balancing scales

**aquajam** · December 17th 2009, 10:13 PM

Imagine the letters(symbols) below to sit on scales.
Scale 1 and 2 are perfectly balanced.

Take D = diamonds H = Heart S = spades.

How many diamonds are needed to balance scale 3?

1. D H ........ S S S S

2. S S D D H ........ H H

3. H S S ........ ?

I get the same answer using both logic and algebra, yet it conflicts with the answer I have been given for this.

Please try!

**Soroban** · December 18th 2009, 08:29 AM

Hello, aquajam!

Imagine the symbols below to sit on balance scales.

Scales 1 and 2 are perfectly balanced.

. . $[1]\quad\begin{array}{ccccc} \diamondsuit\:\heartsuit && \spadesuit\:\spadesuit\:\spadesuit\:\spadesuit\\ \hline & \Delta \end{array}$

. . $[2]\quad\begin{array}{ccccc} \spadesuit\:\spadesuit\:\diamondsuit\:\diamondsuit \:\heartsuit && \heartsuit\:\heartsuit\ \\ \hline & \Delta \end{array}$

. . $[3]\;\;\begin{array}{ccccc} \heartsuit\:\spadesuit\:\spadesuit && ?\,\diamondsuit \\ \hline & \Delta \end{array}$

How many diamonds are needed to balance [3]

From [1], we have: . $D + H \:=\:4S \quad\Rightarrow\quad H \:=\:4S - D\;\;[4]$

From [2]. we have: . $2S + 2D + H \:=\:2H \quad\Rightarrow\quad H \:=\:2S + 2D\;\;[5]$

Equate [4] and [5]: . $4S - D \:=\:2S + 2D \quad\Rightarrow\quad2S \:=\:3D \quad\Rightarrow\quad S \:=\:\tfrac{3}{2}D$

Substitute into [4]: . $H \:=\:4\left(\tfrac{3}{2}D\right) - D \quad\Rightarrow\quad H \:=\:5D$

Substitute into [3]:

. . . . . $H + 2S \;=\;?\,D$

. . $5D + 2\left(\tfrac{3}{2}D\right) \;=\;?\,D$

. . . . . . . $8D \;=\;?\,D$

Therefore, 8 Diamonds are needed.

**Wilmer** · December 18th 2009, 08:56 AM

8 is obviously the answer.

**aquajam** · December 18th 2009, 04:07 PM

Yeah I got 8 as well.

The answer given in the book is 3. Must be a misprint.

**ivankiel23** · July 27th 2012, 07:47 AM

Hahahaha aquajam? You're the man

Everybody did get the answer well done.

Yeah I got 8 as well.

The answer given in the book is 3. Must be a misprint.

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**davidsmith** · August 2nd 2012, 09:06 AM

Bright minds can easily answer this...:-)

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**acabus** · August 2nd 2012, 01:17 PM

If, like some newspaper puzzles I've seen, the force is weight multiplied by distance from the pivot, then the answer is 21, as calculated here.

**hua052011** · August 8th 2012, 11:27 PM

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**hua052011** · August 15th 2012, 07:12 PM

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