5 marks

Question 15 Marks

Seetha 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 a.m, will Seetha catch the train?

Answer

Time of wake up $=5.20 a \cdot m =5$ hour 20 minutes
Time spends $=35$ minutes
Then the time $=5$ hour 55 minutes
Travelling time to reach railway station $=15$ minutes
Now the time will be $=5$ hours 70 minutes
$=5 \text { hours }(60+10) \text { minutes }$
$=5 \text { hours }+(1 \text { hour } 10 \text { minutes })$
$=6 \text { hours } 10 \text { minutes }$
$=6.10 a \cdot m$
The departure time of the train to get ready $=6.0 a \cdot m$.
$\therefore$ She will not be able to catch the train.

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Question 25 Marks

A 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

Answer

Closing time of the post office 5.45 p.m. $=(5+12): 45$ hour
$
=17: 45 \text { hours. }
$
Starting time of the post office $=10 a \cdot m$
$
=10: 00 \text { hours }
$
For a day total working hour without break $=17: 45$ hour $-10: 00$ hours
$
=07: 45 \text { hour }
$
Lunch break $=$ 01:00 hour
Duration of Working hours $=6: 45$ hours
For 6 days working hours $=6: 45$ hour $\times 6$
$
=(6 \text { hours } 45 \text { minutes }) \times 6
$
$=36$ hours 270 minutes
$=36$ hours $+(4 \times 60+30)$ minutes
$=36$ hours + ( 4 hours +30 minutes $)$
$=40$ hours +30 minutes
The total duration of working hours in 6 days $=40$ hours 30 minutes.

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Question 35 Marks

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

Answer

Door:
Width $($ breadth) $=1 m -100 cm$
$\text { height (length) }=2 m 50 cm$
$=250 cm$
Area of the door $=1 \times b$ sq.units
$=250 \times 100 cm ^2$
$=25000 cm ^2$
Bed:
$\text { length = } 2 m 20 cm$
$=220 cm$
width $($ breadth) $=90 cm$
Area of the bed $=I \times b$ sq.units
$=220 \times 90 cm ^2$
$=19800 cm ^2$
We can take the bed into the room.

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Question 45 Marks

A squirrel wants to eat the grains quickly. Help the Squirrel to find the shortest way to reach the grains. (Use your scale to measure length of the line segments)

Answer

The squirrel is in A and the grains are at E
The ways are
(i) A → B → C → D → E
(ii) A → G → F → K → E
(iii) A → H → I → J → E
Distance
(i) AB + BC + CD + DE = 2 cm + 2.5 cm + 2.5 cm + 2 cm
= 9 cm
(ii) AG + GF + FK + KE = (2.6 + 1.7 + 1.8 + 3) cm
= 9.1 cm
(iii) AH + HI + IJ + JE = 3 cm + 2.3 cm + 1 cm + 3.2 cm
= 9.5 cm.
$1^{st}$ way ABCDE is the shortest one.

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Question 55 Marks

Observe and Collect the data for a minute:

i.	Number of times a person breathes	vi.	Number of situps
ii.	Number of times heart beats	vii.	Number of claps
iii.	Number of times the eyes blink	viii.	Number of lines to write
iv.	Distance by walking	ix.	Number of lines to read
v.	Distance by running	x.	Number of Tamil verbs to say

Answer

i.	Number of times a person breathes	15	vi.	Number of situps	5
ii.	Number of times heart beats	50	vii.	Number of claps	100 m
iii.	Number of times the eyes blink	80	viii.	Number of lines to write	10
iv.	Distance by walking	50	ix.	Number of lines to read	200 m
v.	Distance by running	18	x.	Number of Tamil verbs to say	100

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Question 65 Marks

A doctor advised Vairavan to take one tablet every 6 hours once in the $1^{\text {st }}$ day and once every 8 hours on the $2^{\text {nd }}$ and $3^{\text {rd }}$ day. If he starts to take 9.30 a.m first dose. Prepare a time chart to take tablet in railway time

Answer

The first dose is taken at $9.30 a \cdot m =09: 30$ hours
Duration of every dose in $1^{\text {st }}$ day $=6$ hours
Duration of every dose in $2^{\text {nd }}$ and $3^{\text {rd }}$ day $=8$ hours
Time Chart.

	$1^{\text {st }}$ dose	$2^{\text {nd }}$ dose	$3^{\text {rd }}$ dose
I day	09:30 hours	15:30 hours	21:30 hours
II day	05:30 hours	13:30 hours	21:30 hours
III day	05:30 hours	13:30 hours	21:30 hours

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Question 75 Marks

Subtract 10 hours 20 minutes 35 seconds from 12 hours 18 minutes 40 seconds

Answer

$12 \text { hours } 18 \text { min } 40 \text { seconds }$
$=(12 \times 3600)+(18 \times 60)+40 \text { seconds }$
$=43200+1080+40 \text { seconds }$
$=44320 \text { seconds }$
10 hours $20 min 35$ seconds
$=(10 \times 3600)+(20 \times 60)+35 \text { seconds }$
$=36000+1200+35 \text { seconds }$
$=37235 \text { seconds }$
Difference:
$44320-37235$
$=7085$
$7085 \text { seconds }=(1 \times 3600)+3480+5 \text { seconds }$
$=1 \text { hour } 58 \text { minutes } 5 \text { seconds }$

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Question 85 Marks

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

Answer

Before Exams he practiced for 20 days
Days From 10.07.2017 to 31.03 .2018
July - 22 Days ...(From 10.7.2017)
August - 31 Days
September - 30 Days
October - 31 Days
November - 30 Days
December - 31 Days
January 2018 - 31 Days
February - 28 Days
March - 31 Days ...(upto 31.3.2018)
Total - 265 Days
Total number of days practiced $=$ Number of days practiced before exam + Number of days practiced after exam
$
=20+265=285 \text { days }
$
$\therefore$ He practiced for 285 days.

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