For a start, give these problems a try.  A complete explanation will come at the end of the discussion.

1) When positive integer N is divided by positive integer J, the remainder is 14.  If N/J = 134.08, what is value of J?

1. 22
2. 56
3. 78
4. 112
5. 175

2) When positive integer N is divided by positive integer P, the quotient is 18, with a remainder of 7.  When N is divided by (P + 2), the quotient is 15 and the remainder is 1.  What is the value of N?

1. 151
2. 331
3. 511
4. 691
5. 871

3) P and Q are both positive integers.  When P is divided by Q, the remainder is some positive integer D, and when P is divided by (Q + 3), the remainder is also D.  If P/Q = 1020.75 and P/(Q + 3) = 816.6, then which of the following gives the correct set of {D, Q}?

1. {6, 12}
2. {6, 15}
3. {9, 12}
4. {9, 15}
5. {15, 24}

I will discuss those questions at the end of this article.

Division Terminology

Let’s look at division carefully and think about the parts.  Suppose we divide 33 by 4.  Of course, 4 goes into 33 eight times, with a remainder of 1.  Let’s talk about the official names of this cast of characters.

33 = the dividend = the number being divided

4 = the divisor = the number doing the dividing; the number by which you divide

8 = the integer quotient = the integer that results from a whole number of divisions

1 = the remainder

We have to add a caveat here.  Notice: here we are talking about positive integers only —- living in the magical fairyland where the only numbers that exist are positive integers, where skies are not cloudy all day.  Unless you live on a farm where the barnyard animals all sing in unison, you don’t get to stay here forever.

Of course, numbers in the real world aren’t like that, and if you prance through the GMAT Quantitative Section as if it’s a magical fairyland where all the numbers are positive integers, this section will utterly decimate you.  In the real world that involves all possible numbers, this process looks a bit different.  For example, if you type 33/4 into your calculator, your calculator will tell you

33/4 = 8.25

That decimal, 8.25, is the quotient, the actual real-world mathematical quotient.   In this article, for clarity, I will refer to this one as the “decimal quotient.”

Notice, first of all — the integer part of the decimal quotient is exactly equal to the integer quotient.  It has to be.  In fact, we can go a little further:  Let’s look at this process both with words and with numbers:

We divide the dividend (33) by the divisor (4), and we get the decimal quotient.  The integer part of the decimal question is the first piece of the last sum, the integer quotient.  What’s crucially important is — the decimal part of the decimal quotient equals the final fraction:

Virtually any problem on the GMAT that gives you a decimal quotient is relying on this particular formula.  It is crucial for answering #1 and #3 above.

Rebuilding the Dividend

Let’s go back to the integer relationships:

If you are given the divisor, the integer quotient, and the remainder, then you can rebuild the dividend.  In particular, notice that “divisor” is the denominator of both fractions, so we if multiple all three terms by “divisor”, it cancels in two of the three terms:

dividend = (integer quotient)*(divisor) + remainder

That formula is pure gold in questions which give you an integer quotient, a divisor, and a remainder.  Even if one or two of those three are in variable form, it allows us to set up an algebraic relationship we can solve.  This is crucial for answering #2 above.

Practice Questions

It may be, at this point, you want to give those three questions another attempt before reading the solutions.  Here’s yet another practice question:

4) http://gmat.magoosh.com/questions/305

Practice Question Solutions

1) We know that

So 0.08, the decimal part of the decimal quotient, must equal the remainder, 14, divided by the divisor J.

0.08 = 14/J

0.08*J = 14

J = 14/0.08 = 1400/8 = 700/4 = 350/2 = 175

So J = 175, answer = E.

2) Use the rule dividend = (integer quotient)*(divisor) + remainder to translate each sentence.

The first sentence becomes N = 18P + 7.  The second equation becomes N = (P + 2)*15 + 1, which simplifies to N = 15P + 31.  These are ordinary simultaneous equations (http://magoosh.com/gmat/2012/gmat-quant-how-to-solve-two-equations-with-two-variables/).  Since they both equal N already, let’s set them equal and solve for P.

18P + 7 = 15P + 31

3P + 7 = 31

3P = 24

P = 8

 Now that we know P = 8, we can just plug in.  The product 15*8 is particularly easy to do, without a calculator, by using the “doubling and halving” trick.  Double 15 to get 30, and take half of 8 to get 4 —- 15*8 = 30*4 = 120.  So

P = 15(8) + 31 = 120 + 31 = 151

P = 151, answer = A.

3) Here, we have to use

to translate each act of division.  The first one tells us 0.75 = D/Q, and the second one tells us that 0.60 = D/(Q + 3).  These are also two simultaneous equations.  Multiply both to get rid of the fractions.

D = 0.75*Q

D = 0.60*(Q + 3) = 0.60*Q + 1.80

Both are equal to D, so set them equal to each other and solve for Q.

0.75*Q = 0.60*Q + 1.80

0.15*Q = 1.80

Q = 1.80/0.15 = 180/15 = 60/5 = 12

Now that we know Q = 12, we can find D = 0.75*Q = (3/4)*12 = 9.  So D = 9 and Q = 12.  Answer = C.

First of all, consider these three GMAT like questions.  Try them on your own before reading the post.  I will show solutions to these practice questions at the end of this post.

1) Running at the same rate, 8 identical machines can produce 560 paperclips a minute.  At this rate, how many paperclips could 20 machines produce in 6 minutes?

1. 1344
2. 3360
3. 8400
4. 50400
5. 67200

2) Jane can make a handcrafted drum in 4 weeks.  Zane can make a similar handcrafted drum in 6 weeks.  If they both work together, how many weeks will it take for them to produce 15 handcrafted drums?

1. 30
2. 36
3. 70
4. 80
5. 150

3) Machines P and Q are two different machines that cover jars in a factory.  When Machine P works alone, it covers 1500 jars in m hours.  When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours.  In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars?

1. m/(m + n)
2. n/(m + n)
3. mn/(m + n)
4. mn/(m – n)
5. mn/(n – m)

OK, if these problems frustrate the bejeebers out of you, this is the post for you!

Big Idea #1: The “ART” Equation

You may be familiar with the distance equation, D = RT (“distance equals rate times time”), sometimes remembered as the “dirt” equation.  It turns out, that equation is just a specific instance of a much more general equation.  In that equation, R, the rate, is distance per time, but in non-distance problems, rate can be anything over time — wrenches produced per hour, houses painted per day, books written per decade, etc.  In these cases, typical of work problems, we are no longer concerned with “distance” per time, but with the amount of something produced per time.  We use A to represent this amount (the number of wrenches, the number of houses, etc.), and the equation becomes A = RT.  Sometimes folks remember this as the “art” equation.

Here’s a simple mnemonic.  When you travel, you are moving on the Earth, which is made of dirt, so for traveling & distance you use D = RT.  Work problems involve machines, and machines make things —– making is creation, and creation is the essence of art, so use the A = RT equation.  (I know, I know, what comes out of most machines is hardly worthy of aesthetic elevation, but it works for a mnemonic!)

Big Idea #2: Rates are Ratios

The word “rate” and the word “ratio” have the same Latin root: in fact, they also share a Latin root with the “rationality” of our minds, but that’s a discussion that would bring up to our noses into Pythagorean and Platonic philosophies.  The point is: a rate is a ratio, that is to say, a fraction.  Technically, any fraction, any ratio, in which the numerator and the denominator have different units is a rate.  Fuel efficiency (mpg) and price per unit and most baseball fractions (ERA, BA, OBP, SLG, etc.) are rates.  Currency rates and exchanges rates are common financial market rates that, ironically, almost never appear on the GMAT —- go figure!  Most GMAT rates have time in the denominator, and it’s a rate of how fast work is being done or how fast something is being produced or accomplished.

The fact that rates are ratios means: we can solve these problems by setting up proportions and using proportional thinking!  As you will see in the solutions below, that’s an extremely powerful strategy for solution.

Big Idea #3: Add Rates

The vast majority of work problems on the GMAT involve two people or two machines and comparisons of their individual production to their combined production.  Questions #2 and #3 above are of this form.  The questions will often give you information about times and about amounts, and what you need to know is: you can’t add or subtract times to complete a job and you can’t add or subtract amounts of work; instead, you add and subtract rates.

(rate of A alone) + (rate of B alone) = (combined rate of A & B)

Here A and B can be two people, two machines, etc.  The extension of this idea is that if you have N identical machines, and each one works at a rate of R, then the combined rate is N*R.

With just these three ideas, you can unlock any work problem on the GMAT.  At this point, you may want to go back and give another attempt at those three practice questions.  Follow carefully how they are applied in the solutions below.

Practice Problem Solutions

1) “Running at the same rate, 8 identical machines can produce 560 paperclips a minute.”  That 560 is a combined rate of 8 machines —- 560 = 8*R, so the rate of one machine is R = 560/8 = 70 paperclips per minute.

“At this rate, how many paperclips could 20 machines produce in 6 minutes?”  Well, the combined rate of 20 machines would be Rtotal = 20*70 = 1400 pc/min.  Now, plug that into the “art” equation: A = RT = (1400)*(6) = 8400 pc.  Answer = C.

2) Method I: the rates solution

“Jane can make a handcrafted drum in 4 weeks.  Zane can make a similar handcrafted drum in 6 weeks.”  Jane’s rate is (1 drum)/(4 weeks) = 1/4.  Zane’s rate is (1 drum)/(6 weeks) = 1/6.  The combined rate of Jane + Zane is

          R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12

That’s the combined rate.  We need to make 15 drums — we have a rate and we have an amount, so use the “art” equation to solve for time:

          T = A/R = 15/(5/12) = 15*(12/5) = (15/5)*12 =  3*12 = 36

BTW, notice in the penultimate step, the universal fraction strategy: cancel before you multiply (Tip #3: http://magoosh.com/gmat/2012/can-i-use-a-calculator-on-the-gmat/.  Jane and Zane need 36 weeks to make 15 drums.  Answer = B.

Method II: the proportion solution

“Jane can make a handcrafted drum in 4 weeks.  Zane can make a similar handcrafted drum in 6 weeks.”  Let’s find the LCM of 4 and 6 — that’s 12 weeks.  In a 12 week period, Jane, making a drum every 4 weeks, makes three drums.  In a 12 week period, Zane, making a drum every 6 weeks, makes two drums.  Therefore, in a 12 weeks period, they produce 5 drums between the two of them.  If they make 5 drums in 12 weeks, they need triple that time, 36 weeks, to make 15 drums.  Therefore, Jane and Zane need 36 weeks to make 15 drums.  Answer = B.

3) This is a particularly challenging, one because we have variables in the answer choices.  I will show an algebraic solution, although a numerical solution (http://magoosh.com/gmat/2012/variables-in-gmat-answer-choices-algebraic-approach-vs-numerical-approach/) is always possible.

“Machines P and Q are two different machines that cover jars in a factory.  When Machine P works alone, it covers 1500 jars in m hours.  When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours.  In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars? ”

Since the number “1500 jars” appears over and over, let’s arbitrarily say 1500 jars = 1 lot, and we’ll use units of lots per hour to simplify our calculations.

P’s individual rate is (1 lot)/(m hours) = 1/m.  The combined rate of P & Q is (1 lot)/(n hours) = 1/n.  We know

(P’s rate alone) + (Q’s rate alone) = (P and Q’s combined rate)

(Q’s rate alone) = (P and Q’s combined rate) – (P’s rate alone)

(Q’s rate alone) = 1/n – 1/m = m/ (nm) – n/ (nm) = (m – n)/(nm)

We now know Q’s rate, and we want the amount of 1 lot, so we use the “art” equation.

1 = [(m – n)/ (nm)]*T

T = (mn)/(m – n)

Answer = D

Here’s another practice question:

http://gmat.magoosh.com/questions/47

Learn to master this common type of GMAT word problem!

Speed

Rate is another word for speed, and the equation for this is D = RT (which some people remember as the “dirt” equation).  D = distance, R = rate (a.k.a. speed, a.k.a. velocity), and T = time.  One common source of errors is that all three variables have to be in the same units.  If you travel at 30 mph for 10 minutes, you do not go 30*10 = 300 miles!

It’s also helpful to remember, as etymology would suggest, that a “rate” is a “ratio”, which in term is a fraction.  With rates, we can always use ratios and always set up proportions.  One way to find how far one goes at 30 mph in 10 minutes is to say

Cross-multiply, and you get 10 = 2x.  So, x must equal 5 miles.

