# Solved !

SAT question of the day Nov 23 2012

### The third option is to plug in, but in this case since we have n in the equation and (4n) as the answer it is not as an appealing option as in other questions.

Note that the answer on the SAT website follows a correct calculation but marked the wrong answer as the correct one. Don’t be alarmed – this only shows there are humans on the other side of the exam !

Hanan

SAT question of the day Nov 20 2012

### This is a tough question. Not so much because of the calculations involved or because you need to know advanced math in order to solve it. It is difficult because most people will spend a considerable amount of time trying to understand it. In a sense, this is a reading comprehension question rather than a math question.

Lets try and see what would be a good approach to this question. We should also analyze what makes it confusing and work on techniques to reduce the mess.

The questions starts with a statement:

### for all integers

How could we read this information? f() means that if you place a variable inside the function will spit out what ever is written on the other side. We also note that there are limitation on what we can place in the function (also known as the Domain of the function) because we can only place n that are integers (good ! No fractions !).

What does this function do? Lets take an integer n – how about 7? If we calculate f(2 times 7 = 14) we will get 2 times the value of f(7) – whatever that maybe.

OK – not too bad – this defines our function and we understand what it can do.

What other  information are we given?

OK – so when we place into the function the value 4 we get back 4.

That means that f(2 times n)=2 times f(n) and if 2 times n = 4 then n must be 2. So f(2X2)=2f(2)=4 so f(2) should be 2.

Now we are given options to chose from what f() actually does.

Since we combined the two pieces of information and concluded f(2) must give us 2 and we know f(4)=4 we can check which functions actually spit out these values when we plug in 2 or 4.

By elimination only B satisfies both conditions.

Happy Thanksgiving

Hanan

SAT question of the day Nov 17 2012

### Hanan

SAT question of the day Nov 11 2012

### The long way to solve: L times W = original area or 100% 120% of L times 130% of W = 120/100 of L times 130/100 of W = 156/100 of W times L or 156% of W times L. This implies the new area is 156% of the original area. In other words it is equal to the original area + an addition of 56% of the original area.

– Hanan

SAT question of the day Nov 11 2012

### In the figure above, which quadrants contain pairs  that satisfy the condition ?

A.      only

B.      and  only

C.      and  only

D.      and  only

E.     , and

Ha… finally, a question for those who remember their Trig !

and so many ways to solve; such fun!

Well, not really; everyone can answer this one. What are we asked? Where in the Cartesian plain the value of the X coordinate (horizontal location) divided by the value of the Y coordinate (the vertical coordinate) can equal 1?

Well,  X divided by Y = 1 whenever X and Y have the same value.

Now, if they are both positive it is clear that we are in the I quadrant. so I is true (bye bye answer D).

Before you mark A and move on, ask yourself: is there anywhere else their values might be equal?

Yes, actually; when we are in quadrant III both X and Y are negatives and can have the same values. That means III is also true (farewell to answer B). What happens in quadrants II and IV? well either the X is negative and the Y is positive (II) or the X is positive and the Y is negative (IV). We are left with answer C.

For Trig people: x/y = Tan(alpha) and Tan is positive in Quad I and III – easy work for you.

Yet another strategy. As you may be aware by now, following my recommended strategies, it is a good practice to read the answers as part of the reading of the question. Quad I appears in 4 out of the 5 answers  – there is a good chance the statement is correct there. That means that we should start and check if the statement can be correct in Quad II – two advantages – saves time and hoop over answer A – because the question is not “find a Quad where the statement is correct” it is rather “find ALL the Quad where the statement is correct”and if you are on the last part of the growling 3+ hour test you might overlook it.

The SAT website reported 60% success by >100,000 people – which mean it is a mid-level question.

-Hanan

SAT question of the day Nov 08 2012

In a class of  seniors, there are  boys for every  girls. In the junior class, there are  boys for every  girls. If the two classes combined have an equal number of boys and girls, how many students are in the junior class?

### E.

#### First, let’s identify the type of question we have. Reading the question, it is clear we have a ratio problem (boys to girls) with the sub-category of mixing problems – you take one class with a certain ratio, mix it with another class of an unknown size and get a new ratio in the combined mixed class.

