Puys, Can you help me with this one.. Six mobsters have arrived at the theater for the premiere of thefilm "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied? A. 6 B. 24 C. 120 D. 360 E. 720 Thanks in advance.
Big textual matter for a simple Q :) This is what my thought process would be:- 1. There are 5 spots on which Joey can stand 1,2,3,4 and 5. 2. If Joey stands on 1, all other five can stand in any sequence = 5! ways. = 120 3. If Joey stands on 2, Frankie has 4 options, but other 4 can still stand in 4! ways = 4*4! = 96 4. If Joey stands on 3, Frankie has 3 options, but other 4 can still stand in 4! ways. = 3*4! = 72 5. If Joey stands on 4, Frankie has 2 options, but other 4 can still stand in 4! ways. = 2*4! = 48 6. If Joey stands on 5, Frankie has 1 option, but other 4 can still stand in 4! ways. = 1*4! = 24
Total = 120 + 96 + 72 + 48 + 24 = 360
Hence D. Is there any other way, a quick way to solve this....?
Big textual matter for a simple Q :) This is what my thought process would be:- 1. There are 5 spots on which Joey can stand 1,2,3,4 and 5. 2. If Joey stands on 1, all other five can stand in any sequence = 5! ways. = 120 3. If Joey stands on 2, Frankie has 4 options, but other 4 can still stand in 4! ways = 4*4! = 96 4. If Joey stands on 3, Frankie has 3 options, but other 4 can still stand in 4! ways. = 3*4! = 72 5. If Joey stands on 4, Frankie has 2 options, but other 4 can still stand in 4! ways. = 2*4! = 48 6. If Joey stands on 5, Frankie has 1 option, but other 4 can still stand in 4! ways. = 1*4! = 24
Total = 120 + 96 + 72 + 48 + 24 = 360
Hence D. Is there any other way, a quick way to solve this....?
Cheers.
Ok. I kind of figured out another faster way. But this is purely based on options. Steps 1 and 2 will still remain the same as in my previous reply. Now, we know one thing for sure, that the answer has to be greater than 120.
Lets see if this can be 360 or 720. With simple visualisation, we know, we have 4 other combinations left in which we can place Joey. So, even if we assume that for those combinations its the maximum number of ways, i.e. 120 (as we got in our second step), we will still be able to sum everything up to 120*5 = 600. ie. whatever the case be, we can not have number of combinations greater than 600.
I want to get some more perspective on OG11th Edition Questions.....
Q # in OG is 123.
The Price Of Lunch for 15 people was $207.00, including a 15 % gratuity for service. What was t he average price per person excluding the gratuity ?
a) 11.73 b) 12.00 is the correct answer c) 13.80 d) 14.00 e) 15.87
When I saw the question i firstly took 15% of 207 because the gratuity was included in 207.00.this becomes 175.95 and then dividing it by 15 to get the average yields $11.73 which is option A
Now , OG has done some other thing here with the 15 % which i am unable to understand...... if any1 could help me with it , i will be grateful !!
I want to get some more perspective on OG11th Edition Questions.....
Q # in OG is 123.
The Price Of Lunch for 15 people was $207.00, including a 15 % gratuity for service. What was t he average price per person excluding the gratuity ?
a) 11.73 b) 12.00 is the correct answer c) 13.80 d) 14.00 e) 15.87
When I saw the question i firstly took 15% of 207 because the gratuity was included in 207.00.this becomes 175.95 and then dividing it by 15 to get the average yields $11.73 which is option A
Now , OG has done some other thing here with the 15 % which i am unable to understand...... if any1 could help me with it , i will be grateful !!
question says 'including 15% gratuity' which implies
Puys, Can you help me with this one.. Six mobsters have arrived at the theater for the premiere of thefilm Goodbuddies. One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankies requirement is satisfied? A. 6 B. 24 C. 120 D. 360 E. 720 Thanks in advance.
Big textual matter for a simple Q :) This is what my thought process would be:- 1. There are 5 spots on which Joey can stand 1,2,3,4 and 5. 2. If Joey stands on 1, all other five can stand in any sequence = 5! ways. = 120 3. If Joey stands on 2, Frankie has 4 options, but other 4 can still stand in 4! ways = 4*4! = 96 4. If Joey stands on 3, Frankie has 3 options, but other 4 can still stand in 4! ways. = 3*4! = 72 5. If Joey stands on 4, Frankie has 2 options, but other 4 can still stand in 4! ways. = 2*4! = 48 6. If Joey stands on 5, Frankie has 1 option, but other 4 can still stand in 4! ways. = 1*4! = 24
Total = 120 + 96 + 72 + 48 + 24 = 360
Hence D. Is there any other way, a quick way to solve this....?
