SAT Question of the Day May 25th 2013

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.     36

B.     100

C.     136

D.     180

E.     316

This is a “set theory” question – best solved by drawing a Venn diagram.

Write in the left circle dog owners.Write in the right circle cat owners.

In the middle circle are people who own both.

If there are 416 people all together and there are 316+280=596 pets – that means there are 180 who are double counted.

Out of the 316 dog owners 180 have a cat. That implies there are 136 that own only a dog.

45% out of 275,000 got it right.

SAT Question of the Day May 22nd 2013

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

We know 24/15 equals 4/n what is the value of 4n?

4/n=16/4n (expand both top and bottom by 4) – the idea is we will find what is the value of 4n directly.

24/15=16/4n – lets flip both sides 15/24=4n/16 multiply both sides by 16.

15*16/24= 5*3*8*2/(3*8)=10=4n

63% of 236,000 got it right – relatively easy with a calculator.

SAT Question of the Day May 16th 2013

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 limitations do we have with regard to the length of the sides of a triangle?

When we come to draw a triangle we can use any length as our first side. We can then attach to it a second side in any direction with any length we desire. However, when we come to draw the third side we need to close the figure. The third side must be able to close the triangle or in other words any two sides must sum up to be greater in value than the third. A 1,2, and 4 can’t close.

We can imagine that the base is 4 and then we have rods of length 1 and 2 on each side trying to touch one another – they can’t.

49% out of 279,000 got it right.

SAT Question of the Day May 13th 2013

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%

This is a % question. A rectangle can be any rectangle – lets take a 100X100 one.

Now we increase the length by 20% to 120 and the width by 30% to 130.

120X130 = 15600 which is 56% larger than the 10000 we started with.

In letters we can assign X to be the length and Y to be the width. We will write 120/100 X times 130/100 Y equals 1.56XY or 156% of XY.

33% out of 345,000 got it right.

SAT Question of the Day May 10th 2013

5-10-2013-M33370.png

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

We need to identify in which quadrants can we find points that satisfy the condition X/Y=1.

We can re-write this condition as Y=X which is a straight line passing through the origin (0,0) and extends to the I and III quadrants. Another option is to ask ourselves what does it mean X/Y=1? It means the X and the Y have the same absolute values and the same sign. Same absolute values can be in any quadrant but only in quadrants I and III would they have the same signs (+ and +) or (- and -).

For those who took Trig/Algebra II – I will remind you “All Students Take Calculus” and that X/Y is the equivalent of Cotangent in the Unit circle.

58% out of 232,000 got it right.

SAT Question of the Day May 7th 2013

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

We have a ratio problem in front of us.

Lets read it together and try to understand what does the information mean.

The senior class has 80 students. There are 3 boys for every 5 girls. If we divide the class to 8 units (3 boys and 5  girls in each) we have 10 such units. This implies there are 30 boys and 50 girls in this class.

Now, in the Junior class there are more boys than girls. Actually there are 3 boys for every two girls in the junior class. When we combine the classes we have identical number of boys and girls. That means the junior class should have 20 more boys than girls (remember that in the senior class there are 50 girls and only 30 boys). Now, the junior class has 3 boys for every 2 girls. That means that out of every 5 students in the junior class we will have 3 boys and 2 girls or one extra boy. We need 20 such units to obtain those extra 20 boys. We will need 100 students in the junior class.

44% out of 238,000 got it right.

SAT Question of the Day May 4th 2013

If the graph of the function function f in the x 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.

In order to answer this question we need to understand what we are asked.

We have an unknown function ( a relation between two sets of objects with the condition that for each X there is only one Y that is the outcome of f(x)=y ) and are given 3 points (3 X values and their corresponding Y values) that are part of the function.

If we plot these points on a graph paper (during the exam just place them roughly where they should be) we can imagine there are many relations and functions that can contain these points. These functions can have a variety of properties.

However, these points are not on the same line. We can connect each pair but the third point would not be on that line. This is obvious from the slope one can calculate for the lines. Between (3,0) and (0, -9) the slope will be 9/3=3 and between (3,0) and (1, -4) the slope is 4/2=2.

42% out of 260,000 answers were correct.

SAT Question of the Day May 1st 2013

graphic

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

This is a graph+% problem.

What is the % of students who choose action? and the % of the student who choose fiction?

We read from the graph 30% and 14%. Now we know there were 40 more students who choose action. These students represent 16% (30%-14%) of the total student population.

Each % is then 2.5 students and 100% are 250.

51% out of 194,000 got it right.

SAT Question of the Day April 28th 2013

If y equals ((x plus 1) times (x minus 2)) over ((x plus 3) times (x minus 4)), for which of the following values of x is y NOT defined?

 

A.     negative 4

B.     negative 3

C.     negative 1

D.     2

E.     3

In Algebra we are not allowed to divide by 0 because we encounter an undefined expression.

The two values that can make the denominator equal zero are X1=-3 and X2=4 (since we want to avoid X+3=0 and X-4-0).

55% out of 225,000 got it right.

SAT Question of the Day April 25th 2013

If s equals 1 plus (1 over 2) plus (1 over 4) plus (1 over 8) plus (1 over 16) plus (1 over 32) and t equals1 plus (1 over 2) times s, then t exceeds s by

A.     1 over 4

B.     1 over 8

C.     1 over 16

D.     1 over 32

E.     1 over 64

We have the sum of a short sequence of numbers. S = 1 + half of 1 + half of the previous term in the sequence + so on.

Now we want to compare it to t which is 1 + 1/2 of S. Now 1/2 of S will include all the terms in S (because they are one half – so we will lose 1 – because it becomes one half and we will lose 1/32 because it will become 1/64 that is not in the original S).

In t we re-introduce 1 by adding it but we have another term 1/64 that was not there originally.

Only 41% of 262,000 got it right.