii) 600 ml + 400 ml = 1 l
(iii) 6 m 25 cm = 600 cm + 25 cm
(iv) 88 cm < 8 m 8 cm
(v) 55 g > 550 mg
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(i) 800 g + 150g < 3kg
(ii) 600 ml + 400 ml = 1 l
(iii) 6 m 25 cm = 600 cm + 25 cm
(iv) 88 cm < 8 m 8 cm
(v) 55 g > 550 mg
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.7Geetha brought 2 l and 250 ml of water in a bottle. Her friend drank 300 ml from it. How much of water is remaining in the bottle?v

Solution

Quantity of water Geetha brought = 2 l 250 ml
= 2 × 1000 + 250 ml
= 2000 + 250 ml
= 2,250 ml.
Quantity of water her friend drank = 300 ml
Remaining water = 2250 – 300 = 1950 ml. = 1 litre 950 ml.
Remaining water = 1 litre 950 ml.

Answer:

Quantity of water Geetha brought = 2 l 250 ml
= 2 × 1000 + 250 ml
= 2000 + 250 ml
= 2,250 ml.
Quantity of water her friend drank = 300 ml
Remaining water = 2250 – 300 = 1950 ml. = 1 litre 950 ml.
Remaining water = 1 litre 950 ml.

Q.8Thenmozhi’s height is 1.25 m now she grows 5 cm every year. What would be her height after 6 years?v

Solution

Thenmozhi’s present height = 1.25 m
Rate of growth per year = 5 cm
Her growth in 6 years = 5 cm × 6 = 30 cm.
After 6 years her height = 1.25 m + 30 cm
= 1.25 × 100 + 30 cm
= 125 + 30 cm
= 155 cm.
∴ After 6 years Thenmozhi’s height will be 155 cm.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

Thenmozhi’s present height = 1.25 m
Rate of growth per year = 5 cm
Her growth in 6 years = 5 cm × 6 = 30 cm.
After 6 years her height = 1.25 m + 30 cm
= 1.25 × 100 + 30 cm
= 125 + 30 cm
= 155 cm.
∴ After 6 years Thenmozhi’s height will be 155 cm.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.9Priya bought 22\(\frac { 1 }{ 2 }\) kg of onion, Krishna bought 18\(\frac { 3 }{ 4 }\) kg of onion and Sethu bought 9 kg 250 g of onion. What is the total weight of onion did they buy?v

Solution

Priya’s weight = 22 kg 500 g
Krishna’s weight = 18 kg 750 g
Sethu’s weight = 9 kg 250 g
Total weight = 49 kg 1500 g = 49 kg + 1 kg 500 g = 50 kg + 500 g.
Their total weight = 50 kg 500 g.

Answer:

Priya’s weight = 22 kg 500 g
Krishna’s weight = 18 kg 750 g
Sethu’s weight = 9 kg 250 g
Total weight = 49 kg 1500 g = 49 kg + 1 kg 500 g = 50 kg + 500 g.
Their total weight = 50 kg 500 g.

Q.10Maran walks 1.5 km every day to reach the school while Mahizhan walks 1400 m. Who walks more distance and by how much?v

Solution

Distance which Maran walks = 1.5 km = 1.5 × 1000 m = 1500 m
The distance which Mahizhan walks = 1400 m.
Here 1500 > 1400
∴ Difference = 1500 – 1400 = 100 m.
∴ Maran walks more distance = 100 m.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

Distance which Maran walks = 1.5 km = 1.5 × 1000 m = 1500 m
The distance which Mahizhan walks = 1400 m.
Here 1500 > 1400
∴ Difference = 1500 – 1400 = 100 m.
∴ Maran walks more distance = 100 m.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.11In a JRC one day camp, 150 gm of rice and 15 ml oil are needed for a student. If there are 40 students to attend the camp how much rice and oil are needed?v

Solution

Rice needed for one student = 150 g
Rice needed for 40 students = 150 g × 40 = 6000 g. = \(\frac{6000}{1000}\) kg = 6 kg.
Oil needed for one student = 15 ml
Oil needed for 40 students = 15 ml × 40 = 600 ml. = \(\frac{600}{1000}\) l = 0.6 l
∴ For the camp 6 kg of rice and 0.6 l of oil needed.

