When positive integer x is divided by 7 the quotient is q and the rema

What do you mean? Subtracting we get x = 1 ?

You don't get x=1 when you subract, you'd get 0=2q....also when you divide 1 by 10 the remainder is 10..

N=7q+1
N=5q+1

Therefore: N=35X +1

The remeinder could be 1 or 6. for Example if X=0 then the remeinder is 1, but when X=1 then the remeinder is 6.

Precisely what i thought but from Question stem and St 1 we have
but note that it is for the same value of q that we get the remainder 1

x= 7q+1 and x=5q+1 -------> For the same value of q

Also 7q+1=5q+1 --------> 2Q=0 or q=0

Thus the number is 1 and when divided by 10 the remainder is 1 only.

Ans A is correct

if the Question would have stated x= 7q+1 and x=5a+1 then x= 35C+1 and possible value of x = 1,36,71, 106,141.....

then even taking statement 2 the answer would be E. The Question stem needs to mention that x is a 2 digit number.

Hope it helps

For the 2 values highlighted in bold, the remainders will be 6 and 1 respectively.

You need to consider 1 as well the possible value of x....here's why

x=7q+1 when q=0, x=1
q=1,x=8

Similarly x =5q+1, q=0, x=1
q=2,,x=6....and so on

From the question stem, x =7q+1.

F.S 1 states that x = 5q+1. Thus, we know that 7q+1=5q+1 --> q=0.Note that q is also an integer. Thus, x=1.Remainder when 1 is divided by 10 is 1. Sufficient.

F.S 2 for x=8, remainder when divided by 10 is 8.However, when x=15,the remainder when x is divided by 10 is 5. 2 different numerical values, hence Insufficient.

A.



Nope . This is incorrect.

Hi,

I meant if you subtract these equations you get q =0. Now put q =0 in any equation and you will get x =1. Hence the no is 1. Now when you divide x (which is 1) by 10, you will get remainder 1. Sorry for the confusion.

Hi,
Full solution to avoid any confusion-
x= 7q + 1
from 1) x = 5q + 1
Now subtracting these two equations yields q =0. Now put q=0 in any equation and you will get x=1. hence the no is 1. Now if we divide x (which is nothing but 1) we will remainder as 1.

Stmt 1 made it so simple that I assumed there must be typo in the question and solved with x=5p+1 and got C as the answer

if x = 7q+1 and x=5q+1 only value that q can take is ZERO!!! How can stmt 1 be so simple

Hi,
IMO C. Please correct me if I am wrong.

from stmt 1: x = 7q+1 and x=5q+1 which leads to .. x=35Q+36 as the general formula.
hence multiple values of x will satisfy this. 36, 71, 106 .. you cannot say if the remainder will be 1 or 6.

from stmt2 x < 50 .. not sufficient.

from stmt1 and stmt 2 .. x can be only 36. Hence C.

Statement 1: this tells us that when x is divided by either 5 or 7 the quotient is the same and so is the remainder.
That is,
7q+1=5q+1
7q=5q
q=0

This is possible only in the case where X=1. Sufficient.

Statement 2: this gives rise to multiple cases which yield different answers each time. Insufficient.

IMO the answer should be A.

If you thought my answer was useful please hit the kudos button!

From the question we get; x = 7q +1 ---------- (i)
(1) x = 5q + 1 --------- (ii)
Equating (i) and (ii), we get;
7q +1 = 5q +1
7q - 5q = 1-1
2q = 0
q = 0

Substituting q = 0 we can get the value of x. x = 7q +1 ==> x =1
x divided by 10. 1/10. remainder will be 1. (1) is sufficient.

(2) x is less than 50
7q+1<50
7q < 49
q < 7
There are multiple values we for x in this case. Hence (2) is not sufficient.
Answer A...

OS -> x = 7Q + 1
X = 10P + R? What's R?

S1 -> x = 5Q + 1
The statement says Q is the same when the remainder is 1.

x = 1, 6 , 11, 16, 21,.... Plugging in value of Q = 0,1,2,.. in S1
x = 1, 8 , 15, 22, 29,... Plugging in value of Q = 0,1,2,... in the Original Statement

The only value where Q is the same in both cases is when x = 1
=> Can find out R when x is divided by 10.
=> Sufficient.

To clarify, the remainder is also 1 when x = 36, but the quotient changes which is not in line with the given information => it's irrelevant to the question


S2 -> x < 50
Many possible values => Insufficient.

Therefore, the correct answer is A.

IMO E

x=7q+1
x=5q+1

Two value of q 0 and 6 will give different remainders when divided by 10 i.e. 1 and 6

Waiting for the OA and Kudos

x= 7q + 1
from 1) x = 5q + 1
subtracting eqn 1 from given, we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. Sufficient.
stmt 2 is clearly insufficient.
Hence, A

Bunuel the method you mentioned in response to seabhi has been discussed above. i'm curious -- if he's wrong, how were his calculations incorrect? i built out the #s given from the formulas and got the same thing as him (see below):

- x=5q+1, so: 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71
- x = 7q+1, so: 8, 15, 22, 29, 36, 43, 50, 57, 64, 71

Hi,

I think the confusion with this problem is that you neglecting that q in the question stem and in the first statement (1) have to be the same value.

So, from the question stem \(\frac{x}{7}\) = q remainder 1. Or x = 7q+1
which means when q=0 then x = 1. And when q=5, then x =36.

And from statement (1) \(\frac{x}{5}\) = q remainder 1. Or x=5q+1
Thus from this statement if q=0 then x = 1 and if q=7 then x =36

Now, what students are doing incorrectly is assuming since x = 1 and x=36 so they are marking this as insuff. "the quotient is q" in the question stem and "the quotient is q" in statement (1) are of the same value. So when x=1, then q = 0 both in question stem and statement (1). But when x=36, then q = 5 in the question stem and q = 7 in statement (1), so q is not of the same value, and we can ignore x = 36 as a valid solution and only take x = 1 as a valid solution. And of course x=1 divided by 10 leaves remainder of 1.