Official Quant thread for CAT 2013

MBA CAT 2024
19346
@pavimai said:
how many integers between 1000 and 10000 have their sum of the digits equal to 9???
1 x y z 2 x y z 3 x y z 4 x y z 5 x y z 6 x y z 7 x y z 8 x y z 9 x y z

x+y+z=8x+y+z=7x+y+z=6 x+y+z=5x+y+z=4x+y+z=3x+y+z=2x+y+z=1x+y+z=0
10c2 9c28c27c26c25c24c23c2 2c2

45+36+28+21+15+10+6+3+1=165
19347
@pavimai said:
if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C...In this series there will be a point where you will get sum of digits of X=X..Find X
two find sum of digits we have to divide with 9 as substracting 9's from sum will not affect the sum.

8888^8888/9 == 5^8888/9
5/9=5 , 25/9=7 ,125/9=8 ,625/9=4,3125/9=2,15625/9=1,78125/9=5
1 2 3 4 5 6

after 6 cycles it is repeating so 8888/6= 2

There for 7

19348
@IIMAIM said:
1 x y z 2 x y z 3 x y z 4 x y z 5 x y z 6 x y z 7 x y z 8 x y z 9 x y z x+y+z=8x+y+z=7x+y+z=6 x+y+z=5x+y+z=4x+y+z=3x+y+z=2x+y+z=1x+y+z=010c2 9c28c27c26c25c24c23c2 2c245+36+28+21+15+10+6+3+1=165
Rather, treat it as: the four digits are (w+1) x y z
Then w + x + y + z = 8 and hence 11C3 = 165 directly.

Edit: Good morning, world... (goes back to sleep)

regards
scrabbler
19349
@scrabbler said:
I wished to know the OA? I think I interpreted it wrongly...I took the 2/3 as the probability of "knowing the answer" instead of that of "finally getting it right"; I think the latter is more appropriate. I guess the right answer should be 5/6....someone else already solved it while I was updating my answer so I didn't post.

Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?

regardsscrabbler

The correct answer is 8/9 n I saw only u got it, so i asked u to share ur approach. do share it
19350
@DeAdLy said:
The correct answer is 8/9 n I saw only u got it, so i asked u to share ur approach. do share it
Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?

The first sentence should be "probability of knowing the answer is 2/3" to remove ambiguity, then my logic gives a unique answer. My method is mildly unorthodox though...

See, there is a 2/3 and a 1/4 so I took 12 cases (LCM of 3 and 4). So on average in 8 out of 12 cases, he knows the answer and so will get it right. In the remaining 4, he guesses and hence on average in one of those cases he will get it right. So overall there are 9 cases in which he gets it right, and in 8 of them he really knew it. So given that he got it right, his probability of knowing it was 8/9.

(Am still conflicted about the ambiguous wording of the problem though - can see it as 5/6 or as 8/9 depending on how I choose to interpret that first line)

regards
scrabbler
19351
@iLoveTorres said:
" Remove all the cards except aces and kings from a deck. From this, deal two cards to a friend. If she looks at her cards and announces (truthfully) that her hand contains an ace, what is the probability that both her cards are aces? If she announces instead that one of her cards is the ace of spades, what is the probability then that both her cards are aces? "
1) total cases = 4c1*7c1 = 28
favorable = 4c2 = 6
so 3/14
2) 3/7
how many integers between 1000 and 10000 have their sum of the digits equal to 9???
a+b+c+d = 9
here a>=1
==> a'+1+b+c+d = 9
==> a'+b+c+d=8
==> 11c3 = 165
if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C...In this series there will be a point where you will get sum of digits of X=X..Find X
its basically asking for digit sum of A, which is but remainder by 9
8888^8888 mod 9 = 5^8888 mod 9
e(9) = 6
8888mod6 = 2
so 5^2mod9=25mod9=7
hence 7
19352
@scrabbler said:
sir __/\__ , can't tag u for some reason
" Remove all the cards except aces and kings from a deck. From this, deal two cards to a friend. If she looks at her cards and announces (truthfully) that her hand contains an ace, what is the probability that both her cards are aces? If she announces instead that one of her cards is the ace of spades, what is the probability then that both her cards are aces? "
what is the logic behind 3/11? cant figure out

19353
@IIMAIM said:
1 x y z 2 x y z 3 x y z 4 x y z 5 x y z 6 x y z 7 x y z 8 x y z 9 x y zx+y+z=8x+y+z=7x+y+z=6 x+y+z=5x+y+z=4x+y+z=3x+y+z=2x+y+z=1x+y+z=010c2 9c28c27c26c25c24c23c2 2c245+36+28+21+15+10+6+3+1=165
a 4 digit number can be written as : abcd
a+b+c+d = 9
so total ways = 12c3= 220
but this has considered the case in which a=0
so we need to subtract these cases from 220
so b+c+d = 9(as a=0)
total ways 11c2 = 55
required number of ways will be 220-55 = 165...I think this method is faster...or @scrabbler method is better than this..in which he considers the +1 for the first digit...
19354

