Having a problem solving this, I can do basic algebra but stumped with this one. Any help appreciated.
https://ibb.co/GPfVh1R
Maths Problem Help
- Thread starter aumz
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Having a problem solving this, I can do basic algebra but stumped with this one. Any help appreciated.
https://ibb.co/GPfVh1R
from the scond line
2h = 142-t-s
from the third
2s= 154-2h
Therefore
h + t + 2s becomes h + t + (154-2h) which becomes h + t + 154 - (142-t-s) which becomes h + 2t + s +12
You know h + 2t + s = 99 therefore the answer is 99+12 = 111
2h = 142-t-s
from the third
2s= 154-2h
Therefore
h + t + 2s becomes h + t + (154-2h) which becomes h + t + 154 - (142-t-s) which becomes h + 2t + s +12
You know h + 2t + s = 99 therefore the answer is 99+12 = 111
Having a problem solving this, I can do basic algebra but stumped with this one. Any help appreciated.
https://ibb.co/GPfVh1R
https://math.libretexts.org/Bookshe...lities/909:_Solving_Systems_with_Cramers_Rule
You know 2h + 2s = 154 so halving it all
h+s = 77
Taking
h+ 2t+ s = 99 and subbing in h+s = 77 you get
2t+77=99
2t= 99-77
so 2t = 22
triangle = 11
Taking 2h + t + s = 142, subbing in t=11 and h+s = 77 you get
h+77+11 = 142
h = 142 - 77 - 11
heart = 54
h+s=77, subbing in h = 54
54 + s = 77
s = 77 - 54
smiley = 23
so ? =2s + t + h = 46 +11 + 54 = 111
? = 111
Since @Greebo also got 111, I'll assume it's all correct
These days no idea, IIRC we were doing something similar probably between year 9 and year 11, thought that was ~15 years ago now!
h+s = 77
Taking
h+ 2t+ s = 99 and subbing in h+s = 77 you get
2t+77=99
2t= 99-77
so 2t = 22
triangle = 11
Taking 2h + t + s = 142, subbing in t=11 and h+s = 77 you get
h+77+11 = 142
h = 142 - 77 - 11
heart = 54
h+s=77, subbing in h = 54
54 + s = 77
s = 77 - 54
smiley = 23
so ? =2s + t + h = 46 +11 + 54 = 111
? = 111
Since @Greebo also got 111, I'll assume it's all correct
Wouldn't know where to start and not I feel dumb!
What level of maths is this these days? A-Level?
These days no idea, IIRC we were doing something similar probably between year 9 and year 11, thought that was ~15 years ago now!
Soldato
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I did mine similarly to @loftie but not quite so efficiently
2h + 2s = 154
h + s = 77
2t + h + s = 99
2t + 77 = 99
2t = 22
t = 11
h + 2t + s = 99
h + 22 + s = 99
h + s = 77
h = 77 - s
2h + t + s = 142
2h + 11 + s = 142
2h + s = 131
2 (77 - s) + s = 131
-s = 131 - 154
s = 23
h + 2t + s = 99
h + 22 + 23 = 99
h = 99 - 22 - 23
h = 54
Therefore, 2s + t + h = x
46 + 11 + 54 = x
x = 111
I must admit, this took me longer than it should have because I made stupid little errors along the way.
2h + 2s = 154
h + s = 77
2t + h + s = 99
2t + 77 = 99
2t = 22
t = 11
h + 2t + s = 99
h + 22 + s = 99
h + s = 77
h = 77 - s
2h + t + s = 142
2h + 11 + s = 142
2h + s = 131
2 (77 - s) + s = 131
-s = 131 - 154
s = 23
h + 2t + s = 99
h + 22 + 23 = 99
h = 99 - 22 - 23
h = 54
Therefore, 2s + t + h = x
46 + 11 + 54 = x
x = 111
I must admit, this took me longer than it should have because I made stupid little errors along the way.
I think I started off by subbing h= 99 -2t - s or something similarly daft, then I realised that there was 2h + 2s = 154, face palmed and started there! Far easier
Edit : In hindsight, it probably would have been better for us lot to ask you how you'd approached it @aumz and give hints/help rather than just working it all out and showing you
Soldato
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I think I started off by subbing h= 99 -2t - s or something similarly daft, then I realised that there was 2h + 2s = 154, face palmed and started there! Far easier
Edit : In hindsight, it probably would have been better for us lot to ask you how you'd approached it @aumz and give hints/help rather than just working it all out and showing you
How else could we prove how good at basic maths we are if we don't answer for him?
Whilst you can solve these types of problems by setting up a system of simultaneous equations, I don’t think that is the aim of them. I think they are probably more designed to do in your head rather than on a piece of paper.
In this one, you’ve already noticed that there are only 3 pieces of information since many of the rows and columns are duplicates. Let’s call them C1 =142 (column 1), C2 = 99 (column 2) and R3 = 154 (row 3). Notice that R3 has no triangles so we can only create the right number of triangles by using C1 and C2. We want to end up with 1 triangle, so start off with C2 - C1 which leaves us with 1 triangle, -1 hearts and 0 faces and that is equal to 99-142=-43. Now we have no faces, but we need 2 faces so let’s add R3 and we end up with 1 triangle, 1 heart and 2 faces = -43 + 154 = 111. Which is what we wanted.
Of course this is the same as solving the simultaneous equations but you can do it in your head without needing to use a load of algebra.
In this one, you’ve already noticed that there are only 3 pieces of information since many of the rows and columns are duplicates. Let’s call them C1 =142 (column 1), C2 = 99 (column 2) and R3 = 154 (row 3). Notice that R3 has no triangles so we can only create the right number of triangles by using C1 and C2. We want to end up with 1 triangle, so start off with C2 - C1 which leaves us with 1 triangle, -1 hearts and 0 faces and that is equal to 99-142=-43. Now we have no faces, but we need 2 faces so let’s add R3 and we end up with 1 triangle, 1 heart and 2 faces = -43 + 154 = 111. Which is what we wanted.
Of course this is the same as solving the simultaneous equations but you can do it in your head without needing to use a load of algebra.
Associate
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Half a bottle of whiskey and I feel like I've entered the matrix
Wouldn't know where to start and not I feel dumb!
What level of maths is this these days? A-Level?
This was actually from an independant school year 7 entry test.
I think I started off by subbing h= 99 -2t - s or something similarly daft, then I realised that there was 2h + 2s = 154, face palmed and started there! Far easier
Edit : In hindsight, it probably would have been better for us lot to ask you how you'd approached it @aumz and give hints/help rather than just working it all out and showing you
No approach whatsoever as it just baffled me from the start.
Soldato
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111, no need for algebra. Add up right column numbers = 451. Subtract bottom row total to give ?
Year 6/7 depending on kids ability.
As mentioned - these type of questions are often used in 11+ exams.
Ah OK. Simultaneous equations are KS3, which is what I was going by because I fixated on that and completely missed....
^this.
Correct. Some folks are looking at the complicated way, but the total of the numbers on the right will equal the total of the numbers on the bottom, since both totals are just the summation of all the symbols. So take the total of the numbers on the right and subtract the total of the numbers on the bottom and BOOM! Five stars and a pass with honours for spotting that one.
there are 4 answers required, aren't there?
value of each of the symbols and the questionmark
Soldato
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True
..and some folks aren't reading the question. LOL! But I still give extra points to bloodiedathame for working out the ? the simplest way.