Question

Mohan found a box full of 43 coins while searching for treasure. He decides to count the coins by putting them into four piles.
In the first pile, he put some coins.
The second pile has two less than the first pile.
The third pile has 1 less than the last pile.
The last pile had twice as many as the second pile.
How many coins were in each pile?​

Answer 1

Answer:

let x, x-2, 2x-5, 2x-4 be the coins in first pile, second pile, third pile and fourth pile respectively

Step-by-step explanation:

x + x-2 + 2x-5 + 2x-4 = 43

6x -11 = 43

6x = 54

x = 9

therefore,

1. the first pile has 9 coins
2. the second pile has = x-2 = 9-2 = 7
3. the third  pile has = 2x - 5 = 18 - 5 = 13
4. the last pile has = 2x - 4 = 18 - 4 = 14

Answer 2

Step-by-step explanation:

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