SubscribeSaw some comments show up recently on our first scale balance puzzle (hard!), which reminded me how long its been since we put one of these up. With that in mind:
How many triangles are need to balance scale 3?
feel free to submit your answers in the comment section below. Will unmask Thursday, thanks.
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One !
one triangle
Comparing the first two pictures, 1 triangle = 2 circles = 4 squares since 1 circle = 2 squares.
So, the last picture has 2 squares and 1 circle which can be translated to 4 squares.
We need 1 triangle to balance this.
One Triange
______________
X - Triange
Y - Square
Z - Circle
1. X+2Y = 3Z
2. X+Z = 6Y
rearranged to find Triagle
3. X=3Z-2Y
4. X=6Y-Z
Solve for Z
5. 3Z-2y = 6Y-Z
6. 4Z=8Y
7. Z=2Y
Check 7. with scale 1.
1. X+2y = 3Z
2. X = 2Z
SO: X = 2Z= 4Y
Therefore:
Z+2y = ?
Z + Z = ?
2Z = X
QED
The weights are: box 1, ball 2, triangle 4
So one triangle should do the trick. Four boxes or 2 balls would work also.
Triangle= 4
Square = 1
Circle = 2
So on the first balance scale its 6-6
second is 6-6
and on the third its 4-4
So you will need a triangle on the third..
It is one triangle
From second balance one ball is equal to two squares.
So, one Triangle is equal to four squares.
For the third balance,
the total weight in the left hand side is 4 squares which is equal to one triangle.
Let x = Green triangle
y = Blue square
z = Orange circle
x + 2y = 3z -> (1)
x + z = 6y -> (2)
Eqn. (1) + (2) gives
2x + 2y + z = 3z + 6y
Therefore 2x = 2z + 4y
Or x = z + 2y ( Dividing throughout by 2 )
z + 2y is 1 Orange circle and 2 Blue squares
So to balance this we need
1 x or
1 Green triangle
After reading how the previous ones were done I’ve come up with…
In Balance 1, 1triangle (t) and 2 squares (2s) = 3circles (3c)
In Balance 2, 1triange (1) and 1 circle(c) = 6squares (6s)
and in Balance 3, 1 circle and 2 squares = x triangles (x(t))
Re-arranging B1 and B2 to make t the subject…
t = 3c - 2s and t = 6s - c
.: 3c - 2s = 6s - c
.: 4c = 8s
.: c = 2s
B3 can then be expressed as 2s + 2s = x(t)
.: x(t) = 2c
If 1 circle is equal to 2 squares you can take 1 circle from the right of B1 and 2 squares from the left of B1 and it will remain even.
.: 1t = 2c (Or 1 triangle balances with 2 circles)
Match this up with B3 and you get x = 1
.: it will take 1 triangle to balance scale 3
Testing with numbers… if 1t = 2c and 1c = 2s, then t and c must be even numbers. s=1, c=2 and t=4…
Balance 1 is 4+1+1 = 2+2+2 (correct)
Balance 2 is 4+2 = 1+1+1+1+1+1 (correct)
Balance 3 is 2+1+1 = 4 (correct)
Which confirms that 1 triangle will balance scale 3
1 Triangle will balance the scale.
Using the information given for Triangles (T), Squares (S) and Circles (C).
First scale T + 2S = 3C
Second one T + C = 6S which rearranging tells us that C=2S (one circle is 2 squares), and using that info and reinputing in the above that T=4S.
On the left of the 3rd scale there is 1 C + 2 S or also equiv 2S + 2S =4 squares which we calculated was 1 triangle
1 triangle = 2 circles = 4 squares
Therefore, on balance no. 3, 1 circle + 2 squares = 1 triangle
Only one green triangle is required on the rhs of balance 3. Explanation is as follows.
Scale equations are (using the first letters of Green, Blue & Amber):
G + 2B = 3A (B1)
G + A = 6B (B2)
2B + A = nG (B3)
Rearranging:
G = 3A - 2B (B1)
G = 6B - A (B2)
Subtracting these (B1 - B2) & rearranging:
4A = 8B
A = 2B
We can now change the eqns B1 & B2 to eliminate A & express G in terms of B:
3A - 2B = 6B - 2B = 4B = G (B1)
6B - A = 6B - 2B = 4B = G (B2)
(identical, as expected)
Lhs of balance 3 becomes:
2B + A = 2B + 2B = 4B
And we know from above that:
4B = G
So only one G is required on the rhs of B3.
