A car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour. At what speed, in kilometers per hour, is the car traveling?
A. 71.5
B. 72
C. 72.5
D. 73
E. 73.5
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
distinct arrangements, hence permutation..
5 letter code : 4 letter code = 10P5 / 10P4 = / = 6!/5! = 6 :1 ...Ans E
A car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour. At what speed, in kilometers per hour, is the car traveling?
A. 71.5
B. 72
C. 72.5
D. 73
E. 73.5
difference of time is 2 seconds, Hence
1/x - 1/75 = 2/3600
i.e 1800(75-x)=75
simple calculation gives x = 72 km/hr
1) Of the marbles removed, the ratio of the number of red ones to the number of blue ones is 2 : 1.
(2) Of the first 6 marbles removed, 4 are red.
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.
Ans for the above q is option A. Is it correct?
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
Number of 5 letter code is = 10*9*8*7*6---(1)
Number of 4 letter code is = 10*9*8*7---(2)
Eqn 1 divided by Eqn 2 = 6
Hence E
1.If a and b are positive integers such that a - b and b/a are both even integers, which of the following must be an odd integer?
A. a/2
B. b/2
C. (a+b)/2
D. (a+2)/2
E. (b+2)/2
2. What is the two-digit positive integer whose tens digit is a and whose units digit is b?
(1) 2a > 3b > 6
(2) a > 2b > 6.
3.At a party, there were five times as many females as males, and three times as many adults as children. Which of the following could NOT be the number of people at the party?
A. 384
B. 258
C. 216
D. 120
E. 72
someone pls exp these to me
1) Should be E
b/a = 2m (m any positive intger)
so b = 2m*a
(b + 2)/2 = b/2 + 1..--> m*a + 1..always odd integer
2) Combing both the eqns 2a > 3b > 6 and a > 2b > 6.
a > 2b and a> 6 and b > 3
so b can be 4 and then a should be 9
so number is 94
3) then number of males be m and females be f . Also let number of children be c and adults be a ..
So from conditions given f = 5m ----> so total number of ppl at party = f + m = 6m
Also , a = 3c ---> so total number of ppl at party = a + c = 4c
Thus any total should be of 6m or 4c ttpe , which indirectly mens it should be multiple of 12 then only it will be of 4c and 6m type ..
Only option B is not divislble by 12 and hence the answer
Thanks i got it ..
Number of 5 letter code is = 10*9*8*7*6---(1)
Number of 4 letter code is = 10*9*8*7---(2)
Eqn 1 divided by Eqn 2 = 6
Hence E
NOPE d2rock, yours approach is wrong.see the approach of bhavin for the question.
Number of 5 letter code is = 10*9*8*7*6---(1)
Number of 4 letter code is = 10*9*8*7---(2)
Eqn 1 divided by Eqn 2 = 6
Hence E
guy with guts SaysNOPE d2rock, yours approach is wrong.see the approach of bhavin for the question.
nothing wrong with rockstar's approach or the answer ..
Basically his approach is also same as mine ...its the expansion infact ..
for 5 letter code, its 10P5 which is nothing but 10*9*8*7*6.
And for 4 letter code its 10P4 which is 10*9*8*7..
Infact both of us meant the same thing
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
This one is really interesting;
The probability that it will rain in NYC on any given day in July is 40%. What is the probability that it will rain in NYC on exactly 2 days out of 5 in July?
though I got it right, I thought of sharing the question with all Puys.. ;)
The probability that it will rain in NYC on any given day in July is 40%. What is the probability that it will rain in NYC on exactly 2 days out of 5 in July?
though I got it right, I thought of sharing the question with all Puys.. ;)
If f (a + b) = f (a) + f (b), then which of the following could be f (x) for all distinct values of a and b?
A) x^2 +1
B) 2 (x)^1/2
C) x - 4
D)5/x
E) - 7x
I could not comprehend this question.. could someone please help me understand the question.. the use of a and b.. and also share the solution.. ;)
A) x^2 +1
B) 2 (x)^1/2
C) x - 4
D)5/x
E) - 7x
I could not comprehend this question.. could someone please help me understand the question.. the use of a and b.. and also share the solution.. ;)
If f (a + b) = f (a) + f (b), then which of the following could be f (x) for all distinct values of a and b?
A) x^2 +1
B) 2 (x)^1/2
C) x - 4
D)5/x
E) - 7x
I could not comprehend this question.. could someone please help me understand the question.. the use of a and b.. and also share the solution.. ;)
Hey Varun ..
Ans is E
For function problems of these kind, in LHS sub x by (a+b) and verify if is equal to sum when u sub x by a and b resp .
eg 1. (a+b)^2 + 1 is not equal to a^2 +1 + b^2 + 1 ..incorrect
2 . 2root(a+b) is not equal to 2roota +2rootb ..incorrect .
