Official Quant thread for CAT 2013

MBA CAT 2024
9081
@ScareCrow28 Thanks :)
9082
@chillfactor said:
For how many integer values of n, n/(1450 - n) is a perfect square?

n/(1450 - n) = k^2

nk^2 - 1450k^2 + n = 0

n(k^2 + 1) - 1450k^2 = 0

n = 1450k^2/(k^2+1)

k^2 and k^2+1 always coprime


so, for n integer, 1450 should be divisible by k^2+1


we need to find out factors of 1450 that are in the form of k^2+1

1450 = 2*5^2*29

=> 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450

k^2+1 = 1, 2, 5, 10, 50, 145, 290, satisfies

so 7 solutions for n also





9083
@culdip said:
n/(1450 - n) = k^2nk^2 - 1450k^2 + n = 0n(k^2 + 1) - 1450k^2 = 0n = 1450k^2/(k^2+1)k^2 and k^2+1 always coprimeso, for n integer, 1450 should be divisible by k^2+1we need to find out factors of 1450 that are in the form of k^2+11450 = 2*5^2*29=> 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450k^2+1 = 1, 2, 5, 10, 50, 145, 290, satisfiesso 7 solutions for n also
Sir, Is zero (0) a perfect square ??
@chillfactor It's a debated issue counting 0 as a perfect square ..

0 is a null value, so there is no perfect square of something which is nothing.. 0 means lack of anything hence my doubt 😞 ..
9084
@culdip said:
n/(1450 - n) = k^2nk^2 - 1450k^2 + n = 0n(k^2 + 1) - 1450k^2 = 0n = 1450k^2/(k^2+1)k^2 and k^2+1 always coprimeso, for n integer, 1450 should be divisible by k^2+1we need to find out factors of 1450 that are in the form of k^2+11450 = 2*5^2*29=> 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450k^2+1 = 1, 2, 5, 10, 50, 145, 290, satisfiesso 7 solutions for n also
Yup, 7 solutions

n/(1450 - n) = k
n(k + 1) = 1450k

Here k and (k + 1) are coprime, so k + 1 will divide 1450
=> k + 1 will be a factor of 1450

1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450 are factors of 140

As, k is a perfect square, k can be 0, 1, 4, 9, 49, 144, 289

So, n can take 7 values
9085
@nole said:
How many different numbers which are smaller than 2.10^8 can be formed by using digits 1 and 2 only.?I am getting 256 because in the first place 1 is fixed then for the rest 8 places 2^8 ( we can have both 1 and 2 for each place) .But in arun sharma the answer is 766.Where did i go wrong ?
9 digit numbers = 2^8
8 digit number = 2^8
7 digit number = 2^7
6 digit = 2^7
5 digit = 2^5
.
.
.
1 digit = 2^1

total = 766
9086
@nole said:
How many different numbers which are smaller than 2.10^8 can be formed by using digits 1 and 2 only.?

I am getting 256 because in the first place 1 is fixed then for the rest 8 places 2^8 ( we can have both 1 and 2 for each place) .

But in arun sharma the answer is 766.Where did i go wrong ?
nine digit numbers=1*2^8 [1 fixed for first digit]
8 digit numbers=2^8
7---2^7
6--2^6
.
.
.
1-->2
so total =[2^8 + 2.[2^8 -1]]
=766 OA
9087
Solve for integers (x, y) such that:-

x(x + 1) = 4y(y + 1)
9088

"A thief escapes from city A at 2 pm and flees towards city B at 40 kmph. At 3 pm the police realize the escape and star chasing the thief at 50 kmph. Simultaneously, a police team from station B also starts towards station A to apprehend the thief at a speed of 60 kmph. What should be the distance between A and B such that both the police teams nab the thief at the same time?"


options : 200, 240, 400, 440, 480

edit: answer calculated: 440km, right answer
9089
@chillfactor said:
Solve for integers (x, y) such that:-x(x + 1) = 4y(y + 1)
4???
9090
@veertamizhan said:
"A thief escapes from city A at 2 pm and flees towards city B at 40 kmph. At 3 pm the police realize the escape and star chasing the thief at 50 kmph. Simultaneously, a police team from station B also starts towards station A to apprehend the thief at a speed of 60 kmph. What should be the distance between A and B such that both the police teams nab the thief at the same time?"options : 200, 240, 400, 440, 480
If distance between A and B is d, then after 1 hour distance between A and thief will be 40 and distance between thief and B will be (d - 40)

40/(50 - 40) = (d - 40)/(40 + 60)
d = 440
9091

The sum of series -

1/(rt2 +rt1) + 1/(rt2 + rt3) + .......+ 1/(rt120 + rt121) ???
9092
@Tina_angel said:
The sum of series -1/(rt2 +rt1) + 1/(rt2 + rt3) + .......+ 1/(rt120 + rt121) ???
rationalize..
1/(_/2+_/1) * {(_/2-_/1) / (_/2-_/1)} ............simialr way...