Average Speed

Many trickier rate questions ask about “average speed” or “average velocity” (for GMAT purposes, those two are identical).  The formula for average speed is:

For a single trip at one speed, there’s nothing particularly mysterious about this question.  This concept becomes much trickier in two-leg trips, especially trips in which the car travels at one speed in one leg, and at another speed in another leg.  You can never simply average the two velocities given, and that will always be a tempting incorrect choice on the GMAT.  You always need to apply D = RT separately in each leg of the trip, and then you need to add results from the individual legs to find the total distance and the total time.

Practice Questions

For practice, here are some average velocity questions in Two-Leg trips.  The last two are challenging.

1) A car drives 300 miles at 30 mph, and then 300 miles at 60 miles per hour.  What is the car’s average speed, in mph?

1. 36
2. 40
3. 42
4. 45
5. 57

2) A car drives for 3 hours at 40 mph, and then drives 300 miles at 60 mph.  What is the car’s average speed, in mph?

1. 45
2. 47.5
3. 50
4. 52.5
5. 55

3) For the first 150 miles of a trip, a car drives at v mph.  For the next 200 miles, the car drives at (v + 25) mph.  The average speed of the whole trip is 35 mph.  Find the value of v.

1. 20
2. 25
3. 30
4. 35
5. 40

4) A car travels at one speed for 4 hours, and then at twice that speed for 6 hours.  The average velocity for the whole 10 hour trip is 40 mph.  Find the initial speed in mph.

1. 25
2. 35
3. 40
4. 50
5. 60

Practice Question Explanations

1) In order to figure out the average velocity, we need to know both the total distance and the total time.  From the question, we know the total distance is 600 miles.  We need to figure out the time of each leg separately.  In the first leg, T = D/R = 300/30 = 10 hr.  In the second leg, T = D/R = 300/60 = 5 hours.  The total time is 10 + 5 = 15 hours.  The average velocity, total distance divided by total time, is 600/15 = 40 mph. Answer = B.

2) In the first leg, we know time and rate, so find distance: D = RT = (3)*(40) = 120 miles.  In the second leg, we know distance and rate, so find time: T = R/D = 300/60 = 5 hours.  Total distance = 120 + 300 = 420 miles.  Total time = 3 + 5 = 8.  Average velocity = 420/8 = 210/4 = 105/2 = 52.5 mph.  Answer = D.

3) The distance of the first leg is 150 miles, and the rate is v, so the time of the first leg is

The distance for the second leg is 200, and the rate is v, so the time of the second leg is

The total distance was 350 miles, and the average speed was 35 mph, so the total time of the trip must have been T = D/R = 350/35 = 10 hours.  At this point, the algebra becomes hairy, so I will just plug in numbers from the answer choices.

Choice A.  If v = 20 mph, then v + 25 = 45 mph.  The first leg takes 150/20 = 7.5 hours, and the last leg 200/45 takes way more than three hours, so this total time is well over 10 hours.  This choice is not correct.

Choice B.  If v = 25, then v + 25 = 50.  The first leg takes 150/25 = 6 hours.  The second leg takes 200/50 = 4 hours.  The total is 10 hours, which is the correct value, so this is the correct answer choice.  Answer = B.

4) If the average velocity for the 10 hour trip is 40 mph, that means the total distance is D = RT = (40)*(10) = 400 miles.  The distance in the first leg is d₁ = RT = 4v.  The distance in the second leg is d₂ = RT = (2v)*(6) = 12v.  The total distance is the sum, 4v + 12v = 16 v.  Set this equal to the numerical value of the total distance.

So the initial speed is v = 25 mph.  Answer = A.

First, try these practice questions:

1) Every length in a 45°-45°-90° triangle increases by 20%.  By what percent does the area of the triangle increase?

1. 20%
2. 25%
3. 40%
4. 44%
5. 80%

2) At a certain pizza parlor, the diameter of a large pizza is 40% larger than the diameter of a small pizza.  What is the percent increase in total amount of pizza, from a small to a large?

1. 20%
2. 40%
3. 64%
4. 80%
5. 96%

3) For a certain cylinder, the diameter equals the height.  If every length in this cylinder is decreased by 60%, then to the nearest integer, by what percent does the volume decrease?

1. 22%
2. 40%
3. 60%
4. 84%
5. 94%

Do these questions stymie or confuse you?  Then you have found exactly the right post to read!

Scale factor

In Geometry, we often “scale up” and “scale down” shapes — that is to say, we make bigger or smaller copies that have the same “shape”, but in a larger or smaller version.  This process is called dilation, but that’s a word you don’t need to know.  When we want to talk about how much bigger or how much smaller the new shape is, it’s convenient to use the idea of a scale factor.  The scale factor is the factor by which every length in the shape has been multiplied.  If the shape has been scaled “up”, so that the result is larger, then the scale factor is greater than 1.  Conversely, if the shape has been scaled “down”, so that the result is smaller, then the scale factor is less than 1.  I will use the letter k for scale factor.

For example, if k = 3, that means every length in the shape has tripled; and if k = 1/5, that means every length in the shape has been divided by 5 —– of course, that would be an 80% decrease.

Scale factor & percent change

If the appearance of the percent decrease at the end of the last section threw you for a loop, it probably means you need to catch up a little on how percents work.  See this blog.  That blog explains how percent increase and percent decrease can be expressed as a multipliers — that’s one of the most powerful mathematical tricks you can know.  In particular, when a geometric shape scales up or down by a certain percent, the percentage change multiplier is the scale factor!   That’s BIG IDEA #1 of this post.

Change in area

Once you understand this next idea, it is one of the most blessedly easy rules in all of mathematics.  If every length is multiplied by scale factor k, then the area is multiplied by .  That is BIG IDEA #2 of this post.

For example, if I increase each length by 50% — this means, I multiply by 1.5, which in turn means k = 1.5.  From this we can calculate — that means area goes up to 225 percent, which is an increase of 125% above the original 100%.  Thus, area would increase by 125%.

Change in volume

Another one of the easiest mathematical laws imaginable.  If every length is multiplied by scale factor k, then the volume is multiplied by .  That is BIG IDEA #3 of this post.

For example, if each side of a box is tripled (i.e. k = 3), then total volume increases by a factor of .  If each side of a box is multiplied by 5, k = 5, then the total volume is multiplied by .

Summary

At this point, armed with these three “big ideas”, you may want to go back to the practice problems above and give them a second try, before reading the explanations below.

Practice problem solutions

1) The irony of this question is, of course, the specifics of the shape don’t matter at all.  Yes, in general it is a good idea to be familiar with the 45°-45°-90° triangle —- if you’re not, see this post —– but none of that matters here.

We have a 20% increase, so that’s a multiplier of 1.2, which means k = 1.2.  The multiplier for area would be , which is the multiplier for a 44% increase.  That’s the percent increase in area.  Answer = D

2) A real world problem —- yum! —- but again, the specifics of the shape don’t matter.  Diameter increases 40%, so k = 1.4 in this case.  “Amount of pizza” just means the area of the circle, so , which is the multiplier for a 96% increase.  That’s the percent increase in area.  Answer = E

3) Again, the exact dimensions of the cylinder do not matter, as long as every length increases by the same percent or same ratio.  We now have a percent decrease, which means we subtract to construct our multiplier.  A 60% decrease means we have a multiplier of 1 – 0.60 = 0.4, and so k = 0.4 as well.  For volume we need to cube the scale factor: approximately.  That would be the multiplier for a 94% decrease, because 1 – 0.94 = 0.06; therefore, the volume decreases by 94%.  Answer = E.

Practice questions

Let’s start out with a couple practice questions.

2) What is the length of the diagonal of a square with area 48?

The geometry of those questions is not too difficult, but the trick is how to handle those radicals.  If that befuddles you, this is just the post for you.

Distributing radicals

First of all, we need to review some rules for square-roots.  We can distribute a square-root over multiplication or division.  We cannot distribute a square-root over addition or subtraction.  Another way to say that: we can separate a square-root by multiplication or division, but not by addition or subtraction.

Beware of those two on the right: they are very tempting mistakes, and GMAT questions will be designed specifically to elicit that mistake.

Simplifying radicals

Suppose we want to simplify , which means writing it as a product of some positive integer and some much smaller root.  In order to do this, we are going to use the first property given in the previous section: we can separate the square-root by multiplication.  The trick is: for 72, find any factor which is a perfect square.  For example, , and 9 is a perfect square.  Therefore

We have simplified, but not all the way, because 8 still has another factor, 4, which is a perfect square.  To continue:

That last expression is fully simplified, because the number under the radical, 2, has no factor greater than 1 that is a perfect square.  (You always are guaranteed the radical is as simple as it’s going to get if the number under the radical is prime.)  That simplification took a relatively large number of steps.  We would have been more efficient if we had noticed at the outset: .  Then:

It saves you a bit of time if you happen to factor out the largest possible perfect-square factor right at the beginning, but as this process shows, that’s not strictly necessary for arriving at the correct simplification.

Having seen this, you may want to give the practice problems another go before looking at the explanations below.

Practice problems solutions

1) First of all, we use that most amazing theorem, the Pythagorean Theorem.

That’s the easy part.  Now, we have to take a square-root of 80.  Notice that 80 = 16*5 (16 is the largest perfect-square factor of 80).

Answer = B

2) There are a few dozen ways to approach this.  Consider the square:

We know that .  Let’s not simplify that, but just leave that as is.  You see, we are going to use the Pythagorean Theorem again, in the right triangle ACD, so even if we found out the square-root of s, we would have to turn around and square it again anyway.

Again, the meat of the problem involves simplifying this radical.  Notice that .

Answer = C

Attention, mad scientists out there!  Consider these two practice questions.

1) A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution.  How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

1. 200
2. 400
3. 500
4. 600
5. 800

2) A chef mixes P ounces of 60% sugar solution with Q ounces of a 10% sugar solution to produce R ounces of a 25% sugar solution.  What is the value of P?

[Statement #1] Q = 455 mL

[Statement #2] R = 660 mL

Mixtures, solutions, and concentration

First of all, keep in mind: this is relatively rare kind of GMAT Math problem.  There’s a good chance you could take five GMATs in row and never see a mixture problem.  Nevertheless, they do arise occasionally, and it’s good to be familiar with them.

Let’s think for a moment about “concentration” — not the mental quality, but in the chemical solution sense.  What does it mean to say we have a “400 units of a 6% phosphoric acid solution”?  Whatever those units are, the total amount of mixture, of solution, we have is 400 units, and of that, 6% is pure phosphoric acid.  Well, 6% of 400 is 24, so we know we have 24 units of pure phosphoric acid.  That is the amount of concentrate we have in our solution.  (Those “units” could be units of volume or units of mass, but that’s far more detail that you need to know on the GMAT).

I bet “phosphoric acid” sounds highly technical and abstruse: it’s actually one of the ingredients listed on every can of Coca Cola.

The two equations

The secret to any mixture or concentration problem is to use the two equations.  First is the amount of stuff, the total volume.  This is called the volume equation.  The basic idea is:

This makes sense when you think about it: the volume of the resultant solution had to come from the volumes of the two things we mixed.

The second equation is similar.  This concerns, specifically, the amount of concentrate, of whatever the chemical or substances is of which we have a solution.  In problem #1, the concentrate is phosphoric acid, and in problem #2, sugar.   The amount of concentrate that winds up in the resultant solution must come from somewhere.  It must come from the amount of concentrate in the two solutions mixed.

As in the section above, these amounts of concentrate will always be (the concentration percentage) times (the total volume of the solution concerned).

If you had trouble with the problems above before, go back to them, and see if you can set up both of these equations and solve.  Give them another try before reading the explanations below.

Also, here’s another practice question:

3) http://gmat.magoosh.com/questions/29

Practice problems explanations

1) We could backsolve from the numerical answer choices, but let’s use a straight algebra approach.  Let X equal the units of 12% phosphoric acid solution we use, and let Y be the units of 10% sulfuric acid solution that result.

The volume equation is:

400 + X = Y

In the first solution, we have 6% of 400, or 24 units of phosphoric acid.

In the second solution, we have 12% of X = 0.12*X of phosphoric acid.

In the resultant solution, we have 10% of Y = 0.12*X of phosphoric acid.

