The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
59
59
Official Quant Thread for CAT 2012
Quantitative
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Please continue here for all the Quant queries and discussions.
The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
@[521683:harry31]
Post # 7185. See the solution.
Post # 7185. See the solution.
@[482349:sowmyanarayanan] YES MAN ITS 170..PLZZ TELL HOW TO SOLVE ?
170 Kg
Soln to already sabhi logon ne post kar diya hai
Soln to already sabhi logon ne post kar diya hai
1
(a-3) + (b-2) + (c+1) = a+b+c -4 = 9
So, AM >=GM->((a-3)+(b-2)+(c+1))/ 3 >= cuberoot( (a-3)*(b-2)*(c+1))=> 27 is the maximum value of RHS
So, AM >=GM->((a-3)+(b-2)+(c+1))/ 3 >= cuberoot( (a-3)*(b-2)*(c+1))=> 27 is the maximum value of RHS
2
for max.
a-3=b-2=c+1=k
a=k+3
b=k+2
c=k-1
a+b+c=13 => k=3
hence max value=3^3=27
a-3=b-2=c+1=k
a=k+3
b=k+2
c=k-1
a+b+c=13 => k=3
hence max value=3^3=27
1
A-3=b-2=c+1 should give the answer
Which is for A=6,b=5,c=2
So I would go with 27.
Which is for A=6,b=5,c=2
So I would go with 27.
2
If a + b + c = 13, what is the maximum value of (a-3) (b-2) (c+1) ?
a) 26b) 27c) 30d) 19
a) 26b) 27c) 30d) 19
Same to same bhai
Same assumption , same ans
Same assumption , same ans
1
I think there is no single value for this question. The value will depend upon the value of x,y,z and minimum could be 8. (When all are same )
Again if they asked what could the possible value among the choices , then 16 is the answer I guess.
Again if they asked what could the possible value among the choices , then 16 is the answer I guess.
why on earth am i getting 10 as the answer when all the cases i've supposed are real, distinct and positive and sum up to 1.
x=1/6, y=1/3, z=1/2
ans=5*2*1=10 not in the options
x=1/6, y=1/3, z=1/2
ans=5*2*1=10 not in the options
1
ur answer is correct,,i guess d question is nt framed correctly
@[521683:harry31]
More than 1 ans possible
More than 1 ans possible
If x, y, z are three distinct positive real numbers such that x+y+z=1, then the value of (1/x -1) (1/y-1) (1/z-1) is:
a) 16b) 8c) 4d) 1
a) 16b) 8c) 4d) 1
Suppose I say 1,2,3,4 What is the sum ? 10
And it is 2*5 but here we cannot find a series
But we can find a series from 4*(2.5) where 4 will represent the number of numbers in the series and 2.5 will represent the mean or average.
Are we using any fractions ? No. But we are just mod...
And it is 2*5 but here we cannot find a series
But we can find a series from 4*(2.5) where 4 will represent the number of numbers in the series and 2.5 will represent the mean or average.
Are we using any fractions ? No. But we are just mod...
Me too getting <40... if it is given dat(bx-ay)/b = (cy-bz)/c = (az-cx)/a and given that bx ≠ ay , cy ≠ bz, and az ≠ cx , which of the following is true 1) a+ b =c2) a+b+c = 03 ) ab+b c+ca = 04) ab+b c - ca = 0 Leaving for home bhailog... Happy weekend
2
@[577366:chevy] ok. how about my answer convincing?
yaaar shaayad kuch galti kar rha huun mein,,mere toh mre den 1 possible options aa rhe hain
@[521683:harry31] 170kg
1
since gold% is same in both the alloys and is 30% in the new alloy.
therefore, gold%=30, silver %=26; platinum%=44 in the second alloy.
By alligation we have,
z-44/40-z = 3/5=> z=340/8 (z is the %age of platinum in the new alloy)
therefore quantity of platinum = 400*(340/8)*(1/100)=1...
therefore, gold%=30, silver %=26; platinum%=44 in the second alloy.
By alligation we have,
z-44/40-z = 3/5=> z=340/8 (z is the %age of platinum in the new alloy)
therefore quantity of platinum = 400*(340/8)*(1/100)=1...
1
my take would be 39.
Le the Pen's price be Rs. 100 and let A be the portion of pens that were sold at 5% discount.
Then 60*100 < 95*A + (60-A)*110
A<40Hence 39. Am i right?
Le the Pen's price be Rs. 100 and let A be the portion of pens that were sold at 5% discount.
Then 60*100 < 95*A + (60-A)*110
A<40Hence 39. Am i right?
@[278489:subhakimi] Can we include fractions?
@harry31 said: Aman has 60 pens , a part of which he sells at 5%loss and the rest at 10% profit.He earned a profit on the whole . how many pens he sold at 5%loss? 1236482420
Is the question correct ?
