joe has received 11% increase in salary. His new weekly earnings are $222. What was his prrevious weekly salary?
$197.58
2007-10-17 21:25:56 · 11 answers · asked by fana 1 in Science & Mathematics ➔ Mathematics
The answer is actually $200.
Just divide 222 by 1.11.
I see what you did. You took 11% of 222, then subracted it from 222. But if you add 11% to 197.58 you get $219.31, which does not match the weekly earnings.
2007-10-17 21:28:56 · answer #1 · answered by SoulDawg 4 UGA 6 · 0⤊ 1⤋
If the new weekly earnings is $222, divide it by 1.11 to get one hundred percent which was used as the base. This is also his previous salary. Dividing $222 by 1.11 = $200.00, meaning his previous salary was $200. ok?
2007-10-18 04:31:33 · answer #2 · answered by ron 2 · 1⤊ 1⤋
Absolutely right. If you consider e.g. $100 as the 11% increased salary, then the earlier salary was $ 89. Similarly, if $222 is the 11% increased salary, then the earlier salary was $197.58
2007-10-18 04:39:56 · answer #3 · answered by Anonymous · 0⤊ 2⤋
111%<--->$222
100%<--->(100 / 111) x $222 = $200
Previous salary was $200
2007-10-19 02:43:36 · answer #4 · answered by Como 7 · 0⤊ 0⤋
His previous weekly salary was $200.
X_new = X_old + 0.11*X_old
X_new =X_old*(1.11)
222=X_old*(1.11)
thus...
X_old = 222/1.11
X_old = 200
2007-10-18 04:48:19 · answer #5 · answered by cp_exit_105 4 · 0⤊ 1⤋
its $200....
let me tell you how....
Incrsd salary i.e. 111% of his old salary is $222
Old salary (100%) is 222*100/111 = $200
2007-10-18 04:35:44 · answer #6 · answered by ANJUM S 2 · 0⤊ 1⤋
The previous salary is x.
x * 1.11 = $222
Divide both sides by 1.11 and...
x = $200
2007-10-18 04:52:12 · answer #7 · answered by Richard F 6 · 0⤊ 1⤋
umm... my answer is $200
200 * .11 = 22
200 + 22 = 222
=)
2007-10-18 04:31:06 · answer #8 · answered by theresagarfield 3 · 1⤊ 1⤋
Its incredibly close, $197.58 * 1.11 = 219.3138
x*1.11 = 222
x = $200
2007-10-18 04:29:58 · answer #9 · answered by Anonymous · 1⤊ 1⤋
Yes, that's what I got too!!
2007-10-18 04:29:23 · answer #10 · answered by Anonymous · 0⤊ 2⤋