We now have to decide what would be the best strategy to solve. Lets consider the information.

Class I – the senior class – has 80 students with a 3:5 boys to girls. This implies that out of every 8 students (3+5) 3 are boys and 5 are girls. This mean there are 30 boys and 50 girls in this class.

Class II- the junior class – has ? number of students with 2 girls for every 3 boys or 3:2 boys to girls. We don’t know how many students are in the junior class and actually this is the number we are looking for as our answer.

What else do we know? We know that when mixed together the resulting population has a 1:1 ratio of boys to girls or equal number of boys and girls.

There are several possible ways to solve.

One – write an equation – if X is the number of students in the junior class (because that is the number we are looking for) then the number of boys in the class is 3/5 of X and the number of girls is 2/5 of X.

That means that in the combined class the number of all the boys is 30 + (3/5)*X and the number of all the girls is 50+ (2/5)*X. However these two expressions are equal !

We then write an equation: 30 + (3/5)*X = 50+ (2/5)*X or when simplified (1/5)*X=20 and then X=100

Two – plug in. We know that the junior class has a ratio of 3:2 – it will make sense then that the total number of students in the junior class will be a multiplicity of 5 (answers A and C can be ignored). If we plug in 80 we have 48 boys and 32 girls – but we need a difference of 20 to account for the excess girls in the senior class. If we check 120 we get 72 boys and 48 girls – too many. You can see the 100 is the correct answer – plug in is a legitimate option, wise plug in is even better.

Three – If you understand that the junior class should have 20 more boys than girls and that these 20 boys are 1/5 of the class because there are 3/5 boys and only 2/5 girls in the junior class – it is straight forward to find the answer – 100 students in the junior class.

-Hanan

SAT question of the day Nov 5 2012

### E.     The graph of  is a parabola.

Now – let see what this question is really about!

It is a “What can’t be question” because it asks

### “What CANNOT be true?”

Which mean we need to know something about each option.

There are 3 points of a function that are given to us and from their coordinates we need to conclude something about the function.

What happens between the points given? who knows? depends on the function.

It is now wise to remember that in the SAT

### the answers are part of the question !

We know (hopefully) that a line is defined by two points. This seems the most easy option to disprove. If the rise between X=0 to X=1 is 5 =[-4-(-9)] then if we move another two units to the right (to get to X=3 – our run) we should add another 10 units to the rise and get to [10+(-4)]=6 but when X=3 we have 0 – which means these three points are not on a line- QED

Now lets look on the other options – all we need to find or imagine is a case that will allow these points to fulfill the statement – remember the way this question is phrased means we are looking for an answer that is always NOT TRUE.

### A.     The graph of  has a maximum value.

Could be – if it starts at -9, goes up to -4 and continues to 0 – it makes sense that it has a maximum somewhere. Those of you who have taken Calculus can relate to some of the fundamental theorems of Calculus !

So we agree this is not always wrong (in fact it is always true)

### B.      for all points  on the graph of .

Well lets see – we have three y values:  -9, -4, and 0 all are equal to or less than zero. It is possible the graph never crosses the X axes – Does the graph have to cross the X axes and have values greater than zero? NO – so this statement is not always true and thus we can cross it out.

### C.     The graph of  is symmetric with respect to a line.

“What?,” you say? Well lets think what it means. Draw a few points on the bottom half of a piece of paper and connect them. Now fold the paper in the middle and copy the points to the upper part of the paper. Connect the lines on the top part. You just created a symmetric function with respect to the line you folded the paper across.

Now can you imagine you can do the same with the 3 points given? probably yes – so this option goes away.

### D.     The graph of  is a line.

We discussed that this is impossible. and this is the correct answer.

### E.     The graph of  is a parabola.

Well the graph of a parabola y=aX^2+bX+c can be drawn to include these points – just sketch them and see.

Notice we have all the types of answers here – the correct one – an answer that can never be

Answers that are always true and answers that can be true but are not necessarily so.

-Hanan

SAT question of the day Nov 2 2012

In a survey, a group of students from Westville High School were asked about their favorite movie genre. Each student in the group selected exactly one movie genre, and the data collected are summarized in the circle graph above. If 40 more students chose Action than Fiction, how many students were surveyed in total?

