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# Tricky Data Sufficiency Questions: Explanations #1 and #2

Today, we present the explanations for the first two questions in our Tricky Data Sufficiency Questions post. Let’s get started.

## Question #1

1. What is the value of x?

(1) 5

x

+ 3

y

= 15

(2)

y

= 5 – (5/3)

x

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient

(hint)

Since (1) is a linear equation with two variables, it will allow you to solve for x only in terms of y. So that statement is insufficient by itself, eliminate answers A and D. This leaves B, C, and E.

Since (2) is also a linear equation with two variables, it will allow you to solve for x only in terms of y. So that statement is insufficient by itself, eliminate B. This leaves C and E.

You might be tempted to choose C, because (1) and (2) together give you two linear equations with the same two variables. As we’ve seen though, we must first make sure that the two equations are distinct. To see whether they are in fact the same equation in different guises, rewrite one of them—say Statement (1)—in the form that other already has.

So let’s solve for y using (1).

5x + 3y = 15

Subtract 5x from each side.

3y=15-5x

Divide each side by 3.

y = 5 – (5/3)x

So (1) and (2) are in fact the same equation. The correct answer is E.

Rather than checking in this way to see whether the two equations are equivalent, you could instead solve the system of equation by substitution, especially since (2) has already isolated y.

Take the value of y given in Statement and substitute that value for y in Statement (1). Distribute the 3 to clear the parentheses.

5x – 15 + 5x = 15

Subtract 15 to each side to isolate the x terms on the left-hand side of the equation.

5x – 5x = 15 – 15

Simplify.

0=0

That doesn’t give us any information, so the correct answer is E.

## Question #2

2. If y = 2(x – 1) what is the value of x + y?

(1)

x

+ (6/

y

) = 3

(2)

y

=

x

+ 2

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient

(hint)

There’s no obviously useful way to rephrase the question, so we’ll probably need to solve for x and y and then add the values together. Keep an eye open, though, for an opportunity to solve for the sum x + y directly.

You may want to simplify the given equation—the “if” equation—before turning to the statements. Distribute the 2 on the right-hand of the equation to clear the fraction: y = 2x – 2. Now turn to the statements.

You might notice right away that (1) has a variable in a denominator as well as a one outside any denominator. This means that it is not a linear equation and so that information that at first appears sufficient will not be.

Whether you notice that or not, you’ll want to start with the simpler statement (2). Well, that a pleasant surprise! Statement (2) is a linear equation, as is the “if” equation. So unless they’re equivalent equations, (2) should be sufficient. Since both the “if” equation and (2) already isolate y, solving by substitution means setting equal the right hand sides of each equation.

x + 2 = 2x – 2

Subtract x from and add 2 to each side.

x = 4

You could solve for y and add that to 4 to solve x + y, but there’s no need to. If you can see that substituting 4 for x in either of the original equations will yield some constant value for y, then you can conclude that (2) allows you to solve for x + y, and so that (2) is sufficient. Eliminate A, C, and E.

Now let’s turn to Statement (1). A system of a single linear equation (the “if” equation) and a single nonlinear equation (Statement (1)) will not usually yield a unique constant value for each variable, but if we have the time to attempt a solution we should, unless we can rule out the exceptional cases some other way.

Since the “if” equation gives us y in terms of x, we can substitute 2x -2 for y in Statement (1). Simplify the fraction by dividing the numerator and denominator by 2. Multiply all terms by to clear the fraction.

x (x – 1) + 3 = 3 (x – 1)

Distribute to clear the parentheses. Transpose to arrange in the usual quadratic form. Because this is a quadratic but not a quadratic square, it will have two solutions. Eliminate D. The correct answer is B.

Remember to check back on Monday for the explanations to questions #3 and #4. In the meantime, you can check your answers here. Have a good weekend!

### 13 Responses to Tricky Data Sufficiency Questions: Explanations #1 and #2

1. Julia March 1, 2017 at 8:52 pm #

Hi, I believe there is also a typo in question 2, as other students pointed out.

Either the original question was y= 2 (x-1), and there is a typo here, or the solution has a typo and should consider y=2x+2 in the calculations, instead of 2x-2.

• Magoosh Test Prep Expert March 1, 2017 at 11:47 pm #

To Julia and the other students who weighed in– great catch! Thanks for bringing this to our attention! The original question should have read y= 2 (x-1). I’ve edited and corrected this article accordingly.

2. SHUBHESH December 7, 2015 at 8:17 pm #

Hello,

in the second problem:
How does

y = 2(x+1) turn into y = 2x – 2 ?

3. Hardik August 16, 2015 at 3:26 am #

HI Team,

The simpler way would be turn the second equation y=x+2 into y-1=x+1 and then fill the value of x+1 in the equation. So that we can directly say that the answer is B.

4. Kon June 4, 2015 at 3:28 am #

Hello,

in the second problem:
How does

y = 2(x+1) turn into y = 2x – 2 ?

where does the negative come from?

5. faizan May 6, 2015 at 11:23 am #

The question was 5x – 3y = 15, it has since changed to 5x + 3y = 15.
If it is the former then correct answer is C.

• Rita Kreig May 6, 2015 at 4:29 pm #

Thanks Faizan! There was a typo in the initial question, which we fixed. Glad you are able to work it out both ways! 🙂

Cheers,
Rita

6. Harry April 10, 2015 at 1:02 pm #

Distribute the 3 to clear the parentheses.

5x – 15 + 5x = 15

Subtract 15 to each side to isolate the x terms on the left-hand side of the equation.

5x – 5x = 15 – 15

Simplify.

0=0

I cannot understand this,
clearly, 5x+5x=15+15
and so 10x=30
x=3

• Rita Kreig April 10, 2015 at 5:24 pm #

Hi Harry,

Yes, you’re absolutely correct! Thank you so much for pointing out this error. It seems that E is not the correct answer to this problem, as equations (1) and (2) are not equivalent statements (please see my comment to Kmaro for an explanation). Since E isn’t the correct answer, the proof will not work. Your math is 100% correct.

I’m going to forward your comment along to Michael so that he can re-evaluate his explanation. I’m so sorry for the inconvenience! Thank you for sharing your work – you’re helping us help other students! 🙂

Best of luck with your prep, and have a wonderful weekend! I’ll get back to you when we have an updated explanation.

Thank you so much!
Rita

• Rita Kreig April 13, 2015 at 12:25 pm #

Hey Harry,

I just wanted to reach out and let you know that the problem is now fixed. Thank you again for all your help!

Have a great day!
Rita 🙂

7. Kmaro April 10, 2015 at 12:58 pm #

Subtract 5x from each side.

-3y=15-5x

Divide each side by -3.

y=5-5/3х

I`m not sure what i`m missing but i believe if you divide by -3 you should change not only y but also 15/-3 is -5 and -5x/-3 is 5/3x. Thanks in advance

• Rita Kreig April 10, 2015 at 5:17 pm #

Hi Kmaro,

You’re awesome! And absolutely right! Thank you so much for pointing out this typo. I’ll make the change in the post so that we don’t accidentally confuse other students. 🙂

Dividing both sides of the equation by -3 should leave you with y = -5 + (5x/3). Which means that equations (1) and (2) are NOT, in fact, the same equation. I’ll forward your comment along to Michael, so he can replace the explanation. I’ll get back to you when I have an update.

Thank you!!
Rita

• Rita Kreig April 13, 2015 at 12:26 pm #

Hey Kmaro,

I reached out to Michael and we were able to catch the original typo in the problem. The question should work better now. Sorry for the confusion, and thank you so much for all your help!

Cheers! 🙂
Rita

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