SAT Question of the Day January 25th 2013

A jar contains only red marbles and green marbles. If a marble is selected at random from the jar, the probability that a red marble will be selected is 2 over 3. If there are 36 green marbles in the jar, how many red marbles are there in the jar?


A.     18

B.     24

C.     54

D.     72

E.     108


We have here a “reverse” probability question. We know we have only red and green marbles in the jar. The probability to pick, at random, a red one is two thirds. That implies that out of every 3 marbles in the jar 2 are red and 1 is green (thus the ratio red to green is 2:1 and the ratio red to total is 2:3).  Now there are 36 green marbles in the jar. We know there should be twice as many red ones in order to have a ratio of red to (red+green) of 2 to 3. We conclude there must be 72 red marbles in the jar.

47% of about 200,000 got it right.

A nice animation clarifying the concepts can be found at:


SAT Question of the Day January 22nd 2013

If function f of x = x + a times x, and a = 7 over 2, what is function f of (3 over 2) ?

A.     1 over 3

B.     3 over 2

C.     7 over 2

D.     21 over 4

E.     27 over 4


We have a function with a variable X and a constant a. The function operates by taking our input – i.e. X and adding to it a times X. In other words we need to multiply the value of X by (1+a). If a is 7/2 and X is 3/2 we need to multiply 3/2 by (1+7/2) or 3/2 times 9/2 which is 27/4.

function f of (3 over 2) = (9 over 2) times (3 over 2) = 27 over 4

56% got it right of about 200,000 answers. How did you fare?


SAT Question of the Day January 19th 2013

In the xy -plane, the graph of the line with equation y equals a intersects the graph of the quadratic function (f times x) equals x^2 minus (6 times x) plus 8 in exactly one point. What is the value of a?


A.     negative 3

B.     negative 1

C.     1

D.     3

E.     4


We first need to understand the nature of the question.

This is a function “graphing”question. One function is y=a. What type of function is it? It is a line parallel to the X axis (where y=0 for all X values). The second function is a parabola (f times x) equals x^2 minus (6 times x) plus 8

Where would we expect the line to touch the parabola only once? At the vertex – when the X coordinate is equal to -b/(2a) (a, b, and c are the coefficients of the quadratic function). If we plug that value into the parabola we will obtain the y value of the vertex. This value is also the value of a.

b is -6 a is 1 then the X coordinate is +3. If we plug it in we obtain y=9-18+8 or -1.

Only 32% out of ~250,000 answers were correct. Review your Algebra !



SAT Question of the Day Janurary 16th 2013



The scatter plot above shows the number of items purchased at a grocery store by 28 customers and the total cost of each purchase. How many of these 28 customers bought more than 10 items and spent less than $20?


A.     Four

B.     Five

C.     Six

D.     Seven

E.     Eight


Typically, with graph we choose a random point and read its meaning. Lets chose the point above the number 5 on the X axis and opposite the $10 on the Y axis.

How do we read the meaning of this point? This point represents a person who bought 5 items and payed for them $10. Now, there are 28 such points. How many of them represent people who bought more than 10 items (the point to the right of the 10 items mark on the X axis – without the 2 point on this   vertical line. From these people we need to identify those who spent less than $20. Those are represented by all the points below the $20 horizontal line (excluding the 2 points on the line).

Combining these two conditions we identify 4 costumers who fit both criteria.

65$ out of 187,000 – not bad – hope you were one of those who got it right.


SAT Question of the Day January 13th 2013

A train traveling 60 miles per hour for 1 hour covers the same distance as a train traveling 30 miles per hour for how many hours?

A.     3

B.     2

C.     1

D.     1 over 2

E.     1 over 3

If you travel at half the speed it will take you double the time. Mark 2 hours and move on – here is a question without any tricks, deception or issues. Yes, those exist as well. 78% out of 200,000 answers – one of the highest % I’ve seen.

SAT Question of the Day January 10th 2013

If p percent of 75 is greater than 75, which of the following must be true?

A.     p is greater than 100

B.     p is less than 75

C.     p is equal to 75

D.     p is less than 25

E.     p is equal to 25

P is the percent taken out of a number (in our case 75). The result is bigger than 75. 75 is the number from which the percentage is taken from and thus it is our 100%. If P percent out of 75 is bigger it must be larger than 100%. Answer A.

70% got it right out of ~170,000 – an easy question indeed.

SAT Question of the Day January 7th 2012

If a number is chosen at random from the set(negative 10 comma negative 5 comma 0 comma 5 comma 10), what is the probability that it is a member of the solution set of both (3 times x) minus 2 less than 10 and x + 2 greater than negative 8?


A.     0

B.     1 over 5

C.     2 over 5

D.     3 over 5

E.     4 over 5

Let’s read the question.

What are we asked about? we are asked probability – that is what is the chance that something will happen. It is also can be thought about as the ratio of the events we want to happen out of the total possible range of events. Assuming each event can happen at the same likely hood that the other events can, we should find how many events fit our criteria and divide by the total number of events.

We can choose at random one of five numbers (because of that the denominator of all the answers is 5 – yes including 0 since 0/5=0). Now what do we wish would happen? we wish we could get a number that is a member of the solution set of both (3 times x) minus 2 less than 10 and x + 2 greater than negative 8

We then need to solve this inequality.

