SAT Question of the Day December 29th 2012


y = (x^2) - (4 times x) +c

In the quadratic equation above, c is a constant. The graph of the equation in the x y plane contains the points (minus 2 comma 0) and (6 comma 0). What is the value of c?


A.     minus 12

B.     minus 6

C.     4

D.     6

E.     12


We know the function passes through the points (-2, 0) and (6, 0). We also know that the function value, or the y value, is equal to the value of the X squared minus 4 times the value of X plus C.

We can use either point to find the value of C.

We will prefer to use a positive X value (easier to calculate with positives).

6 squared is 36 minus 4 times 6 (24) is 12. 12+C equals the Y value – which is zero when X=6. In this case it is clear C must equal -12.

The same can be done with the other point (-2, 0).

50% out of >140,000 answers – a moderate level of difficulty. If you had problems with understanding the solution – you should review function, quadratics, graphs, and solving equations with one variable.


SAT Question of the Day December 26th 2012

math image


In the figure above, if P Q R S is a quadrilateral and T U V is a triangle, what is the sum of the degree measures of the marked angles?

A.     420

B.     490

C.     540

D.     560

E.     580

Look at the figure – all the angles of the quadrilateral (4 – sum is 360 degrees) and all the angles of the triangle (180 degrees) are marked. +360+180= 540 degrees


70% of >140,000 answers were the correct one – pretty straight forward SAT question of the day.

SAT Question of the day December 23rd 2012

A woman drove to work at an average speed of 40 miles per hour and returned along the same route at 30 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?

A.     30

B.     30 and 1 over 7

C.     34 and 2 over 7

D.     35

E.     40

Well, lets read the question and think about the scenario. If she drove for an hour she traveled a distance that is greater than 30 miles (because she drove part of the time at 40 miles per hour) and lesser than 40 miles (because she drove some of the time at 30 miles per hour).

A and E (and very likely B) are out. Now lets continue thinking a bit. If we knew the average speed all we needed to do is multiply it by the time she drove (an hour) to find the distance. Now we don’t know the distance but we can imagine a case were we drive 10 miles at 10 miles per hour (it will take an hour to get there) and then drive back at 20 miles per hour (it will take half an hour to get back). We drove a total of 20 miles in 1.5 hours. 20 divided by 1.5 is not 15 miles per hour (it is actually less – and will always be less then the mean). So say good-by to answer D. You can circle answer C.

Now if we were to calculate we may take the answers and see which one will give us a total driving time of one hour.

For example – if we take 30 miles as our traveling distance. 30/2= distance to work = 15 miles. Traveling at 30 miles per hour it will take her 1/2 an hour to get there. On her way back she drives faster – thus it will take her 15/40= 0.375 hour or 22.5 minutes to get back home. Thus this answer is wrong.

If we want to solve an equation we can set the distance from home to work as our X


X miles / 30 miles per hour = time travel there

X miles / 40 miles per hour = time travel back

X miles / 30 miles per hour + X miles / 40 miles per hour = 1

Common denominator is 120

4X/120+3X/120= 1

1 = 120/120


X= 120/7

2X (that’s the value we are looking for !)= 240/7=34+2/7

SAT Question of the Day December 20th 2012

y = negative 2 times x^2 + (b times x) + 5

In the xy-plane, the graph of the equation above assumes its maximum value at x = 2. What is the value of b?



A.     –8

B.     –4

C.     4

D.     8

E.     10


y is a parabola. The coefficient of the x square is (-2) – hence it is a parabola opening down and has a maximal value.

This value is obtained when X=2. We know that the X co-ordinate where the parabola obtains its maximal value also equals (-b)/2a were b is the x coefficient and a is the coefficient of the x square.

a in our case equals (-2) and b is the unknown.





Another way is to re-write the parabola in vertex form which will lead to the same results.


SAT Question of the day December 17th 2012

Four distinct lines lie in a plane, and exactly two of them are parallel. Which of the following could be the number of points where at least two of the lines intersect?

Roman numeral 1. Three

Roman numeral 2. Four

Roman numeral 3. Five

A.     Roman numeral 1 only

B.     Roman numeral 3 only

C.     Roman numeral 1 and Roman numeral 2 only

D.     Roman numeral 1 and Roman numeral 3 only

E.     Roman numeral 1Roman numeral 2, and Roman numeral 3

Yet again one can see math questions are, at least in part, reading comprehension questions.

What would make our life easier answering this question? a piece of paper and a pencil? you probably guessed – draw a diagram.

We have a plane – in other word a piece of paper.  Now draw two lines parallel to one another.

As you learned previously they do not meet. Now add a third line. It creates two intersections – one with each of the parallel lines.

At this point we should read carefully the text of the question: “and exactly two of them are parallel.” the last line must then cross all the other lines creating an additional 3 points – all together 5.

Happily we want to circle answer B – however this is after all the SAT. What was the question? “Which of the following could be the number of points” Ha! so it is a could be question. Can we find a case were we don’t have five points? can some of them sometime overlap? – well yes. If the intersection of the two non-parallel lines is exactly overlapping the point where they intersect one of the parallel lines – then we have exactly three points. So both I and III are correct and we choose answer D.





Only 30% of over 220,000 answered correctly – a tough question indeed.

Take care and enjoy your winter break.



