1 st one

21 austr = 12 with exp + 9 without exp

U need to form a group of 3 from 2 without exp and 1 with exp

Number of combinations of 2 people from 9 without exp is
2C9=36

for each of these groups u have 12 austr with exp, so u have
36*12=432 different groups

Q2
In May Mrs. Lee's earnings were 60 percent of the Lee family's total income. In June Mrs. Lee earned 20 percent more than in May. If the rest of the family's income was the same both months, then, in June, Mrs. Lee's earnings were approximately what percent of the Lee family's total income?


May
Lee = 0,6 of total income = 0,6T

June
Lee = 0,6*1,2 = 0,72T

Difference = 0,72T-0,6T=0,12T

so total income in June was 1,12T

0,72T/1,12T = approx. 0,64 or 64%



Q3
A certain meter records voltage between 0 volts and 10 volts, inclusive. If the average (arithmetic mean) value of 3 recordings on the meter was 8 volts, what was the smallest possible recording, in volts?

measures A, B and C have average 8
the summ of measures was 8*3=24

to minimize one u need to maximaze 2 others, let A and B have maximun measures. U know that maximum measure is 10, so
A=10
B=10
C can be only 4 to make the summ 24
Answer 4

thanks u guys

4meonly wrote:Q2
In May Mrs. Lee's earnings were 60 percent of the Lee family's total income. In June Mrs. Lee earned 20 percent more than in May. If the rest of the family's income was the same both months, then, in June, Mrs. Lee's earnings were approximately what percent of the Lee family's total income?


May
Lee = 0,6 of total income = 0,6T

June
Lee = 0,6*1,2 = 0,72T

Difference = 0,72T-0,6T=0,12T

so total income in June was 1,12T

0,72T/1,12T = approx. 0,64 or 64%

Can someone explain this problem again. I am confused why we calculated the difference [Difference = 0,72T-0,6T=0,12T]
Also it is given that " If the rest of the family's income was the same both months" then for both months rest of family should have T as income and for both months it should be 2T

Imagine:
In May Mrs. Lee earned 60$, rest of her family - 40$, total 100
In May Mrs. Lee earned 60$*1,2=72$, rest of her family - 40$ (the rest of the family's income was the same both months), total 112$
72$/112$=64%

4meonly wrote:Imagine:
In May Mrs. Lee earned 60$, rest of her family - 40$, total 100
In May Mrs. Lee earned 60$*1,2=72$, rest of her family - 40$ (the rest of the family's income was the same both months), total 112$
72$/112$=64%

Got it.......
Thanks

Q3
A certain meter records voltage between 0 volts and 10 volts, inclusive. If the average (arithmetic mean) value of 3 recordings on the meter was 8 volts, what was the smallest possible recording, in volts?

i did such silly mistake here - picked 2 as answer and then saw the meter can read between 0 - 10 !!

o heaven ! it hurts !

4meonly wrote:Imagine:
In May Mrs. Lee earned 60$, rest of her family - 40$, total 100
In May Mrs. Lee earned 60$*1,2=72$, rest of her family - 40$ (the rest of the family's income was the same both months), total 112$
72$/112$=64%

So the above makes sense, but I still don't understand the original solution below where Mrs. Lee's earnings in May are subtracted from Mrs. Lee's earning in June and then 1T is added to get 1.12T. Could you explain why those steps are taken?

4meonly wrote:Difference = 0,72T-0,6T=0,12T

so total income in June was 1,12T

aj5105 wrote:Q3
A certain meter records voltage between 0 volts and 10 volts, inclusive. If the average (arithmetic mean) value of 3 recordings on the meter was 8 volts, what was the smallest possible recording, in volts?

i did such silly mistake here - picked 2 as answer and then saw the meter can read between 0 - 10 !!

o heaven ! it hurts !

I forgot the restriction .. 0 - 10

Great explanations for the first two PB.

When I see the type of problem/combination with a large group of people/objects that requires specific qualities I think of each specific group separately:

1. 12 with experience
2. 9 no experience

Now, we need a group of 3 people: we want one with experience and 2 w/o experience

12C1 * 9C2 =12*36=442