SAT Question of the Day November 29th 2012

Miguel is 180 centimeters tall. At 2:00 p.m. one day, his shadow is 60 centimeters long, and the shadow of a nearby fence post is t centimeters long. In terms of t, what is the height, in centimeters, of the fence post?

A.     t + 120

B.     t over 3

C.     3 times t

D.     3 times (square root t)

E.     (t over 3)^2


When we read this question we notice that Miguel is 180 cm tall while is shadow is only 60 cm long (one third of his height). We are then given the length of an object’s shadow (t cm) and assuming it casts it’s shadow with the same RATIO to it’s height we are asked to express the object’s height in terms of t.

Now this might sound intimidating, however we are just asked to find the mathematical operation  that will allow us to calculate the height of an object given the length of the object’s shadow. With Miguel shadow we can calculate we need to multiply the length of his shadow (60 cm) times 3 in order to obtain his height (180 cm). Thus, we conclude that if this ratio is the same for any object, we need to multiply the length of the shadow (t) by 3 – to obtain 3t – which is the correct answer.

We can also look at this question as a ratio problem. 180 to 60 is the same as what to t? and again we conclude what we are looking for is 3t.


SAT Question of the day Nov 26 2012

In a community of 416 people, each person owns a dog or a cat or both. If there are 316 dog owners and 280 cat owners, how many of the dog owners own no cat?

A.  click to choose answer A   36

B.  click to choose answer B   100

C.  click to choose answer C   136

D.  click to choose answer D   180

E.  click to choose answer E   316

A typical time consuming question. Several ways to solve. The one I like uses the Venn-diagram.

On the left there are the dog-only owners (red) on the right are the cat-only owners (yellow) in the middle are those who own both a dog and a cat (orange).

If you add up the dog owners (I have a dog) and the cat owners (I have a cat) you count those who have both a dog and a cat twice (because you count them once when they say “I have a dog” and another time because they will tell you they have a cat). That means the orange population is the sum of dog owners plus cat owners minus the total: (280 +316) – 416 = 316- 416 + 280 = -100+280 = 180

Now – please remember to READ THE QUESTION because we were asked to find how many people own a dog only (red) and not how many own a dog and a cat ! (our X is not the final answer !).

All we have to do is subtract the dog+cat owners from the “have a dog” population: 316-180= 136 people


SAT Question of the Day 23 Nov 2012

If 24 over 15 = 4 over n, what is the value of 4 times n?

A.     6

B.     10

C.     12

D.     30

E.     60


This is a ratio problem. What do we know?

We know that 24 to 15 (24:15) is equal to 4 to n (4:n)

What are we looking for? What is 4 times n or 4n.

Several ways to find:

4 to n is the same as 16 to 4n (expand both top and bottom by 4).

24 to 15 has a common factor of 3 – it is the same as 8 to 5

We can rewrite as 8/5=16/(4n)

Cross multiplying gives:

8 times (4n) = 16 times 5

Divide both sides by 8

(4n)= 16*5/8 or 16/8 * 5 = 2*5 = 10

Another way find the value of n first and then multiply by 4.

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 !


SAT Question of the day Nov 20th 2012

  • function f of (2 times n) = 2 times function f of n for all integers n

  • function f of 4 = 4

If function f is a function defined for all positive integers n, and function f satisfies the two conditions above, which of the following could be the definition of function f?

A.     function f of n = n minus 2

B.     function f of n = n

C.     function f of n = 2 times n

D.     function f of n = 4

E.     function f of n = (2 times n) minus 4


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:

function f of (2 times n) = 2 times function f of n for all integers n

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?

function f of 4 = 4

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



SAT question of the day Nov 17 2012

Which of the following CANNOT be the lengths of the sides of a triangle?

A.     1 comma 1 comma 1

B.     1 comma 2 comma 4

C.     1 comma 75 comma 75

D.     2 comma 3 comma 4

E.     5 comma 6 comma 8


What type of question are we asked to answer?

This is a CANNOT BE question – that is the information in the answer contradicts a fact we should know.

What is the topic? we are dealing with a geometry problem, in particular triangles.

This is a question regarding sides of a triangle. At this point the sentence that states that the combined length of each pair of sides should be greater than the third side of the triangle should surface somewhere in your mind. This  actually make sense – if you imagine the longest side as a line to which the two other sides are connected at the ends, and these two shorter sides are trying to meet one another – their sum should be greater than that of the longest side. It is clear that 1, 2, 4 is a set of lengths that does not do that since 1+2=3<4 and thus this triangle can’t close.