Answer:

Rice needed for one student = 150 g
Rice needed for 40 students = 150 g × 40 = 6000 g. = \(\frac{6000}{1000}\) kg = 6 kg.
Oil needed for one student = 15 ml
Oil needed for 40 students = 15 ml × 40 = 600 ml. = \(\frac{600}{1000}\) l = 0.6 l
∴ For the camp 6 kg of rice and 0.6 l of oil needed.

Q.12In a school, 200 litres of lemon juice is prepared. If 250 ml lemon juice is given to each student, how many students get the juice?v

Solution

Total lemon juice prepared = 200 l = 200 × 1000 ml = 2,00,000 ml.
∴ Quantity of Lemon juice given to one student = 250 ml.
∴ Number of students can get = \(\frac{2,00,000}{250}\) = 800
∴ 800 students can get the lemon juice.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

Total lemon juice prepared = 200 l = 200 × 1000 ml = 2,00,000 ml.
∴ Quantity of Lemon juice given to one student = 250 ml.
∴ Number of students can get = \(\frac{2,00,000}{250}\) = 800
∴ 800 students can get the lemon juice.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.13How many glasses of the given capacity will fill a 2 litre jug? (i) 100 ml ___ (ii) 50 ml ____ (iii) 500 ml ____ (iv) 1 l ____ (v) 250 ml ____v

Solution

2 litre = 2 × 1000 ml = 2000 ml.
(i) 100 ml
\(\frac{2000}{100}=20\)
20 glasses of 100 ml.
(ii) 50 ml
\(\frac{2000}{50}=40\)
40 glasses of 50 ml
(iii) 500 ml
\(\frac{2000}{500}=4\)
4 glasses of 500 ml
(iv) 1 l
\(\frac{2 l}{1 l}=2\)
2 glasses of 1 l.
(v) 250 ml
\(\frac{2000}{250}=8\)
8 glasses of 250 ml can fill the jug.
Objective Type Questions

Answer:

2 litre = 2 × 1000 ml = 2000 ml.
(i) 100 ml
\(\frac{2000}{100}=20\)
20 glasses of 100 ml.
(ii) 50 ml
\(\frac{2000}{50}=40\)
40 glasses of 50 ml
(iii) 500 ml
\(\frac{2000}{500}=4\)
4 glasses of 500 ml
(iv) 1 l
\(\frac{2 l}{1 l}=2\)
2 glasses of 1 l.
(v) 250 ml
\(\frac{2000}{250}=8\)
8 glasses of 250 ml can fill the jug.
Objective Type Questions

Q.149 m 4 cm is equal to …….. (i) 94 cm (ii) 904 cm (iii) 9.4 cm (iv) 0.94 cmv

Solution

(ii) 904 cm
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(ii) 904 cm
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.151006 g is equal to ____v

A. 1 kg 6 g
B. 10 kg 6 g
C. 100 kg 6 g
D. 1 kg 600 g

Solution

(a) 1 kg 6 g

Answer:

(a) 1 kg 6 g

Q.16Every day 150 l of water is sprayed in the garden. Water sprayed in a week is …… (i) 700 l (ii) 1000 l (iii) 950 l (iv) 1050 lv

Solution

(iv) 1050 l

Answer:

(iv) 1050 l

Q.17Which is the greatest 0.007 g, 70 mg, 0.07 cg?v

A. 0.07 cg
B. 0.007 g
C. 70 mg
D. all are equal

Solution

(d) all are equal
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Answer:

(d) all are equal
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

Q.187 km – 4200 m is equal to ………. (i) 3 km 800 m (ii) 2 km 800 m (iii) 3 km 200 m (iv) 2 km 200 mv

Solution

(ii) 2 km 800 m
Posted in Class 6 on January 19, 2025 January 20, 2025
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Answer:

(ii) 2 km 800 m
Posted in Class 6 on January 19, 2025 January 20, 2025
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Copyright © 2026 Samacheer Kalvi

Q.3Convert the following: (i) 20 minutes into seconds (ii) 5 hours 35 minutes 40 seconds into seconds (iii) 3 ½ hours into minutes (iv) 580 minutes into hours (v) 25200 seconds into hoursv

