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If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are

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If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 02:59
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A
B
C
D
E

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Question Stats:

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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 03:15
1
1
IMO B,

3x+4|y|=33 ------(equation)

Putting X = 1,
|Y| = 30/4 =7.5(not integer)

Putting X = 2,
|Y| = 27/4 =6.75(not integer)

Putting X = 3,
|Y| = 24/4 =6(integer)
Therefore y = +-6 -------(1)

Putting X = 4,
|Y| = 21/4 =5.25(not integer)

Putting X = 5,
|Y| = 18/4 =4.5(not integer)

Putting X = 6,
|Y| = 15/4 =3.75(not integer)

Putting X = 7,
|Y| = 12/4 =3(integer)
Therefore Y = -+3 ------(2)

Putting X = 8,
|Y| = 9/4 =2.25(not integer)

Putting X = 9,
|Y| = 6/4 =1.5(not integer)

Putting X = 10,
|Y| = 3/4 =0.75(not integer)

Putting X = 11,
|Y| = 0/4 =0(integer)
Therefore Y = 0 -----(3)

Total values of y = -6,+6,-3,+3,0 (5 values)

Answer should be B

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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 03:17
If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are there?

A. 6
B. 5
C. 4
D. 3
E. 2

Given: x > 0
\(|y| = \frac{3(11-x)}{4}\)

possible values of x,y: [(3,6),(7,3),(11,0)]
we can not go any further as mod can not be less than 0
Ans: D
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 03:27
D.

Given: x > 0 and 3x + 4|y| = 33;

Simplifying the eqn for y; |y| = 3(11-x)/4

Since it given that y is +ve and it is not mentioned that y>0, three values of x satisfies the equation:
(3,6), (7,3), (11, 0)
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 03:34
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\(3x + 4|y| = 33\)

\(|y|=\frac{33-3x}{4}=\frac{3(11-x)}{4}\)

\((11-x)\) is a multiple of \(4\)

So \(x\) can be \(3, 7 or 11\)

When \(x=3\) or \(7\), \(y\) can take two values (+ve/-ve) for each and when \(x=11,y=0\)

(x,y)=(3,6) (3,-6) (7,3) (7,-3) (11,0)

Answer is (B)

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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 04:16
to get an odd answer which is 33, we must add an odd number with an even number and we can get odd number in only multiples of 3 as we can only get even numbers in multiples of 4. We should try with every odd multiples of 3 to add with multiples of 4

Pairs - 21+12 & 24+9

Answer - E
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 06:01
x > 0; x and y are both integers.

3x + 4|y| = 33 --> |y| = 3(11-x)/4
(11-x) has to be a multiple of 4, thus x = 11,7,3

For x=11... y=0
For x=7... y=-1 or y=1
For x=3... y=-2 or y=2
Obviously, there are 5 integer pairs (x, y)

FINAL ANSWER IS (B)

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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 06:44
Quote:
If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are there?

A. 6
B. 5
C. 4
D. 3
E. 2


x>0, then x=(33-4|y|)/3>0 and an integer

(33-4|y|)/3>0, 33-4|y|>0, 4|y|<33,
|y|<33/4, -33/4<y<33/4,
-8≤y≤8

x=11-4/3|y|>0 and an integer
-8≤|y|=multiple of 3≤8
y={-6,-3,0,3,6}=5
y=-6: x=11-4/3*-6=11+8=19
y=6: x=11-4/3*6=11-8=3
both are in range

(x,y) pairs are 5

Ans (B)
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 07:14
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Given x>0, x and you are integers.
3X+4|y|=33
y is integer if x=3 , 7 or 11
So possible ordered pair (x, y) are (3, 6), (3, -6), (7, 3)
(7, -3) &(11,0)i.e total 5 pairs.
So correct ans is (B).

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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 08:18
IMO B

If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are there?

x = 11 - 4/3|y| >0
|y| < 33/4 =8.25
& for integral values, |y| multiple of 3,
So, possible values are |y| = { 0, 3, 6 }
y = { -6, -3, 0, 3 , 6}
So total 5 pairs of (x,y)

B. 5
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If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post Updated on: 29 May 2020, 03:33
given relation of x > 0 and 3x + 4|y| = 33
would be valid at
y=+/-7 ; x= 2
y=+/-6 ; x=3
y=0 ; x= 11
total pairs ; 5 possible
OPTION B

If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are there?

A. 6
B. 5
C. 4
D. 3
E. 2

Originally posted by Archit3110 on 28 May 2020, 10:16.
Last edited by Archit3110 on 29 May 2020, 03:33, edited 1 time in total.
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 15:23
Is it B ?

If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are there?

X > 0 . so 3x + 4|y| = 33 => y = 3/4 (11-x) .. X can be 3 , 7 , 11 and y can be 6 , 3 , 0 respectively .
However as per ques with -ve value of y also the equation will be satisfied .

So total 5 pair of x and y will satisfy the equation.. (3,6) , (3,-6) ,(7,3) , (7,-3) and (11,0)
Hence B , IMO .
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are  [#permalink]

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New post 28 May 2020, 22:06
Ans:C
3x + 4|y| = 33
3x+4y=33..y=(33-3x)/4 as x>0
x=3 y=6
x=7 y=3

3x-4y=33
y=(3x-33)/4
x=3 y=-6
x=7 y=-3
x=15 y=3..no possible as 3*15=45
so total 4 pairs
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Re: If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are   [#permalink] 28 May 2020, 22:06

If x > 0 and 3x + 4|y| = 33, then how many integer pairs of (x, y) are

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