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If 8x + 7y = 30, what is the value of x?

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If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post 26 Nov 2019, 10:34
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

54% (01:26) correct 46% (01:17) wrong based on 46 sessions

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If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post Updated on: 26 Nov 2019, 17:54
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).

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.
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Re: If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post 26 Nov 2019, 10:58
Bunuel wrote:
If 8x + 7y = 30, what is the value of x?

(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
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Re: If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post 26 Nov 2019, 17:30
pk123 wrote:
Bunuel wrote:
If 8x + 7y = 30, what is the value of x?

(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


Is B really sufficient though? You still don't know the actual value of x: could be 1, 2 or 3.
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If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post 26 Nov 2019, 19:16
splcs wrote:
pk123 wrote:
Bunuel wrote:
If 8x + 7y = 30, what is the value of x?

(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


Is B really sufficient though? You still don't know the actual value of x: could be 1, 2 or 3.


B tells us that X and Y are integers. For various integral values of x (shown above), y becomes integer only when x=2
So we get x and y integral value when x =2 and y =2 , the unique (x,y) that satisfies the given equation
Hence B is sufficient
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Re: If 8x + 7y = 30, what is the value of x?  [#permalink]

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New post 26 Nov 2019, 19:38
Bunuel wrote:
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
Multiply by 7 to get 21x+35y=112...(i)
Multiply 8x+7y=30 by 5 to get 40x+35y=150..Subtract (i) from it.
\(40x-21x=150-112.....19x=38...x=2\)
Suff

(2) x and y are positive integers.
\(8x + 7y = 30.....7y=30-8x....y=\frac{30-8x}{7}\)
Check for values of x for which y is an integer.
\(x=1.......y=\frac{30-8x}{7}=\frac{30-8*1}{7}=\frac{22}{7}\)..NO
\(x=2.......y=\frac{30-8x}{7}=\frac{30-8*2}{7}=\frac{14}{7}=2\)....YES
\(x=3.......y=\frac{30-8x}{7}=\frac{30-8*3}{7}=\frac{6}{7}\)..NO
As 8x would go beyond 30 if x>3, we do not require to see further.

D
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Re: If 8x + 7y = 30, what is the value of x?   [#permalink] 26 Nov 2019, 19:38
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