The concentration equation is:

24 + 0.12*X = 0.12*X

Multiply this by 100, to clear the decimals:

2400 + 12X = 10Y

Everything is even, so divide by 2 to simplify:

1200 + 6X = 5Y

We want X, so let’s multiply the volume equation by -5 and add that to this equation we just got:

Answer = E

2) This one is very elegant.  We  have three variables — the amount of 60% sugar solution, the amount of 15% sugar solution, and the amount of the resultant 25% sugar solution.  Three variables.  We have two equations: the volume equation and the concentration equation.  Right now, three variables and two equations: we can’t solve.

Now, look at the statements.  Each statement gives us the value of one of the variables.   If we get the value of one variable, that’s no longer a variable, and thus we are down to two variables with two equations: that’s a situation in which we can find a full solution.  Thus, given the value of either Q or R, we enter a situation in which we can solve for everything, and thus we would know P.  Therefore, each statement, by itself, is sufficient.  Answer = D

First, try this question:

1)

If you find this problem relatively easy, probably most of this post will be for you as well.  If this problem makes your head spin, you have found exactly the post you need.

Monomials and Binomials

First, a review of terms.  In algebra, any product of numbers and variables is called a “monomial” or a “term.”  For example, 2x, , and are all terms.  When we add terms together, we get a polynomial.  A polynomial with two terms is a binomial.  For example, , , and are all binomials.  A polynomial with three terms is a trinomial.  In the question above, each of the five answer choices is a trinomial.

FOIL

The FOIL process is used quite specifically to organize the multiplication of two binomials.  The word FOIL is an acronym, which stands for: First, Outer, Inner, Last.  Those four words tell you which terms of the binomial to multiply together.  Suppose we want to multiply .  Here is a step-by-step guide to the FOIL process for this.

Those purple terms at the end are the individual FOIL terms — we add those together to get the product:

That last expression results from combining the two like terms (–21x and 10x) into a single term.  That is the process of FOILing.  If the question at the beginning of this article was intimidating at first, give it another try now.  Here’s a free Magoosh question for practice.

2)  http://gmat.magoosh.com/questions/137

Practice question explanation

1) ??

FIRST =

OUTER = 6x

INNER = –x

LAST = –3

Answer = E

Master this fundamental algebraic skill which you will need on test day! 

First, try these practice problems.

1)

1. (x – 4)(x + 6)
2. (x + 4)(x – 6)
3. (x – 4)(x – 6)
4. (x – 2)(x + 12)
5. (x + 2)(x – 12)

3) If , then |a| + |b| + |c| + |d| =

1. 10
2. 12
3. 15
4. 18
5. 20

If these are easy for you, you probably have already mastered factoring: kudos to you!  If these confuse you, you have found just the post you need.

Terminology

You don’t need to know any of this terminology for the GMAT, but we need it just to talk about these ideas in words.

A binomial is a polynomial with two terms: all five answer choices to question #1 are the product of two binomials.  A quadratic is a polynomial with three terms whose highest power is x-squared: the stem of question #1 is a quadratic, and the stem of question #2 is a ratio of two quadratics.  To factor a quadratic is to express it as the product of two binomials.  Question #1 is a straightforward “factor the quadratic” problem.  Question #2 involves factoring both quadratics, in the numerator and in the denominator, and then cancelling a common factor.

Technically, any quadratic could be factored, but often the result would be two binomials with horribly ugly numbers — radicals, or even non-real numbers.  You will not have to deal with those cases on the GMAT.  We call a quadratic “factorable” if, when you factor it, the resulting equation has only integers appearing.  You will only have to factor “factorable” quadratics on the GMAT.

Sometimes, factoring quadratics involves quadratic like that in #3, with a leading coefficient (the coefficient of the x-squared term) is an integer other than 1.  These are considerably harder, and you will only see one like this if you are getting most of the math questions correct and the CAT is feeding you one 700-level question after another.

The secret of factoring: FOIL

The best way to understand factoring well is first to understand FOILing well.  Suppose we multiply two binomials:

Notice, if we follow the FOIL process forward, then we see two things.  First, the middle coefficient, the coefficient of the x term, is the sum of the two numbers.  Second, the final term, the constant term with no x, is a product of the two numbers.  Right there, that’s the key of factoring.  If I want to factor any polynomial of the form , then to factor it, we are looking for two numbers that have a product of c and a sum of b.

Things get a little complicated when some negative signs are floating around, so here’s a table, for all cases, for the two numbers, p and q, we need to find to factor the quadratic.  In the table that follows, b is the coefficient of the x-term, c is the constant term with no x, and both of those are given in the quadratic; p & q are the two numbers we need to factor the quadratic into (x + p)(x + q).

BTW, I highly recommend NOT memorizing the above chart, but rather, thinking it through, and doing FOIL examples for each case, to convince yourself of the patterns and to ingrain them into your memory.

So, for example, suppose we want to factor: .  Since c = –18 is negative, that means the p & q we want will have opposite sign: one positive, one negative.  Since b = –7 is also negative, that means which one, p or q, has the larger absolute value, that one will be negative.  We are looking for a bigger negative, and smaller positive, which will have a product of –18 and a sum of ­–7.  One way to think about this is: we need a pair of factors of 18 that have a difference of 7.  These two numbers are clearly 9 & 2.  Make the bigger one negative: –9 & +2.  Those are the numbers we need.  Now, stick these into the factoring format:

Voila!  Having read this, see if you now can figure out questions #1 & #2 above

Advanced topic: dealing with a leading coefficient

Most GMAT test takers will not see this topic.  Only if you anticipating getting the vast majority of questions on the Quant section correct should you even read this section.

Suppose you have to factor something like .  This is tricky, because you are looking for four integers: a, b, c, and d, such that .  The constraints we have are

(i) ac = 8 (both positive)

(ii) bd = –18 (one positive, one negative)

(iii) ad + bc =  –7

This is not quite as methodical and left-brain as factoring in the easy case above.  This involves a certain amount of number sense and a certain amount of pattern matching.  For a & c, the only two possibilities are, in some order, either 2 & 4 or 1 & 8.  For b & d, the possibilities for the absolute values are, in some order, either 1 & 18 or 2 & 9 or 3 & 6 — with the understanding that, in whichever one of those pairs we pick, one must be negative and one positive.  Eventually, after some experimenting and trial-and-error, we find (8)(–2) + (1)(9) = –7 —- this is the combination we need.

If that makes sense, and you feel up to the challenge, try #3.

Summary

Factoring quadratics is such a widely used skill in algebra that you are likely to see something such as Question #1 or #2 on your GMAT.  Here’s another practice question, with its own video explanation.

4) http://gmat.magoosh.com/questions/117

Practice problem explanations

1)

We need two numbers, p and q, which have a product of –24, which means one is negative and one is positive.  Their sum is –10, which means the one with the larger absolute value is negative.  Let’s go through the factor pairs of 24, in each case making the larger one negative

1        –24    sum = –23

2        –12    sum = –10

That’s the one we want!  The pair we want is +2 and –12: they  have a product of –24 and a sum of –10.

Answer = E

2) For this one, we need to factor both the numerator and denominator.  In the numerator, we have x^2 – 4x – 21, so we need two numbers that have a product of –21 and a sum of –4.  After a little trial and error, we find the pair that works is –7 and +3, so:

Now, the denominator, x^2 + 9x + 18.  We need two positive numbers that have a product of 18 and a sum of 9 — these would be 3 & 6.  Thus:

We see that we have a common factor of (x + 3) that will cancel.  Now, let’s put the fraction together:

Answer = B

3) This is the hard one, definitely a 700+ level question.  We need numbers a, b, c, and d such that

This means that ac = 6, bd = –12, and ad + bc = 1.  The a & c pair could be (1, 6) or (2, 3), in some order.  The absolute values of the b & d pair could be (1, 12) or (2, 6) or (3, 4), and of course, in each case, one of the two would have to be negative.  After some trial and error, we find:

Thus, we see:

|a| + |b| + |c| + |d| =  2 + 3 + 3 + 4 = 12

Answer = B

Learn this powerful math trick for GMAT Problem Solving questions. 

First, try these two practice GMAT Quantitative section Problem Solving questions.

1) Bert left the house with N dollars.  He spent 1/4 of this at the hardware store, then $7 at the dry cleaners, and then half of what was left at the grocery store.  When he got home, he had $10 left in his pocket.  What was the value of N?

1. $36
2. $44
3. $52
4. $60
5. $68

2) Ms. Morris invested in Fund A and Fund B.  The total amount he invested, in both funds combined, was $100,000. In one year, Fund A paid 23% and Fund B paid 17%.  The interest earned in Fund B was exactly $200 greater than the interest earned in Fund A.  How much did Ms. Morris invest in Fund A?

1. $32,000
2. $36,000
3. $40,000
4. $42,000
5. $45,000

1. 13
2. 22
3. 33
4. 46
5. 78

A powerful math trick

Do you notice what all three of these Problem Solving questions have in common?  In all three, the five answer choices are numerical.  Whenever an unknown is given in a word problem, and the answer choices are possible numerical values for this unknown, you can plug in.

What does it mean to plug in (also known as “backsolving”)?  Overall, this means that, rather than setting up, say, an algebraic solution to a problem, you will work “backwards”, from the answers first.  I know that, in Algebra Two, your encrusted teacher might have warned you not to solve problems backwards from the answer choices.  On the GMAT, though, time is of the essence, and plugging in is too efficient a strategy to pass up.

Here’s how the plugging in strategy works.  First of all, the GMAT always lists integer answer choices in numerical order, from least to greatest, exactly as these three sample problems do.  Therefore, always start with answer choice (C).  You will pretend that the value of the unknown in the prompt is the value given in (C), and you will work through the “story” with that value, until you get to the end.  At the end, the final value you get will either be too big, too small, or just right.  If it’s just right, that means we were lucky, and (C) is the correct answer.  If choosing (C) leads to a value that is too large, then you know the starting amount must be smaller than (C), so you can immediately eliminate (C) & (D) & (E).  Similarly, if choosing (C) leads to a value that is too small, then you know the starting value must be larger than (C), and you can immediately eliminate (A) & (B) & (C).  Do you realize how powerful a strategy is that, in one fell swoop, allows you to eliminate three of the five answer choices?  If you were pressed for time, you could guess randomly from the remaining to answers, and the odds would be heavily in your favor (this is called “solution behavior“).

Of course, if time is not a huge issue, you just repeat the process.  If (C) is too small, try either (D) or (E) next.  If (C) is too big, then try (A) or (B) next.  By the end of the second plug in, you should know exactly what the answer is.

As an example, I will solve sample problem #1.  This one could be solved with a variety of mean, but to demonstrate the strategy, I will employ backsolving.  Start with answer (C): let’s say that N = 52.  Well, first Fred spends 1/4, that is $13, at the hardware store, which leaves him with $39 dollars.  Then, he spends $7 at the dry cleaners, leaving him with $32.  Then he spends half at the grocery store, which leave him with $16.  This is more than the $10 which he is supposed to have at the end of his trip.  (C) led to a value that is too large, so right away, we know, (C) & (D) & (E) are all wrong.  Now, try (B): let’s say N = $44.  Well, first Fred spends 1/4, that is $11, at the hardware store, which leaves him with $33 dollars.  Then, he spends $7 at the dry cleaners, leaving him with $26.  Then he spends half at the grocery store, which leave him with $13.  This is still more than $10.  (B) is also too big.  This can only mean that (A) is the answer.  Two plug ins, and you are done.  If you were toward the end of the Quantitative section, and had oceans of time left, you could take the extra step of double-checking by backsolving with (A) to verify, but only do that extra step if you really feel that you have more than enough time remaining.

More practice

Now that you’ve read this section, give those other two problems a try with plugging in, before reading the solutions below.  Here’s another Magoosh problem on which you can also practice:

4) http://gmat.magoosh.com/questions/9

Practice problem solutions

1) A

2) This one could be solved with a good deal of hairy algebra, but backsolving is an incredibly efficient alternative.  Let’s start with (C) — we’ll say Ms. Morris invested $40,000 in Fund A, which means she invested $60,0000 in fund B.  In Fund A, she earned 23%, which was (0.23)*40000 = $9200.  In Fund B, she earned 17%, which was (0.17)*60000 = $10200.  Here, the interest earned in Fund B was $1000 more than the interest earned in Fund A, a bigger difference than we want.  To make the difference smaller, we need less money in Fund B and more money in Fund A.  Eliminate answers (A) & (B) & (C).  Try D: We’ll say Ms. Morris invested $42,000 in Fund A, which means she invested $58,000 in Fund B.  In Fund A, she earned 23%, which was (0.23)*42000 = $9660.  In Fund B, she earned 17%, which was (0.17)*58000 = $9860.  The difference is exactly $200, so D is the answer.

