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When positive integer x is divided by 7 the quotient is q
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18 Nov 2013, 08:55
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When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10? (1) When x is divided by 5 the quotient is q and the remainder is 1 (2) x is less than 50  I got the correct answer but took quite a while. I listed out all the possible value of x and realize remainder when x is divided by 10 has to be 1. Any other faster method? 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85 6, 11, 16, 21, 26, 31, 36, 41, 46, 56, 61, 71
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Re: When positive integer x is divided by 7 the quotient is q
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21 May 2014, 06:36
seabhi wrote: 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. No, the correct answer is A. When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?When positive integer x is divided by 7 the quotient is q and the remainder is 1 > x = 7q + 1 > x could be 1, 8, 15, 22, ... (1) When x is divided by 5 the quotient is q and the remainder is 1 > x = 5q + 1 > subtract this from the equation from the stem: 0 = 2q > q = 0 > x = 1. 1 divided by 10 gives the remainder of 1. Sufficient. (2) x is less than 50 > if x = 1, then the remainder upon division 1 by 10 is 1 but if x = 8, then the remainder upon division 8 by 10 is 8. Not sufficient. Answer: A. Hope it helps.
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Re: When positive integer x is divided by 7 the quotient is q
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19 Nov 2013, 03:56
from the ques stem we know x= 7q + 1 from 1) x = 5q + 1 subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient. stmt 2 is clearly insufficient. lots of possible values for x.



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Re: When positive integer x is divided by 7 the quotient is q
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19 Nov 2013, 08:29
zerosleep wrote: from the ques stem we know x= 7q + 1 from 1) x = 5q + 1 subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient. stmt 2 is clearly insufficient. lots of possible values for x. What do you mean? Subtracting we get x = 1 ?



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Re: When positive integer x is divided by 7 the quotient is q
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19 Nov 2013, 10:11
zerosleep wrote: from the ques stem we know x= 7q + 1 from 1) x = 5q + 1 subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient. stmt 2 is clearly insufficient. lots of possible values for x. 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..



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Re: When positive integer x is divided by 7 the quotient is q
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19 Nov 2013, 11:20
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.



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Re: When positive integer x is divided by 7 the quotient is q
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20 Nov 2013, 00:16
bluecatie1 wrote: 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
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Re: When positive integer x is divided by 7 the quotient is q
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20 Nov 2013, 00:21
registerincog wrote: When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?
(1) When x is divided by 5 the quotient is q and the remainder is 1 (2) x is less than 50
 I got the correct answer but took quite a while. I listed out all the possible value of x and realize remainder when x is divided by 10 has to be 1. Any other faster method?
8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85 6, 11, 16, 21, 26, 31, 36, 41, 46, 56, 61, 71 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
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Re: When positive integer x is divided by 7 the quotient is q
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20 Nov 2013, 00:31
registerincog wrote: When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?
(1) When x is divided by 5 the quotient is q and the remainder is 1 (2) x is less than 50
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. AccipiterQ wrote: when you divide 1 by 10 the remainder is 10.. Nope . This is incorrect.
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Re: When positive integer x is divided by 7 the quotient is q
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20 Nov 2013, 06:08
AccipiterQ wrote: zerosleep wrote: from the ques stem we know x= 7q + 1 from 1) x = 5q + 1 subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient. stmt 2 is clearly insufficient. lots of possible values for x. 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.. 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.



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Re: When positive integer x is divided by 7 the quotient is q
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20 Nov 2013, 06:10
registerincog wrote: zerosleep wrote: from the ques stem we know x= 7q + 1 from 1) x = 5q + 1 subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient. stmt 2 is clearly insufficient. lots of possible values for x. What do you mean? Subtracting we get x = 1 ? 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.



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Re: When positive integer x is divided by 7 the quotient is q
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26 Dec 2013, 06:55
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
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Re: When positive integer x is divided by 7 the quotient is q
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21 May 2014, 05:02
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.
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Re: When positive integer x is divided by 7 the quotient is q
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02 Jul 2017, 10:03
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



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Re: When positive integer x is divided by 7 the quotient is q
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08 Aug 2017, 21:26
Bunuel wrote: seabhi wrote: 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. No, the correct answer is A. When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?When positive integer x is divided by 7 the quotient is q and the remainder is 1 > x = 7q + 1 > x could be 1, 8, 15, 22, ... (1) When x is divided by 5 the quotient is q and the remainder is 1 > x = 5q + 1 > subtract this from the equation from the stem: 0 = 2q > q = 0 > x = 1. 1 divided by 10 gives the remainder of 1. Sufficient. (2) x is less than 50 > if x = 1, then the remainder upon division 1 by 10 is 1 but if x = 8, then the remainder upon division 8 by 10 is 8. Not sufficient. Answer: A. Hope it helps. 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



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When positive integer x is divided by 7 the quotient is q
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07 Apr 2018, 12:46
LakerFan24 wrote: Bunuel wrote: seabhi wrote: 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. No, the correct answer is A. When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?When positive integer x is divided by 7 the quotient is q and the remainder is 1 > x = 7q + 1 > x could be 1, 8, 15, 22, ... (1) When x is divided by 5 the quotient is q and the remainder is 1 > x = 5q + 1 > subtract this from the equation from the stem: 0 = 2q > q = 0 > x = 1. 1 divided by 10 gives the remainder of 1. Sufficient. (2) x is less than 50 > if x = 1, then the remainder upon division 1 by 10 is 1 but if x = 8, then the remainder upon division 8 by 10 is 8. Not sufficient. Answer: A. Hope it helps. 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, 71Hi, 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.



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When positive integer x is divided by 7 the quotient is q
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04 Aug 2018, 08:09
Official Explanation from Veritasprep:In quotient remainder problems, number picking is often a valid technique to show patterns and answer the question. For this problem, however, it is not very helpful as the question is really testing a tricky property of division with integers. From statement one, you learn that when x is divided by 5 the quotient is q and the remainder is 1. In the question stem, you learned that when x is divided by 7 the quotient is also q and the remainder is also 1. Most students understand that if you divide a number by two different divisors and get the same remainder, you must be at the LCM (lowest common multiple) of those two numbers plus the remainder (and then at the same location down the number line to infinity). For instance if you divide a number by 5 and 7 and get a remainder of 1, then that can happen at 35+1, 70+1, 105+1, etc. However, if you read carefully you see that the quotient is the same in both operations, a puzzling result. Clearly if you divide 36 by 7 it will give you a different quotient then when you divide 36 by 5. The only way this can happen is if x is equal to 1. Remember that when a smaller integer is divided by a larger integer, the quotient is always 0 and the remainder is the dividend itself. Here when 1 is divided by 7 the quotient is 0 and the remainder is 1 and when 1 is divided by 5 the quotient is also 0 and the remainder is 1. Statement 1 is sufficient as x must be 1 and the remainder when x is divided by 10 is also 1. Statement 2 is clearly not sufficient by itself so the answer is A. This statement is here for people who miss that the quotient is the same and think that x could be any multiple of 35+1. By knowing that x is less than 50 it would lock in the value at 36. However, from the discussion above it is clear that this would be incorrect.
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