A.     100
B.     150
C.     200
D.     250
E.     300

We have in front of us a graph question – we need to be able to read the graph – in this case a pie chart. The data is given in % so we should also be able to work with percents. In addition we have a story – so we better mark up what are we looking for and if there is additional information in the text.

What is the question? “how many students were surveyed in total?”

We will set the number of students surveyed in total as our X if we need to write an equation.

What is a pie-chart? it is a way to describe numerical information graphically – the ration of the area of segment of the pie to the area of the entire pie is equal to the ratio of the group described by that segment to the total population described by the entire pie.

Lets try and read a piece of information. What does Horror 20% mean? It means that 20 out of every 100, or 1/5 of the students surveyed in  Westville High School picked horror movie as their favorite movie genre. If there are 100 student total – then 20 student picked horror, if there are 500 student total 100 picked horror (remember it is a ratio !).

What additional information do we have?

“If 40 more students chose Action than Fiction”
but there are 30% Action and 14% Fiction ! – so we can read this infromation and rewrite it as: there are
16% more students chose Action than Fiction so we come to the conclusion that 16% of the total students (out of X) are actually 40 students.

What is then 4% of the total (1/4 of 16% is 4&) ? 1/4 of 40 is 10 student. How many times does 4% go into 100%? 25 times. We conclude then there are 10X25 or 250 students surveyed in Westville High School.

In equation form:

16% of X= 40 or (16/100)X=40 divide both sides by 16/100 (divide by a fraction is equivalent to multiplication by its reciprocal) and calculate

X= 40*100/16

X=250

-Hanan

SAT question of the day Oct 30 2012

If , for which of the following values of is NOT defined?

A.
B.
C.
D.
E.

This function is written in a form that can be viewed as –  f(x)/g(x).

Both top and bottom are polynomials. The function belongs to a large set of functions that are defined for every X except those that make the bottom equal 0 – that is values of X that zero the denominator are not allowed in the Domain of the function.

We conclude then that we are interested in finding out what numbers, when plugged in instead of X, will make g(x) or the bottom part equal 0.  g(x)= (X+3)(X-4). g(x)=0 will happen when X=4 (then (X-4)=0) or X=(-3) and then X+3=0. 4 is not an option so we are left with (-3).

Hanan

SAT question of the day Oct 27 2012

If  and , then  exceeds  by

A.
B.
C.
D.
E.
I am tagging this question under the “DO YOU UNDERSTAND?” questions category.
What we really need to do is translate the question into math language.
What does “If S=something and T=something else, then T EXCEEDS S BY ” mean?
Well, “Exceeds by” mean “greater then by“. In turn  “greater then by” means you are asked by how much is T bigger than S or in math: Given that S=… and T=…. then T=S+WHAT?
Once you understand the wording of the question in English you can rephrase it in math:
WHAT?=TS
Now, when we understand what we are looking for lets look on T & S. a sequence of numbers, starting with one and getting smaller by one half (a geometric sequence) all the way to 1/32. What is T? T is 1 + one half of S – which mean it equals 1 (like the first term in the sequence making up S) and each member of the S sequence becomes the next member in the sequence because we divide S by a factor of 2 when we translate it into 1/2 times S. That is except the last member – the last one (1/32) becomes 1/64. We actually gained an additional 1/64 – this implies that TS=1/64 which is the WHAT? we were looking for.
We can use a straightforward approach:
T=1+1/2XS
S(1+1/2+1/4+1/8+1/16+1/32)
T=1+1/2X1+(1/2X1+1/2X1/2+1/2X1/4+1/2X1/8+1/2X1/16+1/2*1/32)=1+1/2+1/4+1/8+1/16+1/32+1/64
It is clear then that T is bigger than S by 1/64 – this is assuming we did not do any careless mistakes…
If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).
-Hanan

SAT question of the day Oct 24 2012

Which of the following statements must be true of the lengths of the segments on line above?

.

A.   only
B.   only
C.   only
D.   and only
E.  , , and

This question has two layers of complexity.

One is the question itself, the other is the way the answers are written.