3) For this one, a frontal assault with algebra is likely to be an unproductive disaster.  Backsolving is really the way to go.  Notice that each of the three terms increases as n increase, so we have reason to believe the value of a(n) increase as n increases.  Let’s start with n= 33.  We have to do a little arithmetic to figure out that 33^2 = 1089, but then we can calculate that

That’s much bigger than 500.  Presumably, the very first term larger than 500 would not be that much bigger than 500.  Let’s go down one, to (B).  Suppose n = 22.  Again, we have to do a little arithmetic to calculate that 22^2 = 484, but then we can compute:

Aha!  This is just over 500, and the answers are spaced out enough, that the next one, (A), will undoubtedly be much smaller.  Therefore, answer = B.

Question #1: In the equation,

can you divide both sides by x?

Question #2: In the equation (x – 3)(x + 5) = (2x + 1)(x + 5), can you divide both sides by (x + 5)?

Question #3:

Question #4:

Dividing by a variable or by an algebraic expression

The short answer is: NO.

You see, it’s mathematically illegal to divide by zero, and if you don’t know the value of the variable, then you could be breaking the law without knowing it.  Ask any judge — not knowing that you’re breaking the law generally is not an excuse that holds very well up in court.  In much the same way, not knowing whether you are dividing by zero, because you are dividing by an unknown, is just as bad as dividing by zero directly.

What do you do instead?  Well, there are two alternatives.  One method is: instead of dividing by the variable, factor it out.  For example, with Question #1:

If the produce of two or three or more factors equals zero, this means one of the factors must equal zero.  Here, either x = 0 or (x + 3) = 0, which leads to solutions of x = 0 and x = –3.

The second method is two break the problem into two cases, one in which the variable or expression does equal zero, and one in which it doesn’t.  Treat the two cases separate.  For example, in Question #2:

Case I: let’s consider the case in which (x – 5) = 0.  Well, if this equaled zero, the equation would be true, so this is a solution.  One solution is x = 5.

Case II: let’s consider the case in which (x – 5) ≠ 0, that is, the case in which x ≠ 5.  Well, now we are guaranteed that  (x – 5) ≠ 0 is not equal to zero, so dividing both sides by this expression is now perfectly legal, and this leads to the simple equation x – 3 = 2x + 1, which has a solution of x = –4.  Thus, the overall solutions to this problem are x = 5 and x = –4.

Canceling a variable or expression

Similarly, the blanket answer to the cancelling question is also, NO!, for the same reason.  If there is any possibility that your variable or expression equals zero, then cancelling would be a 100% illegal activity.

For Question #3 — for all values of x other than x = 0, for the entire continuous infinity of numbers on the number line excluding that solitary value, yes, the fraction 2x/5x would equal 2/5.  BUT, when x = 0, that statement is no longer true — it is not even false — it is profoundly meaningless.  It would be like asking whether the number 163 has a flavor — even posing the question implies a profound misconstruing of essential nature of what a number is.  For this one, we would have to say — whatever the question is asking, whatever the question is doing, we have to recognize that x = 0 is not at all a possible value; having eliminating that value, we can proceed with whatever the rest of the problem may be.

Question #4 is a particularly interesting one.  First of all, as with the previous example, we run into major difficulties when the factor-to-be-cancelled equals zero.  As with the other questions, we can’t just do a blanket cancelling with impunity.   As with the previous two questions, we have to consider cases.  If (x + 2) = 0, then the expression on the left becomes 0/0, profoundly meaningless, and any statement setting this equal to anything else would be sheer nonsense.  If (x + 2) = 0, then nothing equals anything else in this problem, so x = –2 is definitely not a legitimate answer.

Now, what happens in the case in which (x + 2) ≠ 0?  Well, in this case, this factor does not equal zero, so it can be cancelled, which leads to:

Now, we have the same expression on both sides of the equation.  This means, these two sides would be equal for all values of x, as long as the expression is defined.  This means the whole continuous infinity of the number line is legal, barring a couple isolated exceptions.  One is x = –4, which makes the denominator equal zero — something divided by zero cannot equal anything, because something divided by zero has already departed from the realm in which any mathematically meaningful statement is possible.  And, of course, as we discovered above, x = –2 cannot be a solution either.  Therefore, the solution consists of all real numbers, the entire continuous infinity of the real number line, except for the values x = –4 and x = –2.

Summary

Don’t divide by variables or by algebraic expressions.  Don’t cancel by variables or by algebraic expressions.  Always consider whether the factor by which you would want to divide could equal zero, and either factor it out or consider the process in separate cases.

A lot of students, especially those away from math for a long time, get lost when trying to divide by a square root. First, consider these three practice questions.

1.  In the equation above, x =

2. Triangle ABC is an equilateral triangle with an altitude of 6.  What is its area?

3.  In the equation above, x =

The second one throws in a little geometry.  You may want to review the properties of the 30-60-90 Triangle and the Equilateral Triangle if those are unfamiliar.  The first one is just straightforward arithmetic.  The third is quite hard.  For any of these, it may well be that, even if you did all your multiplication and division correctly, you wound up with an answers of the form — something divided by the square root of something — and you are left wondering: why doesn’t this answer even appear among the answer choices?  If this has you befuddled, you have found exactly the right post.

Fractions and radicals

When we first met fractions, in our tender prepubescence, both the numerators and denominators were nice easy positive integers.  As we now understand, any kind of real number, any number on the entire number line, can appear in the numerator or denominator of a fraction.  Among other things, radicals —- that is, square-root expressions —- can appear in either the numerator or denominator.  There’s no particular issue if we have the square-root in a numerator.  For example,

is a perfectly good fraction.  In fact, those of you who ever took trigonometry might even recognize this special fraction.   Suppose, though, we have a square root in the denominator: what then?  Let’s take the reciprocal of this fraction.

This is no longer a perfectly good fraction.  Mathematically, this is a fraction “in poor taste”, because we are dividing by a square-root.  This fraction is crying out for some kind of simplification.  How do we simplify this?

Dealing with square roots in the denominator

By standard mathematical convention, a convention the GMAT follows, we don’t leave square-roots in the denominator of a fraction.  If a square-root appears in the denominator of a fraction, we follow a procedure called rationalizing the denominator.

We know that any square root times itself equals a positive integer.  Thus, if we multiplied a denominator of the square root of 3 by itself, it would be 3, no longer a radical.  The trouble is —- we can’t go around multiplying the denominator of fractions by something, leaving the numerator alone, and expect the fraction to maintain its value.  BUT, remember the time-honored fraction trick — we can always multiply a fraction by A/A, by something over itself, because the new fraction would equal 1, and multiplying by 1 does not change the value of anything.

Thus, to simplify a fraction with the square root of 3 in the denominator, we multiply by the square root of 3 over the square root of 3!

That last expression is numerically equal to the first expression, but unlike the first, it is now in mathematical “good taste”, because there’s no square root in the denominator.  The denominator has been rationalized (that is to say, the fraction is now a rational number).

Sometimes, some canceling occurs between the number in the original numerator and the whole number that results from rationalizing the denominator.  Consider the following example:

That pattern of canceling in the simplification process may give you some insight into practice problem #1 above.

Square roots and addition in the denominator

This is the next level of complexity.  Suppose we are dividing a number by an expression that involves adding or subtracting a square root.  For example, consider this fraction:

This is a fraction in need of rationalization.  BUT, if we just multiply the denominator by itself, that WILL NOT eliminate the square root — rather, it will simply create a more complicated expression involving a square root.  Instead, we use the difference of two squares formula, = (a + b)(a – b).  Factors of the form (a + b) and (a – b) are called conjugates of one another.  When we have (number + square root) in the denominator, we create the conjugate of the denominator by changing the addition sign to a subtraction sign, and then multiply both the numerator and the denominator by the conjugate of the denominator.   In the example above, the denominator is three minus the square root of two.  The conjugate of the denominator would be three plus the square root of two.  In order to rationalize the denominator, we multiply both the numerator and denominator by this conjugate.

Notice that the multiplication in the denominator resulted in a “differences of two squares” simplification that cleared the square roots from the denominator.  That final term is a fully rationalized and fully simplified version of the original.

Summary

Having read these posts, you may want to give the three practice questions at the top of this article another try, before reading the explanations below.  If you have any questions on this procedure, please ask them in the comments sections below!

Practice question explanations

1) To solve for x, we will begin by cross-multiplying. Notice that

because, in general, we can multiply and divide through radicals.

Cross-multiplying, we get

You may well have found this and wondered why it’s not listed as an answer.  This is numerically equal to the correct answer, but of course, as this post explains, this form is not rationalized.  We need to rationalize the denominator.

Answer = (D)

2) We know the height of ABC and we need to find the base.  Well, altitude BD divides triangle ABC into two 30-60-90 triangles.  From the proportions in a 30-60-90 triangle, we know:

Now, my predilection would be to rationalize the denominator right away.

Now, AB is simplified. We know AB = AC, because the ABC is equilateral, so we have our base.

Answer = (C)

3) We start by dividing by the expression in parentheses to isolate x.

Of course, this form does not appear among the answer choices.  Again, we need to rationalize the denominator, and this case is a little trickier because we have addition in the denominator along with the square root.  Here we need to find the conjugate of the denominator —- changing the plus sign to a minus sign — and then multiply the numerator and denominator by this conjugate.  This will result in —-

Answer = (A)

First, a few practice problems.

1) The numbers a, b, and c are all positive.  If , then what is the value of ?

Statement #1: a – b = 3

Statement #2: = 7

2) Given that (P + 2Q) is a positive number, what is the value of (P + 2Q)?

Statement #1: Q = 2

Statement #2:

3) In the diagram above, O is the center of the circle, DC = a and DO = b.  What is the area of the circle?

Statement #1:

Statement #2: a + b = 22

4) ABCD is a square with a side y, and JKLM is a side x.  If Rectangle S (not shown) with length (x + y) has the same area as the shaded region above, what is the width of Rectangle S?

(A) x

(B) y

(C) y + x

(D) y – x

(E)

Three important algebra patterns

Doing math involve both following procedures and recognizing patterns.  Three important patterns for algebra on the GMAT are as follows:

Pattern #1: The Difference of Two Squares

Pattern #2: The Squares of a Sum

Pattern #3: The Squares of a Difference

For GMAT Quant success, you need to know these patterns cold.  You need to know them as well as you know your own phone number or address.  The GMAT will throw question after question at you in which you simply will be expected to recognize these patterns.  In such a question, if you recognize the relevant formula, it will enormously simplify the problem.  If you don’t recognize the relevant formula, you are likely to be stymied by such a question.

Memory, not memorizing

You might think I would say: memorize them.  Instead, I will ask you to remember them.  What’s the difference?  Memorization implies a rote process, simply trying to stuff an isolated and disconnected factoid into your head.  By contrast, you strengthen you capacity to remember a math formula when you understand all the logic that underlies it.

Here, the logic behind these formulas is the logic of FOILing and factoring.  You should review those patterns until you can follow each both ways — until you can FOIL the product out, or factor it back into components.  If you can do that, you really understand these, and are much more likely to remember them in an integrated way.

Summary

If these patterns are relatively new to you, you may want to revisit the problems at the top with the list handy: see if you can reason your way through them, before reading the explanations below.  Here’s another practice problem from inside Magoosh:

5) http://gmat.magoosh.com/questions/129

Do you have questions?  Is there anything you would like to say?  Let us know in the comment section at the bottom!

Practice Problem Explanations

1) Let X = , the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get = X – 17.  In other words, if we could find the value of , then we could find the value of X.

Statement #1: a – b = 3

From this statement alone, we cannot calculate , so we can’t find the value of X.  Statement #1, alone and by itself, is insufficient.

Statement #2: = 7

From this statement alone, we cannot calculate , so we can’t find the value of X.  Statement #2, alone and by itself, is insufficient.

Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21.  Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out , and if we know the numerical value of that, we can calculate X and answer the prompt.   Combined, the statements are sufficient.

Answer = C

2) The prompt tells us that (P + 2Q) is a positive number, and we want to know the value of P.  Remember number properties! We don’t know that (P + 2Q) is a positive integer, just a positive number of some kind.

Statement #1: Q = 2

Obvious, by itself, this tells us zilch about P.  Alone and by itself, this statement is completely insufficient.