A standard digital clock displays time in HH:MM (24hour) format. Another digital clock got stuck at a particular time and hence is displaying a particular time.
What is the probability that it is showing the right time in a day?
(Proverb: Even a stopped clock is right twice a day)

19355
@krum said:
sir __/\__ , can't tag u for some reason " Remove all the cards except aces and kings from a deck. From this, deal two cards to a friend. If she looks at her cards and announces (truthfully) that her hand contains an ace, what is the probability that both her cards are aces? If she announces instead that one of her cards is the ace of spades, what is the probability then that both her cards are aces? "what is the logic behind 3/11? cant figure out
Not sure of my answer...

But if you look at it, there at 8 cards. So 8C2 = 28 ways of picking 2.

Ways of picking at least 1 ace = 8C2 - 4C2 (cases where both are kings) = 28 - 6 = 22. This is our sample space.

Ways of picking 2 aces = 4C2 = 6 (Favourable event)

Hence Prob of both being Ace, given that at least one is Ace = 6/22.

Does that make sense?

regards
scrabbler
19356

QA question ;


19357
@krum said:
1) total cases = 4c1*7c1 = 28
My feeling is this step is faulty, in that all the cases with 2 Aces are double counted (4C2 i.e. 6 cases) so actual will be 28-6 = 22.

Point is, if in the 4c1 we pick A(Spades) and in the 7C1 A(Clubs) it is the same as if in the 4c1 we pick A(Clubs) and in the 7C1 A(Spades). But you are counting it as 2 separate cases, right?

Can we have OA please?

regards
scrabbler
19358
@scrabbler said:
Not sure of my answer...But if you look at it, there at 8 cards. So 8C2 = 28 ways of picking 2.Ways of picking at least 1 ace = 8C2 - 4C2 (cases where both are kings) = 28 - 6 = 22. This is our sample space.Ways of picking 2 aces = 4C2 = 6 (Favourable event)Hence Prob of both being Ace, given that at least one is Ace = 6/22.Does that make sense?regardsscrabbler
But I think she has already picked up an ace of spades..so do you really need to do 4c2..
I think it should simply be 3/7...as 1 ace of spades she has picked up...and the probability of getting another ace should be 3/7..as only 7 cards remain and 3 are aces in that...


19359
@iLoveTorres said:
" Remove all the cards except aces and kings from a deck. From this, deal two cards to a friend. If she looks at her cards and announces (truthfully) that her hand contains an ace, what is the probability that both her cards are aces? If she announces instead that one of her cards is the ace of spades, what is the probability then that both her cards are aces? "
Can you provide the OA??
19360
@scrabbler said:
My feeling is this step is faulty, in that all the cases with 2 Aces are double counted (4C2 i.e. 6 cases) so actual will be 28-6 = 22.Point is, if in the 4c1 we pick A(Spades) and in the 7C1 A(Clubs) it is the same as if in the 4c1 we pick A(Clubs) and in the 7C1 A(Spades). But you are counting it as 2 separate cases, right?Can we have OA please?regardsscrabbler
u r right sir,got my mistake , thanks
19361
The avarage age of group decreased by 3 years when a person 56 years old was replaced by another 20 years old . How would the average age have changed, if he were replaced by 2 persons of 32 years each instead???
-JASH
19362
@jashholmes said:
The avarage age of group decreased by 3 years when a person 56 years old was replaced by another 20 years old . How would the average age have changed, if he were replaced by 2 persons of 32 years each instead???-JASH
A=S/N
A-3=(S-36)/N =) N=12
(S+8)/(N+1)=(12A+8)/13

19363

m in !!!

19364
@jashholmes said:
The avarage age of group decreased by 3 years when a person 56 years old was replaced by another 20 years old . How would the average age have changed, if he were replaced by 2 persons of 32 years each instead???-JASH
8/13 years??
19365
@jashholmes said:
The avarage age of group decreased by 3 years when a person 56 years old was replaced by another 20 years old . How would the average age have changed, if he were replaced by 2 persons of 32 years each instead???-JASH
A diff of 36 years induced a diff of 3 years in Avg. Hence, there must have been 36/3=12 people in the original group. After replacing 56 with 2 persons of 32 each, a diff of 8 is induced.
Hence, change = 8/13 in Avg