Class 6 Math Solutions WBBSE Chapter 19 Measurement Of Time Exercise 19
Question 1. Given below is the time chart of Priya on a holiday. Let’s try to get few information.
Solution:
⇒ On that day, Priya woke up: at 6.00 a.m
⇒ Priya studied at: 9.00 a.m
⇒ Priya gossips with friends at: 11.00 a.m
⇒ She had lunch at: 1.00 a.m
⇒ After lunch at: 4.00 p.m
⇒ On that day Priya went to bed: at 10.00 p.m
Question 2. Today I left my house at 10:20 a.m. for school. Returned home at 4:45 p.m. Let’s calculate how long I was out of home.
Solution:
I was out of home = 4:45 p.m – 10:20 am = 6 hrs 25 min
Question 3. Yesterday Deba went to bed at 10: 25 p.m. But today he woke up at 6:10 a.m. Let’s find out how long he slept.
Solution:
Deba slept = 6:10 a.m – 10: 25 p.m = 7 hrs 45 mins
Read and Learn More WBBSE Solutions For Class 6 Maths
Question 4. Today I am going to Puri with my family. The train will leave from Howrah station at 22:35 hours, we will reach Howrah station at 20:44 hrs. Let’s calculate how long we have to wait at the station.
Solution:
Given
Today I am going to Puri with my family. The train will leave from Howrah station at 22:35 hours, we will reach Howrah station at 20:44 hrs.
We have to wait at the station = 22:35 hours – 20:44 hours
= 1 hour 51 mins Hrs min
⇒ \(\begin{array}{lc}
\text { Hrs } & \text { min } \\
22 & 35 \\
20 & 44 \\
\hline 1 & 51 \\
\hline
\end{array}\)
“WBBSE Class 6 Maths Chapter 19 measurement of time solutions”
Question 5. Today I have a history examination. Exam started at 11: 30 a.m. The teacher’s digital wristwatch is showing 13:15. Let me find out how long I had already written the examination. If the exam is to end at 2:30 p.m., how long shall I be able to write the examination, when the exam will be over, let me find out what will the digital watch of the teacher read.
Solution:
Given
Today I have a history examination. Exam started at 11: 30 a.m. The teacher’s digital wristwatch is showing 13:15. Let me find out how long I had already written the examination. If the exam is to end at 2:30 p.m.
⇒ I had already written the examination = 13:15 – 11:30 am = 1 hr 45 min.
⇒ I shalf be able to write the examination = 2:30 p.m. -13:15 = 1 hr 15 min.
⇒ The time of digital watch of the teacher = 13:15 + 1 hr 15 min = 14:30.
Question 6. To reach the father’s office from home, it takes 2 hrs 27 min but while returning home, it takes 2 hrs 51 min. Let me find out how long it takes in total to go to my father’s office and back.
Solution:
Given
To reach father’s office from home, it takes 2 hrs 27 min but while returning home, it takes 2 hrs 51 min.
⇒ The time to go to father’s office and back
= 2 hrs 27 mins + 2 hrs 57 mins
= 5 hrs 24 mins.
