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Sub 505 (Easy)|   Algebra|      
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Bunuel
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 09 Sep 2015, 00:48
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E
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82% (00:48) correct 18% (01:04) wrong based on 129 sessions

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What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


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E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 09 Sep 2015, 01:44
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Bunuel
What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


Kudos for a correct solution.
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Solution :

Solution : For solving for 2 variables we need atleast 2 equations.

Statement1 : 3x + 15y = 35. Only one equation for two variables. Insufficient.

Statement2 : 70 − 6x = 30y ==> 3x + 15y = 35. Only one equation for two variables. Insufficient.

Combined : Both equations are same. Again, only one equation for two variables. Insufficient.

Option, E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 09 Sep 2015, 01:58
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Bunuel
What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


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(1) 3x + 15y = 35: Insufficient. Equation has 2 unknown variables
(2) 70 − 6x = 30y: Insufficient. Equation has 2 unknown variables

Even using both statements, we still have 2 unknown variables as both the equations say the same thing,

3x + 15y = 35
6x+30y=70(This is the same thing as first equation mutilplied by 2)
Answer E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 09 Sep 2015, 08:13
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.



What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


Overall there are 2 variables (x,y) and thus we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), C is likely the answer. Using both 1) & 2) together, 3x+15=35, 6x+30y=70--> 3x+15y=35 thus we have a trivial case. Therefore the answer is E.

Since we skip the actual calculation part, we save some time here. This is the advantage of using variable approach method.
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 09 Sep 2015, 09:09
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From statement 1:
3x + 15y = 35; Insufficient
From statement 2:
70 - 6x = 30y → 6x + 30y = 70 → 3x + 15y = 35
Since both the equations from the two statements are same, the solution is not possible.
Answer: E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 11 Sep 2015, 00:20
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What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y

St 1 --- mulitiple values of x possible for y = 1, 2, 3 ...
and not given interger ...so it can also be a fractional value both negative and positive ,
Hence not possible .
St 1 not sufficient

St 2. 70 − 6x = 30y
divide by and rearrange it becomes
3x + 15y = 35

So same as 1 -- not possible

Together 1 n 2 no new information .

hence E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 14 Sep 2015, 03:10
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Bunuel
What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


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1) We need two equations to find the unique value of x as there are two variables in the equation.

Hence, insufficient.

2) Insufficient for Same reason


Combining the two statements,

Divide equation 1 by 2,
Equation 1 is same as equation 2.

Hence, not sufficient.

Answer is E
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What is the value of x? (1) 3x + 15y = 35 (2) 70 − 6x = 30y 14 Sep 2015, 06:55
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Bunuel
What is the value of x?

(1) 3x + 15y = 35

(2) 70 − 6x = 30y


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PRINCETON OFFICIAL SOLUTION:

This appears to be a straightforward simultaneous equations problem. Neither fact should be enough by itself, but if you combine them then the data should be sufficient, and the answer should be C. But this question is deceptive. The trick here is that the second equation is really the same as the first. It’s just been doubled and the pieces have been rearranged. If you shuffle the pieces around, the second equation becomes 6x + 30y = 70 which is just the first equation multiplied by 2. Thus the second equation doesn’t tell us anything we didn’t already know, and the answer is E.
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