Balancing Problem - Part 4

$1P = 1$ Pentagon

$1C = 1$ Cloud

$1S = 1$ Star

We can write the following equations:

$2P = 2C + 1S$

$1C + 1P + 1S = 1C + 3S$

Now, we can rewrite those equations to:

$2P = 2C + 1S \Leftrightarrow \boxed{4P = 4C + 2S}$

$1C + 1P + 1S = 1C + 3S \Leftrightarrow 1P + 1S = 3S \Leftrightarrow \boxed{1P = 2S}$

Now, let's put the $2S = 1P$ from the second equation into the first equation:

$4 P = 4C + 2S \Leftrightarrow 4P = 4C + 1P \Leftrightarrow 3P = 4C$

Thus, we need $3$ Pentagons to balance $4$ Clouds.

Let pentagon $= p$ , cloud $=c$ and star $=s$

Then, $2p = 2c+s$ ........ (1)

& $c+p+s=3s+c$

$=> p+s=3s$

$=> s=\frac{p}{2}$

put this value in equation (1)

$2p=2c+\frac{p}{2}$

$2c=\frac{3p}{2}$

Multiply by 2

$\boxed{4c=3p}$

x=pentagon , y=cloud ,z=star

Now from figure

2x=2y+z --------> (1)

and

x+y+z=y+3z

therefore, x=2z

.Multiplying equation (1) by 2 and substituting x=2z

we get 4y+x=4x

Thus 4y=3x

Thus Four Clouds Equal to 3 Pentagon :D

let P=1 penta, C=1cloud, S=1star

translate the first and second balance into an eqation; 2P=2C+S (1st balance) S+P+C=C+3S (2nd balance)

combine similar terms in 2nd equation, S+P+C=C+3S P=2S S=P/2 (substitute to the 1st equation)

2P=2C+(P/2) 3P/2=2C 3P=4C

3 Pentagon is needed to balance 4 clouds

We can simply solve it from the condition of the first digram .

weight of 2penta = weight of ( 2cloud + 1 star) this is first condition.

for the the required 4 clouds ,

using condition 1. let 1star = 1cloud .

that is 2penta= 2cloud + 1cloud 2penta= 3clouds

as there are 4 clouds so : adding 1 weight on both sides 1penta + 2penta = 3cloud + 1cloud that is

            3penta = 4clouds

hence penta for 4 clouds = 3 Ans.

Since you need 1 star along with 2 clouds to balance the beam with 2 pentagons on the other side, clouds are LIGHTER! So, for 4 clouds you'll need 3 pentagons.