# simulataneous equation problem

• May 2nd 2009, 01:13 AM
sammy28
simulataneous equation problem
hello,

A climbing centre has 14 fixed staff plus one additional staff member for every six climbers. There are twice as many climbers as staff. Find the number of

a climbers b staff

thanks
• May 2nd 2009, 02:05 AM
Lets say there were 6m climbers

so number of staff members = 14+m

But its given that

2(14+m) = 6m

28 + 2m = 6m

4m =28

Hence m = 7.
Thus
Spoiler:

a = 6m = 42

b = 14+7 = 21
• May 2nd 2009, 04:43 AM
sammy28

if we let a be climbers and b staff then

$a=2b$
$b=14 + \frac{a}{6}$

... i know your answer is correct (because the book says (Hi)), but im confused as to why statements like 'twice as many climbers than staff' gives

$a=2b$

and not

$b=2a$
• May 2nd 2009, 08:22 AM
VonNemo19
if a = climbers and b = staff and there are "twice as many staff as climbers" (keywords of the problem), then if we multiply the number of climbers by 2, then that's how many staff we have. Right? Hence, 2 times the number of climbers = the number of staff. Or better yet a=2b.
• May 5th 2009, 12:15 AM
Consider these things

a is twice of b
.
.

Take an example
6 is twice of 3

6 = 2 * 3

similarly

a = 2 * b

--this might help
try to think taking examples (Smile)