Three children picked 345 apples. Together, the first and second child picked 200 apples, with the second child picking 57 fewer than the third.

1 answer

Let's denote the number of apples picked by the first child as x, the number picked by the second child as y, and the number picked by the third child as z.

We know that:
x + y + z = 345 (equation 1)
x + y = 200 (equation 2)
y = z - 57 (equation 3)

Substitute equation 2 and 3 into equation 1:
x + z - 57 + z = 345
x + 2z = 402
x = 402 - 2z

Substitute x back into equation 2:
402 - 2z + y = 200
y = -202 + 2z (equation 4)

Substitute back y into equation 1:
402 - 2z + (-202 + 2z) + z = 345
402 + z = 345
z = 345 - 402
z = 57

Now, substitute z back into equation 4:
y = -202 + 2(57)
y = -202 + 114
y = 88

Therefore, the number of apples picked by the first child (x) is:
x = 200 - 88
x = 112

So, the first child picked 112 apples, the second child picked 88 apples, and the third child picked 57 apples.