SAT Question of the Day Nov 14th 2012

The length of a rectangle is increased by 20%, and the width of the rectangle is increased by 30%. By what percentage will the area of the rectangle be increased?

A.     25%

B.     36%

C.     50%

D.     56%

E.     60%

In order to solve this question we better have a good understanding of two concepts.

One is area – the other is %.

As usual there are many ways to solve – and I’ll walk you through a few.

AREA of a RECTANGLE – will always be the result of multiplying the length by the width (sometimes it will appear as base times the height perpendicular to that base).

PERCENTAGE – not enough word to describe the suffering students go through with this otherwise “simple” concept. In essence it is a fraction with a denominator of 100. It is always related to something (because of that we always look for % of what?).

The easy way to solve: Assume the length is 10 units – increase it by 20% out of ten and get 12 units as the length of the new rectangle. Assume the width is also 10 units. Increase it by 30% (of the 10) and get 13. 12X13=156 – this is 56 area units bigger than the original or 56% larger than the 100% of the area of the 10X10 original rectangle.

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.


SAT Question of the day Nov 11 2012

SAT Question of the day Nov 11 2012

math image


In the figure above, which quadrants contain pairs (x comma y) that satisfy the condition x over y = 1?

A.     roman numeral 1 only

B.     roman numeral 1 and roman numeral 2 only

C.     roman numeral 1 and roman numeral 3 only

D.     roman numeral 2 and roman numeral 4 only

E.     roman numeral 1, roman numeral 2, roman numeral 3, and roman numeral 4


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 answers 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 (negative over negative is positive). That means III is also true (fair well 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) – that means that even if they have the same absolute value the answer will be NEGATIVE 1 . 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.


SAT question of the day Nov 08 2012

SAT question of the day Nov 08 2012

In a class of 80 seniors, there are 3 boys for every 5 girls. In the junior class, there are 3 boys for every 2 girls. If the two classes combined have an equal number of boys and girls, how many students are in the junior class?


A.     72

B.     80

C.     84

D.     100

E.     120


First lets 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 and get a new ratio.

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.

Several possible ways to solve.

One – write an equation – if X is the number of students in the junior class 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 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.



SAT Question of the day – Nov 05

Here is the SAT question of the day for Nov 05 2012 and my suggested approach to solving it.

If the graph of the function function f in the x times y-plane contains the points (0 comma negative 9), (1 comma negative 4), and (3 comma 0), which of the following CANNOT be true?

A.     The graph of function f has a maximum value.

B.     y less than or equal to 0 for all points (x comma y) on the graph of function f.

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

D.     The graph of function f is a line.

E.     The graph of function f 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 (or I hope you know) 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 – the idea is that a. we can learn something b. we are not always going to be smart and check the option that is the right answer first.

All we need to find or imagine is a case in which these points will 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 function f has a maximum value.

Could be – if it starts at -9 goes up to -4 and continue to 0 – it makes sense 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.     y less than or equal to 0 for all points (x comma y) on the graph of function f.

Well lets see – we have three y values:  -9, -4, and 0 all are equal or smaller 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 false and can be at some instances true and thus we can cross it out.

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

WHAT? well lets think what this statement  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 points on the top part. You just created a symmetric function with respect to the line created by you folding the paper.

Now can you imagine you can do the same with the 3 points given? probably yes – so this option goes away because it can be true and is not always false.

D.     The graph of function f is a line.

We discussed this is impossible. and this is the answer to choose.

E.     The graph of function f is a parabola.

Well, is the graph of a parabola (y=aX^2+bX+c) can be drawn to include these points? yes – we can imagine a part of a parabola passing through any three given 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.

Be ready for this kind of a mix. Read the answers as part of your reading of the question and always start with the one that look the most easy to check.

Hurricane Sandy

In the mid of Hurricane Sandy it seems a little out of place to think of it in terms of math, however someone like me, watching nature’s powers unfold, thinks about the beautiful ways in which physical powers manifest themselves and do so according to laws we can describe using mathematics.

If you want, Mathematics is one language we humans use to describe Nature. One can see this use at work by going to the NOAA website – you can see how math can help predict, using sophisticated mathematical models, how this storm will unfold as well as advanced mathematical modeling for other projects .


Hope this storm will not cause too much damage. Keep safe.