Official Quant thread for CAT 2013

MBA CAT 2024
12722
@anytomdickandhary what if the question would be 5x+12y=25 and what could be the min value of x^2+y^2?

here also will the Multiply first square later
12723

ABC is a three digit number such that ABC = 4 [AB + BC + CA], where AB, BC and CA are all two digit numbers. Find the total number of possible values for the number ABC.
a) 2
b) 1
c) 0
d) 3
e) more than 3

12724
@bhatkushal said:
@anytomdickandhary what if the question would be 5x+12y=25 and what could be the min value of x^2+y^2?here also will the Multiply first square later
yes..... exactly the same thing.

Infact it will hold for any number of pairs.

So identify the pair of numbers and put it in the inequality formula given by cauchy. And after that you might need a bit of tweeking which you can always do to get the answer.

ATDH.
12725
@ravi.theja said:
ABC is a three digit number such that ABC = 4 [AB + BC + CA], where AB, BC and CA are all two digit numbers. Find the total number of possible values for the number ABC.a) 2b) 1c) 0d) 3e) more than 3
definitely zero ..

jus write ABC as 100A + 10 B + C

now 4 [AB + BC + CA] = 44A+44B+44C

compare both eqn , we get 43/56 factor .. its clear that 43 is prime number .. so no numbers are possible

12726
@naga25french said:
definitely zero ..jus write ABC as 100A + 10 B + Cnow 4 [AB + BC + CA] = 44A+44B+44Ccompare both eqn , we get 43/56 factor .. its clear that 43 is prime number .. so no numbers are possible
792
12727
If x, y, z are any real numbers then the minimum possible value of x2 + 2y2 + z2 + 2yz subject of x + 2y + z = €“6 is
a)-6
b)6
c)12
d)None of these
12728
@ravi.theja said:
Find the least five digit number which when divided by 63, 56 and 42 leaves a remainder 1 in each case,A.10082B.10081C.10061D.10071
Lcm(7,14,21) = 42

so smallest 5 digit no. divisible by all 3 = 10038

req. no = 10038+42+1= 10081
12729
@ravi.theja said:
If x, y, z are any real numbers then the minimum possible value of x2 + 2y2 + z2 + 2yz subject of x + 2y + z = €“6 isa)-6b)6c)12d)None of these
6?
12730

Que)How to find the integral solutions that exist in 2x+9y=0 and y>=0, x and y are integers then find the number of solutions.........


I am done making the graph the line will pass through (0,10) and (45,0) now I need to find the integer solutions existing in the shaded area......
Can anybody tell me what is the shortcut way to find these solutions and the explaination......

12731
@bhatkushal u urself have told the asnwer...then what..!??
12732

How many no are coprime to 2304 and lie b/w 1200 and 2200? How to go for it?

12733

what is the answer??@bhatkushal

12734
@ravi.theja said:
Find the least five digit number which when divided by 63, 56 and 42 leaves a remainder 1 in each case,A.10082B.10081C.10061D.10071


Answer: B

the number would be of the form 63k+1, 56k+1, 42k+1
least number which leaves remainder 1 with 63,56 and 42 will be 505

least 5 digit number would be 505+504k... value of k will be 19 and least 5 digit number would be 10081

12735
@krum said:
6?
pls xpln the method sir
12736
@bvdhananjay said:
Answer: B

@bvdhananjay said:
least 5 digit number would be 505+504k... value of k will be 19 and least 5 digit number would be 10081


least 5 digit number would be 505+504k .
How we could conclude that? why 505+504k?
Can u pls elaborate?
Thnkx
12737
@ravi.theja said:
If x, y, z are any real numbers then the minimum possible value of x2 + 2y2 + z2 + 2yz subject of x + 2y + z = €“6 isa)-6b)6c)12d)None of these
given:

x+y+(y+z)=(-6)......(1)

we have to find the min value of :
x^2+2y^2+z^2+2yz

or, x^2+y^2+(y+z)^2...(2)

using cauchy inequalities

[x^2+y^2+(y+z)^2] *[1^2+1^2+1^2]>= [x+y+(y+z)]^2

[x^2+y^2+(y+z)^2]*3>=(-6)^2

[x^2+y^2+(y+z)^2]>=36/3

[x^2+y^2+(y+z)^2]>=12

therefore min value=12
12738
@cadmium said:
least 5 digit number would be 505+504k .How we could conclude that? why 505+504k?Can u pls elaborate?Thnkx
504 is the LCM of 63,56 and 42... so LCM is 504...
Least number leaving remainder 1 will be 504+1=505
Next number that wil satisfy this criteria would be 505+504=1009...

Hence least 5 digit number would be 505+504k
12739
@vbhvgupta said:
How many no are coprime to 2304 and lie b/w 1200 and 2200? How to go for it?
Number of co primes to 2304 and less than 2304 will be..768...
i dont know to make it fall in the range..someone enlighten
12740
@ravi.theja said:
pls xpln the method sir
sry, messed up with sign, 12 it is, u don't wanna know my method

12741
@pavimai said:
Number of co primes to 2304 and less than 2304 will be..768...i dont know to make it fall in the range..someone enlighten
😛 even i struck at the same place