The Official CAT 2007 Quant Thread

[utsavmamoria]

Originally Posted by kondapalli

Which is greater 99^11+100^11 or 101^11 ?

Taking 101^11 common

101^11[(99/101)^11 + (100/101)^11]

Cancelling 101^11 from both sides we are left with


(1 - 2/101)^11 + (1 - 1/101)^11 ............... 1

We can see that the sum of LHS will always be greater than RHS

So LHS > RHS

Konda bhai ..is this approach acceptable

[kekambus]

What for the problem n^2+3n+5 = 121k

Any starts anyone

[sachin2807]

Originally Posted by kondapalli

523abc is divisible by 8,7,9. What is the value of a.b.c?

For no. to be divisible by 9
10+a+b+c=9k
a+b+c=8
a+b+c=17
a+b+c=26(this one is not possible as only no. of this kind can be 998 which is not possible as abc has to be a multiple of
For number to be divisible by 7,
abc-523=7k
From this possible value of abc can be 530,537,544.........656
Now,the number 656 satisfies all above equations (multiple of 8,sum of the digits is equal to 17).
Therefore,value of abc=656(is this correct)

@konda....not able to find out any better approach other than this.

[utsavmamoria]

Originally Posted by kondapalli

How many natural numbers n are there such that n^2+3n+5 is divisible by 121?

n^2 + 3n + 5 = (n+1)(n+2) + 3

n+1 and n+2 are consecutive numbers

The last digit of (n+1)(n+2) is always even

so we need to find p such that

p(p+1) = 121k - 3

So we can say that K has to an odd natural number

Now i tried for a few odd values of k and saw that 121k - 3 does not yield any result of the form p(p+1)

So i conclude that there are no such natural numbers which will satisfy the given condition

Konda ..please tell the correct answer and post the solution

[stayinluv]

wrong post

[reachmonil]

Originally Posted by esh.nil

@konda.. completely agree with the statement..
the answer would be 14c4-1.(14c10-1)
Regarding the cat and dog problem, I am little skeptical abt the soln I gave yesterday.. I will discuss more abt it.. right now little work to catch on, will post when I get time..

Is it (14C4) - 1 or 14C(4-1)? How?

[kondapalli]

Originally Posted by utsavmamoria

n^2 + 3n + 5 = (n+1)(n+2) + 3

n+1 and n+2 are consecutive numbers

The last digit of (n+1)(n+2) is always even

so we need to find p such that

p(p+1) = 121k - 3

So we can say that K has to an odd natural number

Now i tried for a few odd values of k and saw that 121k - 3 does not yield any result of the form p(p+1)

So i conclude that there are no such natural numbers which will satisfy the given condition

Konda ..please tell the correct answer and post the solution

Right dude....there are no such values....

Great thinking....

@Esh,

you are right LHS > RHS.

SOln post kardo....Was that so much obvious to you

[kondapalli]

) Let f(f(x)) = 2f(x) – x for all real x. If f(f(f(f(f(f(7)))))) = 0 then f(7) =

(1) 5/6 (2) 6/7 (3) 1 (4) 7/8 (5) none of these


2) The number of rational points x = p/5 satisfying log(2x-3/4)/logx > 2, where p is an integer and gcd(p, 5) = 1 is/are(1) 2 (2) 3 (3) 5 (4) 1 (5) 4

3) An unlimited number of coupons bearing the letters A, B and C are available. What is the possible number of ways of choosing 3 of these coupons so that they can not be used to spell BAC?(1) 15 (2) 18 (3) 21 (4) 24 (5) 27

4) An integer N, 100 < N < 200, when expressed in base 5 notation has a final (i.e., rightmost) digit of 0. When N is expressed in base 8 notation and in base 11 notation, the leading (i.e., leftmost) digit is 1 in both cases. Then N when divided by 7 gives a remainder of (1) 1 (2) 2 (3) 3 (4) 6 (5) cannot be determined


5) The fraction 2/7 can be written in a unique way as the sum of two unit fractions 1/a + 1/b, where a and b are positive integers with a < b. Then a + b is

(1) 16 (2) 20 (3) 24 (4) 28 (5) 32

6) Number y is defined as the sum of the digit of the number x, and z as the sum of the digits of the number y. How many natural numbers x satisfy the equation x+y+z = 60?
(1) 1 (2) 3 (3) 4 (4) 2 (5) more than 4

[utsavmamoria]

Originally Posted by kondapalli

)
5) The fraction 2/7 can be written in a unique way as the sum of two unit fractions 1/a + 1/b, where a and b are positive integers with a < b. Then a + b is

(1) 16 (2) 20 (3) 24 (4) 28 (5) 32

6) Number y is defined as the sum of the digit of the number x, and z as the sum of the digits of the number y. How many natural numbers x satisfy the equation x+y+z = 60?
(1) 1 (2) 3 (3) 4 (4) 2 (5) more than 4

5) The fraction 2/7 cab be expressed as 1/4 + 1/28

Thus a+b = 4 + 28 = 32

6) The numbers are 47 and 50

47 + 11 + 2 = 60

50 + 05 + 5 = 60

Thus the answers are 5 and 4 respectively

[sachin2807]

Originally Posted by kondapalli

)

4) An integer N, 100 < N < 200, when expressed in base 5 notation has a final (i.e., rightmost) digit of 0. When N is expressed in base 8 notation and in base 11 notation, the leading (i.e., leftmost) digit is 1 in both cases. Then N when divided by 7 gives a remainder of (1) 1 (2) 2 (3) 3 (4) 6 (5) cannot be determined

Number will be 125 and hence when divided by 7 gives 6 as remainder.
(125)(in base 5)=1000

(125)(in base =175
(125)(in base 11)=104