Examples: Properties of Whole Numbers Video Lecture - Class 6

hello friends this video on whole
numbers but 11 is brought to you by exam
fear calm no more fear from exam now
based on all these properties that we
have discussed so far let us now try to
look at some of the questions question
number 1 find the some by suitable
rearrangement so here we have been given
few 3 4 digit numbers and we have to add
them now we can add them normally also
but we need to add them smartly how by
making use of the properties that we
have learnt some time back because that
is the purpose of those properties those
properties are going to make our
calculations easier so let's see how do
we do this first of all what we do is we
rearrange the terms in a proper way
now rearrangement is allowed because we
have learnt about the commutative law
even if you interchange positions it
that overall thing remains the same
because here we are talking about
addition so let us rearrange the terms
so what we do let's keep 1 9 6 2 & 1 5 3
8 together why because both of them are
four-digit as well as when you look at
there once please you see that the 8
plus 2 will add up to form a number with
zero at the end so that will help our
calculations to become easy and we keep
the other two numbers together that is
453 and 647 and we have also put
brackets so that we perform these
operations first so 1 9 6 2 plus 1 5 3 8
is 3 5 0 0 and what about 4 5 3 & 6 4 7
that is 1 1 0 0 so therefore the result
is 4 6 0 0 so you see we could do the
sum in a more simple way by making use
of big properties question number 2 find
the product by suitable rearrangement
125 into 40 into 8 into 25 so here also
we will arrange it in such a way that
our calculations become easier now here
you see 125 into 40 might sound little
more tedious because on one side you
have
three digit number and on the other side
you have a two digit number so to make
it simpler what we do we will bring
eight here and we will take 40 here so
we will just change their positions that
is as per the commutative property so we
will make it 125 into 8 into 40 into 25
so now multiplying this is comparatively
simpler and this comes out to be 1,000
and this comes out to be 1,000 therefore
the result is 10 lakhs question number 3
find the value of the following 8 1 2 6
5 into 1 69 - 8 1 2 6 5 into 69 so
looking at this do you remember any
property is there any property that
comes to your mind
so the the nice thing about it is that
in both the terms you have eight
thousand eighty one thousand 265 as
something common now remember the
distributive property which says that a
into B plus C is equal to a into B plus
a into C now looking at this do you
remember something like this it is of
the form of a into B plus a into C is
just that here you have minus so what we
can do is by distributive property we
can take this common eight one two six
five and then you put inside 169 minus
69 so you see your calculation became
damn easy because otherwise you would
have to multiply these huge terms so
this multiplied it by 169 minus 69 would
be hundred so therefore it would be this
so eighty one lakh twenty six thousand
five hundred would be the answer
question number four find the product
using suitable properties 258 into one
thousand eight so here also we will make
use of distributive property so we can
write 1,000 a times 1000 plus 8 because
the moment we introduced terms with
zeros our calculations become simpler so
by distributive property this would be
258 into 1000 plus 258 into 8 so this
would be equal to 2 lakh 58,000 plus 250
8 into 8 would be 2 0 6 4 so when you
add this 2 you get 2 6 0 0 6 4 that is 2
lakh 60,000 64 question number 5 a
vendor supplies 32 liters of milk to a
hotel in the morning and 68 litres of
milk in the evening if the milk costs
rupees 15 per liter how much money is
due to the vendor per day ok so every
day the milk is delivered to the hotel 2
times so in 1 day once in the morning
and once in the evening so 2 times a day
a milk is being delivered
now what happens during the morning
during the morning 32 liters of milk is
delivered and the cost of 1 liter of
milk is 15 that means the cost of 32
liters of milk would be 32 into 15 now
in the evening 68 litres of milk is
delivered and cost of one liter is 15 so
cost of 68 litres would be 68 into 15
therefore what would be the total cost
for a day so the total cost for a day
would be 15 into 32 plus 15 into 68 can
you think of a way to simplify the
calculation yes by distributive law take
15 comments so 32 plus 68 so 32 plus 68
would be hundred so this can be written
as 15 into hundred so this would be
equal to rupees 1500 so this is the
money which is due per day because in a
day this is the total amount for which
milk is being delivered to the hotel
question number six mash the following
first one for 25 into 136 is equal to
425 into 6 plus 30 plus 100 which law is
this so this is definitely distributive
law because in distributive law we
distribute the number right so we have
broken it into three parts so this can
be written as four twenty five into six
plus four 25 into thirty plus four
twenty five in 200 second one two into
49 into 50 is equal to 2 into 15 to 49
what have we done we have done nothing
we have just exchanged to the position
so wherever we talk about exchange which
I'll talk about commutative property now
this involves multiplication so this is
cumulative 'ti of under multiplication
the last one eighty plus two thousand
five plus 20 is equal to 80 plus 20 plus
two thousand five here also you see that
just the positions have changed so that
is also community but this time it is
addition so computability under addition
thank you please visit exam fee or come
for free quality education you can learn
with a simple four-step learning process
wherein you can watch video lessons you
can ask your questions you can refer
notes and you can take a free online
test we have content for class six to
twelve on physics chemistry mathematics
and biology along with practical videos
so please subscribe to our channel for
daily updates thank you