The remainder of 26 divided by k is k-2, what value is
K?
(1) K> 5
(2) K< 10
Remainder
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- sumgb
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Remainder is 'k-2' when 26 is divided by k. k=?
1. k > 5. we can have k =7 which gives remainder of 5 which is k-2. however, we can also have k = 28 which gives remainder of 26, which is k-2. hence k could be 7 or 28. so Insuff. throw A, D out of window.
2. k < 10. we know k = 7 satisfies the given constraint. But k = 4 also satisfies the given constraints. two values so insuff. cross off B.
Together we know, 5 < k < 10 only k = 7 satisfies given constraints so suff. Answer C
I think answer is C What is OA and OE?
Hope this helps and please post OE, thanks.
1. k > 5. we can have k =7 which gives remainder of 5 which is k-2. however, we can also have k = 28 which gives remainder of 26, which is k-2. hence k could be 7 or 28. so Insuff. throw A, D out of window.
2. k < 10. we know k = 7 satisfies the given constraint. But k = 4 also satisfies the given constraints. two values so insuff. cross off B.
Together we know, 5 < k < 10 only k = 7 satisfies given constraints so suff. Answer C
I think answer is C What is OA and OE?
Hope this helps and please post OE, thanks.
Last edited by sumgb on Sun Aug 14, 2011 4:28 am, edited 2 times in total.
Hi Sumg
Though the OA given is C, I think the answer should be A
Since from the first stmt, the only possible value of K that satisfies the condidition is K=7
From Stmt 2: there are two possible values of K<10 which can be either K=2 or K=4.
Could you please confirm whether my reasoning is correct. Thanks.
Though the OA given is C, I think the answer should be A
Since from the first stmt, the only possible value of K that satisfies the condidition is K=7
From Stmt 2: there are two possible values of K<10 which can be either K=2 or K=4.
Could you please confirm whether my reasoning is correct. Thanks.
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Hi,
26 when divided by k leaves remainder(k-2)
So, 26 - (k-2) = k*p, where p is the quotient
So, 28-k = kp =>k(p+1) = 28 where k>2
So, k can be 4,7,14,28
From(1): k>5
So, k can be 7,14,28
Not sufficient
From(2): k<10
So, k can be 4 or 7
Not sufficient
Both(1) and (2):
k=7
Sufficient
Hence, C
26 when divided by k leaves remainder(k-2)
So, 26 - (k-2) = k*p, where p is the quotient
So, 28-k = kp =>k(p+1) = 28 where k>2
So, k can be 4,7,14,28
From(1): k>5
So, k can be 7,14,28
Not sufficient
From(2): k<10
So, k can be 4 or 7
Not sufficient
Both(1) and (2):
k=7
Sufficient
Hence, C
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
- bblast
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love the algebraic solution by Frank :
Its important to note a lame example - >12/10 = 1 + 2/10
Mapping above to this question :
26/k = P + (k-2)/k
solving above gives :
K(P+1) = 28 where p can be {0,1,3,6...etc- since K is an integer}
Its important to note a lame example - >12/10 = 1 + 2/10
Mapping above to this question :
26/k = P + (k-2)/k
solving above gives :
K(P+1) = 28 where p can be {0,1,3,6...etc- since K is an integer}
Cheers !!
Quant 47-Striving for 50
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Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_