We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
118
4
avatar

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?

 Jan 14, 2019
 #1
avatar+18071 
+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

 Jan 15, 2019
 #2
avatar+18071 
+1

Here is the graph I used     x = days   y = rate

 

 Jan 15, 2019
 #3
avatar+99585 
+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

 

 

cool cool cool

 Jan 15, 2019
 #4
avatar+18071 
+1

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

ElectricPavlov  Jan 15, 2019

6 Online Users

avatar