Official Quant thread for CAT 2013

MBA CAT 2024
18386
@iLoveTorres said:
is the answer 30?
No its not...its 12

Any one pls explain this one

"There are five cities in a state and each of them is to be connected to exactly two other cities using telephone lines. In how many ways can it be done?"

18387
9. The minimum number of real roots of f(x) = |x|^3 + a|x|^2+ b|x| + c,
where a, b and c are real, is
(A) 0,
(B) 2,
(C) 3,
(D) 6
18388
Suppose a1, a2, . . . , an are n positive real numbers with a1a2 . . . an = 1.
Then the minimum value of (1 + a1)(1 + a2). . .(1 + an) is
(A) 2^n
(B) 2^2n
(C) 1,
(D)None of the above
18389
@Exodia said:
x2 ā‚¬ā€œ 3y2 = 1376How many integer solutions exist for the given equation?(a) One (b) Two (c) Four (d) Zero


Plz give answer alongwith explanation.

This could be simplified to the form-
x2/1376-3y2/1376=1
This is analogous to the equation of hyperbola-
x2/a2-y2/b2=1 ;which is having integral solution only if x=a^m and y=b^n,which is not possible for the said values....Hence Zero integral solution is the answer.i am not sure but please tag me in OA.
18390
@Exodia said:
x2 ā€“ 3y2 = 1376How many integer solutions exist for the given equation?(a) One (b) Two (c) Four (d) ZeroPlz give answer alongwith explanation.
@saurabhlumarrai

OA: D) Zero

Explanation: 3y2 = x2 ā€“ 1376
As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3.
There are three possibilities for x in R.H.S:

(i) If x is multiple of 3, so is x2, and R.H.S. will leave a remainder of 1 when divided by 3.
(ii) If x is of the form 3m + 1, x2 will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, nāˆˆN

(iii) If x is of the form 3m + 2, x2 will be of the form 3n + 1 and R.H.S. will leave a remainder of 2.m, nāˆˆN So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists.

Happy CATing
18391
@Exodia said:
@saurabhlumarrai

OA: D) ZeroExplanation: 3y2 = x2 ā€“ 1376As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3.There are three possibilities for x in R.H.S:(i) If x is multiple of 3, so is x2, and R.H.S. will leave a remainder of 1 when divided by 3.(ii) If x is of the form 3m + 1, x2 will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, nāˆˆN(iii) If x is of the form 3m + 2, x2 will be of the form 3n + 1 and R.H.S. will leave a remainder of 2.m, nāˆˆN So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists. Happy CATing
Great Explanation boss !!!
18392
@ishu1991 said:
9. The minimum number of real roots of f(x) = |x|^3 + a|x|^2+ b|x| + c,where a, b and c are real, is(A) 0,(B) 2,(C) 3,(D) 6
puyss help me in this
18393
@Exodia said:
Function f(x) is a continuous function defined for all real values of x, such that f(x) = 0 only for two distinct real values of x. It is also known thatf(6) + f(8) = 0f(7).f(9) > 0f(6).f(10) 0 and f(1) 0III. f(7).f(8) 0(a) 1 (b) 2 (c) 3 (d) 4Plz give answer alongwith explanation.
Answer: b
Explanation:from above given situation,we can derive the result that the graph touches x-axis at two points-
and other observations are-
f(0)>0
f(1)
f(2)
f(3)
f(4)
f(5)
f(6)
f(7)>0
f(8)>0
f(9)>0
f(10)>0
So,option I and II are correct,III is definitely wrong and we cann,t comment on the sitaution of IV.
Altogether,we have 2 correct solutions.Please tag me in OA
18394
@ishu1991 said:
puyss help me in this
is it 0??
18395
@Logrhythm said:
is it 0??
i dont have OA
18396
@ishu1991 said:
i dont have OA
bhai frm whr do u get such ques for which u guys dnt hv an answer...
anyway...took a random equation and checked...take equation |x|^3 + |x|^2 - |x| + 1 = 0 would have no solution...hence my take is 0...
18397
@ishu1991 said:
puyss help me in this
the minimum no of roots will be zero when all a,b,c>0
18398
@leotheking Yes...It is 4500
18399
@vbhvgupta said:
Q 2
snipped
@vbhvgupta said:@chandrakant.k#18604#18604#18606

18400
@chandrakant.k said:
it is given that 18th tap fills the tank in 2 minutes. Lets say its rate =2^aso total capacity of the tank = 2^(a+1)now this means that tank 1 to 17 will take a time of 4 minutesso tank 1 to 16 will take a time of 8 minutestank 1 to 15 will take a time of 16 minutestank 1 to 14 will take a time of 32 mintuesso pipe 15 takes 16 minutesoptions me nahi hai
yar OA 54.....ye Time vale Q..
18401
@chandrakant.k said:
can you post the solution?? btw ye kaha se hai . Time ne maine bhi solve kiya hai lekin ye sab questions
bhai rite now in offc....will post in evening...
18402
@vbhvgupta said:
Q 2
let first tap's rate be k
sum till second tap = 2k, rate of 2nd tap = 2k
sum till third = 3k, rate = 6k
sum till fourth = 9k , rate = 18k
so the rates are in GP, with ratio = 3..
=> 18th tap = 2k(3)^(18-2) 2k(3)^16 = 1/2
=> k = 1/{4(3)^16}
15th tap = 2k(3)^13 = 2(3)^13/4(3)^16 = 1/2(3)^3
since rate = 1/time
hence time = 2(3)^3 = 54 minutes??


18403

solve Q 31

18404
@Logrhythm said:
let first tap's rate be ksum till second tap = 2k, rate of 2nd tap = 2ksum till third = 3k, rate = 6ksum till fourth = 9k , rate = 18kso the rates are in GP, with ratio = 3..=> 18th tap = 2k(3)^(18-2) 2k(3)^16 = 1/2=> k = 1/{4(3)^16}15th tap = 2k(3)^13 = 2(3)^13/4(3)^16 = 1/2(3)^3since rate = 1/timehence time = 2(3)^3 = 54 minutes??
sum till second tap = 2k kaisa?? only k is ther before that na?? and in the question it is given that sum of 1 to (n-1)

sorry twice miss kiya (edited)
18405
@vbhvgupta said:
post the approach??
total units worked each by each tap
tap 1 = x
tap 2 = 2x
tap 3= 2*3x
tap 4= 2*9x=2*3^2x
tap 5= 2*27x=2*3^3x
time taken by tap 18= 2*3^16x
now for 2*3^16x uits are done in 2 mins
in 15 tap will work= 2*3^13x
so time taken = total work done by tap 18 fo 2 mins/work done by 15 tap
2*3^16x * 2/2*3^13x =2*3^3=54mins