Cheers.
Quickest and fastest way for this one :
6 ppl can be arranged in 6 ! ways = 720 ways ... Now, we want one particular person after the other ...
From total, half of the times he is ahead, and the other half he is behind .. Hence, 720/2 = 360 ..
Note: See chart in attachment! The chart lists the sales per month from Jan to June. Which month showed the greatest percentage increase in sales from the prior month?
tiknA_atpuG Says
My Ans: May
May is correct... the ques asks for the greatest percentage increase.. wanna try again?
Q.2. Grocer stacked oranges in a pile. Bottom layer was rectangular with 3 rows of 5 oranges each. In 2nd layer , from bottom , each orange rested on 4 oranges from bottom layer and in the 3rd layer each orange rested on 4 oranges from 2nd layer. Which of the following is max.no. of oranges that could have been in 3rd layer.
Next is a geometry question so i would first try to create a picture in ur mind...
draw pic of two TRIANGLES , all 3 angles in both triangles are in same position. Distanceof base of 1 triangle is s and base of other is S.............
Q. If area of triangle with base S is twice the area of triangle with base s , then in terms of s , S=
a. sq.rt.2/2 *s b. sq.rt. 3/2 *s c. sq.rt.2 *s d. sq.rt.3 *s e. 2s
Q.2. Grocer stacked oranges in a pile. Bottom layer was rectangular with 3 rows of 5 oranges each. In 2nd layer , from bottom , each orange rested on 4 oranges from bottom layer and in the 3rd layer each orange rested on 4 oranges from 2nd layer. Which of the following is max.no. of oranges that could have been in 3rd layer.
Q.2. Grocer stacked oranges in a pile. Bottom layer was rectangular with 3 rows of 5 oranges each. In 2nd layer , from bottom , each orange rested on 4 oranges from bottom layer and in the 3rd layer each orange rested on 4 oranges from 2nd layer. Which of the following is max.no. of oranges that could have been in 3rd layer.
a. 5 b. 4 c. 3 d. 2 e. 1 1st layer has 5 x 3 oranges if each orange in 2nd layer is placed on 4 oranges, no. of rows and columns will each dec by 1 So, 2nd layer shd have 4 x 2 oranges And 3rd layer shd have 3 X1 = 3 oranges
Next is a geometry question so i would first try to create a picture in ur mind...
draw pic of two TRIANGLES , all 3 angles in both triangles are in same position. Distanceof base of 1 triangle is s and base of other is S.............
Q. If area of triangle with base S is twice the area of triangle with base s , then in terms of s , S=
a. sq.rt.2/2 *s b. sq.rt. 3/2 *s c. sq.rt.2 *s d. sq.rt.3 *s e. 2s
Looking at the answer, I'm assuming that the traingles are equilateral...
Q.2. Grocer stacked oranges in a pile. Bottom layer was rectangular with 3 rows of 5 oranges each. In 2nd layer , from bottom , each orange rested on 4 oranges from bottom layer and in the 3rd layer each orange rested on 4 oranges from 2nd layer. Which of the following is max.no. of oranges that could have been in 3rd layer.
a. 5 b. 4 c. 3 d. 2 e. 1
all ur answers are right !!: ) great job !!
can u please explain the oranges question and geometry question more !!!
thankx for ur effort !
For the ques above jus do a lil matrix
1st layer - 5x3 Represented by X X X X X X o o o o X X X X X o o o o X X X X X
So in the 2nd layer if each orange has to be placed on 4 oranges, we can have 4x2 ( Represented by O )
o o o o @ @ @ o o o o
Similarly the in 3rd layer each orange has to be placed on 4 oranges, we'll have 3x1 Represented by @
Hope this helps!!
p.S: the arrangement doesn't show properly in the post... Quote the post, and it'll look just fine..
can u please explain the oranges question and geometry question more !!!
thankx for ur effort !
Next is a geometry question so i would first try to create a picture in ur mind...
draw pic of two TRIANGLES , all 3 angles in both triangles are in same position. Distanceof base of 1 triangle is s and base of other is S.............
Q. If area of triangle with base S is twice the area of triangle with base s , then in terms of s , S=
a. sq.rt.2/2 *s b. sq.rt. 3/2 *s c. sq.rt.2 *s d. sq.rt.3 *s e. 2s
Looking at the answer, I'm assuming that the traingles are equilateral...
Let triangle T1 have base s and area A1 Let triangle T2 have base S and area A2
I meant May is INCorrect...