If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A.20 B.40 C.50 D.80 E.120
IMO: D
Total number of ways to select 3 people out of 10 is : 10C3 Out of which you can have 5C1 ways to have couple and 8C1 ways for the 3rd person.
So number of ways to have 1 couple and one other person on the committee is: 5 * 8 = 40
So number of ways not to have a couple on the committee is: 10C3 - 40 = 80.
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A.20 B.40 C.50 D.80 E.120
hi, there can be these types of committees out of 5 married couples (5 men and 5 women) a) 3 men only no. of ways = 5C3 = 10 b) 3 women only no. of ways = 5C3 = 10 c) 2 men and 1 woman no. of ways = 5C2 * 3C1 = 30 d) 2 women and 1 man no. of ways = 5C2 * 3C1 = 30 so total ways = 10+10+30+30 = 80 hope i am correct...
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A.20 B.40 C.50 D.80 E.120
sorry, couldn't locate the original post therefore picking from here. I think the ans should be A. waiting for the OA before I post explanation.
Quote: Originally Posted by nairpraveenk Fundoo problem -------------------
Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York? (1) Train B arrived in New York before Train A arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) Each statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient.
@Bizzarenym...
could you explain how you got A?? (for the DS question about 2 trains!! )
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A.20 B.40 C.50 D.80 E.120
In a 4 person race, medals are awarded to the fastest 3 runners. The first-place runner receives a gold medal, the second-place runner receives a silver medal, and the third-place runner receives a bronze medal. In the event of a tie, the tied runners receive the same color medal. (For example, if there is a two-way tie for first-place, the top two runners receive gold medals, the next-fastest runner receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a victory circle, how many different victory circles are possible?
Sorry for the delayed response.. here goes the explanation
First, let's consider the different medal combinations that can be awarded to the 3 winners:
1 If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.
2 If there is a 2-WAY tie? for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. **There cannot be a 2-WAY tie for THIRD
3 If there is a 3-WAY tie? then the medals awarded are: GOLD, GOLD, GOLD. **There are no other possible 3-WAY ties.
Thus, there are 4 possible medal combinations:
G, S, B G, G, S G, S, S G, G, G Lets assume there are 4 ppl - A,B C, D
COMBINATION 1: Gold, Silver, Bronze
Using simple permutation - there are 4 possibilities for Gold, 3 for Silver and 2 for Bronze.. ~ 4!
->> 4 x 3 x 2 = 24 different victory circles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.
COMBINATION 2: Gold, Gold, Silver.
Lets find the # of possibilities for 2 Golds : There are 4 candidates, 2 medals.
Thus, applying formula : 4! / = 6 Then the possibilities for the Silver medal : 2
As we need to find possibility for 2 golds AND 1 silver, we multiply : 6*2 = 12 COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, possibility for 1 gold AND 2 silvers : 6*2 = 12 COMBINATION 4: Gold, Gold, Gold.
This is straightfwd: there are 4 posibillities for the 3 golds!!
Either of these combinations will happen, thus, we add the 4.
Thus, there are 24 + 12 + 12 + 4 = 52 unique victory circles.
At a certain food stand, the price of each apple is 40 and the price of each orange is 60. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52? A. 1 B. 2 C. 3 D. 4 E. 5
A+B(orange)=10-- from condition 1 (40A+60B)/10=56--from condition 2 ==>4A+6B=56 or 2A+3B=28
hi, there can be these types of committees out of 5 married couples (5 men and 5 women) a) 3 men only no. of ways = 5C3 = 10 b) 3 women only no. of ways = 5C3 = 10 c) 2 men and 1 woman no. of ways = 5C2 * 3C1 = 30 d) 2 women and 1 man no. of ways = 5C2 * 3C1 = 30 so total ways = 10+10+30+30 = 80 hope i am correct...
@ Raghav I have a doubt :::
c) 2 men and 1 woman no. of ways = 5C2 * 3C1 = 30
Here out of 5 men we choose 2, so 5C2 And out of 5 women we choose 1.. so shouldn't it be 5C1?? Can u pls explain..
Here out of 5 men we choose 2, so 5C2 And out of 5 women we choose 1.. so shouldn't it be 5C1?? Can u pls explain..
You will have to exclude those 2 womens, whose husbands are picked up with 5C2..
in othere words...you pick 2 mans with 5C2...now to make sure that they no married couple be picked up, you need to leave behind those 2 womens who are married to these picked man..