Statement #2:

Now, this may be a pattern-recognition stretch for some folks, but this is simply the “Square of a Sum” pattern.  It may be clearer if we re-write it like this:

This is now the “Square of a Sum” pattern, with P in the role of A and 2Q in the role of B.  Of course, this should equal the square of the sum:

All we have to do is take a square root.  Normally, we would have to consider both the positive and the negative square root, but since the prompt guarantees that (P + 2Q) is a positive number, we need only consider the positive root:

This statement allows us to determine the unique value of (P + 2Q), so this statement, alone and by itself, is sufficient.

Answer = B

3) To find the area of the circle, we need to use Archimedes’ formula, . For that we need the radius, OC.  We are not given this directly, but notice that r = OC = DC – OD = a – b.  If we knew that, we could find the area of the circle.

Statement #1:

A major pattern-matching hit!  This, as written, is the “Square of a Difference” pattern.

In fact, this statement already gives us , so we just have to multiply by pi and we have the area.  This statement, alone and by itself, is sufficient.

Statement #2: a + b = 22

We need a – b, and this statement gives us a value of  a + b.  If we had more information, perhaps we could use this in combination with other information to find what we want, but since this is all we have, it’s simply not enough to find a – b.  This statement, alone and by itself, is insufficient.

Answer = A

4) A tricky one.  First of all, notice that the shaded area, quite literally and visually, is the difference of two square — Area = .  We know from the Difference of Two Squares pattern, this factors into:

Area = = (y + x)(y – x)

Well, if a rectangle had this same area, and it had a length of (y + x), it would have to have to have a width of (y – x) — that would make the area the same.  The width has to be (y – x).  Answer = D

This is a potentially tricky topic.  First, consider these practice questions.

1. Which value(s) of x satisfies the equation above?

I. –1

II. 4

III. 9

1. I
2. III
3. I & II
4. I & III
5. I, II, & III

2. What is the value of y?

A full discussion of these problems will come at the end of this article.

A subtle distinction

Consider the following, very simple algebra equation

Of course, this equation has two solutions, +5 and -5.  Of course, we find that by taking the square-root of both sides, remembering this process involves a ± sign.  Now, by contrast, consider this expression:

Many folks might think this is, in all respects, the same as the solution in the first process — i.e. “taking the square root.”  We need to draw a subtle distinction here, concerning the nature of this sign:

The benighted unfortunately will refer to this as a “square root” sign, but that is a misleading partial name.  The name of this simple is the “principal square root” sign, where the word “principal”, in the sense of “main” or “primary”, here means the positive root only.  Accordingly, the output of this sign is always positive.

The number 25 has two square roots, one positive and one negative, but it has only one principal square root.  A number can have, at most, only one principal square root.  The principal square root of 25 is +5 only.

When a squared algebraic expression appears in a problem, and we ourselves, in the process of problem-solving, find a square root, we need to include all roots, positive and negative (a common mistake is to forget the negative roots).  BUT, when the symbol above, the principle square root symbol, is printed on the page as part and parcel of the given problem, this means its output will always be positive.

Thus, the paradoxical juxtaposition

Solving an equation with a radical

By “equation with a radical”, I will be referring to any algebraic equation in which some of the algebra is under a radical sign, a.k.a a principal square root sign.  The general strategy for such an equation is (a) isolate the radical (i.e. get it alone by itself on one side of the equation); (b) square both sides, thus eliminating the radical; and (c) solve what remains using algebra (often, this will involve factoring a quadratic.)  All well and good, but there’s a catch.

You see, when you square both sides of equation, sometimes that creates solutions that weren’t part of the original equation.  Consider the hyper-simple equation x = 5.  This “equation” has only one solution, positive five.  BUT, if we square both sides, then we get the equation we solved in the previous section, with solutions ±5.  The extra root, x = -5, was not a solution of the original equation, but it became a solution once we squared.  This is an example of an extraneous root —- a number that is not a root of the original equation, but which “becomes” a root when we square both sides.

We have to square both sides to solve an equation with radicals, but doing so introduces the possibility of an extraneous root.  Thus, an essential part of solving any equation with radicals is to check the answers you find, in order to ascertain whether any are extraneous roots.  We verify the roots by plugging them into the original equation —- if the number does not solve the original equation, as given in the problem, then it is not a bonafide solution.

BTW, just as a general point of strategy, regardless of whether radicals are involved, I recommend checking any algebra values you find by plugging them back into the original equation given in the problem, when possible.  It’s just a good habit to check your work.

Extraneous roots play a role in both of the sample problems above.  Having read this post, you may want to go back and give them another attempt before reading the solutions below.  Questions?  Let us know in the comment section at the bottom.

Solutions to the sample questions

1) The radical is already isolated, so square both sides.

(x – 9)(x + 1) = 0

preliminary solutions: x = {+9, –1}

At this point, an unsuspecting student might be tempted to answer (C), the trap answer.  BUT, the problem is: one or both of these answers could be extraneous.  We need to check each by plugging back into the original expression.

Check x = –1

This answer does not check — the left & right sides have different values.  Thus, x = –1 is an extraneous root, not a solution to the problem.

(NB: it’s often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Check x = 9

The value x = 9 checks — it makes the two sides equal, and thus satisfies the original equation.  This is the only solution, so only option III contains a root.

Answer = (B)

2) Using a tried and true DS strategy, start with the easier statement, Statement #2.

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

The radical is already isolated, so square both sides.

Lo and behold!  We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2.  The naïve conclusion would be — this statement says exactly the same thing as the other.  That’s incorrect, though, because we don’t know whether both of these values are valid solutions, or whether one or more is an extraneous root.  We need to test this in the original equation.

Test y = +4

This value checks — y = +4 is a valid solution to the equation

Test y = –2

The two sides are not equal, so this does not check!  This value, y = –2, is an extraneous root.

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question.  This statement, alone and by itself, is sufficient.

Answer = (A)

The GMAT often features quant questions with strange symbols. These symbols should not fluster you too much as long as you remember that they do not represent standard mathematical notation. Instead, the symbols pertain only to the problem and are defined by the GMAT (or whatever prep material you happen to be using).

Let’s have a look at a simple example:

. What is the value of ?

(A) 2
(B) 3
(C) 5
(D) 6
(E) 9

Explanation

To approach strange symbols think of the Q’ as a recipe. To the right of the equals sign are the steps (or the recipe) you have to follow.

Another way of looking at it, whatever we see in place of Q’ we want to plug it into the ‘Q’ in . Therefore . Because the question has two apostrophe signs, we want to repeat this procedure to get, . Answer (D).

This is a basic problem, one that if you saw it on the GMAT, would not bode well. So let’s try a problem that will make you sweat a little more.

. What the value of p in ?
(A) -5
(B) 9
(C) 13
(D) 25
(E) 625

Explanation

Be careful not to fall the trap that switches the order of b and a. Our equation should read: . Solving for p:

.

Answer (E).

For those who are looking to score a Q51, here are two brutally difficult questions. If you think you know the answer, go ahead and post it below with an explanation.

Brutal Question #1

. How many unique sums of x and y result, if x@y is an integer less than 15?

(A) 9
(B) 10
(C) 21
(D) 27
(E) 30

Brutal Question #2

[[x]] is equal to the lesser of the two integer values closest to non-integer x. What is the absolute value of ?

(A)

(B)

(C)

(D)

(E)

First, here are two challenging Quant problems, both involving three equations with three algebraic unknowns.

2a + b + 3c = 6

a – b + 5c = 12

3a + 2b + 2c = 2

1) Given the equations above, what does the product a*b*c equal?

1. 6
2. -6
3. 12
4. -12
5. 24

6x – 5y + 3z = 23

4x + 8y – 11z = 7

5x – 6y + 2z = 12

2) Given the equations above, x + y + z = ?

1. 11
2. 12
3. 13
4. 14
5. 15

Full solutions will follow this article.

Simplifying

In a previous post, I discussed solving two algebra equations with two unknowns.  That’s already a challenging task.  Three equations with three unknowns is even trickier, and something you are quite unlikely to see unless you are already performing brilliantly on the Quant section.

Well, it turns out, solving these, we use a time-honored problem-solving strategy: inside every big problem is a little problem struggling to get out. Yes, that’s playfully stated, but I have found it is often surprisingly apt in all kinds of personal and interpersonal situations.

What’s considerably more pertinent here — this basic idea is the core of much advanced mathematical thinking, at the levels of calculus, analysis, number theory, and other more abstruse topics. At all levels, mathematicians strive to reduce problems which they don’t know how to solve to problems which they do know how to solve.  That can be a hugely valuable perspective on GMAT mathematical strategy.  Among other things, that’s precisely the approach with these problems. I will assume you have read that previous post and are somewhat competent in the two variable/two equation problems — for the purposes of this discussion, I will consider those the problems we do know how to solve. I will also assume you are familiar with substitution and elimination from that previous post.

The approach

Here’s the general strategy for solving three equations with three unknowns.

Step #1: Pick a pair of equations, two of the three, and using either substitution or elimination, eliminate one of the variables.  Most often, elimination is much much easier than substitution!  After this step, we will end up with one equation with two unknowns.

For this one, you have to step back and have your right-brain pattern matching hat on.  You have to think very strategically about what would be the most efficient.  For example, in problem #1 above, if I wanted to pick the first two equations and eliminate c, I could do that, but it would involve multiplying the first equation by 5 and the second equation by (-3), which would lead to some big numbers.  Hmmm.  Not the slickest approach.  Instead, when I look at these equations I notice —- the first equation has a (+b) and the second equation has (-b), so without any fuss, I could add those equations and right away eliminate b.  That’s a considerably more efficient approach.

Step #2: Pick a different pair of equations, and through elimination, eliminate the same variable you eliminated in step #1.  As the result of this step, we now will have two equations with the same two unknowns.

Step #3: At this point, we have reduced the problem we didn’t know how to solve to one we do know how to solve: two equations with two unknowns.  Use those techniques to solve for those two variables.

Step #4: Once you have the numerical values of two of the variable, plug into any of the original equations to solve for the value of the third variable.

I will demonstrate this entire strategy in the solution to #1 below.

Do I have to solve?

For problem #1 above, we have to solve fully, but the coefficients are reasonably small, and as it turns out, the numbers come out nice and neat.  By contrast, #2 is a monster.  The numbers are larger and uglier, and the answer will come up as ugly fractions.   But, as it turns out, we can answer the questions being asked with only a minimum of calculations.

This is where you really have to have your creative, out-of-the-box, right brain cap on.   Question #2 is not asking for the values of individual variables, but for an expression.  As it turns out, there’s an unbelievably simple way to jump directly to the answer with astonishingly little work.  Do you see it?  I will discuss this in the solution below.

The Moral: Don’t automatically assume you always have to slog through the hard work of solving for all the individual variables.   Always keep your antennae up for creative, time-saving shortcuts!

Summary

Once again, these problems are very rare.   You will not see them at all unless you are performing at 700+ level, getting almost everything else right.   Here’s another problem, for further practice:

3) http://gmat.magoosh.com/questions/1009

If you have any questions on what I’ve said here, let me know in the comments sections below!

Solutions to practice problems

1) Here, I will show the full solution outlined above.  First of all, here are the equations, with letter designations.

(P) 2a + b + 3c = 6

(Q) a – b + 5c = 12

(R) 3a + 2b + 2c = 2

(I started later in the alphabet, so these letter-names of the equations wouldn’t be confused with answer choice letters!)  First of all, I notice that lovely (+b) in (P) and (-b) in (Q), so I will add those two.

That is new equation (S), with variable a & c.  Now, in step #2, we want to pick a different pair of equations, and eliminate the same variable, b.  Again, I like the (-b) in (Q) — that won’t be hard to use to cancel (+2b) in equation in (R).  Just multiply (Q) by 2, and add it to (R).

Now, with (S) and (T), we have two equations with the same two unknowns, a & c.  The numbers 8 & 24, the coefficients of c, have a LCM of 24.  Multiply (S) by 3 and (T) by (-2).

Thus, a  = -2.  Plug this value into (S).

Plug these two values into (P).

Thus, {a, b, c} = (-2, 1, 3), and their product a*b*c = -6.  Answer = (B).

2) As with the last problem, I will begin by giving letter names to the equations.

(P) 6x – 5y + 3z = 23

(Q) 4x + 8y – 11z = 7

(R) 5x – 6y + 2z = 12

Here, it would be a colossal waste of time to solve for the individual values of x & y & z separately.  We want to find the value of x + y + z.  Notice, first of all, that the x-coefficient of (R) is one higher than that of (Q); unfortunately, their y-coefficients are 14 units apart from each other, not close at all.  Now, notice that the x-coefficient of (P) is one higher than that of (R); also, the y-coefficient of (P) is one higher than that of (R); also, the z-coefficient of (P) is one higher than that of (R)!  BINGO!  The difference, (P) – (R), equals the expression we seek!

