 Question

Dear All,

How to solve below problem simply with model diagram?

Boxes A, B and C contain 360g of sand altogether. 1/6 th of sand in BOX A is poured into Box B.

1/3 rd Box B is poured into Box C.

1/5th of Sand in Box C is poured into Box A.

There was an equal amount of sand in each box in the end. Find the amount of sand in each box at first.

Concept: Total amount of sand in the 3 boxes did not change after each transfer of sand from one box to another.

Strategy: Working backwards

At the end (After Transfer from C to A):

Each box –> 360g / 3 = 120g

Before Transfer from C to A:

4 units out of 5 units of C –> 120g

Amount of sand transferred to A –> 120g / 4 = 30g

Amount of sand in A –> 120g – 30g = 90g

Amount of sand in C –> 120g + 30g = 150g

Before Transfer from B to C:

2 units out of 3 units of B –> 120g

Amount of sand transferred to C –> 120g /2 = 60g

Amount of sand in C –> 150g – 60g = 90g

Amount of sand in B –> 120g + 60g = 180g

At first (Before Transfer from A to B):

5 units out of 6 units –> 90g

Amount of sand transferred to B –> 90g / 5 = 18g

Amount of sand in B –> 180g – 18g = 162g

Amount of sand in A –> 90g + 18g = 108g

Answer: A: 108g, B:162g and C: 90g

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Thank you so much

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360/3 = 120
Working backwards, before :
1/5 of sand in Box C is poured into Box A,
1 – 1/5 = 4/5 ——-> 120
5/5 ——-> 150 (Box C)
120 – (150 – 120) = 90 (Box A)

1/3 of Box B is poured into Box C,
1 – 1/3 = 2/3 ——-> 120
3/3 = 180 (Box B)
150 – (180 – 120) = 90 (Box C at first)

1/6 of sand in Box A is poured into Box B,
1 – 1/6 = 5/6 ——-> 90
6/6 ——-> 108 (Box A at first)
180 – (108 – 90) = 162 (Box B at first)

Ans : Box A — 108g; Box B — 162g and Box C — 90g.

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