1. h is amount that Harry spent, then 2h+h+2h+40 = 3(h+60)
5h+40=3h+180
2h=140
h=70
2. the prob. that no man selected is 6/10*5/9*4/8=120/720
the prob at least one man selected is 1-120/720=600/720=5/6
1.Tom, brik and Harry agree to chip in to buy a television. Tom agrees to spend twice as much as Harry, and brik agrees to spend $40 more than Tom. If they had all chipped in the same amount, Harry would have had to put in $60 more than he does. How much does Harry spend on the television?
A) $50
B) $60
C) $70
D) $80
E) $100
2.Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least one of the people chosen is a man?
A) 1/6B)1/3 C)2/3 D) 4/5E) 5/6
Thanks apo!!
Need some light on the below prob as well.Any shortcut to solve?
a merchant sells 3 diff sizes of canned tOMATOES. A large can costs as much as 5 medium cans or 7 small cans. If a customer buys an equal number of small and large cans of tomatoes for th eexact amount of money that would buy 200 medium cans, how many small cans will she buy??
a. 35
b. 45
c. 72
d. 199
e. 208
Last edited by abhishek1787; 06-27-2012 at 06:20 AM. Reason: correction
little algebra helped me to get the answer.
Let L be the cost of large can
M=number of medium sized can.
S=number of small sized can.
then 5M=7S=L
200 medium sized can costs 5*200=$1000.(where 5 is cost of each medium sized can)
deriving the equation which says -"customer buys an equal number of small and large cans of tomatoes for th eexact amount of money that would buy 200 medium cans".
5M+7S=1000;solve for M and S by plugging in the given options for S.
S=35 is the answer.Let me know if i am correct.
a is the number of large cans
b is the number of medium cans
c is the number of small cans
pa is the price of a large can
pb is the price of a medium can
pc is the price of a small can
Then,
pa=5pb
pa=7pc
pc=5pb/7
a*pa+a*pc=200*pb
then,
a*5pb+a*5pb/7=200*pb
a*5+a*5/7=200
40*a/7=200
a/7=5
a=35
Wow, this is a tough question. I got 56. Is that right?
This was my method:
Just to get them out of the way: 5, 50, 500, and 5000 = 4 possibilities.
Then:
2 digits: Use the pairs {2,3} and {1,4}. Each has 2 orders = 4 possibilites
3 digits: Use the sets:
{2,3,0} = 4 possibilities (don't count 0 as first digit)
{4,1,0} = 4 possibilities (don't count 0 as first digit)
{1,1,3} = 3 possibilities (don't count the two 1s as different)
{2,2,1} = 3 possibilities (don't count the two 2s as different)
= 14 possibilites
4 digits: Use the sets:
Use the sets:
{2,3,0,0} = 6 possibilities (no 0 as first digit and don't count two 0s as different)
{4,1,0,0} = 6 possibilities (no 0 as first digit and don't count two 0s as different)
{1,1,3,0} = 9 possibilities (no 0 as first digit and don't count two 1s as different)
{2,1,1,1} = 4 possibilities (don't count three 1s as different)
{2,2,0,1} = 9 possibilites (no 0 as first digit and don't count two 2s as different)
= 34 possibilites
So adding these all up that's 4 + 4 + 14 + 34 = 56 total possibilities
Did I forget any?
Last edited by thepillow; 08-03-2012 at 09:51 PM.
Actually, I think I just wasted a whole lot of time. After seeing that the answer is 56 it occurred to me that 56 is the same as 8c3, which makes sense:
___ ___ ___ ___ ___ ___ ___ ___
There are 8 empty slots above and we're looking for a number (4 digits or less) where all the digits together add to 5. If we consider each of the 8 slots to have a value of 1 then all we need to do is choose 3 out of those 8 to cross out and we'll be left with 5. We can do that in 8c3 = 56 ways. Depending on which slots we cross out, the left over slots will be grouped together in various ways that will all be equivalent to 4 digits or less and sum to 5.
For example:
___ ___ ___ ___ X X X ___ = 41
X ___ X ___ X ___ ___ ___ = 113
___ ___ X ___ ___ ___ X X = 23
___ ___ ___ ___ ___ X X X = 5000
etc.
To me this seems like a much much faster solution but also much harder to conceptualize initially.
That's beautiful. I never would have looked at it that way.
Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the digits are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56.
In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed.
And I would have gotten the correct answer very quickly, if I hadn't forgotten about the 0122 combinations. Drat.
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