SAT Question of the Day July 18th 2013

In a certain lawn-mower factory, 0.06 percent of all mowers produced are defective. On the average, there will be 3 defective mowers out of how many produced?

A.     500

B.     1800

C.     5000

D.     18000

E.     50000

We have a percentage problem on our hands.

We recall that 1% means 1 out of a hundred. If we had 0.1% – that implies that 0.1% is 0.1 out of a 100 and if we have 0.01% that means 0.01% is 0.01 out of a 100. How many times does 0.01 fit in 100? 100/0.01=10,000.

Now we have 0.06% or six times that number. This in-turn tells us we can expect to see 6 defective lawn mowers out of every 10,000. We observed only 3 such defective lawn mowers so we probably checked only half of 10,000 or 5,000 lawn mowers.

Another way to approach this question is to set up the equation were X stands for the number of lawn mowers.

100 (3/X) = 0.06% which is a single variable algebraic equation.

47% out of 212,000 got it right.

SAT Question of the Day July 15th 2013

A 6-sided number cube, with faces numbered 1 through 6, is to be rolled twice. What is the probability that the number that comes up on the first roll will be less than the number that comes up on the second roll?

A.     1 over 4

B.     1 over 3

C.     5 over 12

D.     7 over 12

E.     1 over 2

We have a probability question on our hands.

There are a few ways to attack this problem and get to the right answer.

The easiest way, in my opinion, to solve this question is unfortunately the most time demanding.

We have 6X6=36 possible outcomes – lets write them down:

In the table we’ll make we will place the number that can come up in the first roll and we will mark the desired outcomes of the second roll (with a value less than and not including the value we got on the first roll) in parentheses.

First roll   Second roll              number of options that satisfy the criteria

1                 1, 2, 3, 4, 5, 6                                                 0

2                 (1), 2, 3, 4, 5, 6                                               1

3                 (1, 2), 3, 4, 5, 6                                               2

4                 (1, 2, 3), 4, 5, 6                                               3

5                 (1, 2, 3, 4), 5, 6                                               4

6                 (1, 2, 3, 4, 5), 6                                               5

Summing up the desired outcomes we obtain 15 out of 36 or 5/12.

This is one of those problems that can be solved theoretically using formulas but lend themselves to be solved by “brute force” which also allows for us to check our work.

47% out of 184,000 got it right.

SAT Question of The Day July 12th 2013

Triangle A B C is isosceles, and the measure of angle A is 74 degrees. What must be the measure of another angle of this triangle?

A.     32 degrees

B.     42 degrees

C.     53 degrees

D.     74 degrees

E.     It cannot be determined from the information given.

If a triangle is an isosceles two of its sides are equal and two of his angles are equal.

We are given one angle with a measure of 74. It could be that this is one of the base angles and thus the other base angle is equal to it and the head angle is 32. Another option is that 74 is the head angle and then each of the base angles is 53. Without additional information we can’t tell.

We must therefore choose E.

38% out of 184,000 considered both options and choose the correct third option.

SAT Question of The Day July 9th 2013

If a minus b = b minus 3 = 6, what is the value of a?

A.     negative 3

B.     6

C.     9

D.     15

E.     18

Yet again, in the SAT it is useful to think backwards.

If b-3=6 it means that b=9

If a-b=6 and b=9 then a must be 15.

Simplify and solve – this question tries to confuse by a non standard layout of the mathematical information.

Clean it up and you have it.

61% out of 171,000 did it.

SAT Question of The Day 6th July 2013

3611,18, …

The first four terms of a sequence are shown above. Which of the following could be the formula that gives the nth term of this sequence for all positive integers n?

A.     2 times n

B.     (2 times n) + 1

C.     3 times n

D.     (n^2) + 1

E.     (n^2) + 2

One approach is plug in – n stands for the position of the term in the sequence.

If we plug in 1 to the formula instead of n we should obtain 3, when we plug in 2 we will get 6, 3 to give 11 and 4 to obtain 18.

Only n^2+2 delivers each time.

52% of 172,000 got it right.

P.S. in questions where plugging in seems to be the best option start from the last answer.

SAT Question of The Day July 3rd 2013


In the figure above, all intersecting sides of the polygon meet at right angles. What is the area of the polygon?

A.     108

B.     104

C.     102

D.     98

E.     96

My approach is to close the rectangle (12X9=108 units) and then subtract two small rectangles. One is  2X2 and the other is 2X3 or 4 and 6 units that are taken out – giving 98 units.


61% of 164,000 got it right

SAT Question of The Day June 30th 2013

Ten cars containing a total of 32 people passed through a checkpoint. If none of these cars contained more than 4people, what is the greatest possible number of these cars that could have contained exactly 2 people?

A.     One

B.     Two

C.     Three

D.     Four

E.     Five

This is a “logic” question.

We have 10 cars and 32 people. None of them contain more than 4 people. We need to find the MAXIMUM number of cars that contain exactly 2 people.

Lets try and construct the ideal condition. If we had 10 cars with 2 people in them we will have the maximum number of cars. However, this scenario enables only 20 people to ride in the cars and 12 are left out. Now, if each car contains no more than 4 people we can sit another 2 people in each car until we sit the 12 people we left out. We need 6 cars to sit these people with 4 people in each of these cars. We are then left with 4 cars with 2 people in each.

There are other legitimate logical approaches to this question – I shared the one that works best for me.

57% out of 162,000 got it right.

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





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 !