I cot the answer but i am not able to show my child how I got the answer

Question

Mrs. Carlyle bought a bag of peanuts for her children. child 1 took 1/3 of the bag. Child 2 took 1/4 of remaining peanuts. Child 3 took 1/2 of the remaining peanuts. Child 4 took 10 peanuts. there were 71 peanuts remaining in the bag. How many peanuts were originally in the bag?

Answer 1

Hi Joann,

              If we let P=the original number of peanuts then we can express what happened as:

 

Child 1 took 1/3P so 2/3P remained

Child 2 took 1/4of2/3P or 2/12P so 2/3P-2/12P or 1/2P remained

Child 3 took 1/2of1/2P or 1/4P so 1/2P-1/4P or 1/4P remained

Child 4 took 10 of the remaining and there were 71 left so 1/4P-10=71

 

Solving for P gives P=324 which is the original number of peanuts in the bag.

 

The key here is to account for the remaining peanuts at each step.

 

Jim

 

Answer 2

Say there are x peanuts in the bag.

 

Child 1 gets= x/3

Child 2 gets =(2x/3)*1/4= x/6

Child 3 gets =(x-x/3-x/4)*1/2=x/4

Child 4 gets = 10

x/3+x/6+x/4+10=3x/4+10

x-3x/4-10=x/4-10

x/4-10=71

x/4=81

x=324

No of peanuts in the bag= 324