Does anyone have the OG Quant and willing to share, please let me know. I'll om my id to him.
Also, can anyone please tell me about the difficulty level of Quant in the actual test as in is OG enough or should we prepare on similar lines as CAT, etc.
Thanks in advance 😃
Six mobsters have arrived at the theater for the premiere of the film 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?
6
24
120
360
720
try downloading from torrentz.com ......OG is definitely there not sure about OG Quant
Six mobsters have arrived at the theater for the premiere of the film "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?
6
24
120
360
720
suppose there is no constraint. Then the total number of ways of arranging 6 people=6!=720 ways
Now apply the constraint. Among this 720, in half of them joey will come ahead of frankie and in the other half, frankie will come ahead of joey.
So Ans:720/2 = 360
Is x an even integer?
(1) is divisible by 4.
(2) is divisible by 16.
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 to answer the question asked, and additional data are needed.
Answer is D
Six mobsters have arrived at the theater for the premiere of the film 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?
6
24
120
360
720
A B C D E F be the six mobsters.
Let C and D be the Joey and Frankie.
Suppose C stands in 1st position, then D has 5 choices, rest of four has to be arranged in 4 position taken 4 at a time i.e. 4!
1*5*4! = 5!
C stands in 2nd position, then D has 4 choices -
1*4*4!
So (5+4+3+2+1)*4! = 15*4! = 15*24 = 360 😃
Need help in identifying simplest technique to solve problems like this
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37
B and D are out of the question; they repeat same digits (3 and 6 respectively)
Made a guess of E :-(. Need a way to solve this problem without guess 😃
Guys,
I have started a thread
GMAT study centre in Kolkata
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There are 26 students who have read a total of 56 books among them. The only
books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only
read Aye, and 8 students have read only Cod and Dee, what is the smallest number
of books any of the remaining students could have read?
I found this in MGMAT Prep Materials.
Explanation not understood.
Anyone with a understandable one please.
OA is 2
Thanks
Venkat, go about it in this way,
10 students have read Aye
8 students have read Bye
& 8 students have read Cod
Total books read = 10+8+8 = 26 by 18 (10+8 ) students.
Remaining No of books = 56-26 = 30
Remaining Students = 26 - 18 = 8
Now, each remaining student can, at the most, read 4 books each (A,B,C or D)
So if we assume that 7 students read all the 4 books, 7*4 = 28, the last student will have to read a minimum of 2 (30-28 = 2) books.
There are 26 students who have read a total of 56 books among them. The only
books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only
read Aye, and 8 students have read only Cod and Dee, what is the smallest number
of books any of the remaining students could have read?
I found this in MGMAT Prep Materials.
Explanation not understood.
Anyone with a understandable one please.
OA is 2
Thanks
I dont know of a generic way to solve these kind of problems.
But this one can be treated easily.
A,B,D are all out of question, you dont need to do any calculation to figure that out.
Remaining C&E; can be compared, but if I am running short of time, I would have made a calculated guess and marked E since any 2 digit number when divided by 99 will not give a decimal which repeats more than twice.
Need help in identifying simplest technique to solve problems like this
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37
B and D are out of the question; they repeat same digits (3 and 6 respectively)
Made a guess of E :-(. Need a way to solve this problem without guess :)
Venkat, go about it in this way,
10 students have read Aye
8 students have read Bye
& 8 students have read Cod
Total books read = 10+8+8 = 26 by 18 (10+8 ) students.
Remaining No of books = 56-26 = 30
Remaining Students = 26 - 18 = 8
Now, each remaining student can, at the most, read 4 books each (A,B,C or D)
So if we assume that 7 students read all the 4 books, 7*4 = 28, the last student will have to read a minimum of 2 (30-28 = 2) books.
Sausi, venkat one more approach...
There are 4 books, and total books students read are 56, so each of 4 books must repeat.
Let all students read all books then we would have 26*4 = 104 books, so all students couldn't have read 4 books, lets say all students read three books 26*3 = 78 still above 56, 26*2=52 less than 58; so each student must have read at least 2 books
In short 56/26=2+6/56 so take quotient 2
Your opinion?
OFF TOPIC on this forum, but can anyone suggest me if they know someone who takes online classes for quant???
Thanks
If the remainder is 7 when positive integer n is divided by 18, what is the remainder when n is divided by 6 ?
pratyasha1990 SaysIf the remainder is 7 when positive integer n is divided by 18, what is the remainder when n is divided by 6 ?
It will be 1.
Plug numbers like 25, 43, 61 etc........
What is the sum of the remainders of the first 40 positive integers when didvided by the nummber 6.
itsvenkathere SaysWhat is the sum of the remainders of the first 40 positive integers when didvided by the nummber 6.
SHould be 100........because on dividing by 6 a number can give 0-5 as remainder........
Thus 1 will give 1, 2 will give 2 etc...............36 will give 0, 37 will give 1, 38 will give 2........40 wil give 4.....
Thus sum = ((1 + 2 + 3+ 4 +5 ) * 36/6) + (1 + 2 + 3 + 4) = 100
pratyasha1990 SaysIf the remainder is 7 when positive integer n is divided by 18, what is the remainder when n is divided by 6 ?
Remainder of 7 divided by 6.
How? n divided by 18 gives remainder of 7 and say quotient of x
so number is 18x+7
divide this by 6, 3x+7/6 = 3x+1 with remainder of 1