⇒ \(\begin{array}{lll}
\text { Hrs } & \min & \\
2 & 27 & \\
2 & 57 & \\
5 & 24 &
\end{array}\)
⇒ \(\begin{array}{rcc}
7 . \mathrm{hr} & \min & \mathrm{sec} \\
8 & 32 & 41 \\
+18 & 42 & 25 \\
& 74 & 66 \\
\hline \mathrm{hrs} & \min & \mathrm{sec} \\
\hline
\end{array}\)
Solution:
1. \(\begin{array}{rrr}
\mathrm{hr} & \mathrm{min} & \mathrm{sec} \\
8 & 32 & 41 \\
+18& 42 & 25 \\
26 & 74 & 66 \\
\hline 27 \mathrm{hrs} & 15 \mathrm{~min} & 6 \mathrm{sec} \\
\hline
\end{array}\)
2. \(\begin{array}{ccc}
\text { 2 hr } & \text { min } & \text { sec } \\
\text { (8) } & 60+11 & 60+37 \\
9 & 12 & 37 \\
-3 & 38 & 41 \\
\hline 5 \mathrm{hr} & 33 \mathrm{~min} & 56 \mathrm{sec} \\
\hline
\end{array}\)
“WBBSE solutions for Class 6 Maths measurement of time”
Question 8. Let’s add and subtract
1. (4 hr 33 min 20 sec) + (9 hr 52 min 25 sec)
Solution:
⇒ \(\begin{array}{ccc}
\text { Hrs } & \text { Mins } & \text { Sec } \\
4 & 33 & 20 \\
+9 & 52 & 25 \\
13 & 85 & 45 \\
\hline=14 \mathrm{Hrs} & 25 \mathrm{~min} & 45 \mathrm{sec} . \\
\hline
\end{array}\)
2. (6 hr 42 min 2 sec) – (2 hr 55 min 42 sec)
Solution:
⇒ \(\begin{array}{ccc}
\text { Hrs } & \text { Min } & \text { Sec } \\
5 & 60+41 & 60+2 \\
6 & 42 & 2 \\
-2 & 55 & 42 \\
\hline 3 & 46 & 20
\end{array}\)
= 3 hrs 46 min 20 Sec
3. (18 hr 19 min 15 sec) + (9 hr 55 min 48 sec)
Solution:
⇒ \(\begin{array}{rcc}
\text { Hrs } & \text { Min } & \text { Sec } \\
18 & 19 & 15 \\
9 & 55 & 48 \\
\hline 27 & 74 & 63 \\
\hline
\end{array}\)
= 28 hrs 15 min 3 sec
4. (23 hr 7 min) – (19 hr 29 min 18 sec)
Solution:
⇒ \(\begin{array}{ccc}
\text { Hrs } & \min& \text { Sec } \\
22 & 60+6 & 60+0 \\
23 & 7 & 00 \\
9 & 55 & 48 \\
\hline 13 & 11 & 12 \\
\hline
\end{array}\)
= 13 hrs 11 min 12 Sec.
Class 6 WBBSE Math Solutions Chapter 19 Measurement Of Time Exercise 19.1
Question 1. Debubabu built a new house. He will paint the two windows of his house. Each window has two panes. If he takes 2 hr 15 min to paint 1 window pane, let’s find how long he will take to colour 2 windows.
Solution:
Given
Debubabu built a new house. He will paint the two windows of his house. Each window has two panes. If he takes 2 hr 15 min to paint 1 window pane,
⇒ Time required to paint 1 window pane = 2 hrs 15 m
∴ Time required to paint 2 window panes = 2 hr 15 min x 2 = 4 hrs 30 min
“Class 6 Maths WBBSE Chapter 19 time measurement exercises”
Question 2. In 11 hrs 36 min, Phani Da can make 4 exactly similar clay statues. Let’s find out how long he will take to make one such statue. He takes equal time to make each statue.
Solution:
Given
In 11 hrs 36 min, Phani Da can make 4 exactly similar clay statues.
⇒ Time required to make 4 clay statues = 11 hrs 36 min
∴ Time required to make 1 clay statue = 11 hr 36 min ± 4
= 2 hrs 54 mins.
Question 3.
1. 3 hrs 26 min x 4 = How many hours and minutes?
Solution:
⇒ 3 hrs 26 min x 4 = 12 hr 104 min = 13 hrs 44 min
2. 7 hrs 13 min x 12 = How many hours and minutes?
Solution:
⇒ 7 hrs 13 min x 12 = 84 hrs 156 min = 86 hr. 36 min
3. 3 hrs 27 min ÷ 9 = How many minutes and seconds?
Solution:
4. 15 hrs ÷ 12 = How many hrs and mins?
Solution:
5. 6 hrs 18 sec ÷ 9 = How many min, sec?
Solution:
6. 5 hrs10 min 4 sec ÷ 4 = How many hours, min, sec?
Solution:
“Step-by-step solutions for measurement of time Class 6 WBBSE”
7. 2 hrs 32 min 41 sec÷ 3 = How many hours, min, secs?