SAT Question of the Day December 14th 2012

A geologist has 10 rocks of equal weight. If 6 rocks and a 10-ounce weight balance on a scale with 4 rocks and a 22-ounce weight, what is the weight, in ounces, of one of these rocks?

A.     4

B.     5

C.     6

D.     7

E.     8


Lets see – you have 6 rocks on one side and 4 rocks on the other – that means there are 2 extra rocks on one side (lets say the left). Then there are 10 ounce on the left side and 22 on the right – so an extra 12 ounce on the right. If both sides are equal – two rocks must way 12 ounce – so one rock weigh 6 ounce.



SAT Question of the Day December 11th 2012

A, B, C, and D are points on a line, with the midpoint of segment line B C. The lengths of segments line A B, line A C, and line B C are 10, 2, and 12, respectively. What is the length of segment line A D?

A.     2

B.     4

C.     6

D.     10

E.     12

When you approach this question you need to keep in mind two things: 1. READ 2. DRAW

What do we know? we know that there is a line with 4 points scattered on the line.

We also know that D is exactly in the middle between B and C.

Now draw a line and place a point B and a point C somewhere to it’s right. Then mark another point, D, in between them (estimate the middle). If you decided to mark C to the left of B, that’s fine – just reverse the instructions.

Line Segments

BC is 12 units long. that means that BD=6=DC

Point A can be anywhere, as much as we know, it can be on the left side of B, it can be between B and D, between D and C, or on the right side of C.

How would we know where exactly point A is marked? We know that points A & B are 10 units apart. That means either point A is between D and C (10 units out of the 12) if A is on the right side of B, or it is 10 units to the left of B. Now A and C are only 2 units apart and not 22 – so we conclude option one – A is somewhere in between D and C is correct. Now what are we looking for? we want to know how far are points A and D. If we count from point C we need to move 2 units to point A and then another 4 to point D (6 units apart) – we choose then 4 as our answer.

Line Segment

Only 53% got it right ! hope you were one of them.


SAT question of the Day Dec 8th 2012

Milk costs x cents per half-gallon and y cents per gallon. If a gallon of milk costs z cents less than 2 half-gallons, which of the following equations must be true?

A.     x minus (2 times y) + z = 0

B.     (2 times x) minus y plus z equals 0

C.     x minus y minus z equals 0

D.     (2 times x) minus y minus z equals 0

E.     x + (2 times y ) minus z = 0

Here we use our understanding that it is wise to read the answers as part of the question.

Look at the answers – they all have zero on the right side of the equation. That mean we just need to write an equality something equals something else and then subtract one side from the other to get the equation. Two half gallon containers cost 2X cents. One gallon container cost Y cents. If this gallon container costs Z cents less than the two half gallon containers then if we add Z cents to Y cents we would get 2X. In math terms we would write: 2X=Y+Z

Now lets subtract Y from both sides and then lest subtract Z from both sides.

We would obtain: 2X-Y-Z=0 which is the correct answer.

55% got it right out of 140,000 responses. A intermediate level question. You will probably get it right just be attentive of the time you spend answering it.



SAT question of the Day Dec 5th 2012

math image


In the triangles above, 3 times (y minus x) =


A.     15

B.     30

C.     45

D.     60

E.     105

We have an algebra question that uses your Geometry knowledge in order to obtain the values of X and Y.

Y is an angle in a triangle in which all three angles are Y and hence equal – this is thus an equilateral triangle and each of the Y’s equal 50 degrees.

Similarly, the X is an angle in a right triangle where the other angle is X as well. This makes it an isosceles triangle in which the two equal angles sum is equal to 90 degrees. This mean each is 45 degrees.

Now we can solve the equation. 3 times (60-45) is 3 times 15 or 45 degrees.


This was a relative easy question with 72% correct answers out of ~160,000 responses.


SAT Question of the Day Dec 2nd 2012

The stopping distance of a car is the number of feet that the car travels after the driver starts applying the brakes. The stopping distance of a certain car is directly proportional to the square of the speed of the car, in miles per hour, at the time the brakes are first applied. If the car’s stopping distance for an initial speed of 20 miles per hour is 17 feet, what is its stopping distance for an initial speed of 40 miles per hour?


A.     34 feet

B.     51 feet

C.     60 feet

D.     68 feet

E.     85 feet


The key to this problem is reading. The words “directly proportional” should be underline while you are reading this problem. Why? because they tell you that when the speed of the car is zero the stopping distance is also zero. However, this does not tell you the shape of the function -do not assume it is linear without further evidence.

What else do we know about the link between the two variables (stopping distance as function of speed)?  We know it is linked through an increase that is proportional to the “square of the speed of the car“. This is a parabola that passes through (0,0). If the stopping distance when traveling at 20 mils per hour is 17 feet and the stopping distance is directly proportional to the square of the speed of the car, when you increase the speed by a certain factor – the stopping distance will increase by the same factor squared. That implies that if we double the speed to 40 miles per hour the stopping distance will increase by 2 square or by a factor of 4 to 68 feet. Note that directly proportional does not always mean a linear relation.

The success rate answering this question was 38% out of ~200,000 submissions – you can see that reading is a critical part of math. I assume most of the wrong answers were 34 feet as this is “directly proportional” but linear and not quadratic.