1 triangle
triangle = t
square = s
circle = c
1t + 2s = 3c
1t + 1c = 6s
1c + 2s = ??*t
(1) => 1t + 2s = 3c => t = 3c - 2s (a)
(2) => 1t + 1c = 6s (a),(2)=> 3c - 2s +1c = 6s => 4c = 8s
so c = 2s (a) => t = 3c - 2s = 3*2s -2s = 4s
(3) 1c + 2s = 2s + 2s = 4s
and 1t = 4s
so the solution is 1 triangle
I got one triangle to balance the circle and two squares
t=triangle, s=square and c=circle
so from scale 1 we get the formula t+2s=3c and from scale 2 we have t+c=6s. From there we make t the subject of both formulas and get (Scale 1) t=3c-2s and (Scale 2) t=6s-c
joining these two equations i get
3c-2s=6s-c
3c+c=6s+2s
4c=8s
c= 2s and conversely s=c/2
if you then resubstitute these into the orginal scale 1 and scale 2 formulas you get
Scale 1
t+2s=3c
t+2s=3(2s)
t+2s=6s
t=4s or s=t/4
Scale 2
t+c=6s
t+c=6(c/2)
t+c=3c
t=2c or c=t/2
Scale 3 is c+2s, which is (in terms of t) = t/2+2(t/4)= t/2+t/2 = t
I’m sure i could’ve simplified that, couldn’t I?
2 squares equal 1 circle
2 circles equal 1 triangle
the answer —
1 triangle is needed to balance scale 3
Adding scale 1 & 2:
2 Triangles + 2 squares + 1 circle = 6 squares + 3 circles
=> 2 Triangles = 4 squares + 2 circles
=> 1 Triangle = 2 square + 1 circle ! Scale 3
I this this can be an answer:
From the first 2 scales: 1 circle= 2 squares;
1 triangle= 4 squares–> 1 triangle= 2 circles
So in the thid scale: 1 circle + 2 squares= 4 squares—–>
1 triangle, that I think is the answer
One triangle.. unless I made a mistake in my algebra
1
one ball equels 2 squares
one triangle equels 4 squares
therefore 2 squares and one ball(2 squares) equels one triangle
one triangle.
I just did a quick trial and error with the sequences of shapes and came up with this code:
2 Blue squares = 1 Orange circle
2 Orange circles = 1 green triangle
1 Green triangle = 4 Blue squares
3.)
2 blue squares = 1 orange circle;
Which leaves two orange circles, which equal one green triangle.
Therefore to balance the other side of the scale, place one green triangle.
1 triangle.
circles->c, triangles->t, and squares->s
3c=1t+2s and c+t=6s. Combining these and removing t, we get that 8s=4c, or 2s=c so we can consider the left side of scale three to be two circles. Plugging that into t+2s=3c, we get t+c=3c, which yields t=2c. That’s exactly how much we have on the left, so one triangle will balance it out.
One triangle is needed to balance scale 3.
Start by adding two triangles to the left-hand side. Then we have a triangle-plus-circle set and a triangle-plus-two-squares set. From scale one, we can add three circles to the right-hand side; from scale two, we can add six squares to the right-hand side, so scale three is balanced. Next, “cancel out” one circle and two squares from both sides. We’re left with two triangles balancing two sets of circle-plus-two-squares. So one triangle balances one circle plus two squares.
1
http://img510.imageshack.us/im.....no3nw1.jpg
Okay, I’ve come up with ..
1Triangle=2 Circles
1Square=.5Circles
1Circle=2Squares
1Triangle=4Squares
So, 1 Triangle would balance?
One Triangle.
1T + 2S = 3C
1T + 1C = 6S
1C + 2S = 1T
Only 1 Triangle.
1T + 2S = 3C
-1T + 6S = 1C
8S = 4C => 2S = 1C
1T + 1C = 6S => 1T = 4S
2S + 1C = 1T
Scale 1: 2 x S + T = 3 x C
Scale 2: T + C = 6 x S
Solving for C and S wrt T:
C=T/2
S=T/4
Scale 3: 2 x S + C = T/2 + T/2 = T
So one triangle balances scale 3
1 triangle.
1 triangle = 4 squares = 2 circles
1 Triangle = Answer. LOTS of good explanations above for this one, check them out
jasc,
That’s a neat way to look at it!
I got the same answer as you did, but had to do all the algebra with pencil and paper.
I’ll have to look at simpler ways to do things in the future. A picture really says a 1,000 words!