Lets deal with this question by breaking it down.

We are given a drawing – as usual unless otherwise stated it is not to scale and is there to confuse us.

We have a line and segments – lets call AB=X BC=Y and CD=Z and since they are segments they are non-zero positive numbers.

This is an always true question – that is if we can come up with one example that is not true that answer is wrong.

Now  if we can disproof I we cross out three answers where it appears as true – and if it is true we cross out the other two answers and tell them “bye-bye” – it is clear it is the first answer to examine.

Then lets rethink the answers as:

I. AB+CD=AD as X+Y=X+Y+Z it is obvious that Z is missing from the left side and so this will never be true (cross out answers A, D and E)

Now it is between II and III or answers B and C.

II is AB+BC=AD-CD X+Y=(X+Y+Z)-Z that is always true and correct.

Well, lets see why III is wrong…

AC-AB=AD-CD  (X+Y)-X=(X+Y+Z)-Z or Y=X+Y again this will never be true – since we have the extra Y.

Another way to address this question is to replace the segments with numbers (note my choice of numbers – think why this is a wise choice and a set worth using?).

AB=1, BC=3, CD=5 and then AC=4 BD=8 and AD=9

Plug in and see which statement/s are true:

Note that I chose these numbers because it is very hard to combine them (add or subtract) to get an addition or subtraction of another set that will not be true (unlike the typical 1,2,3..).

A third option is to actually deal with the question as segments – that option is good if you have a good spatial vision.

A mid-level question – not too hard but it is unlikely you’ll save time on it.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Oct 21 2012

If  is  more than , then

A.
B.
C.
D.
E.

A classical three step math question. What do we have to find out? X-Y – that means neither X or Y are of interest – only the difference between them (this also make sense since we have only one equation). Now step two. What do we know? the information is given in words and we need to translate it into math. “X+2X” – isn’t that just 3 times X? yes ! so let write 3X. is 5 more than – that means the 3X is bigger by 5 units then something else – that means that if we SUBTRACT 5 from 3X we will have the same as – or equal amount as the other side: we rewrite this as 3X-5= Now equal what? “Y+2Y” – didn’t we just do something similar? isn’t it 3Y?

OK, the hard part is done we wrote 3X-5=3Y and we remember we are looking to see what (X-Y) is.

To do that we better have all the Xs and Ys at the same side and what ever they are equal to, on the other. Lets subtract 3Y from both sides -> 3X-3Y-5=0 and then add 5 to both sides = 3X-3Y=5

Now we can observe we have 3 times X-Y since 3(X-Y)=3X-3Y so lets divide both sides by 3 and get (X-Y)=5/3

A hard one – mainly due to the need to translate, calculate, and lack of straight forward alternative ways to solve.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credit !).

-Hanan

SAT question of the day Oct 18 2012

A manager estimates that if the company charges dollars for their new product, where , then the revenue from the product will be dollars each week. According to this model, for which of the following values of would the company’s weekly revenue for the product be the greatest?

A.
B.
C.
D.
E.

Several ways to solve. I like these type of questions due to the versatility of the paths to get to the correct answer.

Option one – plug-in – brute force but works – since you have a calculator it is not hard to find the correct answer.

Option two – this is a second degree polynomial (quadratic) with a negative coefficient preceding the X^2. This means it has a max. This fits the question which looks for what P will produce the largest value of  r(P)? Now the equation is 2000p-10p*p – we can factor this into 10(200-p)p which mean the X intercepts are at p=0 and p=200. The vertex is exactly in the middle of the curve – p=100 and will yield the maximal value.

If you took calculus you can take the first derivative – 2000-20p and equate it to zero – which will also yield p=100.

I think this is a moderate SAT math question – and as you can see different routes will yield the same answer however – the more on-target you are, able to categorize the question (max of a quadratic) and use the mathematical tools to simplify the problem the faster you’ll get to the correct answer.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Oct 15 2012

If and , then

A.
B.
C.
D.
E.

Not a hard nut to crack. However there is what to learn about the way we do things, how we save time, and why we pay attention.

