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When x is divided by 3, the remainder is 2, and when y is divided by 7
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27 Apr 2016, 00:06
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When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21? (1) x^2 divided by 7 leaves a remainder of 4. (2) y – 4 is divisible by 3.
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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27 Apr 2016, 02:01
Bunuel wrote: When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4. (2) y – 4 is divisible by 3. X can be 2,5 and y can be 18 so reminder is different in both..SO E
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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29 Jul 2017, 10:07
Bunuel wrote: When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4. (2) y – 4 is divisible by 3. can someone give a solution with equations ?



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When x is divided by 3, the remainder is 2, and when y is divided by 7
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29 Jul 2017, 10:38
Takdir wrote: Bunuel wrote: When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4. (2) y – 4 is divisible by 3. can someone give a solution with equations ? Hi, if you are looking for solution with equation..the main statement tells us.. A) x=3a+2.... x could be 2,5,8,11,14,17.... B) y=7b+4.... y could be 4,11,18,25...statements.. I) \(x^2=7c+4\)..... \(x^2\) could be 4,11,18,25..... so x can be \(\sqrt{4},\sqrt{11},\sqrt{18},\sqrt{25}\)... or \(2,\sqrt{11},\sqrt{18},5\)... not required to go beyond this as we see ATLEAST two values of x possible 2,5...insuff II. it will give you some value of y.. nothing about x insuff combined.. whatever be the value of y.. y+2 and y+5 will surely give different REMAINDER when div by 21..hence insuff E
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7
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09 Nov 2018, 16:58
Hi, I guess Answer is C. x when divided by 3, Remainder is 2. x = 3a + 2 y when divided by 7, remainder is 4. y = 7b + 4 Remainder when x + y divided by 21? x + y = 3a + 2 + 7b+ 4 x + y = 3a + 7b+ 6. If a is divisible by 7 and b is divisible by 3 then Remainder = 6. A. x2 / 7 , Remainder = 4. x = (3a + 2). x2 = 9a2 +6a + 4. a is divisible by 7, so that remainder is 4. B. y  4 is divisible by 3 y = 7b + 4 ; y4 = 7b 7b is divisible by 3, so b is divisible by 3. C. a = divisible by 7, b = divisible by 3. x + y = 3a + 7b + 6 when divided by 21 should give remainder 6. The OA is E, Please can someone tell why C is incorrect?
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7
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09 Nov 2018, 17:48
Akshay_Naik wrote: Hi, I guess Answer is C.
x when divided by 3, Remainder is 2. x = 3a + 2
y when divided by 7, remainder is 4. y = 7b + 4
Remainder when x + y divided by 21? x + y = 3a + 2 + 7b+ 4 x + y = 3a + 7b+ 6. If a is divisible by 7 and b is divisible by 3 then Remainder = 6.
A. x2 / 7 , Remainder = 4. x = (3a + 2). x2 = 9a2 +6a + 4. a is divisible by 7, so that remainder is 4.
B. y  4 is divisible by 3 y = 7b + 4 ; y4 = 7b 7b is divisible by 3, so b is divisible by 3.
C. a = divisible by 7, b = divisible by 3. x + y = 3a + 7b + 6 when divided by 21 should give remainder 6.
The OA is E, Please can someone tell why C is incorrect? Try the numbers x=5 and y=4 it gives a remainder of 9 when divided by 21 But when you take x=23 and y=4, it gives a remainder of 6 when divided by 21



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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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31 Dec 2018, 00:31
Bunuel chetan2uSo what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong: Stem: \(x=3q+2\) and \(y=7p+4\) \(–>\) \(x+y=3q+7p+6\) Hence, our question becomes: is \(3q+7p=21k\). This is the case whenever \(3q\) is divisible by \(7\) and \(7p\) is divisible by \(3\). Statement I: \(x^2=9q^2+12q+4=7L+4 –> q=7L\) Still, Not sufficient as no information is given about yStatement II: \(y4=3m\). Now from the stem, we know that \(y=7p+4\) hence, this statement tells us that \(7p=3m\) Still, Not sufficient as no information is given about xStatement I&II combined: From I) we know that \(3q\) is divisible by \(7\), hence, \(3q\) must be a multiple of \(21\) From II) we know that \(7p\) is divisible by \(3\), hence, \(7p\) must be a multiple of \(21\) Thus, \(x+y=(3q+7p)+6= 21b+6\) Please let me know where I went wrong!
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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04 Jan 2019, 03:55
no one has any idea why above solution is incorrect?
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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04 Jan 2019, 05:03
ghnlrug wrote: Bunuel chetan2uSo what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong: Stem: \(x=3q+2\) and \(y=7p+4\) \(–>\) \(x+y=3q+7p+6\) Hence, our question becomes: is \(3q+7p=21k\). This is the case whenever \(3q\) is divisible by \(7\) and \(7p\) is divisible by \(3\). Statement I: \(x^2=9q^2+12q+4=7L+4 –> q=7L\) Still, Not sufficient as no information is given about yStatement II: \(y4=3m\). Now from the stem, we know that \(y=7p+4\) hence, this statement tells us that \(7p=3m\) Still, Not sufficient as no information is given about xStatement I&II combined: From I) we know that \(3q\) is divisible by \(7\), hence, \(3q\) must be a multiple of \(21\) From II) we know that \(7p\) is divisible by \(3\), hence, \(7p\) must be a multiple of \(21\) Thus, \(x+y=(3q+7p)+6= 21b+6\) Please let me know where I went wrong! You are wrong here.. Statement I: \(x^2=9q^2+12q+4=7L+4 –> q=7L\) Still, Not sufficient as no information is given about y \(x^2=9q^2+12q+4=7L+4 –>9q^2+12q=7L......3q(3q+4)=7L\).. When q is multiple of 7, 3q(3q+4)=21a(21a+4)=7L.... possible Here 3q is divisible by 7When 3q+4 is multiple of 7, say q is 1....3*1(3*1+4)=3*(7).... possible Here 3q=3 and it is not divisible by 7
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7
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