SAT Question of The Day June 27th 2013

y = (x^2) minus (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.     negative 12

B.     negative 6

C.     4

D.     6

E.     12

Well – this is a nice question as you are actually given too much information.

We have the equation that correlates between X values and Y values.

This relation is quadratic. We are missing the value of a number that is replaced by the letter C.

It is clear that if the relation between X and Y is true then C must assumes a certain value.

The graph of the equation passes through 6,0 that implies that when X=6 Y=0

0=6^2-4*6+C or 0=36-24+C

That in turn means C=-12

You could have come to the same conclusion using the other point – however – if you don’t like to work hard try to avoid negative numbers.

52% out of 143,000 got it right.

SAT Question of The Day June 24th 2013

It costs d dollars to ship a package that weighs p pounds if and only if 6 is less than or equal to p is less than or equal to 10. If Alan’s package weighs p pounds and costs d dollars to ship, which of the following must be true?

A.     (absolute value of p minus 6) is less than or equal to 2

B.     (absolute value of p minus 10) is less than or equal to 2

C.     (absolute value of p minus 8) is less than or equal to 2

D.     (absolute value of p plus 6) is less than or equal to 10

E.     (absolute value of p plus 10) is less than or equal to 6

The wording of this question is horrible. It tells you about Allan who shipped a package and it cost him d Dollars to ship.

It wastes your time to try and understand what d has to do with the question (hint: NOTHING).

You have a package that weighs anything between 6 and 10 pounds. Shipping packages that weigh in between 6 and 10 pounds cost the same (d) regardless of the actual weight of the package.

That means Allan’s package must weigh something between 6 and 10 pounds.

You are asked to find the expression that best describe the relation between the data above and the weight (p) of this specific package.

All the answers have absolute value in them – that is they strip the expression inside of them of it’s sign (be it positive or negative) and place a positive sign instead.

Now Allan’s package can weigh 6, 7.5, 7.8, 9.9 or 10 pounds – all are OK.

The answer, in this question, must always be true (think why). You can plug in 6, 8, and 10. This will nullify A, B, D, and E.

Another way to think about it is that with absolute value expression we are measuring the distance (positive) from a mid point. In our case the difference between 10 and 6 is 4 and the mid-point is 8. If we measure the difference between the actual weight of Allan’s package and 8 – it’s absolute value must always be smaller or equal to 2.

50% out of 145,000 got it right.

SAT Question of the Day June 21 2013

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

Let’s start with the obvious – it is not going to be 30 miles (because she drove part of the time faster and it took her an hour) and it is not 40 (because she drove for an hour and part of the time at lower speed than 40 miles per hour). Now, you might be tempted to answer 35. This is almost correct but it is not the right answer. If she drove for half an hour at 30 miles per hour and for half an hour at 40 miles per hour than the average speed would be 35 miles per hour – THIS ASSUMES THE TIME TRAVELED IS IDENTICAL !

In our case the distance traveled is identical – driving this distance at 30 miles per hour takes a little more than 1/2 an hour and driving the same distance at 40 miles per hour takes a little less – together these two times add up to an hour.

This means that if I had to guess I would choose 34 and 2/7 of a mile.

The woman drove from home to work – a distance of X miles – it took her then to get to work X/40 hours or (X/40)*60 minutes.

On the way back she drove the same distance – X at 30 miles per hour. This trip took her a longer time period to complete – it took her X/30 of an hour or (X/30)*60 minutes.

Together, back and forth, it took her an hour or 60 minutes.

(X/40)*60+(X/30)*60=60 minutes

1.5X+2X=60

3.5X=60

7/2X=60

7X=120

X=120/7 or 17 and 1/7th of a mile.

We are looking on the entire distance or twice as many miles – 34 and 2/7th of a mile.

41% out of 155,000 got it right !

