Official Quant thread for CAT 2013

MBA CAT 2024
4181
@rachit_28 said:
bhai maine x = 0 and y = 1 maan liya, usse k= 1 ho gayathen x^2 + y^2 = 1 ho gaya.isme kya galat ho gaya ?
Nothing wrong in dis.. 😛 Ans "k" ki form me nikalna hai bas..Your answer is just one of the cases.. kyuki its given that x+y=k and k can be any number.. so answer should be in terms of "k" only.. Tumara answer is one of the cases only..nothing wrong 😃
4182
@ayushnasa said:
if 3 will hit( and this itself is wrong...we cant assume half will hit when the number of outcomes is so low...) Then we have a 100% prob of it destroying the target in 6 hits flat..I gave 6 as the answer on the same lines, but was pretty sure it was wrong..
I am still thinking the same which you thought but couldn't follow further.
Can you please make my life simple ?
4183
@ScareCrow28 said:
Let k be a fixed odd positive no. find minimun value of x^2 +y^2, where x, y are nonnegative interger and x+y =k ? Mai options nahi dunga
dy/dx=-1
2x-2y=0
x=y
but this not possible as k is odd so
minimum value of x and y =1 and 2
so 5 ???
4184
@ScareCrow28 said:
Let k be a fixed odd positive no. find minimun value of x^2 +y^2, where x, y are nonnegative interger and x+y =k ? Mai options nahi dunga
k^2/2....bolo bhai..

4185
@ScareCrow28 said:
Nothing wrong in dis.. Ans "k" ki form me nikalna hai bas..Your answer is just one of the cases.. kyuki its given that x+y=k and k can be any number.. so answer should be in terms of "k" only.. Tumara answer is one of the cases only..nothing wrong
aap tabhi options nhi dete :P
4186
@nick_baba said:
k^2/2....bolo bhai..
4187
@techsurge said:
dy/dx=-12x-2y=0x=ybut this not possible as k is odd so minimum value of x and y =1 and 2so 5 ???
Calculate in terms of "k" ..It depends on"k"
4188
@rachit_28 said:
I am still thinking the same which you thought but can't follow.Can you please make my life simple ?
Humari wrong thinking ke according out of 6 shots, 3 will miss and 3 will hit as probability is 0.5 . But we cannot assume half of the shots hit if the number of shots is so low. If the number of shots were, say, 10000. Then we can safely assume 5000 hit and 5000 miss.

So short me, when we say probability of hitting is 0.5 we do not mean that if 6 shots are fired, 3 will hit. We mean that if one shot is fired, the probability of it getting hit is 0.5
4189
@rachit_28 ..Ap padho..Edit kia last ka :P
4190
@rachit_28 said:
I am still thinking the same which you thought but can't follow.Can you please make my life simple ?
@ScareCrow28 , @shadowwarrior
probability that target is destroyed in 3 shots = 2c2(1/2)^3=1/8 =0.125
probability that target is destroyed in 4 shots = 3c2(1/2)^3*(1/2)^1=4/16=0.1875
probability that target is destroyed in 5 shots = 4c2(1/2)^3*(1/2)^2=10/32=0.1875
probability that target is destroyed in 6 shots = 5c2(1/2)^3*(1/2)^3=10/64=0.15625
probability that target is destroyed in 7 shots = 6c2(1/2)^3*(1/2)^4=15/128=0.1171875
probability that target is destroyed in 8 shots = 7c2(1/2)^3*(1/2)^5=21/256=0.08203125
probability that target is destroyed in 9 shots = 8c2(1/2)^3*(1/2)^6=0.0546875


0.125+0.1875+0.1875+0.15625+0.1171875+0.08203125+0.0546875=0.91015625

OA bhai log?

4191
@techsurge said:
2k^2+ 2k+1minimum 0
Minimum can't be 0..since k is an odd number.. minimum is "1" as calculated by @rachit_28 But, since we don't know what "k" is..hence we get the answer in terms of "k"..We can simply put values of "k" and get the answer
4192
@ScareCrow28 said:
Let k be a fixed odd positive no. find minimun value of x^2 +y^2, where x, y are nonnegative interger and x+y =k ? Mai options nahi dunga
2k^2 +1/2
x=k-1/2
y=k+1/2
4193
@ScareCrow28 said:
Minimum can't be 0..since k is an odd number.. minimum is "1" as calculated by @rachit_28 But, since we don't know what "k" is..hence we get the answer in terms of "k"..We can simply put values of "k" and get the answer
edited
posted new answer
4194
@ScareCrow28 said:
Minimum can't be 0..since k is an odd number.. minimum is "1" as calculated by @rachit_28 But, since we don't know what "k" is..hence we get the answer in terms of "k"..We can simply put values of "k" and get the answer
min value won't be 1 dost..

4195
@krum said:
@ScareCrow28 , @shadowwarrior probability that target is destroyed in 3 shots = 2c2(1/2)^3=1/8 =0.125probability that target is destroyed in 4 shots = 3c2(1/2)^3*(1/2)^1=4/16=0.1875probability that target is destroyed in 5 shots = 4c2(1/2)^3*(1/2)^2=10/32=0.1875probability that target is destroyed in 6 shots = 5c2(1/2)^3*(1/2)^3=10/64=0.15625probability that target is destroyed in 7 shots = 6c2(1/2)^3*(1/2)^4=15/128=0.1171875probability that target is destroyed in 8 shots = 7c2(1/2)^3*(1/2)^5=21/256=0.08203125probability that target is destroyed in 9 shots = 8c2(1/2)^3*(1/2)^6=0.05468750.125+0.1875+0.1875+0.15625+0.1171875+0.08203125+0.0546875=0.91015625OA bhai log?
bhai..OA to 7 tha..we also got the answer as 9 but naga sir got 7..you can check the post here http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-44541/p-1674757/r-1722099?page=374
4196
@nick_baba said:
min value won't be 1 dost..
Yes it will be dost Put x=0 and y=1
4197
@ScareCrow28 said:
Probability that I hit a target is 1/2. The target is destroyed if 3 shots hit the target. What is the minimum number of shots that I should try, so that the probability that I destroy the target is more than 0.9 ?1) 62) 73) 84) 9
(nc0 + nc1 + nc2)/2^n

min n = 9
4198
@ScareCrow28 said:
Yes it will be dost Put x=0 and y=1
and put x=y=1/2..still k will be odd integer and x^2 + y^2 will be 1/2..
4199
@ScareCrow28 said:
Yes it will be dost Put x=0 and y=1
whats the answer
is it not 2k^2 +1/2 ?
4200
@nick_baba said:
and put x=y=1/2..still k will be odd integer and x^2 + y^2 will be 1/2..
x and y are non negative integers :sneaky: