CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[satanica]

Originally Posted by implex

...
for part 2)
as a+b+c=5 sum of roots is 5 and ab+bc+ca=3( product of roots 2 at a time ) is 3

now product abc is not given we assume it to be t where t is positive
f(m)=m^3-5m^2+3m-t=0
f'(m)=3m^2-10m+3
f'(m)=0 has two roots 1/3 and 3
so f(m) has its three roots, one less than 1/3 other between 1/3 and 3 and the the last greater than 3
this can be seen from rolle's theorem..
from the other values let a=b then we get a quadratic where a=1/3 is one of the roots
we can find c=5-2/3

How do you justify your taking a=b=(1/3) ?
As far as Rolle's Theorem could take you, you got to one root less than equal to 1/3 and other between 1/3 and 3 and the third greater than equal to 3. Now you'll have to prove that one root being (1/3 + δ) and other being (1/3 - δ) will not increase the value of 'c'.

[implex]

Originally Posted by satanica

How do you justify your taking a=b=(1/3) ?
As far as Rolle's Theorem could take you, you got to one root less than equal to 1/3 and other between 1/3 and 3 and the third greater than equal to 3. Now you'll have to prove that one root being (1/3 + δ) and other being (1/3 - δ) will not increase the value of 'c'.

there is no such compulsion in rolle's

the roots can be equal as well.
one root less than 1/3 and other between 1/3 and 3
does not mean that both of them can't be 3.

in rolle's theorem when we identify the intervals, its closed at both ends!!

[BORADESA]

Originally Posted by kharesaurabh84

Let x and y be positive real numbers such that x^3 + y^3 = 4x^2. A is the maximum value of x + y.


Let a, b, c be real such that a+b+c = 5 and ab + bc + ca = 3. B is the largest possible value of c.

Then A + B lies in the range



(1) [7, (2) [8, 9) (3) [9, 10) (4) [10, 11) (5) none of these


Soln.
x^3 + y^3 =4x^2
(x+y)^3-3xy(x+y)=4x^2
=>replacing x+y by A and y by A-x
(4-3a)x^2 + 3A^2.x - A^3 = 0
for real values of x
D>0
9A^4+4A^3(4-3A) > 0
A (- [0,16/3]
Amax 16/3

a+b+c = 5 and ab + bc + ca = 3
replace a by (5-b-c) in second eq. and form a quadratic in b
=>b^2 + (c-5)b + (c^2-5c+3) = 0

again for real b
D>0
-3c^2+10c+13>0
C (- [1,13/3]
Cmax=13/3
Amax + Cmax =29/3=9.66

option (3) is correct

thats the best approach
Thanx

[pavanpadekal]

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Quantitative Question # 014
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Let x be the number of base systems in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number. Then x in base 7 is a


(1) odd but not prime (2) prime (3) even and perfect square (4) even but not perfect square (5) none of these

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[pavanpadekal]

N=abc
(3 digit number in base 6)
N=a*6^2+b*6^1+c*6^0
To find the maximum value for this
Maximum value of N = 5*6^2+5*6^1+5*6^0=180+30+5=215
Hence the largest 3 digit perfect square
i.e.,14^2=100+80+16=196
N=5*6^2+2*6^1+4*6^0=180+12+4=196=524 base 6
N=pq=p*l^1+q*l^0=196(2 digit number in an arbitrary base l)

Obviously l<=196 BUT >=38 as 196/5=37.2
Hence x=196-38+1=197-38=159
Expressing in base 7
159=3*7^2+1*7^1+5*7^0
x=375 to base 5
Hence
(1) odd but not prime

Obviously l<197 BUT >14
Hence x=196-14=182
Expressing in base 7
182=3*7^2+5*7^1+0*7^0
x=350 to base 7
Hence (4) even but not perfect square
A strong feeling that I have not understood the problem!

[nbangalorekar]

Originally Posted by pavanpadekal

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Quantitative Question # 014
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Let x be the number of base systems in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number. Then x in base 7 is a


(1) odd but not prime (2) prime (3) even and perfect square (4) even but not perfect square (5) none of these

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the square which will be the largest in base 6 is 196.
now 196 can be expressed as a 2 digit number in (196-14)=182 ways
182 in base 7 will give 350 which is even but not a perfect square.
so answer is option '4'

[slam]

Originally Posted by pavanpadekal

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Quantitative Question # 014
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Let x be the number of base systems in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number. Then x in base 7 is a


(1) odd but not prime (2) prime (3) even and perfect square (4) even but not perfect square (5) none of these

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good to be back in the morning session. Here's how I approached this:
max 3 digit number in base 6 = (555)in base 6 = (215) inbase10
=> largest perfect square that can be accomodated is 196

Now, let the base of the number system that can accomodate 196 in two digits be y.
=> (y-1)y + (y-1) >= 196 (the largest two digit nbr in base y)
=> y^2 >= 197 which means y >= 15 (since y is a positive integer)

Also,
y(1) + 0 <= 196 (the smallest two digit number in base y)
=> y <= 196

=> number of values of y that may satisfy this = 196 - 15 + 1 = 182

=> x = (182) in base 10
= (346) in base 7

x = 182 in base 10 => even but not a perfect square => answer option (4).
(I think we should be checking odd/even and prime/composite properties of x in base 10 itself, as these will be the same no matter which base you express the number in. Is this correct?)
Please correct me if required.

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slam.

[slam]

Originally Posted by nbangalorekar

the square which will be the largest in base 6 is 196.
now 196 can be expressed as a 2 digit number in (196-14)=182 ways
182 in base 7 will give 350 which is even but not a perfect square.
so answer is option '4'

196 should not be included here as in base 196 it will be a 1 digit number rather than a 2 digit number.right?

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slam.

[nbangalorekar]

Originally Posted by slam

196 should not be included here as in base 196 it will be a 1 digit number rather than a 2 digit number.right?

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slam.

196 in base 196 will be 10. definitely a 2 digit no.

[shankey_595]

a largest number in base 6 is 555
so d largest square which can be represnted in base 6 is 441
so d equivalent of 441 in base 10 is 169.....
let x be d numbers in certain bases where 169 can be represnted as a two digit number
so at max base could be 169 and base should be greater den 13
so d number of x= 169-14+1=156
though i have a doubt in dis in base 16 can we take A9 as a two digit number?



converting 156 into base 7whic is equal to 312 hence option 4