|
Question 1180388: Robert and Caleb each had a collection of rocks. Caleb had 2/9 as many rocks as Robert. When Robert lost 40 rocks, Caleb then had twice as many as Robert. How many rocks were in Robert’s collection at first?
Found 3 solutions by mananth, MathTherapy, ikleyn:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Robert and Caleb each had a collection of rocks. Caleb had 2/9 as many rocks as Robert. When Robert lost 40 rocks, Caleb then had twice as many as Robert. How many rocks were in Robert’s collection at first?
Let Robert have initially x rocks
Celeb will have 2x/9 rocks
When Robert lost 40 rocks he had x-40 rocks
Caleb then had twice as many as Robert
2x/9*2 = (x-40)
4x/9 = x-40
4x = 9x -360
5x =360
x= 72
Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website!
Robert and Caleb each had a collection of rocks. Caleb had 2/9 as many rocks as Robert. When Robert lost 40 rocks, Caleb then had twice as many as Robert. How many rocks were in Robert’s collection at first?
That PERSON is WRONG, AGAIN, as usual!
Why is this person allowed to continue to respond with RUBBBISH answers in this forum? Can't this be addressed?
Correct answer: Initial number in Robert's collection: 
Answer by ikleyn(52884)  (Show Source):
You can put this solution on YOUR website! .
Let x be the number of rocks Robert has initially.
Then Caleb's amount is  rocks.
After Robert lost 40 rocks, you have THIS EQUATION
= 2*(x-40)
Simplify and solve
2x = 9*2*(x-40)
2x = 18*(x-40)
2x = 18x - 720
720 = 18x - 2x
720 = 16x
x = 720/16 = 45.
ANSWER. Initially, Robert had 45 rocks.
Solved, answered and carefully explained.
-------------
@mananth setup the problem INCORRECTLY, so both his/her equation and the answer are WRONG.
For your safety, ignore his/her post.
|
|
|
| |