Solution:
⇒ 2 hr 32 min 4 sec x 3
⇒ 2 hr 32 min 4 sec x 3
⇒ 6 hr 96 min 12 sec
= 7 hr 36 min 12 sec
Class 6 Maths Solutions WBBSE Chapter 19 Measurement Of Time Exercise 19.1
Question 1. 1st February of 2010 was Monday. Let’s calculate 1st March 2010 and 1st April 2010 will be which days.
Solution:
⇒ 1 st February 2010 was monday and February has 28 days.
∴ 1st March 2010 was Monday and March has 31 days.
∴ 3 days after Monday is Thursday.
∴ 1st April 2010 is Thursday.
Question 2. 01/02/2012 was Wednesday, then let’s calculate which days were the following dates: 01/03/2012, 01/04/2012, 01/05/2015,04/06/2012
Solution:
⇒ 1st February of 2012 was Wednesday and February 2012 had 29 days.
∴ 1 day after Wednesday is Thursday.
∴ . 1st March 2012 was Thursday.
⇒ March has 31 days.
⇒ 3 days after Thursday is Sunday.
∴ 1st April 2012 was Sunday.
⇒ April has 30 days.
⇒ 2 days after Sunday is Tuesday.
∴ 1st May 2012 was Tuesday.
⇒ May has 31 days, 6 days after Tuesday is Monday.
⇒ 4th June 2012 is Monday.
“WBBSE Class 6 Maths Chapter 19 important questions and answers”
Question 3. 1st January of 1996 was Sunday. Let us calulate which day 1st January of 1997 will be.
Solution:
⇒ 1st January 1996 was Monday.
⇒ Since 1996 was a leap year so 1996 had 366 days.
⇒ 2 days after Monday is Wednesday.
∴ 1st January 1997 was Wednesday.
Question 4. 1st March of 2004 was Monday, which day will be 1st April of 2005.
Solution:
⇒ 1st March 2004 was Monday 2005 year had 365 days.
⇒ 1 day after Monday is Tuesday.
⇒ 1st March 2005 was Tuesday
⇒ 1st March 2005 had 31 days.
⇒ 3days after Tuesday is Friday.
∴ 1st April 2005 was Friday.
Question 5. 1st June of 2008 was Tuesday. Let’s calculate which day was 1st June of 2006.
Solution:
⇒ 1st June 2008 was Sunday. 2008 was a leap year.
⇒ So, 2008 had 366 days.
⇒ 2006 and 2007 had 365 days.
⇒ 4days before Sunday is Thursday.
⇒ 1st June 2006 was Thursday.
Question 6. Independence day of 2013 is Thursday, let’s find which day will be Independence day of 2016.
Solution:
⇒ Independence day of 2013, 15th August 2013 was Thursday.
⇒ 2013 had 365 days and 2016 is a leap year.
⇒ So 2016 has 366 days.
⇒ 4 days after Thursday is Monday.
∴ Independence day of 2016 will be Monday.
Question 7. From the calendar of 2013 let us write which day of the week are the following and without looking into calender, let’s find which days of the week these days were in the year 2011 — Children’s day, Teachers’ day, Gandhi birthday, Republic day, Netaji Jayanti, World environment day (5th of June).
Solution:
1. Children’s Day (14th November)
⇒ 14th November 2013 was Thursday.
⇒ 2014 had 365 days and 2012 had 366 days.
⇒ 3 days before Thursday is Monday.
2. Teachers’ day (5th September)
⇒ 5th September of 2013 was Thursday.
⇒ 2011 had 365 days and 2012 had 366 days.
⇒ 3 days before Thursday is Monday.
∴ Teachers’ day – 5th September 2011 was Monday.
“Solved examples of time measurement WBBSE Class 6”
3. Gandhi birthday – 2nd October.
⇒ 2nd October 2013 was Wednesday.
⇒ 2011 had 365 days and 2012 had 366 days.
3 days before Wednesday Sunday.
∴ Gandhi birthday (2nd October) 2011 was Sunday.
4. Republic day (26th January)
⇒ 26th January of 2013 was Saturday.
⇒ 2011 had 365 days and 2012 had 366 days.