u will get _/121- 1 after cancelling out the terms....
9093
@chillfactor said:
Solve for integers (x, y) such that:-x(x + 1) = 4y(y + 1)
x(x + 1) = 4y(y + 1)
y^2+y=(1/4) * x(x+1)
f(x)=1/4*x(x+1)
f(1)=1/2, f(2)=3/2...6/2,10/2
nth term will be (1/4)(n*(n+1)...
y^2+y=(1/4)(n*(n+1)

y=[-1+/- sqrt(1+n*(n+1)]/2
Hence for x=n; y=[-1+/- sqrt(1+n*(n+1)]/2

Integer value of y is possible if n=0, or n=-1
Case n=0
x=0, y=[-1+/-sqrt(1)]/2= -1,0
Case n=-1
x=-1 y=0,-1
Hence four pairs
9094
@veertamizhan said:
"A thief escapes from city A at 2 pm and flees towards city B at 40 kmph. At 3 pm the police realize the escape and star chasing the thief at 50 kmph. Simultaneously, a police team from station B also starts towards station A to apprehend the thief at a speed of 60 kmph. What should be the distance between A and B such that both the police teams nab the thief at the same time?"options : 200, 240, 400, 440, 480edit: answer calculated: 440km, right answer
A----A'--------------B
At 3pm Thief(T) will be at A' , AA'=40(As speed is 40km/hr)
Relative speed of Pa and T = 50-40=10; Time=40/10=4hr
So according to question Time(A'B)=4hr
Relative speed of Pb and T=40+60=100
A'B/100=4; A'b=400; AB=AA'+A'B=40+400=440
9095
@Tina_angel said:
The sum of series -1/(rt2 +rt1) + 1/(rt2 + rt3) + .......+ 1/(rt120 + rt121) ???
Rationalise it

1/{√(n + 1) + √n} = {√(n + 1) - √n}/{(n + 1) - n} = √(n + 1) - √n

So, sum will be:-
S = √2 - 1 + √3 - √2 + √4 - √3 + .... + √121 - √120 = √121 - 1 = 10

Given that a + b = 5 and ab = 3, then what is the value of a⁴ + b⁴ ?
9096
@chillfactor said:
Rationalise it1/{√(n + 1) + √n} = {√(n + 1) - √n}/{(n + 1) - n} = √(n + 1) - √nSo, sum will be:-S = √2 - 1 + √3 - √2 + √4 - √3 + .... + √121 - √120 = √121 - 1 = 10Given that a + b = 5 and ab = 3, then what is the value of a⁴ + b⁴ ?
is it 343 ?
9097
@chillfactor said:
Rationalise it1/{√(n + 1) + √n} = {√(n + 1) - √n}/{(n + 1) - n} = √(n + 1) - √nSo, sum will be:-S = √2 - 1 + √3 - √2 + √4 - √3 + .... + √121 - √120 = √121 - 1 = 10Given that a + b = 5 and ab = 3, then what is the value of a⁴ + b⁴ ?
a^4+b^4=(a^2+b^2)^2-2*a^2*b^2=((a^2+b^2)^2-18---1
a^2+b^2=(a+b)^2-2ab=25-6=19--2
a^4+b^4=19^2-18=343
9098
@chillfactor said:
Given that a + b = 5 and ab = 3, then what is the value of a ´ + b ´ ?
a^2 + b^2 = 25 - 6 =19
a^4 + b^4 = 361 - 18 = 343.

9099
@Brooklyn said:
4???
@Torque024 said:
x(x + 1) = 4y(y + 1)
Hence for x=n; y=[-1+/- sqrt(1+n*(n+1)]/2Integer value of y is possible if n=0, or n=-1Case n=0x=0, y=[-1+/-sqrt(1)]/2= -1,0Case n=-1x=-1 y=0,-1Hence four pairs
Yup, 4 is correct.

x(x + 1) = 4y(y + 1)
4x² + 4x = 16y² + 16y
(4y + 2)² - (2x + 1)² = 3
(4y - 2x + 1)(4y + 2x + 3) = 3 = 1*3 = 3*1 = (-1)(-3) = (-3)(-1)

So, we will get 4 solutions
@rachit_28 said:
is it 343 ?
@Torque024 said:
a^4+b^4=(a^2+b^2)^2-2*a^2*b^2=((a^2+b^2)^2-18---1a^2+b^2=(a+b)^2-2ab=25-6=19--2 a^4+b^4=19^2-18=343
@akansh_1 said:
a^2 + b^2 = 25 - 6 =19a^4 + b^4 = 361 - 18 = 343.
Yup, 343 is correct.

Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?
9100
@chillfactor said:
Solve for integers (x, y) such that:-x(x + 1) = 4y(y + 1)

4 pairs of integers,i.e
(0,0), (0,-1),(-1,0),(-1,-1)