Today, we’re going to review the last two questions from the Tricky Data Sufficiency Questions practice post. If you’re just tuning in, give the questions a shot, check your answers here, and then review the explanations to questions #1-2, #3-4, and #5-6.
Ready? Let’s move along.
Question #7
7. If p + q = 2r + 3, what is the value of p?
(1) p – r = 6
(2) r = (q/2) + 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)
Whenever the question stem presents you with an equation with multiple variables and asks you to solve for one of them, you should immediately rewrite the equation to find a more useful question. Specifically, you should isolate the variable you’re asked to solve for and then focus on the expression equal to that variable.
Here’s what that would look like for this problem:
p + q = 2r + 3
Subtract q from each side of the equation to isolate p.
p = 2r – q + 3
Because this is Data Sufficiency, ask, “What do I need to know to know the value of 2r – q + 3?” The answer to that question is simply “2r – q.”
Statement (1) includes p (which we don’t want) but not q (which we do want), so it is not sufficient. Eliminate A and D.
Turn to Statement (2) and try to isolate 2r – q.
r = (q/2) + 2
Multiply each side by 2 to clear the fraction.
2r = q + 4
Subtract q from each side to isolate 2r – q.
2r – q = 4
Statement (2) is sufficient. The correct answer is B.
Question #8
8. What is the value of x?
(1) x – (4/y) = 2
(2) x + 2y = 8
(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)
Because both (1) and (2) involve y as well as x, each may allow you to solve for x in terms of y, but neither by itself will yield a unique constant value for x. Eliminate A, B, and D.
Are the two statements sufficient together? Probably not, since (1) is not linear equation, and a system of a single linear equation with a single nonlinear equation doesn’t ordinarily yield a solution. Just in case this an exception, though, let’s try to solve this system for x.
(If we’re very clever we might notice that solving for y is just as good as solving for x, since it leads to a value for x. Solving for y might also be easier here. Let’s suppose that we missed that shortcut, though, and just solve for x.)
Solving the system for x begins with solving for y in terms of x in the second equation.
x + 2y = 8
Subtract x from each side.
2y = 8 – x
Divide each side by 2.
y = 4 – (x/2)
Substitute the expression 4 – (x/2) for y in the first equation.
x – 4/(4 – (x/2)) = 2
Rewrite the expression 4 – (x/2) as (8/2) – (x/2) or simply as (8 – x)/2.
x – (4/((8 – x)/2) = 2
Simplify the compound fraction.
x – (8/8 – (x)) = 2
Multiply each term by 8 – x to clear the fraction.
x(8 – x) – 8 = 2(8 – x)
Distribute to clear the parentheses.
8x – x^2 – 8 = 16 – 2x
Transpose to arrange in the usual quadratic form
–x^2 + 10x – 24 = 0
Multiply through by -1.
x^2 – 10x + 24 = 0
At this point you might recognize that this is a quadratic but not a perfect quadratic square, and so must have two solutions. If you don’t recognize that, go ahead and solve it.
(x – 4)(x – 6) = 0
x = {4, 6}
Since the two statements together don’t yield a unique constant value, they are not sufficient. The correct answer is E.
Hi Rita and Michael
Even after fixing the typo. After this step “Distribute to clear the parentheses.
8x – x^2 – 8 = 16 – 2x” and transposing
It should be X^2-10X+24 and soln for x are 6 and 4 and not 2 and 4.
But I am glad that I am not the only one doing such mistakes. it happens to best of us. 🙂
Dear Harsh,
This is Mike McGarry. Thank you very much for pointing this out: I just corrected the algebra above. 🙂 Best of luck to you, my friend!
Mike 🙂
Michael, thank you for your explanation. I think there is a little error in the solution of Question #8.
you jump from this
x + 2y = 8
into this:
2y = x + 8
and I think you meant to say 2y=8-x
and when you plug this into statements 1 & 2 together you end up with the roots being x=4 and x=6, so the solution is E.
Hi Javi,
Thank you so much for letting us know about this error! I fixed the typos, and the explanation should make more sense now. Thanks for teaching us a thing or two, and for looking out for other students!
Cheers!
Rita