Percentage Help

Lets approach this with nice numbers:

Of the families in City X in 1994, 40 percent owned a personal computer:

lets say there are a total of 100 families and 40 of them own a computer.

in 1998 that num was 30 percent greater than it was in 1994, or 130% of 40
1.3*40=52


total number of families in City X was 4 percent greater in 1998 than it was in 1994 or 104% of our 100 families = 104 families

52 of those owned a computer which is 50% Answer A

Many thanks

yup ! thats how i got it

Just a thought - could there be any other approach whatsoever???

bhumika.k.shah wrote:yup ! thats how i got it

Just a thought - could there be any other approach whatsoever???

usually putting numbers is an easier approach in these kind of problems but I've one more

say X was population in '94 and Y in '98

so in '94 people having PC = 40/100*x = 0.4x

in '98 population was 4% greater than that in '94 so Y = X + 4/100*x = 1.04X

and in '98 number of PC owners was 30% greater than that in '94 which will be .40X + 30/100(.40*x) = .40X + .12X = .52x

now % of PC owners in '98 in terms of population in '98 = [.52X/1.04 x] * 100 = 1/2 * 100 = 50%

xilef wrote:Lets approach this with nice numbers:

Of the families in City X in 1994, 40 percent owned a personal computer:

lets say there are a total of 100 families and 40 of them own a computer.

in 1998 that num was 30 percent greater than it was in 1994, or 130% of 40
1.3*40=52


total number of families in City X was 4 percent greater in 1998 than it was in 1994 or 104% of our 100 families = 104 families

52 of those owned a computer which is 50% Answer A

That's how I did it and is the easiest way I could think of.