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17 Nov 2018, 10:13
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C
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24% (02:16) correct
76%
(02:34)
wrong
based on 172
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How many ordered pairs of real numbers (x,y) satisfy the following system of equations?
x+3y =3
||x|-|y|| = 1
A) 1
B) 2
C) 3
D) 4
E) 8
17 Nov 2018, 12:12
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x + 3y = 3, so, x = 3 - 3y, substituting this in second equation we get:
||3-3y| - |y|| = 1, which gives, |3-3y| - |y| = +1 or -1 which gives,
3 - 4y = +1 or -1
3 - 2y = +1 or -1
-3 + 4y = +1 or -1
-3 + 2y = +1 or -1
We can see that 3 - 4y = 1 is the same as -3 + 4y = -1, with the same logic we can eliminate 4 equations as they are all the same. Finally we get the following four equations:
3 - 4y = 1 which gives y = 0.5
3 - 4y = -1 which gives y = 1
3 - 2y = 1 which gives y = 1
3 - 2y = -1 which gives y = 2
We get three distinct values of y, that is 0.5, 1 and 2, corresponding, values of x will be 1.5, 0, -3.
Hence we get 3 possible solutions.
Option C
||3-3y| - |y|| = 1, which gives, |3-3y| - |y| = +1 or -1 which gives,
3 - 4y = +1 or -1
3 - 2y = +1 or -1
-3 + 4y = +1 or -1
-3 + 2y = +1 or -1
We can see that 3 - 4y = 1 is the same as -3 + 4y = -1, with the same logic we can eliminate 4 equations as they are all the same. Finally we get the following four equations:
3 - 4y = 1 which gives y = 0.5
3 - 4y = -1 which gives y = 1
3 - 2y = 1 which gives y = 1
3 - 2y = -1 which gives y = 2
We get three distinct values of y, that is 0.5, 1 and 2, corresponding, values of x will be 1.5, 0, -3.
Hence we get 3 possible solutions.
Option C
General Discussion
17 Nov 2018, 10:57
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clearly, there are 4 sets of solutions ((y=0.5,x=1.5), (y=1,x=0),(y=1,x=0),(y=2,x=-3)) but only 3 are distinct(y=1,x=0 repeated). So, IMO Option c- 3 ordered pairs.
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GMATbuster's Weekly Quant Quiz#9 Ques 7 answer.pdf [742.01 KiB]
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