Good Morning everyone! 😀
We are back to the normal maths challenges today! Although, if you enjoyed yesterday’s challenge please tell me and I can post some more of those sometimes!
Well done for having a go at yesterday’s challenge, it was a tricky one! You all did so well to get so close to the target number of 683 and some of you even got exactly 683!
This was my answer:
6 x 3 = 18
18 + 2 = 20
25 + 9 = 34
34 x 20 = 680
680 + 3 = 683
……………………⭐️⭐️⭐️……………………
A big well done to these children for reaching the number 683:
Albert, Sophie, Martin, Ajay, Mayowa, Rafael, Vincent, Clare, Pablo, Holly, William, Paolo and Tijne.
However, some of you did not use all of the numbers: 25, 3, 6, 9, 2 and 3.
The first person to reach 683 AND use all of the numbers was…
Martin!
……………………⭐️⭐️⭐️……………………
And now for today’s maths challenge:
Good luck! 😀
From Miss Lee
Pile B = x
Pile A = x + 4
Pile C = x + 1
Pile D = 2x
Altogether, they add up to 20 coins:
x + x + 4 + x + 1 + 2x = 20
5x + 5 = 20
5x = 15
x = 3
Therefore,
Pile A = 7 coins
Pile B = 3 coins
Pile C = 4 coins
Pile D = 6 coins
Good morning Miss Lee,
The answer to these questions are:
pile 1 has 7 , pile 2 has 3
pile 3 has 4 , pile 4 has 6 🙂
First pile, 7
Second pile, 3
Third pile, 4
Forth pile, 6
The first pile has 7 coins, the second pile has 3 coins, the third pile has 4 coins and the fourth pile has 6 coins.
First pile-7 coins
Second pile-3 coins
Third pile-4 coins
Fourth pile-6 coins
My answers are :
The first pile had 7 coins.
The second pile had 3 coins.
The third pile had 4 coins.
The fourth pile had 6 coins.
Does anybody have a different answer?
7,3,4,6
a=b+4
b=c-1
c
d=bx2
if c=4 then
b=4-1=3
a=3+4=7
d=3×2=6
We liked the yesterday’s one!
1st pile has 10 coins
2nd pile has 2 coins
3rd pile has 4 coins
4th pile has 4 coins
That is great to hear Martin!
https://nrich.maths.org/6499
On this link you can have a go at a similar activity!
Good morning miss Lee,
Here’s my answer.
Pile 1 = 7 coins
Pile 2 = 3 coins
Pile 3 = 4 coins
Pile 4 = 6 coins
This makes a total of 20 coins.
Thanks, Tijne
Hi Miss Lee! My answer is:
Pile #1 has 7 coins, pile #2 has 3 coins, pile #3 has 4 coins and pile #4 has 6 coins ?. This is because :
3+4=7
3+1=4
3×2=6
3+7+4+6=20
(She has loads of money ?????????!)
Good morning Miss Lee ?
Here is my answer:
7+3+4+6=20
Thank you
Morning,
Pile 1= 7 coins ( 4 more than pile 2 which has 3 )
Pile 2=3 coins ( 1 coin less than pile 3 which has 4 )
Pile 3=4 coins
Pile 4=6 coins ( twice as much as the second pile 3X2=6)
Total =20 coins
Hi Miss Lee!
Here are my answers…
pile 1. 7 coins
pile 2. 3 coins
pile 3. 4 coins
pile 4. 6 coins
Hope it is correct??
Pile 1 = 7 gold coins
Pile 2 = 3 gold coins
Pile 3 = 4 gold coins
Pile 4 = 6 gold coins
7+3+4+6=20
Good afternoon Miss Lee,
My answer for
Queen Edmerelder:
1st
=7 coins
2nd
=3 coins
3rd
=4 coins
4th
=6 coins
The first pile has 7 coins
And the second pile has 3
The third pile has 4
The fourth pile has 6 piles
Hello Miss Lee and classmates(year 4)!
My answer for today’s math challenge is…
in pile 1 we have 7 coins
and then in pile 2 we have 3 coins
and last but not least, in pile 4 Esmerelda has 6 coins!!!
🙂 thank you miss I miss you guys so
much!!
I am going to explain every pile using the third one that I will call T:
First = T-1+4
Second = T-1
Third = T
Fourth = T-1+T-1
There are 5 T and zero numbers therefore T=20/5=4
First = 7
Second = 3
Third = 4
Fourth = 6
7+3+4+6=20