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registerincog
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:01) correct
47%
(02:15)
wrong
based on 532
sessions
History
Date
Time
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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
(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
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
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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.
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.
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.
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.
seabhi
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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Bunuel
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...
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
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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Bunuel
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
Posts: 30
LakerFan24
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.
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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.
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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Bunuel
x is of the form x = 7*q + 1
x = 1, 8, 15, 22, 29, 36, .......
1. when x is divided by 5, it leaves a reminder of 1
This leaves us with 1, 36, 106, .............
5*0 + 1 = 1
5*7 + 1 = 36
5*21 + 1 = 106
Not sufficient
2. Not sufficient on it's own
1&2 combined -
We can have x = 1 or x =36
Not sufficient
Where am I wrong? Can someone help
venkycs98
You're missing the point that the quotient obtained by dividing by 7 in the stem and by 5 in the first statement is the same, q.
Hope this helps.
P.S. Worth noting here that pure algebraic questions, such as the one above, are no longer part of the Data Sufficiency syllabus of the GMAT.
DS questions in GMAT Focus encompass various types of word problems, such as:
- Word Problems
- Work Problems
- Distance Problems
- Mixture Problems
- Percent and Interest Problems
- Overlapping Sets Problems
- Statistics Problems
- Combination and Probability Problems
While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."
Check GMAT Syllabus for Focus Edition
You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.
So, you can ignore this question.
Hope it helps.
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