⇒ 3 days before Saturday is Wednesday.
∴ Republic day (26th January, 2011) was Wednesday.
5. Netaji Jayanti (23rd January)
⇒ 23rd January 2013 was Wednesday.
⇒ 2011 had 365 days and 2012 had 366 days.
⇒ 3 days before Wednesday is Sunday.
∴ Netaji Jayanti (23rd January) 2011 was Sunday.
6. World Environment Day (5th June)
⇒ 5th June 2013 was Wednesday.
⇒ 2011 had 365 days and 2012 had 366 days.
⇒ 3 days before Wednesday is Sunday.
⇒ World Environment Day 5th June, 2011 was Sunday.
“Best guide for Class 6 Maths WBBSE measurement of time”
Question 8.
- Let’s write the leap years between 1895 to 1915.
- Let’s write the leap years between 2010 to 2030.
Solution:
- Leap years between 1895 to 1915 = 1896; 1900; 1904,1908, and 1912.
- Leap years between 2010 to 2030 = 2012, 2016, 2020, 2024, and 2028.
Question 9. I stayed for 4 years in a house at College Ghat Road, from 2010 to 2013. Let’s calculate for how many days I stayed in that house.
Solution:
⇒ In 2010 – Number of days = 365 In 2011 — Number of days = 365
⇒ In 2012 — Number of days = 366 In 2013 — Number of days = 365 Total days in 4 years = 365 + 365 = 366 + 365 = 1461
Question 10. My birthday is on 15th December. In 2013, my birthday was on Sunday. Which day of the week will my birthday be in the years 2014, 2015 and 2016?
Solution:
My birthday is on 15th December 2013, which was Sunday. In 2014 had 365 days
⇒ 1 day after Sunday is Monday.
∴ 15th December 2014 was Monday.
⇒ In 2015 had 365 days.
∴ 1 days after Monday is Tuesday.
∴ 15th December, 2015 is Tuesday.
⇒ 2016 is of 366 days.
∴ 2 days after Tuesday is Thursday.
∴ 15th December 2016 will be Thursday.
Question 11. After independence, how many leap years have passed till 2014?
Solution:
⇒ Independence day was 15th August 1947.
⇒ The leap years that have passed till 2014 are 1948,1952,1956,1960,1964,1968 1972,1976,1980,1984,1988,1992 1996, 2000, 2004,2008 and 2012.
WBBSE Class 6 Maths Solutions Chapter 19 Measurement Of Time Exercise 19.2
Question 1. My date of birth is 19 -11 -1975, i.e., 19th November 1975. Let us find my
Solution:
⇒ \(\begin{array}{lll}
\text { Years } & \text { Months } & \text { Days } \\
& 12+9 & 30+10 \\
2010 & 10 & 10 \\
-1975 & 11 & 19 \\
\hline 24 & 10 & 21 \\
\hline
\end{array}\)
= 24 years 10 months 21 days.
Question 2. The construction of our main road began in summer on 6/6/2010. It took 1 yr 3 months 18 days to finish the work. Let’s calculate to find on which date the road construction work was completed.
Solution:
⇒ \(\begin{array}{ccc}
\text { Years } & \text { Months } & \text { Days } \\
2010 & 06 & 06 \\
+\quad 1 & 03 & 18 \\
\hline 2011 & 09 & 24 \\
\hline
\end{array}\)
⇒ The date of road construction work was completed on 24/09/ 2011.
Question 3. My present age is 11 yrs 7 months 10 days. Let me calculate after how many years I shall be eligible to cast my vote.
Solution:
⇒ \(\begin{array}{cll}
\text { Years } & \text { Months } & \text { Days } \\
17 & 12+0 & 30 \\
(-) 11 & 07 & 10 \\
\hline 6 & 08 & 20 \\
\hline
\end{array}\)
= 6 yrs 04 months 20 days.
“How to measure time accurately Class 6 WBBSE”
Question 4. The age of my father is 52 years 8 months 20 days. My uncle is 3 years 10 months 26 days older than my father. Let me find the age of my uncle.
