for 1> all the numbers 38,63,88 satisfied the condition. So more than 1 ans. Insufficient.
2>23,43,63,83 all satisfy the condition. So insufficient.

Combining 1 and 2 only 1 answer i.e. 63. So we need both .

I hope I am clear.

P.S. Sorry I am not Stuart

nailyad wrote:
When the positive integer n is divided by 25, the reminder is 13.What is the value of n?

1. n<100

2.When n is divided by 20 the reminder is 3.

My way of solving this:

n<100, there is only one number that can fit into this condition-63. So sufficient, right?

2. Again the same way, only one number.

1. n<100, I do not agree with the statement in red you made at all. 13, 38, 63 and 88 satisfy this. I think you overlooked other possibilities than 63.

2.When n is divided by 20 the reminder is 3. Note that if we consider only this statement we do not have any upper limit on n, 63 leaves a remainder of 13 when divided by 25 and leaves a remainder of 3 and when divided by 20. But, 163 also does the same and if you think about it 263 will also do the same. So, there are infinite number of possibilities of n 63,163,263..... Clearly insufficient


When you combine statement 1 and 2, there is only one value of n possible that is n=63
Hence C

Stmnt I : 13 and 63 both are possible , infact all such numbers of form 25p+13 ( where p is any integer from 0 onwards) , so you have many 13,38,63,89 --So insufficient

Stmnt II : Again number could of form 20q+3 ( where q is any integer from 0 onwards)
So you can have 3 , 23,43 , etc -so not sufficient

You know for sure only when both stmnts are combined

Let the quotient be q when we divide n by 100.
We can write
25q+13=n -------(1)
Also n is a positive integer,hence its possible values will be 13,38,63,88,113,138,163...........
I. n<100
Hence,n=13,38,63,88
So,Statement I is insufficient as it fails to give us a unique value of n.
II.When n is divided by 20 the reminder is 3.
If quotient in this case is Q,we can write
20Q+3=n
or, 100 Q+15=5n-------(2)
Multiply eqn (1) by 4,we get
100q+52=5n --------(3)
(3) - (2) gives,
100(q-Q) -37 =n
Hence possible positive solutions to above equation can be 63,163,263,363..........
So,Statement II is also insufficient as it fails to give us a unique value of n.
Combining I and II, n=63
Hence,option C i.e. both statements are needed.

The no is of the form n = 25x+13 cos when divided by 25 it leaves a remainder 13
(25x+13)/20 leaves a remainder 3
(25x+13) = (20x+5x+10+3) of this expression 20x will be divisible by 20
take 5x+10+3 which leaves a remainder 3 so (5x+10) will be divisible by 20
5(x+2)/20 = (x+2) /4 so when x = 2, 6, 10 so on it will be divisible by 4 , Possible values but not definite. So Insuff.

only when the value of n is restricted by n<100 we can get a definite value
x=2 , n= 50+13/25 = 2 13/25, 63/20 = 3 3/20
x= 6 , n = 150+13= 6 13/25, 163/20 = 8 3/20

so n= 63,
IMO C

THANK YOU ALL GUYS!!!!!