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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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When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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New post 27 Apr 2016, 00:06
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C
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E

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Question Stats:

35% (03:10) correct 65% (02:42) wrong based on 135 sessions

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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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New post 27 Apr 2016, 02:01
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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  [#permalink]

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New post 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  [#permalink]

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New post 29 Jul 2017, 10:38
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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  [#permalink]

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New post 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 ; y-4 = 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  [#permalink]

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New post 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 ; y-4 = 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  [#permalink]

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New post 31 Dec 2018, 00:31
Bunuel chetan2u
So 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 y

Statement II: \(y-4=3m\). 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 x

Statement 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  [#permalink]

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New post 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  [#permalink]

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New post 04 Jan 2019, 05:03
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ghnlrug wrote:
Bunuel chetan2u
So 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 y

Statement II: \(y-4=3m\). 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 x

Statement 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 7
When 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   [#permalink] 04 Jan 2019, 05:03
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