If when x is divided by z, the result is y remainder q, then

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Source: Princeton Review

If when \(x\) is divided by \(z\), the result is \(y\) remainder \(q\), then which of the following must be true?


A. \(z(y+q)=x\)

B. \(\frac{x}{z}−y=\frac{q}{z}\)

C. \(xz−q=y\)

D. \(\frac{x}{z}=y+q\)

E. \(\frac{x}{z}=yz+q\)

The OA is B

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by Jay@ManhattanReview » Tue Mar 12, 2019 8:54 pm
BTGmoderatorLU wrote:Source: Princeton Review

If when \(x\) is divided by \(z\), the result is \(y\) remainder \(q\), then which of the following must be true?


A. \(z(y+q)=x\)

B. \(\frac{x}{z}−y=\frac{q}{z}\)

C. \(xz−q=y\)

D. \(\frac{x}{z}=y+q\)

E. \(\frac{x}{z}=yz+q\)

The OA is B
Let's take this question as: when 13 is divided by 3, the result is 4 and the remainder is 1.

Thus, 13 = 3*4 + 1

Coming to the question:

x = yz + q

There is no option as x = yz + q. Dividing by z, we get x/z = y + q/z => x/z - y = q/z

The correct answer: B

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Wed Mar 13, 2019 8:22 am
BTGmoderatorLU wrote:Source: Princeton Review

If when \(x\) is divided by \(z\), the result is \(y\) remainder \(q\), then which of the following must be true?


A. \(z(y+q)=x\)

B. \(\frac{x}{z}−y=\frac{q}{z}\)

C. \(xz−q=y\)

D. \(\frac{x}{z}=y+q\)

E. \(\frac{x}{z}=yz+q\)

The OA is B
When it comes to remainders, there's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

GIVEN: when x is divided by z, the result (aka quotient) is y remainder is q
Applying the above rule, we get: x = zy + q .
Check the answer choices . . . x = zy + q is not there.
Looks like we need to rewrite the expression

Take: x = zy + q
Subtract yz from both sides to get: x - zy = q
Divide both sides by z to get: x/z - zy/z = q/z
Simplify to get: x/z - y = q/z

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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