Q3: Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.) A. 20 B. 92 C. 300 D. 372 E. 476
You will have to exclude those 2 womens, whose husbands are picked up with 5C2..
in othere words...you pick 2 mans with 5C2...now to make sure that they no married couple be picked up, you need to leave behind those 2 womens who are married to these picked man..
hence 5-2 = 3
therefore..3C1
Ohh yess!! silly me... Got it now!! thanks a bunch
Puys, Please provide explanations for the following. Please note that Std dev is a pain area for me :)
1. a set of data consists of the foll 5 nos:{0,2,4,6,8}. what 2 nos if added to create a set of 7 numbers will result in a std deviation for the set with just the original 5 nos? A. -1 and 9 B.4 and 4 C.3 and 5 D. 2 and 6 E.0 and 8 OA:D 2. In a game of chess, the moves of white and black pieces alternate, woth white pieces having the first move .During a certain chess tournament, the white pieces have made 2319 moves altogether, while the black pieces have made a total of 2315 moves. In every game , either the white side wins , black side wins , or there is a draw. If it is impossible for the losing to make the last move in a game, which of the following could be true about the tournament? I. The black side lost 5 games. II. The black side won more games than the white sides III. All games ended in a draw. A.III only B.I and II only C.I and III only D. II and III only E. I, II and III OA:D
Puys, Please provide explanations for the following. Please note that Std dev is a pain area for me :)
1. a set of data consists of the foll 5 nos:{0,2,4,6,8}. what 2 nos if added to create a set of 7 numbers will result in a std deviation for the set with just the original 5 nos? A. -1 and 9 B.4 and 4 C.3 and 5 D. 2 and 6 E.0 and 8 OA:D 2. In a game of chess, the moves of white and black pieces alternate, woth white pieces having the first move .During a certain chess tournament, the white pieces have made 2319 moves altogether, while the black pieces have made a total of 2315 moves. In every game , either the white side wins , black side wins , or there is a draw. If it is impossible for the losing to make the last move in a game, which of the following could be true about the tournament? I. The black side lost 5 games. II. The black side won more games than the white sides III. All games ended in a draw. A.III only B.I and II only C.I and III only D. II and III only E. I, II and III OA:D
First has to be D)...without any calculations...
Average of the current set = 4..
if you add 2 numbers whoese sum = 8, average will not be change...so all of them qualifies..
Standard daviation = the davitaion from the average and because the current set is uniform across the average 4, the new 2 numbers need to be within this range..to keep the set uniform ( If SD need not to be changed)
Now SD = (AVG{SUM(difference from avergae ) ^ 2 })1/2
so in order to keep the difference same, the difference between the numbers from the average need to be agreeing the the current difference = 2
w,x,y,z are integers such that w>x>y>z>0, is y a common devisor of w, x??
1) w/x = z^-1 + X^-1 2) w^2-wy-2w = 0.?
IMO: Option B
1. does not say much about the relationship with y. w=(z+x)/z is the best we can deduce here. NOT SUFF
2. w^2 -wy-2w = 0 we get either w=0 or w=y+2. Since w cannot be 0, we have to hold w = y+2 true. Which means w,x,y are consecutive integers greater than 0. In which case, y can never be a common factor for w and x. SUFF
p,q,r.s are +ve integers, if p/q = r/s, is r devisible by 5?
1) p is a multiple of 140 2) Q = 7^x
IMO: C
1. Given p/q = r/s Lets pick numbers here, p=140, q = 10, r = 14, s = 1. Suffices p/q=r/s, but r is not divisible by 5.
Au Contraire': 140,7,20,1 suffices p/q=r/s and r is divisible by 5. So, NOT SUFF
2. q = 7^x, => q can be 1, 7,49,.... does not help by itself. NOT SUFF
1 and 2 together: p/q will be 140 X n / 7^x. for n = 1 and x = 1 lets say, p/q will be 140. For any combination of n and x, there will always be a number r which will be a multiple of 5 as 140 is 7 X 20. So Together SUFF.
1. Given p/q = r/s Lets pick numbers here, p=140, q = 10, r = 14, s = 1. Suffices p/q=r/s, but r is not divisible by 5.
Au Contraire': 140,7,20,1 suffices p/q=r/s and r is divisible by 5. So, NOT SUFF
2. q = 7^x, => q can be 1, 7,49,.... does not help by itself. NOT SUFF
1 and 2 together: p/q will be 140 X n / 7^x. for n = 1 and x = 1 lets say, p/q will be 140. For any combination of n and x, there will always be a number r which will be a multiple of 5 as 140 is 7 X 20. So Together SUFF.
1. does not say much about the relationship with y. w=(z+x)/z is the best we can deduce here. NOT SUFF
2. w^2 -wy-2w = 0 we get either w=0 or w=y+2. Since w cannot be 0, we have to hold w = y+2 true. Which means w,x,y are consecutive integers greater than 0. In which case, y can never be a common factor for w and x. SUFF
its not B)...let others try, then will post the OA