3. a+b-4 is not equal to a-4 +b-4 ...incorrect
4. 5/(a+b) is not equal to 5/a + 5/b
5. -7(a+b) = -7a-7b ...correct ....Ans E
As a general rule , squares, roots, adding or subtracting constants to variables does not hold true ..
Single variate function without any additional operation would hold true ..
Hope that helps
This one is really interesting;
The probability that it will rain in NYC on any given day in July is 40%. What is the probability that it will rain in NYC on exactly 2 days out of 5 in July?
though I got it right, I thought of sharing the question with all Puys.. ;)
Binomial distribution does really come in handy out here ..
p(r) = nCr*p^r*q^(n-r) where p is success and q = 1-p is failure
Hence p(exactly 2 out 5 days ) = 5C2 * (2/5)^2 * (3/5)^3
= 216/625 ...Ans
[LEFT]A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?[/LEFT]
[LEFT](A) 3510
(B) 2620
(C) 1404
(D) 700[/LEFT]
(E) 635
Total possible committees = Total possible committees without restriction - Total committees with restriction
i.e Selecting a team of 6 as:
(2 Men & 4 Women OR 3 Men and 3 women ) -
(2 Men & 4 Women OR 3 Men and 3 women considering both the men who refuse to serve together)..
i.e [(8C2 * 5C4) + (8C3 * 5C4)] - [(2C2 * 5C4 + 2C2 * 6C1 * 5C3)]
= [140 + 560] - [ 5 + 60 ]
= 700 - 65
=635 ...Ans
Else, we could just figure out that without any restriction there are 700 ways to select a team of 6 ..so added restriction is going to reduce the outcome , only one answer option is below 700, so it had to be 635
Binomial distribution does really come in handy out here ..
p(r) = nCr*p^r*q^(n-r) where p is success and q = 1-p is failure
Hence p(exactly 2 out 5 days ) = 5C2 * (2/5)^2 * (3/5)^3
= 216/625 ...Ans
Hi can you explain the funda?
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
let one of the two man,refused to serve, is selected-
1.three man+three woman=7c3+5c3=35+10=45
2.Two man+four w=7c2+5c4=21+5=26
so the case will be similiar for the another man, so total combinations=2*(45+26)=142
neither of the two man selected
1.three man+three woman=6c3+5c3=20+10=30
2.Two man+four wom=6c2+5c4=15+5=20
So total=142+30+20=192
Answer dosent match.Bhavin,can you tell me whr i went ulta!!!
let one of the two man,refused to serve, is selected-
1.three man+three woman=7c3+5c3=35+10=45
2.Two man+four w=7c2+5c4=21+5=26
so the case will be similiar for the another man, so total combinations=2*(45+26)=142
neither of the two man selected
1.three man+three woman=6c3+5c3=20+10=30
2.Two man+four wom=6c2+5c4=15+5=20
So total=142+30+20=192
Answer dosent match.Bhavin,can you tell me whr i went ulta!!!
Hey ...error on multiple grounds ..let me attempt to correct them ..red highlighted portion is incorrect ..
1) And refers to multiplication , OR refers to addition ..
3M & 3W is not addition of selection of men and women, its multiplication ..
2) when u select 1 or none of the 2 unwanted men, u need to segregate the total no of available men as a team of 6 favorable and 2 unfavorable men first ...we cannot randomly exclude 1 man, and select 3 from remaining 7 as 7C3 ...we dont know whom to exclude !!..
So going your way maybe the approach could be ,
1 out of 2 unfavorable men & 2 out of 6 fav men & 3 women = 2C1*6C2*5C3 = 300
Similarly , phrasing out the exact 4 condition mentioned by u ...
2) 2C1*6C1*5C4 =60
3) 2C0 *6C3*5C3 = 200
4) 2C0 *6C2*5C4 = 75
Hence, 300+60+200+75 = 635 ...exact same answer
..Hope this helps u correct your mistake !!
Cheers,
Bhavin
guy with guts SaysHi can you explain the funda?
Hello ....
Very rarely a GMAT problem expects us to know the binomial distribution ..It is basically used when any process can have only 2 possible outcomes ...one is termed as success (anything that we want it to happen) and another is failure (anything that we do not want)
By usual notations, p is associated with success and q with failure ..
Since for any probability, sum of prob of all discrete outcomes is always 1 , we get p+q = 1
Flipping of coins to generate a desired no of heads / tails follows binomial distr. It is a special case, since success = failure = 0.5 and hence we can solve even without having any clue about binomial distribution !!
Hope this helps .
U can also read about binomial distribution here : Binomial distribution - Wikipedia, the free encyclopedia