Welcome back to our series on GMAT quant rate problems. In this final article, we’ll explore problems that have algebraic expressions as answer choices. You have a few options whenever you encounter one of these problems, but this article will focus on solving them by applying the knowledge you’ve acquired by reading through this series.

## GMAT quant rates problems: Problems involving algebraic expressions

Time to put it all to work!

### Let’s return to machines and production rates for our first official problem:

**Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in ***x*** hours. Working alone at its constant rate, Machine A produces 800 nails in ***y*** hours. In terms of ***x*** and ***y***, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?**

**(A)*** x*** / (***x*** + ***y***)**

**(B) y**

**/ (**

*x***+**

*y***)**

**(C) x**

**y**

**/**

**(**

*x***+**

*y***)**

**(D) xy**

**/ (**

*x***–**

*y***)**

**(E) xy **

**/ (**

*y***–**

*x***)**

A well-trained GMAT test-taker always assesses the answer choices before diving into solving. Hopefully the choices here, especially choice C, remind you of something. We haven’t seen it since the third article of this series, but it’s the “new and improved” combined work formula! The combined time for two machines, working together, to complete a job is the product of their independent times divided by the sum of their independent times.** In this problem, ***x* is the combined time and *y*** is the time for Machine A. Let’s use ***z* to represent the time for machine B:

*x* = (*y* * *z*) / (*y* + *z*)

**Because the variable z is so “tangled up” with the variable y in this form of the equation, it will be difficult to isolate. But remember that this form is equivalent to the “original” form of the equation:**

1/*x* = 1/*y* + 1/*z*

**This looks more solvable for z:**

1/*x* = 1/*y* + 1/*z*

1/*z* = (1 / *x*) – (1 / *y*)

1/*z* = (1 / *x*)(*y */ *y*) – (1 / *y*)(*x */ *x*)

1/*z* = (*y */ *xy*) – (*x */ *xy*)

1/*z* = (*y* – *x*) / *xy*

*z* = *xy */ (*y* – *x*)

We used the familiar “common denominator-finding” method of multiplying the fractions (1 / *x*) and (1 / y) by (*y* / *y*) and (*x* / *x*), respectively. Hopefully, this whole solution reminded you of the derivation of the improved combined work formula from article 3. The solution of the equation for *z* matches** the correct answer choice E.**

### Let’s try another official GMAT problem involving travel speed instead of production rate:

**A man drove his automobile ***d***1**** kilometers at the rate of ***r***1**** kilometers per hour and an additional ***d***2**** kilometers at the rate of ***r***2**** kilometers per hour. In terms of ***d***1****, ***d***2****, ***r***1****, and ***r***2****, what was his average speed, in kilometers per hour, for the entire trip?**

**(A) (d****1**** + d****2****) / [(d****1**** / r****1****) + (d****2**** / r****2****)]**

**(B) (d****1**** + d****2****) / (r****1**** + r****2****)**

**(C) [(d****1**** + d****2****) / (r****1**** + r****2****)] / (d****1**** + d****2****)**

**(D) [(d****1**** / r****1****) + (d****2**** / r****2****)] / (d****1**** + d****2****)**

**(E) It cannot be determined from the information given.**

The answer choices here might look overwhelming, and** it would be tempting to select answer choice E and avoid thinking about the relationships between all those d’s and r’s.** But a solid grasp of the speed, distance, and time relationships can make this problem relatively painless.

**The average speed for the entire trip will be the total distance divided by the total time.** The total distance will be the sum of the independent distances *d*1 and *d*2, so the numerator we’re looking for is (*d*1 + *d*2). **This narrows our search to answer choices A and B.** So what expression for total time

The total time will be the sum of the times for each “leg” of the trip. **In answer choice B,** the denominator is simply the sum of the speeds *r*1 and *r*2. **This can’t be right, so the answer should be A.** But we can prove it. Since time = distance / speed, the time for each leg of the trip will be the distance covered on that leg divided by the speed at which that distance was covered, or, in terms of the variables in the problem, (*d*1 / *r*1) and (*d*2 / *r*2). We see the addition of these expressions in the denominator of **choice A, so that’s our answer.**

### Here’s a final official GMAT rate problem for the article and for the series:

**If Car A took ***n*** hours to travel 2 miles and Car B took ***m ***hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?**

**(A) 10***nm*** / (3***n*** + 2***m***)**

**(B) (3***n*** + 2***m***) / 10(***n*** + ***m***)**

**(C) (2***n*** + 3***m***) / 5***nm*

**(D) 10(***n*** + ***m***) / (2***n*** + 3***m***)**

**(E) 5(***n*** + ***m***) / (2***n*** + 3***m***)**

Since time = distance / speed, the time for Car C to travel 5 miles is the distance 5 divided by the arithmetic mean of the speeds of Car A and Car B. All our work will be modeling this arithmetic mean.

To start, we can say this:

mean = (speed A + speed B) / 2

**With this framework, we can model the independent speeds A and B and then plug them in. Since speed = distance / time, speed A = (2 / n), and speed B = (3 / m**

**).**

Let’s plug these in:

mean = (speed A + speed B) / 2

mean = [(2 / *n*) + (3 / *m*)] / 2

**To add the fractions (2 / n) and (3 / m), we can use the same method as in our derivation of the improved combined work formula: multiply (2 / n) by (m / m) and (3 / m) by (n / n):**

mean = [(2 / *n*) + (3 / *m*)] / 2

mean = [(2 / *n*)(*m* / *m*) + (3 / *m*)(*n* / *n*)] / 2

mean = [(2*m* / *nm*) + (3*n* / *nm*)] / 2

mean = [(2*m* + 3*n*) / *nm*] / 2

**This is the speed of Car C, so since time = distance / speed, the distance 5 divided by this expression represents the time for Car C to travel 5 miles.**

time = 5 / {[(2*m* + 3*n*) / *nm*] / 2}

time = 10 / [(2*m* + 3*n*) / *nm*]

time = 10*nm* / (2*m* + 3*n*)

And here, again, are our answer choices:

**(A) 10***nm*** / (3***n*** + 2***m***)**

**(B) (3***n*** + 2***m***) / 10(***n*** + ***m***)**

**(C) (2***n*** + 3***m***) / 5***nm*

**(D) 10(***n*** + ***m***) / (2***n*** + 3***m***)**

**(E) 5(***n*** + ***m***) / (2***n*** + 3***m***)**

The order of the addends 3*n* and 2*n* in the denominator is reversed, but this doesn’t matter. **The correct answer is choice A.**

We’ve learned a lot about rates and speeds in this series. Hopefully, you were able to draw on your knowledge to tackle the challenging problems in this final article. If not, review and keep practicing! With repeated application of the fundamentals, you’ll develop the skills to confidently solve rate problems on GMAT quant.

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**Contributor: ***Elijah Mize (Apex GMAT Instructor)*