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26 Nov 2019, 10:34
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If 8x + 7y = 30, what is the value of x?
(1) 3x + 5y = 16
(2) x and y are positive integers.
(1) 3x + 5y = 16
(2) x and y are positive integers.
mykrasovski
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Updated on: 26 Nov 2019, 17:54
Originally posted by mykrasovski on 26 Nov 2019, 10:52.
Last edited by mykrasovski on 26 Nov 2019, 17:54, edited 1 time in total.
Last edited by mykrasovski on 26 Nov 2019, 17:54, edited 1 time in total.
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Statement 1: sufficient. We have two equations and two unknowns, we can get both x and y.
Statement 2: x > 0 and y > 0, both are integers.
8x + 7y = 30 can be rewritten as 7y = 30 - 8x = 2(15 - 4x). In other words, (1) 7k = 15 - 4x and (2) y = 2m.
Let's look at (1) closely. Only one value of x, x = 2, will satisfy the conditions that are given in Statement 2. So, we can find x.
Answer (D).
Statement 2: x > 0 and y > 0, both are integers.
8x + 7y = 30 can be rewritten as 7y = 30 - 8x = 2(15 - 4x). In other words, (1) 7k = 15 - 4x and (2) y = 2m.
Let's look at (1) closely. Only one value of x, x = 2, will satisfy the conditions that are given in Statement 2. So, we can find x.
Answer (D).
26 Nov 2019, 10:58
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Bunuel
If 8x + 7y = 30, what is the value of x?
(1) 3x + 5y = 16
(2) x and y are positive integers.
(1) 3x + 5y = 16
(2) x and y are positive integers.
Option A, sufficient
8x+7y=30 & 3x+5y=16 are 2 equations with 2 variable can be solved
24x+21y=90, multiplied given equation by 3
24x+40y=128, multiplied option A by 8
Subtracting we get 19y=38, hence y=2
Substituting y=2 x is 2
Option B
x, y> 0 and Integers
Say x=1, y=(30-8x)/7 hence y= 22/7
x=2, y = 14/7=2
x=3, y=6/7
x=4, y<0 not possible
hence even Option B is sufficient
Both the options can yield unique value of x
