Official Quant thread for CAT 2013

MBA CAT 2024
10801
@Shray14 said:
A music shop sells cassettes of English music, Hindi film songs and Telegu film songs at a ratio of 2:4:5. If in the month of June the shop sold 452 cassettes of Hindi film songs, the number of cassettes of Telegu film songs is??
565??
10802
@Shray14 said:
A music shop sells cassettes of English music, Hindi film songs and Telegu film songs at a ratio of 2:4:5. If in the month of June the shop sold 452 cassettes of Hindi film songs, the number of cassettes of Telegu film songs is??
565
10803
@Torque024 said:
A1=5555, A2=5551, A3=5511; E=Select 2 5rs coins out of 4; P(M)=probablity of selecting 1rs coin
P(EA1)=1, P(EA2)=3C2/4C2=1/2, P(EA3)=2C2/4C2=1/6
P(A1)=1/(1+1/2+1/6)
P(A2)=1/2/(1+1/2+1/6)
P(A3)=1/6/(1+1/2+1/6)
P(THAT A COIN DRAWN IS OS 1rs)=P(A1)*P(M/A1)+P(A2)*P(M/A2)+P(A3)*P(M/A3)
=3/5*0+(3/10)*1/4+(1/10)*1/2=
1/8

@Torque024 bhai thoda detail mein explain kar do

didnt understood what is P(EA1) and how did u use combination here ??? :banghead:

i am getting 1/4 from my approach as

3 cases A=5555 B= 5511 C=5551

selecting a cases 1/3

no.of one in those cases

1/4*1/3 + 2/4+1/3 +0*1/3 =3/12=1/4

10804
@Torque024 said:
A1=5555, A2=5551, A3=5511; E=Select 2 5rs coins out of 4; P(M)=probablity of selecting 1rs coin
P(EA1)=1, P(EA2)=3C2/4C2=1/2, P(EA3)=2C2/4C2=1/6
P(A1)=1/(1+1/2+1/6)
P(A2)=1/2/(1+1/2+1/6)
P(A3)=1/6/(1+1/2+1/6)
P(THAT A COIN DRAWN IS OS 1rs)=P(A1)*P(M/A1)+P(A2)*P(M/A2)+P(A3)*P(M/A3)
=3/5*0+(3/10)*1/4+(1/10)*1/2=
1/8

@Torque024

Bhai please explain kaise kiya ... not getting what is F(EA1)... and how u used combination here

Pls tell what i did wrong

3 possible cases after 55 were recovered 5555 , 5511 , 5551

Now of these 3 cases getting number of 1s =1/4*1/3 +1/3*2/4 +0*1/3 =1/4

what did i miss

10805
@soumitrabengeri ya got it...
10806
@zealouszestful said:
In the image below: - AB and AD are tangent to the circle - BC and AD are parallelWhat is the length of AC?
bhai what is OA
10807
@gautam22 said:
sir 63 aa raha hai aapka kya aaya hai?
yar mere dimag ki bhujia ban gayi...nhn hua orally..firse try karunga dinner ke baad

In triangle ABC, AB = 13, BC = 15 and CA = 14. Point D is on BC with CD = 6. Point E is on BC such that ˆ BAE = ˆ CAD. Given that BE = p/q where p and q are relatively prime positive integers, find q.
OPTIONS

1) 338
2) 469
3) 463
4) 336


10808
@sujamait said:
yar mere dimag ki bhujia ban gayi...nhn hua orally..firse try karunga dinner ke baadIn triangle ABC, AB = 13, BC = 15 and CA = 14. Point D is on BC with CD = 6. Point E is on BC such that ∠BAE = ∠CAD. Given that BE = p/q where p and q are relatively prime positive integers, find q.OPTIONS1) 338 2) 469 3) 463 4) 336
463

angle BAE =angle CAD
Hence, angle BAD= angle CAE
Using Sine rule,

CD/AC = sin CAD/sin ADC = sin BAE/sin ADC
And, CE/AC = sinCAE/sinAEC=sinBAD/sin AEC
Multiplying both equations
CD*CE/AC^2 = sinBAE*sinBAD/sinADC*sinAEC
= sinBAE*sinBAD/sinADB*sinAEB
=BE*BD/AE^2
Hence CD*CE/AC^2 = BE*BD/AE^2
From here, BE = 2535/463
Hence q=463