SAT Question of the Day June 18 2013

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.     negative 8

B.     negative 4

C.     4

D.     8

E.     10

We have a quadratic equation (X to the 2nd). The X coordinate of the vertex (where the quadratic equation obtained its max or min value) is determined by X=(-b)/(2a) where b is the number coefficient of the X and a is the number coefficient of the X^2

b=? a=(-2) and we know X=2

plugging in we have: 2=(-b)/[2*(-2)]

or 2=(-b)/(-4)

-8=-b

or b=8

You can also use your TI calculator to plug numbers in and find, using the min/max functions which value satisfies the conditions – but this is much more difficult.

36% out of 145,000 got it right.

SAT Question of the Day June 15 2013

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

In a math question concentrated on Geometry I recommend to make a little diagram.

We have 4 lines in the plane. Exactly two of them are parallel – implying they never cross.

Now lets call these lines A and B and lets draw them parallel to the bottom of the page.

————————————————————————

————————————————————————

The remaining two lines (C and D) are not parallel to one another or to A or B.

This mean we can imagine they are forming a gigantic

\                /

   \           /

      \     /

         X

       /    \

    /          \

/                 \

Now lets move this X up.

It is clear C and D meet exactly once. C intersect A once and intersect B once. The same is true for D.

We have an option of 1+2+2 intersections.

However, if we continue to move the X up and the intersection between C and D is now overlapping one of the parallel lines we reduced two points.

We conclude there are either 5 or 3 points were the lines intersect.

30% out of 218,000 got it right – a surprisingly difficult question.

SAT Question of the Day June 12 2013

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

All we need to do is to set an equation – one rock will weigh X ounces.

6X+10=4X+22

It is clear we have two more rocks on one side and 12 ounce more on the other.

This implies each rock weighs 6 ounces.

71% out of 128,000 got it right – indeed a relatively simple question.

SAT Question of the Day June 09 2013

A, B, C, and D are points on a line, with D 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

This is a reading comprehension question as much as it is a mathematical question.

Notice that the assumption that the points are in order from A to B to C to D is challenged by the statement that D is actually exactly in the middle between B and C.

Now, what we want to do is create a diagram that will help us solve the important question – what is the distance between A and D?

Lets start with what we know to be true – D is in the middle between B and C.

Draw a line and two points on it and name them B and C (for the sake of simplicity let’s call the left one B and the right one C). Mark the middle of the segment as point D.

Now point A can be either on the left side of B, between B and D, between D and C, or on the right side of C.

(A)—B—(A)—D—(A)—C—(A)

Now, let us look on the distances between the points: BC is 12 – so BD=DC=6 We know AB is 10 – which places A either on the left of B or between D and C (10 is not far enough from B to be on the other side of C). We also know that the distance between A and C is 2 units – this leaves us only one option –  A is between D and C – and is located 4 units from D.

B——D—-A–C

46% out of 150,000 got it right.

SAT Question of the Day June 6th 2013

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

We can buy milk either in one or half gallon containers.

We pay Y cents per one gallon and X cents per half gallon.

If we buy two half gallons of milk we pay another Z cents  than if we buy a one gallon container.

In Math we will write 2X=Y+Z or 2X-Y-Z=0

You can try it with numbers: 10 cents for a gallon 6 cents for each half gallon and 2 cents for the difference between buying a gallon or buying two half gallons.

57% out of 141,000 got it right.

SAT Question of the Day June 3rd 2013

2013-M37917.png

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

A.     15

B.     30

C.     45

D.     60

E.     105

The triangle on the left is an isosceles right triangle (two legs are equal and one angle is 90 degrees). This means X=45

The triangle on the right is equilateral – each angle is 60 degrees.

y-x=15 so 3(Y-X)=3 times 15 = 45

74% out of 160,000 got it right.

SAT Question of the Day May 31st 2013

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

Lets read the question and highlight the critical words:

The stopping distance of a certain car is directly proportional to the square of the speed.

That is – the distance equals a constant times the speed squared.

If we double the speed the distance of 17 feet would be quadrupled (2 square is 4) and will equal 68 feet.

38% out of 258,000 got it right.