Answer = A

More formally, these are called “rational expressions” in mathematics — “rational” in the sense of “having do with a ratio”, as the word is used in the phrase “rational numbers“.   These rational expression appear in some of the most challenging Quant problems on the GMAT.   Here are a few practice questions.

1. –2
2. –1
3. 0
4. 1
5. 2

1. zero
2. one
3. two
4. a finite number greater than two
5. infinitely many

5. As y increases from y = 247 to y = 248, which of the following decreases?

1. I only
2. II only
3. III only
4. II and III only
5. I, II, and III

Solutions will follow this article.

Thoughts on rational algebraic expression

First of all, to understand this stuff, you should be clear on the basic rules of fractions: how to add, subtract, multiply, and divide them.  If you can’t do this basic arithmetic with numerical fractions, it will be very hard to do it with algebraic rational expressions!   Some further tips:

1.  Suppose you have an equation involving one or more algebraic rational expressions.  Suppose you are asked to solve for values of the variable.  It’s important to note that any value of the variable that makes any individual denominator equal to zero cannot possibly be a solution of the equation.   This can be a powerful tool in “which of the following could be a solution” question, because usually you can immediately eliminate a few answers right away, which sets you up very well for backsolving or solution behavior.

2. When adding or subtracting rational expressions, as when adding or subtracting ordinary fractions, we must find a common denominator to combine.  We must do precisely the same thing with rational expressions.  Here are a couple examples of this process.

Example #1

Example #2

3.  Whenever you have just one algebraic fraction on one side of the equation equal to just one algebraic fraction on the other side of the equation, then you can cross-multiply. If either side has more than one fraction, added or subtracted, you would have to combine them, via the previous hint, before you are ready to cross-multiply.

4.  If the whole equation has only one or two denominators, you also can simply multiply every term on both sides by the denominators.  That can be a very efficient way to get rid of all the fractions in one fell swoop.  For example, the equation:

can be simplified by multiplying each term by (x – 2) —- with the fraction, it cancels the denominator, simply leaving the numerator.

x(x – 2) = 2(x – 2) + 1

If we were to multiply this all out, we would get a quadratic that we could solve.

5.  One could always use a direct algebraic solution: that may be efficient or that may take several steps and be time-consuming, even if you know you are doing. Remember that backsolving may be quicker. For a compound fraction (a big fraction with a little fraction in the numerator or denominator), it may well be quicker to step back and perform a more holistic solution, looking at what must be true about each piece: I demonstrate this in the solution for practice problem #2 below.

Summary

If this article gave you any insights, you may want to give the practice problems another look before jumping into the explanations below.  Here’s another practice question from inside Magoosh.

6. http://gmat.magoosh.com/questions/137

If you would like to express anything or ask for clarification, please let us know in the comments section below.

Explanations to the practice problems

1) We could multiply by the three denominators, (x – 2), then (x – 1), then (x + 1), get three quadratics, and simplify.  That’s a lot of work.

We could find a common denominator on the right, combine them into a single ugly fraction, then when we had a single fraction on each side, we could cross multiply.  That’s also a lot of work.

The simple solution for which this problem is crying out is backsolving.  Notice that three of the answer choices — (A) & (D) & (E) — are illegal because each makes one of the denominators zero.  We can immediately eliminate all three of those.  Now, it’s just a matter of plugging in the other two answer choices.

The two sides are not equal, so (B) can’t be the right answer.   At this point, we pretty much know that (C) must be the answer, but it’s always good to verify that it works.

Both sides are equal, so x = 0 satisfies this equation.  Therefore, answer = (C).

2) Rather than do a ton of algebraic re-arranging, let’s think about this.  We have 3 divided by (something) equals 1/2.  This means, the “something” must equal 6.   That immediately produces the much simpler equation:

Answer = (A)

3) Multiply all three terms by x and we get

This equation is unfactorable.  It is not a perfect square.  Think about its graph, which is a parabola:

When x = 0, y is negative, and when x = 2, y is positive.  Therefore, the parabola intersects the x-axis twice, which means the equation has two real solutions.

Answer = (C).

BTW, this is a special mathematical equation.  One solution is the Golden Ratio, and the other solution is the negative reciprocal of the Golden Ratio.

4) This is an easy one to solve.  Subtract 2 from both sides:

Now, add 3/y to both sides.  Because the two fractions have the same denominator, y, we can just add the numerators:

Answer = (E)

5) As y gets larger, what happens to each one of these?

For statement I, as y gets larger, the 2y gets larger.  Since the subtracting 100 stays the same as the value of y changes, that makes no difference.  This one increases as y increases, so it is not a correct choice.

For statement II, as the denominator of a fraction increases, the value of the fraction overall decreases.   When y increases, 50/y has to decrease.  Again, adding 80 remains the same as y changes, so this doesn’t make any difference.  This is a correct choice.

For statement III, as long as y > 3, then y^2 – 3y will increase as y increase.  That means the entire fraction decrease.  We are subtracting 100 minus the fraction, and if the fraction gets smaller, then we are subtracting something smaller and therefore are left with more.   This means the entire expression, the difference, gets bigger as y increase.   This one increases as y increases, so it is not a correct choice.

Answer = (B)

Muchos estudiantes, especialmente los que están lejos de las matemáticas durante mucho tiempo, se pierden al tratar de dividir por una raíz cuadrada. Primero, revisemos estas tres preguntas de práctica.
 

 
1. En la ecuación anterior, x =
 

 

2. El triángulo ABC es un triángulo equilátero con una altitud de 6. ¿Cuál es su área?
 

 

 
3. En la ecuación anterior, x =
 

 
En la segunda pregunta necesitas un poco de geometría. Es posible que desees revisar las propiedades del triángulo 30-60-90 y del triángulo equilátero si no estás familiarizado con ellas. El primer ejercicio es aritmética simple. La tercera pregunta es bastante difícil. Para cualquiera de estos, bien puede ser que, incluso si multiplicaste y dividiste correctamente, tu respuesta sea del tipo: algo dividido por la raíz cuadrada de algo y te preguntarás “¿por qué mi respuesta no aparece entre las opciones?”. Si esto te confunde, has encontrado el post correcto.

Fracciones y radicales

Cuando nos enseñan las fracciones por primera vez, en nuestra tierna pubertad, tanto los numeradores como los denominadores eran fáciles números enteros positivos. Como sabemos ahora, cualquier número real, puede aparecer en el numerador o en el denominador de una fracción. Entre otras cosas, los radicales, o sea, las expresiones con raíces cuadradas, pueden aparecer tanto en el numerador como en el denominador. No hay ningún problema en particular si tenemos la raíz cuadrada en un numerador. Por ejemplo,
 

 
Es una fracción fácil. De hecho, aquellos que alguna vez han estudiado trigonometría podrían incluso reconocer este tipo de fracción. Supongamos ahora que tenemos una raíz cuadrada en el denominador: ¿qué ocurre entonces? Tomemos el recíproco de esta fracción.
 

 
Esto ya no es una fracción fácil. Matemáticamente, esta es una fracción “de mal gusto”, porque estamos dividiendo por una raíz cuadrada. Esta fracción nos pide a gritos algún tipo de simplificación. ¿Cómo simplificamos esto?

Cómo lidiar con raíces cuadradas en el denominador

Por convención matemática estándar, una convención que se sigue en el GMAT, no dejamos raíces cuadradas en el denominador de una fracción. Si una raíz cuadrada aparece en el denominador de una fracción, seguimos un procedimiento llamado racionalización del denominador.
 
Sabemos que al multiplicar la raíz cuadrada por sí misma obtenemos un entero positivo. Así que, si multiplicamos un denominador con la raíz cuadrada de 3 por sí mismo, el resultado sería 3, que ya no es un número radical. El problema es que no podemos ir multiplicando el denominador de de una fracción por cualquier número sin multiplicar también el numerador porque la fracción perdería su valor original. PERO, recuerda el el viejo e infalible truco de multiplicar la fracción por uno; siempre podemos multiplicar una fracción por A / A, por un número cualquiera sobre sí mismo, porque esta nueva fracción sería igual a 1, y la multiplicación por 1 no cambia el valor de la fracción.
 
Así, para simplificar una fracción con la raíz cuadrada de 3 en el denominador, ¡multiplicamos por la raíz cuadrada de 3 sobre la raíz cuadrada de 3!
 

 
Esa última expresión es numéricamente igual a la primera expresión, pero a diferencia de la primera, ahora es “de buen gusto” para las matemáticas, porque no hay ninguna raíz cuadrada en el denominador. El denominador ha sido racionalizado (es decir, la fracción es ahora un número racional).
 
A veces, se produce alguna cancelación entre el número del numerador original y el número entero que resulta de racionalizar el denominador. Considera el siguiente ejemplo:
 

 
Ese patrón de cancelación en el proceso de simplificación puede darte una idea de la solución al ejercicio # 1 de arriba.

Raíces cuadradas y suma en el denominador

Este es el siguiente nivel de complejidad. Supongamos que estamos dividiendo un número por una expresión que implica sumar o restar una raíz cuadrada. Por ejemplo, revisemos esta fracción:
 

 
Esta es una fracción que necesita ser racionalizada. PERO, si simplemente multiplicamos el denominador por sí mismo, eso NO eliminará la raíz cuadrada, sino que simplemente creará una expresión aún más complicada con una raíz cuadrada. En lugar de esto, utilizamos la fórmula de diferencia de cuadrados, = (a + b)(a – b). Los factores de la forma (a + b) y (a – b) se llaman conjugados entre sí. Cuando tenemos (número + raíz cuadrada) en el denominador, creamos el conjugado del denominador cambiando el signo de suma a un signo de resta, y luego multiplicamos tanto el numerador como el denominador por el conjugado del denominador. En el ejemplo anterior, el denominador es tres menos la raíz cuadrada de dos. El conjugado del denominador sería tres más la raíz cuadrada de dos. Para racionalizar el denominador, multiplicamos tanto el numerador como el denominador por este conjugado.
 

 
Observa que la multiplicación en el denominador resultó en una simplificación mediante la “diferencia de dos cuadrados” que eliminó las raíces cuadradas del denominador. La expresión final es una versión completamente racionalizada y simplificada de la original.

Resumen

Después de leer este post, puede ser que quieras intentar resolver las tres preguntas de práctica de arriba antes de leer las siguientes explicaciones. Si tiene alguna pregunta sobre este procedimiento, ¡por favor házla en las sección de comentarios de abajo!

Explicaciones a los ejercicios de práctica

1) Para encontrar el valor de x, comenzaremos con una multiplicación cruzada.

 
Observa que lo anterior ocurre porque en general, podemos multiplicar y dividir con radicales.
 

 
Al realizar una multiplicación cruzada, obtenemos
 

 
Es posible que hayas obtenido este resultado y te hayas preguntado por qué no aparece entre las opciones de respuesta. Esto es numéricamente igual a la respuesta correcta, pero por supuesto, como explicamos en este post, esta forma no está racionalizada. Necesitamos racionalizar el denominador.
 

 
Respuesta = (D)
 
2) Sabemos la altura del triángulo ABC y necesitamos encontrar la medida de su base. Bueno, la altitud BD divide el triángulo ABC en dos triángulos de 30-60-90 grados. De las proporciones de un triángulo de 30-60-90 grados, sabemos:
 

 
Ahora, yo trataría de racionalizar el denominador de inmediato.
 

 
Ahora, AB se debe simplificar. Sabemos que AB = AC, porque el triángulo ABC es equilátero, por lo que obtenemos la medida de su base.
 

 
Respuesta = (C)
 
3) Comenzamos dividiendo por la expresión que está entre paréntesis para despejar a x.

 
Por supuesto, esta expresión no aparece entre las opciones de respuesta. Una vez más, necesitamos racionalizar el denominador, y este caso es un poco más complicado porque tenemos una suma en el denominador además de la raíz cuadrada. Aquí tenemos que encontrar el conjugado del denominador: cambiar el signo más a un signo menos y luego multiplicar el numerador y el denominador por este conjugado. Esto nos dará como resultado:
 

 
Respuesta = (A)
 
 
Este post originalmente apareció en inglés en el blog Magoosh y fue traducido por Brenda Cabrera.

The post Matemáticas para el GMAT: cómo dividir por una raíz cuadrada appeared first on Magoosh GMAT Blog.