10809
@gautam22 said:
sir B aur D ko join kar dofir aa jayega 28/sin2Θ=49/sinΘ...fir easily ho jayega agar 63 hi ans hai to.......aur sir doosre ka 463 hai.....ye vala ques pehle bhi ho chuka hai iss saal vali quant thread pe ya pata nahi pichle saal vali pe tha
apka woh samjha nhn yar BD wala...kya kiya hai age...
iska toh sahi hai 463.
10810
@gautam22 said:
sir AB=AD therefore ABD=ADB=Θ....fir alternate angle segment theorem lagao.....angle CDE=Θ aa jayega(E is a point wen we extend AD).....fir (AD+CDcosΘ)^2+(CDsinΘ)^2 =AC^2......Θ aapka aa jayega triangle ABD mein sine rule lagane se
E is point ? where how to get it ? not getting..:P

Find the remainder when 'a^p' is divided by 4. ('a' and 'p' are positive integers)
I. 'a' is a prime number greater than 126 but less than 148.
II. a = 2p +1.

10811
@sujamait said:
bhai what is OA
bhai oa to mujhe nahi pata. bade aascharya ki baat hai ki ye question mere se solve nahi hua, hours dimag marne ke bavzood.
10812
@gautam22 said:
sir AD(D ki side se) extend kar do...uspe ek point lelo E .....sir doosre ka CBD hai kya?
ABD me sine rule lagakar ? angle kaise aayega yar ?
uska OA nhn hai
10813
Look at this series: 8, 22, 8, 28, 8, ... What number should come next?

A. 9
B. 29
C. 32
D. 34
10814
@mbajamesbond said:
Look at this series: 8, 22, 8, 28, 8, ... What number should come next?A. 9 B. 29 C. 32 D. 34
32??
8 - 22(8*3 - 2)
8 - 28 (8*4 - 2^2)
8 - 30(8*5 - 2^3)
i know this is the worst logic anyone can think of 😛 😛
10815
@chandrakant.k said:
32??8 - 22(8*3 - 2)8 - 28 (8*4 - 2^2)8 - 30(8*5 - 2^3)i know this is the worst logic anyone can think of
Logic looks correct to me but OA is 34.
10816

nah....lemme come up with even more stupid logic......
8 seems to be constant in odd positions and in the even positions, numbers seem to be increasing with d=6....so i will go with (d)34...

@mbajamesbond is this correct by any chance...?

10817
Each root of the equation ax^3 – 7x^2 + cx + 231 = 0 is an integer. One of the roots is - 1/2 times the sum of the other two roots. What is the sum of all the possible values of a?

a17
b–7
c–17
dNone of these
10818
@shadowwarrior said:
nah....lemme come up with even more stupid logic......8 seems to be constant in odd positions and in the even positions, numbers seem to be increasing with d=6....so i will go with (d)34...@mbajamesbond is this correct by any chance...?
yes wasent complex...
10819
@mbajamesbond said:
yes wasent complex...
i thought more than necessary
10820
@gautam22 said:
sir AB=AD therefore therefore ABD=ADB=Θ....fir alternate angle segment theorem lagao.....angle CDE=Θ aa jayega..........now sir BC|| AD....therefore angle CDE=BCD=Θ and angle CDE=CBD(alternate angle theorem)....therefore BCD=CBD.....=>CD=BD......now in triangle ABD............ BD/sin(180-2Θ)= AB/sinΘ..........angle BAD=180-2Θ,AB=AD=49,ABD=ADB=Θ ,BD=28......hope it is clear nowsir vo doosre vale ka ans CBD hai kya?
yar OA nhn hai ..wohi lag rh ahai..
hmm was missing 1 condition..BD=DC..my eye cudn catch it.. 😛 thanks.