Mrs. Carlyle bought a bag of peanuts for her children. when Philip Joy Brent and Preston came home from school and they each took some peanuts from the bag. Phillip took 1/3 of peanuts from the bag Joy took 1/4 of the remaining peanuts. Brent 1/2 of the remaining peanuts. Preston took 10 peanuts there 71 peanuts remaining in the bag. how many peanuts were originally in the bag? how many peanuts in each child take?

Answer 1

Let's denote the original number of peanuts in the bag as "x".

Philip took 1/3 of x = x/3 peanuts.
The remaining peanuts in the bag after Philip took his share is (x - x/3) = 2x/3 peanuts.

Joy took 1/4 of the remaining peanuts, which is (1/4) * (2x/3) = x/6 peanuts.
The remaining peanuts in the bag after Joy took her share is (2x/3 - x/6) = 4x/6 - x/6 = 3x/6 = x/2 peanuts.

Brent took 1/2 of the remaining peanuts, which is (1/2) * (x/2) = x/4 peanuts.
The remaining peanuts in the bag after Brent took his share is (x/2 - x/4) = 2x/4 - x/4 = x/4 peanuts.

After Preston took 10 peanuts, there were (x/4 - 10) peanuts remaining in the bag.

We are given that there were 71 peanuts remaining, so we can set up the following equation:
x/4 - 10 = 71

Adding 10 to both sides of the equation:
x/4 = 81

Multiplying both sides of the equation by 4:
x = 81 * 4 = 324

Therefore, originally there were 324 peanuts in the bag.

Now, let's find out how many peanuts each child took:
- Philip took 1/3 of 324 = (1/3) * 324 = 108 peanuts.
- Joy took 1/4 of the remaining peanuts (216) = (1/4) * 216 = 54 peanuts.
- Brent took 1/2 of the remaining peanuts (162) = (1/2) * 162 = 81 peanuts.
- Preston took 10 peanuts.

Therefore:
Philip took 108 peanuts,
Joy took 54 peanuts,
Brent took 81 peanuts, and
Preston took 10 peanuts.