Applying alligations we can say that for profit the pens at 5% loss should be at max twice as the number of pens sold at 10% profit. And there we have no profit
So number of pens sold at loss is 40 and our answer should be less than 40 but there are 4 options.
If the question were what cannot be the number of pens sold at 5% Loss then 48 is the answer.
Applying alligations we can say that for profit the pens at 5% loss should be at max twice as the number of pens sold at 10% profit. And there we have no profit
So number of pens sold at loss is 40 and our answer should be less than 40 but there are 4 options.
If the question were what cannot be the number of pens sold at 5% Loss then 48 is the answer.
This too shall pass ...
@[278489:subhakimi] Can we include fractions?
@[521683:harry31]
a+b = 60....eq 1
a...sold at lossb...sold at profit
no loss no gain condition gives:0.95a = 1.1b
So, a = 22/19 b
Put this in eq
b = 27.74
So b = 28 (it cannot be fractional)Any numbers of pens sold above 28(inclusive) out of 60 will yield a profit.
5% lo...
a+b = 60....eq 1
a...sold at lossb...sold at profit
no loss no gain condition gives:0.95a = 1.1b
So, a = 22/19 b
Put this in eq
b = 27.74
So b = 28 (it cannot be fractional)Any numbers of pens sold above 28(inclusive) out of 60 will yield a profit.
5% lo...
See for 24*25 you need not to take 25 numbers on each side you just need 12 numbers on each side.
Like from 12 to 36 .
In a similar way for 75*8 it is actually not coming from this form but from 37.5*16 where 37.5 is the mean and 16 is the number of numbers in the series , viz, 30 to 45.<...
Like from 12 to 36 .
In a similar way for 75*8 it is actually not coming from this form but from 37.5*16 where 37.5 is the mean and 16 is the number of numbers in the series , viz, 30 to 45.<...
but how did you arrive at 24? totally lost!!!
See for 24*25 you need not to take 25 numbers on each side you just need 12 numbers on each side.
Like from 12 to 36 .
In a similar way for 75*8 it is actually not coming from this form but from 37.5*16 where 37.5 is the mean and 16 is the number of numbers in the series , viz, 30 to 45.<...
Like from 12 to 36 .
In a similar way for 75*8 it is actually not coming from this form but from 37.5*16 where 37.5 is the mean and 16 is the number of numbers in the series , viz, 30 to 45.<...
is it 24???
24*.95+ 36*1.1 >60,,,so profit
my take : 24???
i guess mre den i answers possible,,true for 12 also,,, :/
24*.95+ 36*1.1 >60,,,so profit
my take : 24???
i guess mre den i answers possible,,true for 12 also,,, :/
1
@[592846:YouMadFellow] Yes thanks ,silly mistake,therefore answer should be 4 ryt?
170 kg?
1
I guess in case of 24*25 .. the numbers will be from 12 to 36 .. (-1) is not in the picture yet.
In the last case.. the least number will be (8-37) and not (8-75)... 8 is the middle term of the AP.
In the last case.. the least number will be (8-37) and not (8-75)... 8 is the middle term of the AP.
2
@[521683:harry31] 12??
@[577366:chevy] please look at my answer if i am wrong.
there are two alloys of gold , silver and platinum. the first alloy is known to contain 40%of platinum and the second alloy 26% of silver. the percentage of gold is same in both the alloys .Having alloyed 150kg of the first alloy and 250kg of the second, we get a new alloy that contains 30% of go...
@[278489:subhakimi] In how many ways can 600 be represented as the sum of two or more consecutive natural numbers ?
What can be the general formula of these type of sums ?Even i want general approach for questions like this ,i guess it is something to do with no being multiple of 4n and 4n+2.B...
What can be the general formula of these type of sums ?Even i want general approach for questions like this ,i guess it is something to do with no being multiple of 4n and 4n+2.B...
1
Aman has 60 pens , a part of which he sells at 5%loss and the rest at 10% profit.He earned a profit on the whole . how many pens he sold at 5%loss?
1236482420
1236482420
600 = 2^3*3*5^2
no of odd factors = 2*3=6
no. of ways in which 600 be represented as the sum of two or more consecutive natural numbers= 6-1=5???
no of odd factors = 2*3=6
no. of ways in which 600 be represented as the sum of two or more consecutive natural numbers= 6-1=5???
Let 7 neighbors sit in any way, 7 sisters can eat with same set of neighbors till 7 days
I didnt get this part.. Why will they dine for 7 days? P.S. thanks for the Editing suggestion. :)
I didnt get this part.. Why will they dine for 7 days? P.S. thanks for the Editing suggestion. :)
600 = 3*2^3*5^2
600/3 , 600/5 , 600/25 , 600/16 , 600/15 = 5 ways
don't know any formula :(
600/3 , 600/5 , 600/25 , 600/16 , 600/15 = 5 ways
don't know any formula :(
in this case, it should have been three going by this formula,
(13,14)
(8,9,10)
(2,3,4,5,6,7).