Solution

(i) 20 minutes into seconds:
1 min = 60 seconds
20 min = 20 × 60 seconds
= 1200 seconds
(ii) 5 hours 35 min 40 seconds into seconds
1 hour = 60 min
1 min = 60 seconds
1 hour = 3600 seconds
5 hours = 5 × 3600 seconds
= 18000 seconds
35 min = 35 × 60 seconds
= 2100 seconds
5 hours 35 minutes 40 seconds
= (18000 + 2100 + 40) seconds
= 20140 seconds
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2
(iii) 3 ½ hours into minutes
1 hour = 60 minutes
3 ½ hours = 3 hours + 30 min
= (3 × 60 + 30) min
= (180 + 30)min
= 210 min
(iv) 580 minutes into hours
1 hour = 60 min
580 min
= \(\frac{580}{60}\) hours
= \(\frac{290}{30}\) hours
= \(\frac{29}{3}\)
= 9 \(\frac{2}{3}\)
= 9 hours 40 min
(v) 25200 seconds into hours:
25200 seconds = \(\frac{25200}{3600}\)
= \(\frac{126}{18}\) hours
= \(\frac{63}{9}\) hours
= 7 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

(i) 20 minutes into seconds:
1 min = 60 seconds
20 min = 20 × 60 seconds
= 1200 seconds
(ii) 5 hours 35 min 40 seconds into seconds
1 hour = 60 min
1 min = 60 seconds
1 hour = 3600 seconds
5 hours = 5 × 3600 seconds
= 18000 seconds
35 min = 35 × 60 seconds
= 2100 seconds
5 hours 35 minutes 40 seconds
= (18000 + 2100 + 40) seconds
= 20140 seconds
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2
(iii) 3 ½ hours into minutes
1 hour = 60 minutes
3 ½ hours = 3 hours + 30 min
= (3 × 60 + 30) min
= (180 + 30)min
= 210 min
(iv) 580 minutes into hours
1 hour = 60 min
580 min
= \(\frac{580}{60}\) hours
= \(\frac{290}{30}\) hours
= \(\frac{29}{3}\)
= 9 \(\frac{2}{3}\)
= 9 hours 40 min
(v) 25200 seconds into hours:
25200 seconds = \(\frac{25200}{3600}\)
= \(\frac{126}{18}\) hours
= \(\frac{63}{9}\) hours
= 7 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.4The duration of electricity consumed by the farmer for his pump set on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity.v

Solution

The total duration of electricity consumed on both days
= 7 hours 20 min 35 sec + 3 hours 44 min 50 sec
= (7 + 3) hours (20 + 44) min (35 + 50) sec
= 10 hours 64 min 85 seconds
= 11 hours 5 min 25 seconds

Answer:

The total duration of electricity consumed on both days
= 7 hours 20 min 35 sec + 3 hours 44 min 50 sec
= (7 + 3) hours (20 + 44) min (35 + 50) sec
= 10 hours 64 min 85 seconds
= 11 hours 5 min 25 seconds

Q.5Subtract 10 hours 20 min 35 seconds from 12 hours 18 min 40 seconds.v

Solution

12 hours 18 min 40 seconds
= (12 × 3600) + (18 × 60) + 40 seconds
= 43200 + 1080 + 40 seconds
= 44320 seconds
10 hours 20 min 35 seconds
= (10 × 3600) + (20 × 60) + 35 seconds
= 36000 + 1200 + 35 seconds
= 37235 seconds
Difference:
44320 – 37235
= 7085
7085 seconds = (1 × 3600) + 3480 + 5 seconds
= 1 hour 58 minutes 5 seconds
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

12 hours 18 min 40 seconds
= (12 × 3600) + (18 × 60) + 40 seconds
= 43200 + 1080 + 40 seconds
= 44320 seconds
10 hours 20 min 35 seconds
= (10 × 3600) + (20 × 60) + 35 seconds
= 36000 + 1200 + 35 seconds
= 37235 seconds
Difference:
44320 – 37235
= 7085
7085 seconds = (1 × 3600) + 3480 + 5 seconds
= 1 hour 58 minutes 5 seconds
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.6Change the following into 12 hour format (i) 02:00 hours (ii) 08:45 hours (iii) 21:10 hours (iv) 11:20 hours (v) 00:00 hoursv

Solution

(i) 2 am
(ii) 08:45 am
(iii) 9:10 pm
(iv) 11:20 am
(v) 12 midrid

Answer:

(i) 2 am
(ii) 08:45 am
(iii) 9:10 pm
(iv) 11:20 am
(v) 12 midrid

Q.7Change the following into 24-hour format. (i) 3.15 am ( ii) 12.35 pm (iii) 12.00 noon (iv) 12.00 mid nightv

Solution

(i) 03.15 hours
(ii) 12.35 hours
(iii) 12.00 hours
(iv) 24.00 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

(i) 03.15 hours
(ii) 12.35 hours
(iii) 12.00 hours
(iv) 24.00 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.8Calculate the duration of time (i) from 5.30 am to 12.40 pm (ii) from 1.30 pm to 10.25 pm (iii) from 20.00 hours to 4.00 hours (iv) from 17.00 hours to 5.15 hoursv

Solution

(i) from 5.30 a.m. to 12 .40 p.m.
Duration of time from 5.30 a.m. to noon = 12 : 00 – 5 : 30 = 6 : 30 i.e 6 hours 30 minutes
From noon to 12.40 p.m the duration = 00 hours 40 minutes
Total duration = 6 hours 30 minutes + 00 hours 40 minutes
= 6 hours 70 minutes
= 6 hours + (60 + 10) minutes
= 6 hours + 1 hr 10 minutes
= 7 hours 10 minutes
∴ Duration of time from 5.30 am to 12.40 pm = 7 hours 10 minutes
(ii) From 1.30 pm to 10.25 pm
= (1.30 pm to 10.00 pm) + 25 min
= 8 hrs 30 min + 25 min
= 8 hrs 55 min
(iii) From 20.00 hours to 4.00 hours
= (20.00 hrs to 24.00 hrs) + (24.00 hrs to 4.00 hrs)
= 4 hrs + 4 hrs
= 8 hours
(iv) From 17.00 hrs to 5.15 hours
= (17.00 hrs to 05.00 hrs) + 15 min
= 12 hours + 15 min
= 12 hours 1.5 min
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

(i) from 5.30 a.m. to 12 .40 p.m.
Duration of time from 5.30 a.m. to noon = 12 : 00 – 5 : 30 = 6 : 30 i.e 6 hours 30 minutes
From noon to 12.40 p.m the duration = 00 hours 40 minutes
Total duration = 6 hours 30 minutes + 00 hours 40 minutes
= 6 hours 70 minutes
= 6 hours + (60 + 10) minutes
= 6 hours + 1 hr 10 minutes
= 7 hours 10 minutes
∴ Duration of time from 5.30 am to 12.40 pm = 7 hours 10 minutes
(ii) From 1.30 pm to 10.25 pm
= (1.30 pm to 10.00 pm) + 25 min
= 8 hrs 30 min + 25 min
= 8 hrs 55 min
(iii) From 20.00 hours to 4.00 hours
= (20.00 hrs to 24.00 hrs) + (24.00 hrs to 4.00 hrs)
= 4 hrs + 4 hrs
= 8 hours
(iv) From 17.00 hrs to 5.15 hours
= (17.00 hrs to 05.00 hrs) + 15 min
= 12 hours + 15 min
= 12 hours 1.5 min
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.10Manickam joined a chess class on 20.02.2017 and due to an exam, he left practice after 20 days. Again he continued to practice from 10.07.2017 to 31.03.2018. Calculate how many days did he practice?v

Solution

From the date of joining = 20 days From 10.07.2017 to 31.03.2018
July – 22
Aug – 31
Sep – 30
Oct – 31
Nov – 30
Dec – 31
Jan – 31
Feb – 28
Mar – 31
Total – 265
Total no of practice days = 265 + 20 = 285 days

Answer:

From the date of joining = 20 days From 10.07.2017 to 31.03.2018
July – 22
Aug – 31
Sep – 30
Oct – 31
Nov – 30
Dec – 31
Jan – 31
Feb – 28
Mar – 31
Total – 265
Total no of practice days = 265 + 20 = 285 days

Q.11A clock gains 3 minutes every hour. If the clock is set correctly at 5 am, find the time shown by the clock at 7 p.m?v

Solution

Time gained for 1 hour = 3 min
Time duration from 5 am to 7 pm = 14 hours
Time gained for 14 hours = 1 4 × 3 minutes
= 42 minutes
So, at 7 pm, the clock shows 7 hrs 42 minutes
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

Time gained for 1 hour = 3 min
Time duration from 5 am to 7 pm = 14 hours
Time gained for 14 hours = 1 4 × 3 minutes
= 42 minutes
So, at 7 pm, the clock shows 7 hrs 42 minutes
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.12Find the number of days between Republic day and Kalvi Valarchi Day in 2020.v

Solution

In 2020 Republic Day will be celebrated on 26th January and Kalvi Valarchi Day will be celebrated on 15th July.
Number of days between 26.01.2020 and 15.07.2020
January – 6 Days (from 26.01.2020)
February – 29 Days (2020 is a leap year)
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 15 Days (upto 15.07.2020)
Total – 172 Days.
∴ Total number of days = 172

Answer:

In 2020 Republic Day will be celebrated on 26th January and Kalvi Valarchi Day will be celebrated on 15th July.
Number of days between 26.01.2020 and 15.07.2020
January – 6 Days (from 26.01.2020)
February – 29 Days (2020 is a leap year)
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 15 Days (upto 15.07.2020)
Total – 172 Days.
∴ Total number of days = 172

Q.13If the 11th of Jan 2018 is Thursday, what is the day on 20th July of the same year?v

Solution

Jan – 21
Feb – 28
Mar – 31
April – 30
May – 31
June – 30
July – 19
Total – 190 days
190 days = 27 weeks + 1 day
The required day is the first day after Thursday.
Therefore 20th July 2018 is Friday.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

Jan – 21
Feb – 28
Mar – 31
April – 30
May – 31
June – 30
July – 19
Total – 190 days
190 days = 27 weeks + 1 day
The required day is the first day after Thursday.
Therefore 20th July 2018 is Friday.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.14(i) Convert 480 days into years. (ii) Convert 38 months into yearsv

Solution

(i) 480 days = \(\frac{480}{365}\)
= 1 year 115 days
= 1 year 3 months 25 days
(ii) 38 months = \(\frac{38}{12}\)
= 3 years 2 months

Answer:

(i) 480 days = \(\frac{480}{365}\)
= 1 year 115 days
= 1 year 3 months 25 days
(ii) 38 months = \(\frac{38}{12}\)
= 3 years 2 months

Q.15Calculate your age as on 01.06.2018v

Solution

My date of birth 20.11.1999
Convert in the format yyyy/mm/dd
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2 4
My age is 18 yrs 6 months 11 days

Objective Type Questions

Answer:

My date of birth 20.11.1999
Convert in the format yyyy/mm/dd
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2 4
My age is 18 yrs 6 months 11 days

Objective Type Questions

Q.162 days = _____ hours.v

A. 38
B. 48
C. 28
D. 40

Solution

(b) 48

Answer:

(b) 48

Q.173 weeks = ……… days (i) 21 (ii) 7 (iii) 14 (iv) 28v

Solution

(i) 21
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

(i) 21
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.18The number of ordinary years between two consecutive leap years is _____.v

A. 4 years
B. 2 years
C. 1 year
D. 3 years

Solution

(d) 3 years

Answer:

(d) 3 years

Q.19What time will it be 5 hours after 22:35 hours? (i) 2:30 hours (ii) 3:35 hours (iii) 4:35 hours (iv) 5:35 hoursv

Solution

(ii) 3:35 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Answer:

(ii) 3:35 hours
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.2

Q.202\(\frac { 1 }{ 2 }\) years is equal to ______ months.v

A. 25
B. 30
C. 24
D. 5

Solution

Answer:

Q.1Two pipes whose lengths are 7 m 25 cm and 8 m 13 cm joined by welding and then a small piece 60 cm is cut from the whole. What is the remaining length of the pipe?v