Muitos estudantes, especialmente aqueles que não mexem com matemática há um bom tempo, ficam perdidos quando tentam fazer divisões com raízes quadradas. Primeiro, considere essas três questões práticas.
 

1. Na equação acima, x =

 

 
2. O triângulo ABC é um triângulo equilátero com altura 6. Qual é a sua área?

 

 
3. Na equação acima, x =

 
Para segunda questão, é necessário um pouco de geometria. Revise as propriedades do Triângulo 30-60-90 (Triângulo Retângulo) e do Triângulo Equilátero se elas não são familiares para você. A primeira questão é apenas aritmética simples. A terceira é um pouco difícil. Para qualquer uma dessas, pode ser que, mesmo que tenha feito todas suas multiplicações e divisões corretamente, você acabe com uma resposta do tipo – alguma coisa dividido pela raiz quadrada de outra – e e perguntando: por que esta resposta nem aparece nas opções de escolha? Se isto lhe deixa atordoado, você achou o post certo.

Frações e Radicais

Quando nós começamos a mexer com frações, na nossa doce pré-adolescência, tanto os numeradores quanto os denominadores eram lindos inteiros positivos. Agora nós sabemos qualquer tipo de número real, qualquer número na linha dos inteiros, pode aparecer no numerador ou denominador de uma fração. Dentre outras coisas, radicais – ou seja, expressões de raiz quadrada – podem aparecer tanto no numerador quanto no denominador. Não há nenhum problema quando elas aparecem no numerador. Por exemplo,
 

 
é uma fração perfeitamente boa. De fato, quem já estudou trigonometria deve até reconhecer essa fração especial. Suponha, então, que temos uma raiz quadrada no denominador: e agora? Vamos pegar o recíproco dessa fração.
 

 
Esta não é mais uma fração perfeitamente boa. Matematicamente, esta fração é de “mau gosto”, porque estamos dividindo por uma raiz quadrada. Esta fração está implorando por algum tipo de simplificação. Como simplificamos isto?

Lidando com raiz quadrada no denominador

Pela convenção matemática padrão, seguida pelo GMAT, nós não deixamos raízes quadradas no denominador de uma fração. Se uma raiz quadrada aparece de tal forma, nós seguimos um procedimento chamado racionalização do denominador.
 
Nós sabemos que qualquer raiz quadrada multiplicada por ela mesma equivale a um número inteiro. Então, se nós multiplicarmos o denominador da raiz quadrada de 3 por ela mesma, daria 3, que não é mais um radical. O problema é – nós não podemos sair por aí multiplicando o denominador de frações por algo, deixando o numerador sozinho, e esperar que a fração se mantenha com o mesmo valor. MAS, lembre-se daquele antigo truque de frações – nós podemos sempre multiplicar uma fração por A/A, por um número sobre ele mesmo, porque a fração seria igual a 1, e multiplicá-la por 1 não muda o valor de nada.
 
Então, para simplificar a fração com a raiz quadrada de 3 no denominador, nós multiplicamos pela raiz quadrada de 3 sobre a raiz quadrada de 3!
 

 
A última expressão é numericamente igual à primeira, mas ao contrário desta, é agora matematicamente aceita, porque não há nenhuma raiz quadrada no denominador. Ele foi racionalizado (isto é, a fração é agora um número racional).
 
Às vezes, alguns cancelamentos acontecem entre o número do numerador original e o número inteiro que resulta da racionalização do denominador. Considere o seguinte exemplo:
 

 
Este padrão de cancelamento no processo de simplificação pode fornecer algum discernimento para o problema prático nº 1 acima.

Raiz quadrada e adição no denominador

Este é o próximo nível de complexidade. Suponha que estamos dividindo um número por uma expressão que envolva adição ou subtração de uma raiz quadrada. Por exemplo, considere esta fração:
 

 
Esta é uma fração que precisa ser racionalizada. MAS, se apenas multiplicamos o denominador por si mesmo, isto NÃO IRÁ eliminar a raiz quadrada – em vez disso, irá apenas criar expressão mais complicada envolvendo raiz quadrada. Nós usamos a fórmula da diferença de dois quadrados, = (a + b)(a – b). Fatores da forma (a + b) e (a – b) são chamados conjugados um ao outro. Quando temos (número + raiz quadrada) no denominador, nós criamos o conjugado do denominador mudando o sinal de adição para um sinal de subtração, e então multiplicamos numeradores e denominadores pelo conjugado do denominador. No exemplo acima, o denominador é três menos a raiz quadrada de dois. O conjugado do denominador seria três mais a raiz quadrada de dois. A fim de racionalizar o denominador, nós multiplicamos tanto o numerador quanto o denominador por este conjugado.
 

 
Note que a multiplicação no denominador resultou em uma simplificação da “diferença de dois quadrados” que tirou a raiz quadrada do denominador. O termo final está completamente racionalizado e é uma versão totalmente simplificada da original.

Resumo

Após ler esse post, tente resolver as três primeiras questões práticas do começo do artigo antes de ler as explicações abaixo. Se tiver quaisquer dúvidas no processo, por favor, faça-as nos comentários na seção abaixo!

Explicações das questões práticas

1) Para resolver x, nós começamos fazendo a multiplicação cruzada. Note que
 

 
porque, em geral, nós podemos multiplicar e dividir através de radicais.
 

 
Fazendo a multiplicação cruzada, nós temos
 

 
Você poderá ter encontrado isso e se perguntar porque não está listado na resposta. Isto é numericamente igual à resposta correta, mas é claro, como este post explica, esta forma não é racionalizada. Nós precisamos racionalizar o denominador.
 

 
Resposta = (D)
 
2) Nós sabemos a altura do triângulo ABC e precisamos encontrar sua base. Bem, a altura BD divide o triângulo ABC em dois triângulos retângulos. Pelas proporções em um triângulo retângulo, nós sabemos:
 

 
Agora, meu trabalho é racionalizar o denominador logo.
 

 
Agora, AB está simplificado. Sabemos que AB = AC, porque o triângulo ABC é equilátero, então nós temos nossa base.
 

 
Resposta = (C)
 
3) Nós começamos dividindo a expressão entre parênteses para isolar x.
 

É claro, esta forma não aparece entre as opções de respostas. Novamente, nós precisamos racionalizar o denominador, e nesse caso é um pouco mais complicado, porque nós temos uma adição no denominador junto com uma raiz quadrada. Aqui, nós precisamos encontrar o conjugado do denominador – mudando o sinal de mais para um sinal de menos – e então multiplicar o numerador e o denominador por este conjugado. Isto irá resultar em –
 

 
Resposta = (A)
 
 
Esta postagem apareceu originalmente em inglês no Magoosh blog e foi traduzida por Jonas Lomonaco.

The post Matemática GMAT: Como Dividir por Uma Raíz Quadrada appeared first on Magoosh GMAT Blog.

Here are fifteen challenging questions for GMAT math practice, with explanations below.  Can you keep the GMAT Quant pace, doing these in under 90 seconds each?   As always, no calculator!

1) Let abcd be a general four-digit number and all the digits are non-zero.  How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?

(A) 6

(B) 7

(C) 24

(D) 36

(E) 42

2) Let abcd be a general four-digit number.  How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?

(A) 4

(B) 6

(C) 12

(D) 24

(E) 36

3) There are 500 cars on a sales lot, all of which have either two doors or four doors.  There are 165 two-door cars on the lot.  There are 120 four-door cars that have a back-up camera.   Eighteen percent of all the cars with back-up cameras have standard transmission.  If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?

(A) 18

(B) 27

(C) 36

(D) 45

(E) 54

4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4^th or 5^th grade.   There are 320 students in the 4^th grade, and there are 210 girls in the 5^th grade.  Fifty percent of the 5^th graders and 40% of the 4^th graders take Mandarin Chinese.   Ninety 5^th grade boys do not take Mandarin Chinese.  The number of 4^th grade girls taking Mandarin Chinese is less than half of the number of 5^th grade girls taking Mandarin Chinese.  Which of the following could be the number of 4^th grade boys in Mandarin Chinese?

(A) 10

(B) 40

(C) 70

(D) 100

(E) 130

5)  A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube.  These four are not touching each other.  All outward faces are painted and all inward faces are not painted.  These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square.  The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom.  How many individual faces will have to be painted to accommodate the requirements of this new design?

(A) 0

(B) 5

(C) 9

(D) 16

(E) 27

6) Twelve points are spaced evenly around a circle, lettered from A to L.  Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices.   What is the value of N?

(A) 48

(B) 52

(C) 60

(D) 72

(E) 120

7)  Theresa is a basketball player practicing her free throws.  On her first free throw, she has a 60% chance of making the basket.  If she has just made a basket on her previous throw, she has a 80% of making the next basket.  If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket.  What is the probability that, in five throws, she will make at least four baskets?

8) Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other.  For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not.  How many “Secret Pair” numbers are there?

(A) 720

(B) 1440

(C) 1800

(D) 1944

(E) 2160

9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6).  What is the area of the part of this circle in the first quadrant?

10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84).  Which of the following could be the slope of line L?

(A) 0

(B) 1/2

(C) 1/4

(D) 2/5

(E) 6/7

11) At the beginning of the year, an item had a price of A.  At the end of January, the price was increased by 60%.  At the end of February, the new price was decreased by 60%.  At the end of March, the new price was increased by 60%.  At the end of April, the new price was decreased by 60%.  On May 1^st, the final price was approximately what percent of A?

(A) 41%

(B) 64%

(C) 100%

(D) 136%

(E) 159%

12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan.   Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs.  These then are sold to an American company, where plastic seats & backs will be affixed to these frames.  If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?

13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year.  Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year.  What is the possible percentage range for students taking both English and German in the same year?

(A) 0% to 70%

(B) 0% to 100%

(C) 10% to 70%

(D) 10% to 100%

(E) 40% to 70%

14) On any given day, the probability that Bob will have breakfast is more than 0.6.  The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5.  The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7.  Let P = the probability that, on any given day, Bob will have a sandwich for lunch.  If all the statements are true, what possible range can be established for P?

(A) 0 < P < 0.6

(B) 0 ≤ P < 0.6

(C) 0 ≤ P ≤ 0.6

(D) 0 < P < 0.7

(E) 0 ≤ P < 0.7

(A) – 64

(B) – 7

(C) 38

(D) 88

(E) 128

Explanations for this problem are at the end of this article.

More Practice

Here are twenty-eight other articles on this blog with free GMAT Quant practice questions.  Some have easy questions, some have medium, and few have quite challenging questions.

1) GMAT Geometry: Is It a Square?

2) GMAT Shortcut: Adding to the Numerator and Denominator

3) GMAT Quant: Difficult Units Digits Questions

4) GMAT Quant: Coordinate Geometry Practice Questions

5) GMAT Data Sufficiency Practice Questions on Probability

6) GMAT Quant: Practice Problems with Percents

7) GMAT Quant: Arithmetic with Inequalities

8) Difficult GMAT Counting Problems

9) Difficult Numerical Reasoning Questions

10) Challenging Coordinate Geometry Practice Questions

11) GMAT Geometry Practice Problems

12) GMAT Practice Questions with Fractions and Decimals

13) Practice Problems on Powers and Roots

14) GMAT Practice Word Problems

15) GMAT Practice Problems: Sets

16) GMAT Practice Problems: Sequences

17) GMAT Practice Problems on Motion

18) Challenging GMAT Problems with Exponents and Roots

19) GMAT Practice Problems on Coordinate Geometry

20) GMAT Practice Problems: Similar Geometry Figures

20) GMAT Practice Problems: Variables in the Answer Choices

21) Counting Practice Problems for the GMAT

22) GMAT Math: Weighted Averages

23) GMAT Data Sufficiency: More Practice Questions

24) Intro to GMAT Word Problems, Part I

25) GMAT Data Sufficiency Geometry Practice Questions

26) GMAT Data Sufficiency Logic: Tautological Questions

27) GMAT Quant: Rates and Ratios

28) Absolute Value Inequalities

Summary

These are hard problems.  When you read the solutions, don’t merely read them passively.  Study the strategies used, and do what you can to retain them.  Learn from your mistakes!

