CAT 2007: Quantitative Questions a Day 134 till the end - The Discussions

[Aarav]

This will be the last thread of this season for QQAD. Hope you make the maximum from this

Good luck to all.

[Aarav]

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Quantitative Question # 134
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Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

Part (B)

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

[Rohit .Sinha]

Originally Posted by Aarav

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Quantitative Question # 134
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Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

Part (B)

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

1.

(x^2 + y^2)^2 - 2x^2*y^2=b
((x+y)^2 - 2xy)^2 - 2x^2*y^2 =b
or
(a^2 - 2xy)^2 - 2x^2*y^2 =b
a^4 + 4x^2*y^2 - 4x*y*a^2 - 2x^2*y^2 =b
a^4 + 2x^2y^2 - 4x*y*a^2 -b=0

let this be quadratic in xy

since they are real,

16a^4 >= 8(a^4 -b)
a^4 +b >= 0

Hence (5)
------------------------
Rohit

[Rohit .Sinha]

Originally Posted by Aarav

------------------------------------------------------
Quantitative Question # 134
------------------------------------------------------

Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

Part (B)

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

part (2)

only n = 3,2

Hence answer(1)
-------------
Rohit

[frigid sapphire]

part b choice(1)

for 10+x/110+x =20+y/120+y =30+z/130+z

x-y=10, y-z=10

so the values of x, y z cud be

21, 11,1 22,12,2 23,13,3........

now 130+z/30+z should be a positive int.
therfore (30+z)n =130+z
z=130-30n/n-1
n=2,n=3 satisfy.
so ans is 2.

[kondapalli]

Originally Posted by Aarav

------------------------------------------------------
Quantitative Question # 134
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Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

Part (B)

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

x = 10(11-n)/(n-1); y = 10(12-2n)/(n-1) ; z = 10(13-3n)/(n-1)

clearly 3n<13 => n<=4; hence n can be 2,3,4. But 4 doesn't satisfy z.

Hence n = 2 and n = 3 satisfy. Hence 2 distinct solutions.

[prade]

Originally Posted by Rohit .Sinha

1.

(x^2 + y^2)^2 - 2x^2*y^2=b
((x+y)^2 - 2xy)^2 - 2x^2*y^2 =b
or
(a^2 - 2xy)^2 - 2x^2*y^2 =b
a^4 + 4x^2*y^2 - 4x*y*a^2 - 2x^2*y^2 =b
a^4 + 2x^2y^2 - 4x*y*a^2 -b=0

let this be quadratic in xy

since they are real,

16a^4 >= 8(a^4 -b)
a^4 +b >= 0

Hence (5)
------------------------
Rohit

if x and y are complex and x = m+in and y =m-in
then x.y = m^2-n^2 is real. so taking xy as real will not be the necessary and sufficient condition.

[Tabloid]

134 Part A.
x^4 + (x-a)^4 = b
let x^2 = z
2z^2 - 2a^2z +a^4-b = 0

for real roots 2b >=a^4

Ans. 5

134 Part b
applying componendo and dividendo
100/120+2x = 100/140+2y = 100/160+2z = n-1/n+1
60+x=70+y=80+z=50(n+1)/(n-1)=a
x,y,z can take many values for this equation.
but for all of them to be positive integer together.
(n+1)/(n-1) can be integer only for n=2,3
giving out a= 100,150 and integer values of x,y,z

Ans 1

[prade]

Originally Posted by Aarav

------------------------------------------------------
Quantitative Question # 134
------------------------------------------------------

Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

x^4-ax^3 +3a^2x^2 -a^3x + (a^4-b)/2 = 0 = (x^2+px+q)(x^2+mx+n)
p+m = -a
n+q+pm = 3a^2
qm+pn = -a^3
qn = (a^4-b)/2

for real roots p^2-4pq >= 0
and m^2-4mn > = 0

aarav help needed...

[prade]

Originally Posted by Aarav

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Quantitative Question # 134
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Part (B)

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

n=2,3 are the only possible solutions option 1