First what are we looking for? X-Y=? Now we know X=12 so it is actually 12-Y=? Now what other piece of information do we have? X/Y=3 – and we remember X=12. Plug in and 12/Y=3 then Y must be 4.  Now don’t rush to mark 4 if that was a possible answer – because this is not what you were asked ! We are looking for 12-Y=? Back to the question (remember finding X or Y does not mean you answered the question!) 12-4=8

Notice – we work our way backwards (one of the techniques the person who wrote the question tries to waste your time – working from end to start can simplify things). We did not stop when we found the variable – because we stated the question as the first thing. We know the variable is only an intermediate solution and not our final answer  Lastly, we can see how easy it is to make careless mistakes (especially if you are on the last part and just to go home) – it is not done and we don’t turn off our guard until we turn in the exam !

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Oct 12 2012

If , then

A.
B.
C.
D.
E.

This one is as simple as it gets. I actually think most people who get this one wrong say to themselves “it can’t be this easy”. Well, it can – as there are hard questions there are easy ones.

If you remember that a times b equal b times a (also known as the commutative property of multiplication) then you must have concluded that N equals 7/9.

The take home lesson is – YES there are places where answering a questions takes 5 seconds. It is worth making sure this is not a trick question and be careful but do not over complicate things – there are enough hard questions to deal with.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Oct 9 2012

If and , then

A.
B.
C.
D.
E.

Two ways to solve this one – the fast (but you need to memorize and identify a basic formula) or the slow.

The easy one is if you identify it is a formula (a+b)(a-b)=a^2-b^2 – all you need to do is multiply the two values (5 times 3 is 15). If you are at it, it is good to memorize (a+b)^2= as well as (a-b)^2= and for the brave (a^3-b^3)= and (a^3+b^3)=

Now, for the rest, it becomes solving two equations with two variables – add the two values on one side (5+3=8) and the variables on the other (x+y)+(x-y)=2x – divide both by two and get X=4 Now X+Y=3 so Y must be negative one (-1). Square of 4 is 16 square of (-1) is +1. 16-1=15

The two ways are leading to the same answer but the first one will get you there so much faster. It makes sense to memorize those basic identities, expressions, and equations. The time you put into it will pay off big time during the exam.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Oct 6 2012

On the line above, if , which of the following must be true?

A.
B.
C.
D.
E.
Note – the most important thing in this question – FIGURE NOT DRAWN TO SCALE
That means it is just there to confuse you – don’t trust it and don’t try measuring it !
Now what do we have here ? It is a simple inequality question pretending to be a Geometry question !
Is it an always true question ? YES – that mean some of the answers can be sometime true but only one of the answers is ALWAYS true. If it is easier for you, you can plug in variables Instead of AB<BC<CD<DE – for example you can write W<U<X<Y<Z and look at it as an inequality question.
Now, one way to solve it is to cross out wrong answers: lets look at answer A –  AC is composed of two segments AB and BC – each is smaller than CD but is it always true they are smaller than CD together ? for example 10<11<12 does 10+11<12 ? NO. Note that if you used 1,2,3 you’ll get equal which is also not true but if you use 1,2, 5 you run into a “true”answer – remember that your strategy calls in this type of questions to find one incorrect answer – that will disqualify this option. The same is true for answers C, D and E – which mean B is the correct answer.
Another option is to try and grasp what the answer mean – for (B) – AC<CE you can break it to AB+BC<CD+DE. AB is always smaller than CD and BC is always smaller than DE. SMALL+SMALL will always be smaller than BIG+BIG.
Both the elimination and looking for the TRUE answer routes will work.
Note that this is an “ALWAYS TRUE” question – which mean that if you are plugging in numbers they should be aimed at crossing out wrong answers and not at finding the right one !
If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).
-Hanan

SAT question of the day Oct 3 2012

If , what is the value of ?