Practice Problem Explanations

1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d.  There are seven possibilities:

1. a) {1, 2, 3}, sum = 6
2. b) {1, 2, 4}, sum = 7
3. c) {1, 2, 5}, sum = 8
4. d) {1, 3, 4}, sum = 8
5. e) {1, 2, 6}, sum = 9
6. f) {1, 3, 5}, sum = 9
7. g) {2, 3, 4}, sum = 9

For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits.  Thus, for each item on that list, there are six different possible four-digit numbers.  The total number of possible four-digit numbers would be 7*6 = 42.  Answer = (E)

2)  The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd.  That limits the choices significantly.  We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product—there are no zeros in abcd.  We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.

Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}.  We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:

1. use {3, 7}, product = 21, abcd could be 3721 or 7321
2. use {3, 9}, product = 27, abcd could be 3927 or 9327
3. use {7, 9}, product = 63, abcd could be 7963 or 9763

Those six are the only possibilities for abcd.

Answer = (B)

3) Total number of cars = 500

2D cars total = 165, so

4D cars total = 335

120 4D cars have BUC

“Eighteen percent of all the cars with back-up cameras have standard transmission.”

18% = 18/100 = 9/50

This means that the number of cars with BUC must be a multiple of 50.

How many 2D cars can we add to 120 4D cars to get a multiple of 50?  We could add 30, or 80, or 130, but after that, we would run out of 2D cars.  These leaves three possibilities for the total number with BUC:

If a total of 150 have BUC, then 18% or 27 of them also have ST.

If a total of 200 have BUC, then 18% or 36 of them also have ST.

If a total of 250 have BUC, then 18% or 45 of them also have ST.

Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”

40% = 40/100 = 2/5

This means that number of cars with both back-up cameras and standard transmission must be divisible by 5.  Of the three possibilities we have, only the third words.

Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)

18% or 45 of these also have ST.

40% of that is 18, the number of 2D cars with both BUC and ST.

Thus, the number of 4D cars with both BUC and ST would be

45 – 18 = 27

Answer = (B)

4) 700 student total

4G = total number of fourth graders

5G = total number of fifth graders

We are told 4G = 320, so 5G = 700 – 320 = 380

5GM, 5GF = fifth grade boys and girls, respectively

We are told 5GF = 210, so 5GM = 380 – 210 = 170

4GC, 5GC = total number of 4^th or 5^th graders, respectively taking Chinese

We are told

5GC = 0.5(5G) = 0.5(380) = 190

4GC = 0.4(4G) = 0.4(320) = 128

4GFM, 4GMC, 5GFC, 5GMC = 4^th/5^th grade boys & girls taking Chinese

We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do.  Thus 5GMC = 80.

5GMC + 5GFC = 5GC

80 + 5GFC = 190

5GFC = 110

We are told:

4GFM < (0.5)(5GFC)

4GFM < (0.5)(100)

4GFM < 55

Thus, 4GFM could be as low as zero or as high as 54.

4GMC = 4GC – 4GFM

If 4GFM = 0, then 4GMC = 128 – 0 = 128

If 4GFM = 54, then 4GMC = 128 – 54 = 74

Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128.  Of the answer choices listed, the only one that works is 100.

Answer = (D)

5) The single cube has paint on all six sides.  Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces).  In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint).  In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces.  This chart summarizes what we have:

For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).

We could use either the single total box or any of the 24 corner boxes for the four corners of the square.  That leaves 21 of these, and 35 edge boxes, more than enough to cover the 32 edges of the square.  The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center.  The only boxes that will need to be painted, one side each, are the 9 interior boxes.  Thus, we have 9 sides to paint.

Answer = (C)

6) Here’s a diagram.

First, let’s count the equilateral triangles.  They are {AEI, BFJ, CGK, DHL}.  There are only four of them.

Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex.  We could have BAL, CAK, DAJ, and FAH.  All four of those have a line of symmetry that is vertical (through A and G).  Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation.  That’s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.

Answer = (B)

7) There are five basic scenarios for this:

Case I: (make)(make)(make)(make)(any)

If she makes the first four, then it doesn’t matter if she makes or misses the fifth!

Case II: (miss)(make)(make)(make)(make)

Case III: (make)(miss)(make)(make)(make)

Case IV: (make)(make)(miss)(make)(make)

Case V: (make)(make)(make)(miss)(make)

Put in the probabilities:

Case I: (0.6)(0.8)(0.8)(0.8)

Case II: (0.4)(0.4)(0.8)(0.8)(0.8)

Case III: (0.6)(0.2)(0.4)(0.8)(0.8)

Case IV: (0.6)(0.8)(0.2)(0.4)(0.8)

Case V: (0.6)(0.8)(0.8)(0.2)(0.4)

Since all the answers are fractions, change all of those to fractions.  Multiply the first by (5/5) so it has the same denominator as the other products.

Case I: (3/5)(4/5)(4/5)(4/5)(5/5) = 960/5^5

Case II: (2/5)(2/5)(4/5)(4/5)(4/5) = 256/5^5

Case III: (3/5)(1/5)(2/5)(4/5)(4/5) = 96/5^5

Case IV: (3/5)(4/5)(1/5)(2/5)(4/5) = 96/5^5

Case V: (3/5)(4/5)(4/5)(1/5)(2/5) = 96/5^5

Add the numerators.  Since 96 = 100 – 4, 3*96 = 3(100 – 4) = 300 – 12 = 288.

288 + 256 + 960 = 1504

P = 1504/5^5

Answer = (E)

8) There are three cases: AABC, ABBC, and ABCC.

In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C.  9*9*8 = 81*8 = 648.

In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

48*3 = (50 – 2)*3 = 150 – 6 = 144

3*648 = 3(600 + 48) = 1800 + 144 = 1948

Answer = (D)

9)

We know that the distance from A (0,6) to B (0, – 6) is 12, so triangle ABO is equilateral.  This means that angle AOB is 60°.   The entire circle has an area of

A 60° angle is 1/6 of the circle, so the area of sector AOB (the “slice of pizza” shape) is

The area of an equilateral triangle with side s is

Equilateral triangle AOB has s = 12, so the area is

If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.

Again, that’s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV.  We just want the part in Quadrant I, which would be exactly half of this.

Answer = (C)

10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those.  Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!

What is the slope from (50, 70) to (100, 90)?  Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5.  A line with a slope of 2/5 could pass just above these points.

Now, what about the third point?  For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points.  Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83).  This means it would pass under the third point, (80, 84).  A slope of 2/5 works for all three points.

We don’t have to do all the calculations, but none of the other slope values works.

Answer = (D)

11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.

Let’s say that the A = $100 at the beginning of the year.

End of January, 60% increase.  New price = $160

End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.

10% of $160 = $16

40% of $160 = 4(16) = $64

That’s the price at the end of February.

End of March, a 60% increase: that’s a increase of 60% of $64.

10% of $64 = $6.40

60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40

Add that to the starting amount, $64:

New price = $64 + $38.40 = $102.40

End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.

At this point, we are going to approximate a bit.  Approximate $102.40 as $100, so 40% of that would be $40.  The final price will be slightly more than $40.

Well, what is slightly more than $40, as a percent of the beginning of the year price of $100?  That would be slightly more than 40%.

Answer = (A)

12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan.  The German company must pay this amount.

Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan.  That’s the amount that the Germans pay to the Chinese.

That is the German company’s outlay, in euros.  Now, they make N metal chairs, and sell them, making a gross profit of P euros.

That must be the total revenue of the German company, in euros.  This comes from the sale to the American company.  Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.

That must be the total dollar amount that leaves the American company and goes to the German company.  This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.

Answer = (A)

13) First, we will focus on the least, the lowest value.  Suppose the minimum of 70% take English, and the minimum of 40% take German.  Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English.  Thus, 10% is the minimum of this region.

Now, the maximum.  Both the German and English percents are “at least” percents, so either could be cranked up to 100%.  The trouble is, though, that both can’t be 100%, because some folks have to take Italian, and nobody can take three languages at once.  The minimum taking Italian is 30%.  Let’s assume all 100% take German, and that everyone not taking Italian is taking English: that’s 70% taking English, all of whom also would be taking German.  Thus, 70% is the maximum of this region.

Answer = (C)

14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch.  The problem tells us that:

P(A) > 0.6

P(A and B) < 0.5

P(A or B) = 0.7

First, let’s establish the minimum value.  If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7.  All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.

Now, the maximum value.  Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region.  See the conceptual diagram.

The top line, 1, is the entire probability space.  The second line, P(A or B) = 0.7, fixes the boundaries for A and B.  P(A) is the purple arrow, extending from the right.  P(B) is the green arrow extending from the left.  The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.

Let’s say that P(A) is just slightly more than 0.6.  That means the region outside of P(A), but inside of P(A or B) is slightly less than 1.  That’s the part of P(B) that doesn’t overlap with P(A).    Then, the overlap has to be less than 0.5.  If we add something less than 1 to something less than 5, we get something less than 6.  P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.

Thus, 0 ≤ P(B) < 0.6.

Answer = (B)

15)

Answer = (E)

The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.

Question #1: In the equation,

can you divide both sides by x?

Question #2: In the equation (x – 3)(x + 5) = (2x + 1)(x + 5), can you divide both sides by (x + 5)?

Question #3:

Question #4:

Dividing by a variable or by an algebraic expression

The short answer is: NO.

You see, it’s mathematically illegal to divide by zero, and if you don’t know the value of the variable, then you could be breaking the law without knowing it.  Ask any judge — not knowing that you’re breaking the law generally is not an excuse that holds very well up in court.  In much the same way, not knowing whether you are dividing by zero, because you are dividing by an unknown, is just as bad as dividing by zero directly.

What do you do instead?  Well, there are two alternatives.  One method is: instead of dividing by the variable, factor it out.  For example, with Question #1:

If the produce of two or three or more factors equals zero, this means one of the factors must equal zero.  Here, either x = 0 or (x + 3) = 0, which leads to solutions of x = 0 and x = –3.

The second method is two break the problem into two cases, one in which the variable or expression does equal zero, and one in which it doesn’t.  Treat the two cases separate.  For example, in Question #2:

Case I: let’s consider the case in which (x + 5) = 0.  Well, if this equaled zero, the equation would be true, so this is a solution.  One solution is x = –5.

Case II: let’s consider the case in which (x + 5) ≠ 0, that is, the case in which x ≠ –5.  Well, now we are guaranteed that  (x + 5) ≠ 0 is not equal to zero, so dividing both sides by this expression is now perfectly legal, and this leads to the simple equation x – 3 = 2x + 1, which has a solution of x = –4.  Thus, the overall solutions to this problem are x = –5 and x = –4.

Canceling a variable or expression

Similarly, the blanket answer to the cancelling question is also, NO!, for the same reason.  If there is any possibility that your variable or expression equals zero, then cancelling would be a 100% illegal activity.

For Question #3 — for all values of x other than x = 0, for the entire continuous infinity of numbers on the number line excluding that solitary value, yes, the fraction 2x/5x would equal 2/5.  BUT, when x = 0, that statement is no longer true — it is not even false — it is profoundly meaningless.  It would be like asking whether the number 163 has a flavor — even posing the question implies a profound misconstruing of essential nature of what a number is.  For this one, we would have to say — whatever the question is asking, whatever the question is doing, we have to recognize that x = 0 is not at all a possible value; having eliminating that value, we can proceed with whatever the rest of the problem may be.

Question #4 is a particularly interesting one.  First of all, as with the previous example, we run into major difficulties when the factor-to-be-cancelled equals zero.  As with the other questions, we can’t just do a blanket cancelling with impunity.   As with the previous two questions, we have to consider cases.  If (x + 2) = 0, then the expression on the left becomes 0/0, profoundly meaningless, and any statement setting this equal to anything else would be sheer nonsense.  If (x + 2) = 0, then nothing equals anything else in this problem, so x = –2 is definitely not a legitimate answer.

Now, what happens in the case in which (x + 2) ≠ 0?  Well, in this case, this factor does not equal zero, so it can be cancelled, which leads to:

Now, we have the same expression on both sides of the equation.  This means, these two sides would be equal for all values of x, as long as the expression is defined.  This means the whole continuous infinity of the number line is legal, barring a couple isolated exceptions.  One is x = –4, which makes the denominator equal zero — something divided by zero cannot equal anything, because something divided by zero has already departed from the realm in which any mathematically meaningful statement is possible.  And, of course, as we discovered above, x = –2 cannot be a solution either.  Therefore, the solution consists of all real numbers, the entire continuous infinity of the real number line, except for the values x = –4 and x = –2.

Summary

Don’t divide by variables or by algebraic expressions.  Don’t cancel by variables or by algebraic expressions.  Always consider whether the factor by which you would want to divide could equal zero, and either factor it out or consider the process in separate cases.

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