sureshbala's posts

sureshbala
replied to Number System - Questions & Discussions

@bpolagani Remainder when 2^1100 is divided by 101Since the divisor 101 is prime make use of the Fermat's rule. Fermat's theorem says that if p is prime and p does not divide a, then a^(p-1) when divided by p the remainder is 1.Hence, 2^100 divided by 101, the remainder is 1 => 2^1100 when divid...

1
1

sureshbala
replied to percentages and profit,loss and interest

gains 0.5%(X) and loses 0.25%(X) is as good as loosing 0.25%(X)Hence in two weeks the clock looses 0.25%(7x24x60) = 25.2 minutes = 25 mins 12 secThus, after two weeks the clock shows 12:00 noon - 25 mins 12 sec = 11:34:48

2

sureshbala
replied to Permutations & Combinations - Questions & Discu...

With oranges i have 5 chances namely -> picking one orange or two oranges or three oranges or four oranges and not picking up an orange at all i.e. 5 chancesSimilarly with mangoes 4 chances and with apples 3 chances.

Total chances = 5X4X3 = 60. But I cannot leave out all of them since i need ...

Total chances = 5X4X3 = 60. But I cannot leave out all of them since i need ...

2

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

Same here as well...

Even if you consider only integral percentages it comes down to 98

Even if you consider only integral percentages it comes down to 98

1

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

Two cuts will give us three pieces

If both the cuts are on the same half of the stick, one resulting piece will have a length more than or equal to half of the length of stick thus violating the rule that the sum of two sides must be greater than the third side.

Thus if both the cuts are on...

If both the cuts are on the same half of the stick, one resulting piece will have a length more than or equal to half of the length of stick thus violating the rule that the sum of two sides must be greater than the third side.

Thus if both the cuts are on...

10

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

Number is 42857

Let the 5 digit number be n

Adding 1 at the beginning, the resultant number can be written as 10^5+n

Adding 1 at the end, the resultant number can be written as 10n+1

Given 3(10^5+n) = 10n+1

=> 7n = 299999 => n = 42857

Hence sum = 26

Bed Time...

Let the 5 digit number be n

Adding 1 at the beginning, the resultant number can be written as 10^5+n

Adding 1 at the end, the resultant number can be written as 10n+1

Given 3(10^5+n) = 10n+1

=> 7n = 299999 => n = 42857

Hence sum = 26

Bed Time...

1

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

29786

850

850

-------

31486

850

850

-------

31486

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

2012 = 2^2 * 503

Let a = 2^x1 * 503^y1 and b = 2^x2 * 503^y2

(x1,x2) => 5 chances

(y1,y2) => 3 chances

Total 5*3 = 15 numbers

Let a = 2^x1 * 503^y1 and b = 2^x2 * 503^y2

(x1,x2) => 5 chances

(y1,y2) => 3 chances

Total 5*3 = 15 numbers

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

23/29

s12/s15 = 12^2/15^2

=>12/2/15/2= 144/225

=>d = 2a

T12/T15 = a+11d/a+14d = 23a/29a = 23/29

s12/s15 = 12^2/15^2

=>12/2/15/2= 144/225

=>d = 2a

T12/T15 = a+11d/a+14d = 23a/29a = 23/29

1

sureshbala
replied to Official Quant Thread for CAT 2012 [part 2]

31 I just made use of the choices