Mrs. Carlyle bought a bag of nuts for her children. Phillip, Joy, Brent, and Preston took some from the bag. Phillip took 1/3. Joy took 1/4 of the remaining nuts. Brent took 1/2 of the remaining peanuts. Preston took 10. There were 71 remaining in the bag. How many peanuts were oringinally in the bag? How many peanuts did each child take?

Originally there was 324 nuts in the bag.

Phillip took 108, Joy took 54, Brent took 81 and Preston took 10.

Step-by-step explanation:

Let's call the total amount of nuts N, and the number of nuts each child took by the initial of their name (Phillip will be P1 and Preston will be P2).

So we can write the following equations:

P1 = N/3

After removing N/3, the remaining nuts is (N - N/3):

J = (N - N/3)/4 = (2N/3)/4 = N/6

After removing N/6 from (N - N/3), the remaining nuts is (N - N/3 - N/6):

B = (N - N/3 - N/6)/2 = (N/2)/2 = N/4

P2 = 10

In the final there were 71 nuts remaining, so we have that:

N - N/3 - N/6 - N/4 - 10 = 71

N - N/3 - N/6 - N/4 = 81

N/4 = 81

N = 324 nuts

The amount of nuts took by each child is:

P1 = N/3 = 108 nuts

J = N/6 = 54 nuts

B = N/4 = 81 nuts

P2 = 10 nuts

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