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?
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.
To solve this problem, let's work through it step by step.
Step 1: Determine how many peanuts Preston took.
Preston took 10 peanuts.
Step 2: Determine how many peanuts were left after Preston took some.
After Preston took 10 peanuts, there were 71 peanuts remaining in the bag.
Step 3: Determine the fraction of peanuts that Brent took.
Brent took 1/2 of the remaining peanuts.
Step 4: Determine the total number of peanuts remaining before Brent took some.
Let X be the total number of peanuts remaining before Brent took some.
We have:
X - (1/2)X = 71 (since Brent took half of the remaining peanuts)
Simplifying the equation, we get:
(1/2)X = 71
X = 71 * 2
X = 142
Therefore, there were 142 peanuts remaining before Brent took some.
Step 5: Determine the fraction of peanuts that Joy took.
Joy took 1/4 of the remaining peanuts.
Step 6: Determine the total number of peanuts remaining before Joy took some.
Let Y be the total number of peanuts remaining before Joy took some.
We have:
Y - (1/4)Y = 142 (since Joy took one-fourth of the remaining peanuts)
Simplifying the equation, we get:
(3/4)Y = 142
Y = 142 * (4/3)
Y = 189.33
Therefore, there were approximately 189 peanuts remaining before Joy took some.
Step 7: Determine the fraction of peanuts that Philip took.
Philip took 1/3 of the peanuts.
Step 8: Determine the total number of peanuts remaining before Philip took some.
Let Z be the total number of peanuts remaining before Philip took some.
We have:
Z - (1/3)Z = 189.33 (since Philip took one-third of the peanuts)
Simplifying the equation, we get:
(2/3)Z = 189.33
Z = 189.33 * (3/2)
Z = 283
Therefore, there were 283 peanuts remaining before Philip took some.
Step 9: Determine the original number of peanuts in the bag.
The original number of peanuts in the bag is the sum of the peanuts taken by each child and the remaining peanuts:
Original number = Philip's peanuts + Joy's peanuts + Brent's peanuts + Preston's peanuts + Remaining peanuts
Original number = (1/3)Z + (1/4)Y + (1/2)X + 10 + 71
Original number = (1/3)(283) + (1/4)(189.33) + (1/2)(142) + 10 + 71
Original number ≈ 191.67 + 47.33 + 71 + 10 + 71
Original number ≈ 391
Therefore, there were approximately 391 peanuts originally in the bag.
Step 10: Determine how many peanuts each child took.
Philip took (1/3)(283) ≈ 94 peanuts.
Joy took (1/4)(189.33) ≈ 47 peanuts.
Brent took (1/2)(142) ≈ 71 peanuts.
Preston took 10 peanuts.
So, each child took the following number of peanuts:
Philip: 94 peanuts
Joy: 47 peanuts
Brent: 71 peanuts
Preston: 10 peanuts