Solution

Total length = 7 m 25 cm + 8 m 13 cm = 15 m 38 cm (or) 1538 cm
length detatched = 60 cm
Remaining length = 14 m 78 cm
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Answer:

Total length = 7 m 25 cm + 8 m 13 cm = 15 m 38 cm (or) 1538 cm
length detatched = 60 cm
Remaining length = 14 m 78 cm
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Q.2The saplings are planted at a distance of 2 m 50 cm in the road of length 5 km by Saravanan. If he has 2560 saplings, how many saplings will be planted by him? how many saplings are left?v

Solution

Distance between two saplings = 2 m 50 cm = 250 cm
Total length of the road = 5000 m = 500000 cm

Answer:

Distance between two saplings = 2 m 50 cm = 250 cm
Total length of the road = 5000 m = 500000 cm

Q.4Make a calendar for the month of February 2020. (Hint: January 1st, 2020 is Wednesday)v

Solution

February 2020 is a leap year
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 3

Answer:

February 2020 is a leap year
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 3

Q.6A squirrel wants to eat the grains quickly. Help the Squirrel to find the shortest way to reach the grains. (Use your scale to measure the length of the line segments) Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 5

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 5

v

Solution

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 6
The Shortest way to reach the grains is the AGFKE path.

Answer:

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 6
The Shortest way to reach the grains is the AGFKE path.

Q.7A room has a door whose measures are 1 m wide and 2 m 50 cm high. Can we make a bed of 2 m and 20 cm in length and 90 cm wide into the room?v

Solution

Measures of the door length 2 m 50 cm width and lm (100 cm)
(i) Measures of the bed 2 m 20 cm width and 90 cm
Measures of the bed < measures of the door
∴ Yes, we can take the bed into the room.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Answer:

Measures of the door length 2 m 50 cm width and lm (100 cm)
(i) Measures of the bed 2 m 20 cm width and 90 cm
Measures of the bed < measures of the door
∴ Yes, we can take the bed into the room.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Q.8A post office functions from 10 a.m. to 5.45 p.m. with a lunch break of 1 hour. If the post office works for 6 days a week, find the total duration of working hours in a week.v

Solution

Working hours in a day = 6 hrs 45 min
= (6 × 60 min) + 45 min
= (360 + 45) min
= 405 min
Total duration of working hours in a week
= 6 × 405 min
= 2430 min
= \(\frac{2430}{60}\)
= \(\frac{810}{20}\)
= 40 \(\frac{1}{2}\)
= 40 hours 30 minutes
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Answer:

Working hours in a day = 6 hrs 45 min
= (6 × 60 min) + 45 min
= (360 + 45) min
= 405 min
Total duration of working hours in a week
= 6 × 405 min
= 2430 min
= \(\frac{2430}{60}\)
= \(\frac{810}{20}\)
= 40 \(\frac{1}{2}\)
= 40 hours 30 minutes
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Q.9Seetha wakes up at 5.20 a.m. She spends 35 minutes to get ready and travels 15 minutes to reach the railway station. If the train departs exactly at 6.00 am, will Seetha catch the train?v

Solution

No, Seetha will not catch the train
Time at Seetha wakes up = 5.20 am
Time is taken for getting ready = 35 min
Travelling time to station = 15 min
Reporting time = 5.20 am + 50 min = 6.10 am
But, the train departs exactly at 6.00 am
So, Seetha will not catch the train.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Answer:

No, Seetha will not catch the train
Time at Seetha wakes up = 5.20 am
Time is taken for getting ready = 35 min
Travelling time to station = 15 min
Reporting time = 5.20 am + 50 min = 6.10 am
But, the train departs exactly at 6.00 am
So, Seetha will not catch the train.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3

Q.10A doctor advised Vairavan to take one tablet every 6 hours once on the 1st day and once every 8 hours on the 2nd and 3rd day. If he starts to take 9.30 am the first dose. Prepare a time chart to take the tablet in railway time.v

Solution

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 6
Posted in Class 6 on January 22, 2025 January 23, 2025
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Answer:

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.3 6
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
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Copyright © 2026 Samacheer Kalvi