Question

Four balances are shown where the two sides of each contain exactly the same amount of weight. The first balance has 4 squares, 1 triangle and 1 ball on the left side, right side has 2 diamonds. The second balance has 1 triangle and 1 square on the left side, 1 ball on the right side. The third balance has 1 triangle and 1 diamond on the left side, 1 ball and 2 squares on the right side. The fourth triangle has 1 diamond, 1 triangle and 1 square on the left side. How many balls are on the right side? Explain/show how this was determined.

Answer 1

balance 1: 4S+1T+1B=2D
balance 2: 1S+1T=1B
banace 3: 1T+1D=1B+2S
balance 4: 1S+1T+1D= XXXX B

solve for XXXX

now, rearranging the four equations in order of BDST

a) 1B-2D+4S+T=0
b) -1B+0D+1S+T=0
c) -1B+D-2S+1T=0
d) XXXB+D+S+T=0

subtract d from b
(1-xx)B+D=0
add a + C
-D+2S+2T=0
now add these two equations
(1-xx)B+2S+2T=0 , now subtract 2*b from this
(xx-1+2)B =0

which means xxx=1

check my work on this, it is easy to make an error.

Answer 2

I got 3B on the last scale