#Yr 4
81 messages · Page 1 of 1 (latest)
3 and 4 numbering are not visible
the two questions there
Grandchildren one and the rectangle paper one these are 3 and 4 right?
Yes
Ok let's start with 3 first
Let's see the 2nd part
i have as many sons and daughters together as one eighth of my age
number of sons = x
Number of daughters= y
Her age = 88
So
x+y = 1/8×(88)
x+y = 11```
Is this part clear?
yupp
Now first part
I have half as many grandchildren as I have daughter
And
Three fifth as many as I have sons
So we can write
Number of grandchildren=3/5(no of sons)=3x/5```
okyup
So we can write y/2 = 3x/5
Because both are same and denote no of grandchildren
So
y = 6x/5
okie
Substitute y=6x/5 in this equation
See this
3 x 5/5
15/5
3
Wait, why arent we substitutin the other equation?
What other equation?
Wait jokes we just simplified it
So we can sub 5 into any of the son and daughter equaton
5 is no of sons that is x
No of daughters is y, we did not find that because it's not require
But as x+y = 11
You can find y = 6
aHH YTEAH That makes sense
ID HAVE to write it liek this for the kid
okk
cooliesss
Lets do the next one
Unfortunately my work will start now so I have to go, do helpers ping and someone will assist you
Ok1! NO worries. Thank you aso much for the help :))
@alpine leaf You there ?.
yES
HGELO
Let's do 2nd one
area of small rectangle is 1/5 of that of original paper
Width of original paper=15
So area = 15×20
Area of small rectangle= 1/5(area of original)
1/5×(15×20) = 60```
Clear till this point?
Yup easy
Now for area to be 60 , you need to find all pairs of length and breadth such that length×breadth = 60
yupp
So pairs are
(1,60) (2,30) (3,20) (4,15) (5,12) (6,10)
Yep
Now as we know the smaller rectangle is made of the bigger paper , it's length and breadth can either be equal to bigger paper or less, but can't be greater than either length or breadth
Understood?
Yehh
Now we are left with these 4 (3,20) (4,15) (5,12) (6,10)```
okay
Now you gotta visualise a bit because as a teacher I need a blackboard or whiteboard, but try to understand
I think what's confusing is how do you know that 3x20 and 4x15 can fit 5 innit
Let's take the first pair (3,20), see the length is same as the original rectangle paper
So we can take how many small rectangle
(3,20)
(3,20)
(3,20)
(3,20)```
These will be stacked like
=====
=====
=====
=====```
No, that's for all small rectangle because all of them has area of 60
Yes but we need to find the pairs for which 4 of them can fit and 1 more can be of different size we don't care
Let me find a paper , maybe you will understand better that way
Wait Yes it makes sense
so thats why they chose 5 x 12 and 6 x10 because ther would be 4 full squares that can fit and the leftovers
Yessssss
4 size of same dimensions and leftover
the 3x20
For 3,20 and 4,15 it's 5 of same dimensions
Where does it come from
Our question restrict only 4 of same dimensions