
By: shirley wee (offline) Wednesday, February 29 2012 @ 07:37 AM CST (Read 910 times)



shirley wee 
A normal chick has two legs, A lame chick has one leg. A sitting chick has no legs. 606 chicks have a total of 764 legs. The number of sitting chicks is half of the total number of normal chicks and lame chicks. How many chicks of each type are there?
Can anybody help me to solve this problem? Please show the steps on how do you solve it. Thanks!

Chatty
Registered: 02/28/12 Posts: 68





By: tlk (offline) Wednesday, February 29 2012 @ 08:58 AM CST



tlk 
Refer attached. I used guess & check method.

Newbie
Registered: 10/21/11 Posts: 14





By: shirley wee (offline) Wednesday, February 29 2012 @ 09:37 AM CST



shirley wee 

Chatty
Registered: 02/28/12 Posts: 68





By: w13158 (offline) Thursday, March 01 2012 @ 07:25 PM CST



w13158 
Sitting chicks > 1u
Normal and lame chicks >2u
3u>606
1u>202 (sitting chicks)
2u>404 (normal and lame chicks)
404 x 2= 808 (assume all chicks are normal, the no. of legs = 808)
808764=44 (take the assume no. of leg minus the total legs given, will you the no. of lame chicks)
40444=360 (this will give you the total no. of normal chicks with 2 legs)
Ans: sitting chicks 202
normal chicks 360
lame chicks 44

Newbie
Registered: 08/01/11 Posts: 8