Lets simplify each inequality, draw it on the number line and see which of our answers is a member of this set.

(3 times x) minus 2 less than 10 can be simplified by adding two to both sides of the equation and then dividing both side by 3 to obtain

x + 2 greater than negative 8 can be simplified by subtracting 2 from both side to obtain 

It is now clear that only -5 and 0 fit our criteria

and thus the answer is 2 over 5


52% out of ~190,000 responses – another intermediate level question

SAT Question of the Day January 4th 2013

A machine can insert letters in envelopes at the rate of 120per minute. Another machine can stamp the envelopes at the rate of 3 per second. How many such stamping machines are needed to keep up with 18 inserting machines of this kind?

A.     12

B.     16

C.     20

D.     22

E.     24

WOW – seems complicated isn’t it?

We have machines, minutes, seconds and then we need to keep up with things.

What should we do? First breath deeply – yes we can! It is a great slogan and in our case true.

Let’s follow the usual way we address every SAT question.

We have a machine that can inset letters into envelopes (so you can get all the bills and junk mail etc.). Each individual machine inserts 120 such letters into envelops every minute (in a minute we have sixty seconds). We can think of this machine inserting TWO letters into envelops every second.

Now we need to stamp these envelopes before we deliver them. Each stamping machine can stamp 3 envelops every second. It is clear that one such machine can easily cope with one letter inserting machine. However, we have 18 such inserting machines. Each one of them inserts 2 letters into envelops and together they insert 2X18=36 letters every second. Now how many stamping machines do we need in order to keep up with the flow of envelopes?  36/3=12 and that is the answer.

Note that we could have translated the rate of work done into minutes. However, this is not recommended since it cause an inflation of the numbers (they are much bigger and thus harder to handle). We could also write work equations to describe the situation or work with ratios – this is not necessary here due to the relative simplicity of the problem.

56% got this question right out of ~200,000 – intermediate level question


SAT Question of the Day – January 1st 2013

What a start to the new year !

Lets – read and try to understand the question.


In the figure, the slope of the line through points P and Q is three over two. What is the value of k?

A.     4

B.     5

C.     6

D.     7

E.     8

If we draw a line connecting the two points – the slope of the line is 3/2 – that is for every 2 units we move to the right on the X axis (run) we need to move up 3 units (rise).

We know the X and Y values of one point – X=1 Y=1. We are looking for the X coordinate of the other point (point Q) but we know its Y coordinate – Y=7. That means we have a rise of 6 units (7-1) which mean a run of 4 – so we need to move from X equal 1 (point P) 4 units to the right X=5 (Q).

Trig Identities January 01 2013

Trigonometric identities are, maybe, the most difficult trig topic to master.

It is, on the surface, an easy task – all you need to show is that the expression on one side of the equation is equal to the expression written on the other side.

You don’t need to solve anything – just rewrite !

However, due to the multitude of options, formulas, and sometimes lack of time or will to explain this topic in an adequate way, many students find themselves reluctant to solve problems involving trigonometric identities.

The way I’m describing here is by no means the fastest, most elegant, or the simplest – however the principles to follow are straightforward and comprehensive. They are general enough to enable you to use them each time, and with practice, these guidelines will help you overcome even the most daunting trig identities problem.


Many students assume that since they have taken the Algebra HSA (Montgomery county) and are done with Algebra I and Algebra II the material covered during the study of these topics is no longer needed – as you will see they are wrong.

An important first step in mastering trig identities is mastering FOILing and unFOILing as well as memorizing and being able to identify on sight several identities.

The most common ones are:

and the most important to identify is the third one (when A=X and B=1) which take the form of:

 which can and should be expanded to

2. Change the trigonometric expressions appearing on both left and right to expressions containing SIN and COS only – second from the left column in the table below.

If all terms are single multiples of the variable (x, t, alpha, etc) don’t bother to copy the variable throughout the steps – saves time and less messy.

3. Remember to check after each step what is the expression you are trying to obtain (usually the one on the right side).

This will help you determine which additional steps are needed to expand, rewrite, or manipulate the intermediate form you have.

4. Trig identities expressions are separated by an equal sign – that implies you can transition from left to right as well as from right to left. If you get stuck – start with the other side. All the time the expressions you write are identical somewhere in the middle you demonstrated they are equivalent.

Left side expression                            =                                            Right side equation

Left side expression –>step 1 —> step 2 —> …..last step —> Right side equation

Left side expression <– last step…. <— step 2 <— step 1 <— Right side equation

Left side expression –> step 2 —> step 3—-> last step <—- step 4 <— step 1 <— Right side equation

5. If one of the sides has an expression with dividend or numerator and a divisor or denominator and somewhere along the way you obtained either one – don’t change it. Work on the one you don’t have and try to make it the same as in the final expression.

6. If worse comes to worse and you are stuck with a numerator and a denominator – you can always on one side and the expression on the other side has a different numerator and a different denominator and you can’t see a way out – you can expand one side by the other expression denominator (multiply and divide). The numerator multiplied by the other side denominator and divided by the original denominator should then be simplified to give the desired numerator.

7. Most important of all :