 # Question

There were 156 more silver balloons than gold balloons at a party. After 5/6 of the silver balloons and 3/4 of the gold balloons burst, there were 106 balloons left. How many balloons were there altogether at first ?

This is a Simultaneous Equation question.

 Bofore R1 left R2 R2 × 4 R1  + R2 Silver 6 1 4 10 Gold – 4 1 4 0 Balloon 156 106 424 580

10u = 580

1u = 580 ÷ 10 = 58

6u = 58 × 6 = 348 (silver)

348 – 156 = 192 (gold)

348 + 192 = 540 balloons

There are more Silver than Gold balloons.

So 6 silver – 4 gold = 156

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silver balloon : (12 units + 156) (at first)
(5/6) x (12 units + 156) = (10 units + 130) (burst)
(12 units + 156) – (10 units + 130) = 2 units + 26 (left)

gold balloons : 12 units (at first)
(3/4) x 12 units = 9 units (burst)
12 units – 9 units = 3 units (left)

2 units + 26 + 3 units = 5 units + 26 = 106
1 unit = (106 – 26)/5 = 16
12 units + 156 + 12 units = 24 units + 156 = 24 x 16 + 156 = 540

Ans : 540 balloons.

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of the 156 “more silver balloons”, 26 silver balloons are left after 5/6 burst.

106-26 = 80 balloons

of the remaining silver balloons, 1/6 are left (or 2/12).

of the gold balloons, 1/4 are left (or 3/12).

the sum of these two fractions = 5/12

Since 5/12 corresponds to 80 balloons, 12/12 corresponds to 192 balloons

(80 balloons divided by 5 multiplied by 12).

silver balloons = 192 + 156

gold balloons = 192

total balloons = 192 + 156 + 192 = 540 balloons

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