YEAR 1:
ART 50,000
ARI 25,000 + 27,500 = 52,500
ARI made 2,500 more
YEAR 2:
ART 60,000
ARI 30,000 + 32,500 = $62,500
ARI made 2,500 more this year as well
YEAR 3:
ART 70,000
ARI 35,000 + 37,5000 = 72,500
ARI made 2,500 more than ART this year
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YEAR 1:
ART 50,000
ARI 25,000 + 27,500 = 52,500
ARI made 2,500 more
YEAR 2:
ART 60,000
ARI 30,000 + 32,500 = $62,500
ARI made 2,500 more this year as well
YEAR 3:
ART 70,000
ARI 35,000 + 37,5000 = 72,500
ARI made 2,500 more than ART this year
There.Quote:
Originally Posted by 3fingersalute
3 points plus a bonus for difficulty. ;)
bloody hell, i always had this problem with brain teasers. read the entire bloody question.
:thumbs2:Quote:
Originally Posted by a d e p t
Coolness............now I need to stop for a :drink: after work cause my head hurts from thinking so much.
of course I made a mistake, these are their saleries per yearQuote:
Originally Posted by OskarTheSmurf
so after year 1:
Base Pay:
Art made 50000 Ari made 50000 + 1250 from the raise (was in middle of year) so Ari made more in year one by 1250
Year 2:
Base pay
Art 60000 Ari 55000
final: Art made 60000 Ari made 56250 (again half a year of the 2500 a year raise) so Art made more by 3750 in year 2.
Year 3:
Base Pay
Art 70000 Ari 60000
Art made 70000 this year, Ari made 66250 giving a difference of 8750
Am I at least getting closer?
Sort of.
But 3fingersalute already got it... Sorry.
Suppose you have an open topped box that is 5cm x 5cm x 5cm. (capacity: 125 cubic centimetres)
Inside the box is a steel ball-bearing that is 25 cubic centimetres in size.
Next to the box is a two litre pail filled with mercury.
How many cubic centimetres of mercury would you have to pour into the box to completely submerge the ball bearing?
__________
Mercury is denser than steel and the ball would float atop the liquid metal.
That depends...Quote:
Originally Posted by a d e p t
Does a steal ball bearing float in mercury?
This question is not one where we can ask for clues, sorry bro.
volume is 25cm cubed so the radius is 1.8cm (diameter 2.9cm therefore the vloume of mercury required to submerge the bearing would be 5X5X2.9 - 25 = 72.5 cm cubed. although mercury should be refered to in mL not centimeters, and like 3fs said, assuming the ball bearing has a greated density than mercury.
Psst.
That was RIOT, not 3fingersalute. ;)
my bad
Alright then, the ball bearing will float in the mercury and therefore, can never be covered.Quote:
Originally Posted by a d e p t
Quote:
Originally Posted by RIOT
Cool.
3.
Hey technical objection :p - I knew for sure that steel was less dense than mercury & wouldn't have had to try for clues ('cos he got one despite your objection) one quick squizz at the periodic table on the back of my 'junior chemist' ruler told me all I needed to know.. :DQuote:
Originally Posted by a d e p t