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Mixture, Series,and Combination Problems

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Mixture, Series,and Combination Problems

by Onigbogi Tosin » Sun Sep 27, 2009 8:54 am
1. According to the directions on a can of frozen orange juice concentrate. 1 can of concentrate is to be mixed with 3 cans of water to make orange juice. How many 12-ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?
A 25
B 34
C 50
D 67
E 100

2. If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?
A 38
B 39
C 40
D 41
E 42

3. A certain law firm consists of 4 senior partners 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different?.
A 48
B 100
C 120
D 288
E 600
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by fiercepoint » Mon Sep 28, 2009 5:28 am
1. 25
The ratio of concentrate to water is 1:3. The total amount of orange juice is 6 x 200 = 1200 ounces. So if you divide this total amount into four parts, it is one part concentrate and three parts water which means you need 300 ounces of concentrate. If one can is 12 ounces then 25 cans are needed to get a total of 300 ounces.
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by mp2437 » Mon Sep 28, 2009 6:53 am
Agree with fiercepoint on #1.

#2. Note that 77 = 11 * 7, and you know 350 = 50 * 7

So, you would need 50 - 11 = 39 "7's", and 1 77, which is a total of n = 39 + 1 = 40. Choice C.

Check: 39 * 7 + 77 = 273 + 77 = 350. Works.

#3. Choice B. I did it the longer way to verify, however, not doing all the math, I've identified a quick way to solve this.

The way I thought about this was as follows: You have 10 partners total, and you need to choose 3, so 10C3 = 120. Now, you know that you have additional constraints, therefore, the total number of combinations must be less than 120. Eliminate C,D,E.

Now, I just did the first set of combinations by myself (choose 1 senior member, and 2 juniors).

Your combinations for the first senior partner is as follows: {S1,J1,J2},{S1,J1,J3},{S1,J1,J4},{S1,J1,J5},{S1,J1,J6},{S1,J2,J3},{S1,J2,J4},{S1,J2,J5},{S1,J2,J6},{S1,J3,J4},{S1,J3,J5},{S1,J3,J6},{S1,J4,J5},{S1,J4,J6},{S1,J5,J6}, a total of 15 combinations for 1 senior partner.

There are 4 senior partners, so there are 4 * 15 = 60 possible combinations JUST by using 1 senior partner and 2 juniors.

That means you can eliminate A, and you're left with choice B.
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by sanjana » Tue Sep 29, 2009 12:54 am
For problem 3 I have a simpler solution

Whenever the question asks for atleast 1,I always use the probability rule of
Atleast 1 = 1 - none
In our case 1 would be choosing 3 members from a group of 10 people without any restriction

10C3 = 120

Now choosing 3 such that none is a Senior partner is equivalent to choosing 3 from the 6 junior members
6C3 = 20

Therefore required number = 120-20 = 100

Hope this helps.
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by Onigbogi Tosin » Thu Oct 01, 2009 8:32 am
sanjana wrote:For problem 3 I have a simpler solution

Whenever the question asks for atleast 1,I always use the probability rule of
Atleast 1 = 1 - none
In our case 1 would be choosing 3 members from a group of 10 people without any restriction

10C3 = 120

Now choosing 3 such that none is a Senior partner is equivalent to choosing 3 from the 6 junior members
6C3 = 20

Therefore required number = 120-20 = 100

Hope this helps.
I love you style, It's so simple and straight forward
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