A.     16
B.     32
C.     64
D.     128
E.     256
Square root looks for the number that when multiplied by itself will give the argument under the square root.
If Sqrt of X is 16 – X must be 16 times 16 or 256. Plug this into the other equation – 256 times four is 1024. Which of the answers, when multiplied by itself can yield 1024? the only number that can produce 1024 when multiplied by itself is 32 (remember the one digit is determined by the one digits of the numbers multiplied – for example xxx4 times xxx4 will always end with 6, xxx7 times xxx7 will always end with a nine, etc. ).
Another way is to break down the sqrt (that is if you remember the rules) sqrt(4x)=sqrt(4)*sqrt(x)=sqrt(4) is two and sqrt(x) is 16 – that is 2 times 16 or 32.
Yet again, multiple ways to find the solution – practice and proficiency save time – so you can crack the harder questions and have time to check your work.
If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).
-Hanan

SAT question of the day Sep 30 2012

If , which of the following expresses  in terms of ?

A.
B.
C.
D.
E.

Like in many SAT questions it is good to find out first “What are we asked?” – it is apparent we need to express a in terms of b based on the information provided. Now lets see what is this information – 2 to the power of a equals 4 to the power of b.

If you reviewed your power laws you know that on the left side is an expression that has 2 multiplied by itself a times. On the right you have four multiplied by itself b times.

Each time you multiply by 4 is equivalent to multiplying by 2 two times. That means there should be twice as many 2s on the left side as 4s on the right. This implies a=2b

Another way to address this problem is to rewrite both side as 2 to the power: 2 ^a = 4 ^b = [(2)^2]^b = 2^2b

Thus, a=2b

Lastly, if this seems too complicated – you can use a simple SAT rule – if it is equal – it is always equal. YES – you can plug numbers in. 2^2=4^1 , 2^4=16=4^2 and so on – again leading to the same conclusion: a=2b

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan

SAT question of the day Sep 27 2012

The histogram above shows the distribution of 31 black cherry trees, by height. For example, the leftmost bar represents the black cherry trees that are at least 60 feet, but not less than 65 feet, in height. Based on the histogram, which of the following can be the average (arithmetic mean) height of the 31 black cherry trees?

70 feet, 72 feet, 74 feet, 78 feet, or 80 feet ?

How do you solve this question?

Lets translate the question. What are we exactly asked? “which of the following can be” that means one of the answers can be but is not necessarily the “average” and the other ones can never be the answer.

Now that we understand we are tasked to find a number that can be the answer we understand we are actually looking for the range of possible averages.

One SAT strategy is not to calculate but to save time and come up with an educated guess. If you are short on time, identified this is a hard to crack problem and need to come up with an answer (something is better than nothing). The key here – is the word EDUCATED which means you need to think. Now can the average be 90? can it be 60? No it can’t – this is a weighted average or an average problem -so it needs to be near the middle – which leave 72, 74, and 78 as possible guesses. 72 and 78 need the average to be “pushed”to one side by a very unlikely combination of tree heights (and as you will see below – an impossible event) so we are left with a reasonable guess of 74 – which is the right answer as shown below mathematically.

How can we find the possible range the average tree height is bounded in? we need to identify the end points – highest and lowest possible.

Now it is time to use the “Simplify” command – lets see what is the possible range of average tree heights for the trees between 60 ft and 70 ft. There are 3 trees in each group. The trees in the leftmost bar can be all 60 ft, all just shy of 65 feet, or any number between them will be the average height of these three . The 3 trees on the second bar can be all 65 feet, all shy of 70 or any number between 65 and 70. That mean the average of height of these six trees can obtain any value between (3*60+3*65)/6 and (3*65+3*70)/6.

Pleased with our understanding of the miniature problem we should expand this weighted average to the other bars: (3*60+3*65+8*70+10*75+5*80+2*85)/31 = a little over 72 – that mean 70 and 72 are not possible averages since we calculated the lowest possible one.

Since this is a SAT question you can pretty much mark 74 as the right answer and move on (think why?)

However, if we want to fully solve the problem we should find the upper limit. We can either do the calculation again and find it is a little below 78 (I told you you could have marked 74 already !) or use our understanding of how weighted averages behave when you add a constant to all terms (which is what we do when we look at the upper bound – we add 5 feet to each item as the intervals are 5 feet each).

We can see that this relatively hard SAT question requires a bit of thinking, engagement, and skill to solve. It is also a time sink – so use proper time-management strategy when working on this question.

If you have another way, that works better for you and make sense to you, please share it with me and I’ll post it to the website (with credits !).

-Hanan