Annie, Ben and Claire are training for MATHCOUNTS. On January 1st, they each started to solve a large packet of problems that their coach gave them. Each of them worked at a constant rate. Annie solved six problems each day and finished four days after Ben did. Claire solved three more problems each day than Ben did and finished two days before he did. How many problems were in the packet their coach gave them?

Guest Jan 14, 2019

#1**+2 **

I had to solve this one graphically...

72 questions

Anne rate = 6 days = 12

Ben rate = 9 days = 8

Claire rate = 12 days = 6

ElectricPavlov Jan 15, 2019

#3**+1 **

Thanks, EP

Here's an algebraic solution :

Let the number of problems Ben solved per day = P

Let the number days that Ben takes = D

So Anne solved 6( D + 4) problems = 6D + 24 problems (1)

And Ben solved PD problems (2)

And Claire solved ( P + 3) ( D - 2) problems (3)

So using (2) and (3)

PD = (P + 3) (D - 2)

PD = PD + 3D - 2P - 6

3D = 2P + 6

P = [ 3D - 6 ] / 2 (4)

And subbing (4) into (2) and equating to (1) .....we have that

[ 3D - 6 ] ( D) / 2 = 6D + 24

[ 3D - 6 ] * D = 12D + 48

3D^2 - 6D = 12D + 48

3D^2 - 18D - 48 = 0

D^2 - 6D - 16 = 0 factor

(D - 8) ( D + 2) = 0

Since D is positive.....then D = 8

So....the number of problems is 6(8) + 24 = 72

CPhill Jan 15, 2019

#4**+1 **

Thanx, Chris.... those are the equations I used in my graph....but I got bogged down in the algabraic solution of them......

ElectricPavlov
Jan 15, 2019