If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
10p5 to 10p4 = 5 to 1
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
i feel its 6:1....
no. of alphabets = 10
hence, no. of possible 5 letter codes = 10*9*8*7*6
no. of possible 4 letter codes = 10*9*8*7
thus, answer is E...6:1
i feel its 6:1....
no. of alphabets = 10
hence, no. of possible 5 letter codes = 10*9*8*7*6
no. of possible 4 letter codes = 10*9*8*7
thus, answer is E...6:1
right . I did not calculate.
Hi All,
Note: I had posted a similar question earlier. Please note however that the questions in this attachment are different from the earlier ones.
Attached are some of the questions from GMATPrep that I got wrong. The options marked with the bold black dot in the attachment are mine and are the wrong answers. The options marked by a square blue frame are correct as these are the correct answers marked by the software. However the software does not explain why a particular choice is correct and others are wrong. I will be very grateful if some of you can help justify why the correct choice is actually correct!
Many thanks.
- TheToastedBread
Hi It would be better if you post the question directly here. instaed of winzip. may one question per post.
otherwise it is difficult to read answer and question together
hey can anyone guide me to any forum or a thread wherein i can find some material on CR/SC.....
thanxx
Hi folks,
Does anyone have the kalpan 800 software for GMat...if so can u please upload the software......My friend is in need of that software...He is going to write GMat in the nxt month....if u are done with this u are helpful to him a lot.....Thnx in advance.
Regards,
Varun.
Hi folks,
Does anyone have the kalpan 800 software for GMat...if so can u please upload the software......My friend is in need of that software...He is going to write GMat in the nxt month....if u are done with this u are helpful to him a lot.....Thnx in advance.
Regards,
Varun.
Dude go to this thread
http://www.pagalguy.com/discussions/gmat-prep-material-25024311
Vikram
But where can i find this material there are so many pages...is it necessary to search all the pages i have checked through the search option but unable to find it.....
varunvanama SaysBut where can i find this material there are so many pages...is it necessary to search all the pages i have checked through the search option but unable to find it.....
Just post your query on that thread .. and people will get back to you ....
You need not go thru all the pages in that thread
ok...thank you...i will post there....
ok...thank you...i will post there....
Hi
Can somebody help with this probability question
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?
a)5
b)6
c)7
d)14
e)21
Hi
Can somebody help with this probability question
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?
a)5
b)6
c)7
d)14
e)21
hey, first of all, its nt a probability question...its a P&C; one...see, u hv to find the no. of combinations in which both Ben and Ann are not there...so that leave 5 ppl wid us from which we hv to select 4...so u can select 4 ppl frm a group of 5 ppl in 5C4 = 5 ways...
answer is A
Question from the GMAT prep test 1.
For how many integers is 2^n = n^2
The answer is TWO integers. One of them is '2', I couldn't figure out the other.
JJ
Question from the GMAT prep test 1. For how many integers is 2^n = n^2 The answer is TWO integers. One of them is '2', I couldn't figure out the other. JJthe other number is 4.... 😃
ajayreddy Saysthe other number is 4.... :)
Bingo..! the other is 4..!
Hi .. Here goes a question.. !Since we are talking of numbers.. lets try this one..
find the total number of possible numbers between 300-400 such that the sum of the cube of its digits is equal to the number itself.
options
a. 0
b.1
c.2
d.4
Hi .. Here goes a question.. !Since we are talking of numbers.. lets try this one..
find the total number of possible numbers between 300-400 such that the sum of the cube of its digits is equal to the number itself.
options
a. 0
b.1
c.2
d.4
Number wil start with 3 3^3=27.
8 or 9 can not be in the numbe. becos 8^3 and 9^3 exceed 400. 7^3=343 , 37x is a possible anwwr. 3^3=7^3=370 if units is 1 371 is one possible answer.
now if the other 2 digits are 5 5^3+5^3= 250 and the hundred digit is 3. sum will be below 300. So one of the digits should be atleast 6. It can be in units or tens digits. 3^3+6^3=243. the other digit can be 4 or 5 only. becos if it is 3 or below the sum will be below 300. 6 or above sum will be above 400.
possible anwers is 346,356.364,365. none of them meet the other condition(sum of cubes)
So there is only one number (371 ) meets the criteris. Answer is B
Number wil start with 3 3^3=27.
8 or 9 can not be in the numbe. becos 8^3 and 9^3 exceed 400. 7^3=343 , 37x is a possible anwwr. 3^3=7^3=370 if units is 1 371 is one possible answer.
now if the other 2 digits are 5 5^3+5^3= 250 and the hundred digit is 3. sum will be below 300. So one of the digits should be atleast 6. It can be in units or tens digits. 3^3+6^3=243. the other digit can be 4 or 5 only. becos if it is 3 or below the sum will be below 300. 6 or above sum will be above 400.
possible anwers is 346,356.364,365. none of them meet the other condition(sum of cubes)
So there is only one number (371 ) meets the criteris. Answer is B
You just missed it! The answer is not B.. but C.. u did consider 371 but you did not take another no into consideration..! think.. !