Solution:
⇒ \(\begin{array}{ccl}
\text { Years } & \text { Months } & \text { Days } \\
52 & 8 & 20 \\
+3 & 10 & 26 \\
\hline 56 & 7 & 16 \\
\hline
\end{array}\)
= Age of my uncle = 56 years 7 months 16 days.
Question 5. Let us find values
1. \(\begin{array}{rcc}
\text { Year } & \text { Months } & \text { Days } \\
9 & 10 & 27 \\
+\quad 5 & 8 & 21 \\
\hline 14 & 18 & 48 \\
\hline
\end{array}\)
Solution: 15 years 7 months 18 days.
2. \(\begin{array}{ccc}
\text { Year } & \text { Month } & \text { Days } \\
29 & 11 & 19 \\
5 & 9 & 25 \\
+6 & 3 & 13 \\
\hline 40 & 23 & 57 \\
\hline
\end{array}\)
Solution: 42 years 0 months 27 days
3. \(\begin{array}{ccl}
\text { Year } & \text { Month } & \text { Days } \\
& 12+3 & 30 \\
11 & 3 & \mathrm{x} \\
-6 & 10 & 28 \\
\hline 5 & 5 & 02 \\
\hline
\end{array}\)
Solution: 5 years 5 months 2 days.
4. \(\begin{array}{|c|c|c|}
\hline \text { Year } & \begin{array}{c}
\text { Month } \\
12+6
\end{array} & \begin{array}{l}
\text { Days } \\
30+19
\end{array} \\
\hline 11 & 6 & 19 \\
\hline-6 & 10 & 21 \\
\hline 5 & 8 & 28 \\
\hline
\end{array}\)
Solution: 5 years 8 months 28 days.
Question 6. (1) 8 yrs 8 months 28 days + 11 yrs 8 months 18 days = How many yr, months and days?
Solution:
⇒ \(\begin{array}{rcl}
\text { Years } & \text { Months } & \text { Days } \\
8 & 8 & 28 \\
+11 & 8 & 18 \\
\hline 19 & 16 & 46 \\
\hline
\end{array}\)
= 20 yrs 5 months 16 days.
2. 20 yrs 11 months and days? -10 yrs 8 months 23 days = How many yr, month
Solution:
⇒\(\begin{array}{ccc}
\text { Years } & \text { Months } & \begin{array}{c}
\text { days } \\
30+0
\end{array} \\
20 & 11 & 0 \\
10 & 89 & 23 . \\
\hline 10 & 2 & 7 \\
\hline
\end{array}\)
= 10 years 2 months 7 days
3. 8 yrs 7 months 21 days x 9 = How many yr, month and days?
Solution:
= 77 yrs.9 months 9 days.
Question 7. My age is □ yrs □ months □ days. The age of my friend is □ yrs □ months □ days. What is the sum of our age and who is older than the two and by how much? Let us find out.
Solution:
My age is 15 years 11 months 26 days. The age of my friend is 14 yrs 9 months 20 days.
⇒ \(\text { Sum }=\begin{array}{ccc}
\text { Year } & \text { Month } & \text { Days } \\
15 & 11 & 26 . \\
14 & 9 & 20 \\
\hline 29 & 20 & 46 \\
\hline
\end{array}\)
= 30 yrs 9 months 16 days.
I am older than my friends.
Difference
⇒ \(\begin{array}{lll}
\text { Years } & \text { Month } & \text { Days } \\
15 & 11 & 26 \\
14 & 9 & 20 \\
\hline 1 & 2 & 6 \\
\hline
\end{array}\)
= 1 yr 2 months 6 days.
“Understanding time conversion and measurement Class 6 WBBSE”
Question 8. My date of birth is □. Today my age is □ yrs □ months □ days.
Solution:
My date of birth is 04.10-2001
Today my age is — (19-11 -2014)
⇒ \(\begin{array}{rll}
\text { Year } & \text { Month } & \text { Days } \\
2014 & 11 & 19 \\
-2001 & 10 & 04 \\
\hline 13 & 1 & 15 \\
\hline
\end{array}\)
Today my age is 13yrs 1month 15days