(13,14)
(8,9,10)
(2,3,4,5,6,7).
1
I dont know what is I here.. Had seen this sum a few pages back and was not solved. So pasted it again.
Okay , wrong post . No need to answer. Bit rusty :D
1
typical application of AP and factor of number.
Only the odd factors of 600 can be represented that way
eg . 600/3 = 200 , so we can have 199,200,201 .
Only the odd factors of 600 can be represented that way
eg . 600/3 = 200 , so we can have 199,200,201 .
600=2^3*3*5^2
no of odd factors = 6
hence no of ways in which this can be expressed as sum of consecutive numbers = no of odd factors -1 = 6-1=5
no of odd factors = 6
hence no of ways in which this can be expressed as sum of consecutive numbers = no of odd factors -1 = 6-1=5
1) total cases = 22C2 = 231 favorable cases = 8C2
28/231 = 4/33
2) 5C2/22C2 = 10/231
28/231 = 4/33
2) 5C2/22C2 = 10/231
@[278489:subhakimi]
Numbers of factors of 600:
600 = 2^3*3*5^2=> no of factors = 4*2*3 = 24
As sum has to be consecutive we neglect even powers of 600 i.e. factors left are only multiples of 3 and 5 = 2*3 = 6 in numbers
As it has to be 2 or more numbers total possibilities = 6-1 = 5
Numbers of factors of 600:
600 = 2^3*3*5^2=> no of factors = 4*2*3 = 24
As sum has to be consecutive we neglect even powers of 600 i.e. factors left are only multiples of 3 and 5 = 2*3 = 6 in numbers
As it has to be 2 or more numbers total possibilities = 6-1 = 5
1)
8/22 * 7/21 = 4/33
2)
5/22 * 4/21 = 10/231
8/22 * 7/21 = 4/33
2)
5/22 * 4/21 = 10/231
@[517487:shashikhawani] Is the answer is 9/8?
In how many ways can 600 be represented as the sum of two or more consecutive natural numbers ?
What can be the general formula of these type of sums ?
What can be the general formula of these type of sums ?
@[517487:shashikhawani] What is I?? Variable or imaginary i?? (root of -1)
1. 8C2/22C2 = 4/33
2. 5C2/22C2 = 10/231
2. 5C2/22C2 = 10/231
2
Let 7 neighbors sit in any way, 7 sisters can eat with same set of neighbors till 7 days
Now if we make any neighbor sit, then her 2 neighbors can be selected in 6c2 = 15 ways. so 15 such sets
15*7 = 105 days.. Not 100 % sure
Reg editing : hover to the top right corner, you will find a...
Now if we make any neighbor sit, then her 2 neighbors can be selected in 6c2 = 15 ways. so 15 such sets
15*7 = 105 days.. Not 100 % sure
Reg editing : hover to the top right corner, you will find a...
7 sisters dine at a round table. they dine together till each f them dine with different neighbors i.e. they donot like to dine with the same neighbors in any two arrangments. in a year how many days they dine together? Please explain this in detail. P.S. I am not able to edit the post in new...
Q) if one root of the eqn (I-m)x^2 + Ix +1=0 is double of the other and is real,find the greatest value of m a)9/8 b)8/7 c)8/6 d)7/5 e)5/7
In a bag, there are 10 black, 8 white, 5 red balls. Three balls are chosen at random and one is found to be black. 1. The probability that the rest two are white is a. 8/23 b. 4/33 c. 4/23 d. 5/23 2. The probability that the rest two are red is a. 10/231 b. 12/231 c. 12/363 d. 14/231 ...
16*9*x/(4*9)= 0.5*x*h
=>h= 8cm
=>h= 8cm
R = abc/4*area
9 * 4 * (1/2)*(a*c* sinB) = a*c*9
=> sinB = 1/2 = (h)/AB => h = 8
9 * 4 * (1/2)*(a*c* sinB) = a*c*9
=> sinB = 1/2 = (h)/AB => h = 8
@[317134:the.lordhimself]
Area =abc/4R
1/2 *B C*AD=16*9*BC/(4*9)
Solve it AD=8 CM
Area =abc/4R
1/2 *B C*AD=16*9*BC/(4*9)
Solve it AD=8 CM
d(d+10) + 17 mod (13 + d)
put d = 1
so , total number of students are = 14
put d = 1
so , total number of students are = 14
1
In Triangle ABC , AB=16cm and AC=9cm.What is the length of the perpendicular drawn from the vertex A to the side BC if the circumradius of the triangle is 9cm?