# Help

• Aug 13th 2005, 11:19 PM
UKMUG
Help
:eek: I am having trouble with this one any help would be apprieciated.....

Two companies merge. Each company has high speed test machines as well as other test machines. One company has 400 chip test machines of which 20% are high speed. The combined company will have 44% high speed test machines. If the second company has 60% high speed test machines, then how many machines are there in the second company in total?
• Aug 14th 2005, 03:41 AM
rgep
You're told that company A has 400 machines of which 20% are high-speed: that is 80. Company B has X machines in total (this is the quantity you're being asked to determine) of which 60% are high-speed : that is, 60X/100. The total is 400+X machines of which 44% are high-speed: that is, 44(40+X)/100. But the total number of high-speed machines is also 80 (from company A) plus 60X/100 (from company B). So we have an equation 44(X+400)/100 = 80 + 60X/100. You now have to solve for X .
• Aug 20th 2005, 07:36 AM
UKMUG
Quote:

Originally Posted by rgep
You're told that company A has 400 machines of which 20% are high-speed: that is 80. Company B has X machines in total (this is the quantity you're being asked to determine) of which 60% are high-speed : that is, 60X/100. The total is 400+X machines of which 44% are high-speed: that is, 44(40+X)/100. But the total number of high-speed machines is also 80 (from company A) plus 60X/100 (from company B). So we have an equation 44(X+400)/100 = 80 + 60X/100. You now have to solve for X .

I still don't understand this one I keep getting lost when I try to solve
• Aug 20th 2005, 07:57 AM
44(X+400)/100 = 80 + 60X/100

Taking LCM on the right hand side gives you
( 8000+60X)/100

NOw since 100 is common on both sides,it can be cancelled out leaving you with

44(X+400)=8000 + 60X

Dividing throughout with 4

11(x+400) = 2000+15X

11X + 4400 = 2000+15X

15X-11X= 4400-2000

4X= 2400

X= 600