# Official Quant Thread for CAT 2012 [Part 1]Quantitative Report

Hi puys,
CAT 2011 season is finally over [smiley] [smiley]
Let start a new season of Quants thread for CAT 2012.. [smiley] [smiley]
Part-8
http://www.pagalguy.com/forum/quanti...at-2011-a.html (Official Quant Threa...
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Bhai sorry, I just read my post, it sounded ambiguous. What i meant was. All cubes leave a remainder of +1, -1 or 0 with 7.
Case 1 -> Remainder is 1.
4*1 + 1 + 5 = 10.
10 mod 7 = 3.
Case 2 -> Remainder is -1.
4*1 + (-1) + 5 = 9.
9 mod 7 = 2.
Case 3 -> Remainder is 0.
4*0...
2 2
A)0????
@SUMEET-YES,HAVE PUT ON THE SPECS
underlined part thoda samjhaoge.....
e.g. 15^3 leaves remainder 1 when divided by 7
A train starts from Delhi at a : b o'clock (i.e. 'b' minutes after a o'clock). It reaches Chandigarh on the same day at b : c o'clock after taking exactly 'c' hours and 'a' minutes. How many different values...
My take is 0.
x^3 will always leave a remainder of -1 with 7.
and x^6,ie, (x^3)^2 mod 7 wld be +1.
so, x^6 + x^3 + 5 = 5 (mod 7).
1 1
u overlooked that question says digit at ten's place should be 7
so it can't be 484 or 121
hota hai kabhi kabhi
1 1
sorry!gone blind
Koi bi element subset mein hoga ya toh ni hoga. So har 10 element ke pas 2 options hai. Hence 2^10.
but sir there is 121 and 484 as well
for 484 we get 2 as false rem(484)/13=3 for 121 all r false!
Oa is option 3.........
For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 is divisible by 7?
a. 0
b. 4
c. 11
d. 36
(A7A)^17 is a perfect square means A7A is also a prfct sq.
26^2 is 676 (remember this).
676 = 26*26 which is divisible by 13.
So 3 is false.
my take is opt2
sood bold part samjhaiyo !!
There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How man...
Square can be 676. Just out of memory as it is better to keep all sqaures till 30 in mind.
(A7A)^17 = (A7A)^16 * (A7A)
=> (A7A) is a perfect sqaure
=> Remainder by 13 is 0 and not 3.
2 2
OA is A < B.....
I think one of the ways to be sure is to take the first 3-4 terms and test.
N = (A7A)^17 is a perfect square. Which of the following statement is FALSE?
(1) A is an even digit.
(2) A is divisible by 3
(3) When N is divided by 13 we get remainder 3.
(4) None of t...
Here you go: