Aug 04

## SAT Question of the Day 21th July 2013

### How many integers satisfy the inequality above?

### A. None

### B. One

### C. Two

### D. Three

### E. Five

### We have an inequality on our hands (it has a < or a > sign) and also an absolute value hidden inside. We know that when we take an integer and subtract five of that number and change the sign of the answer to a positive we obtain a number that is smaller or equal to 2.

### How many non-negative integers are there that are smaller or equal to 2? well there is 2 (equal), 1 (smaller), and 0 (a non negative). We obtain 0 by X=5, we obtain 1 by X=6 OR by X=4 (remember that absolute value function turns negative to positives), and the same for 2 – we will obtain it by X=7 or X=3.

### We conclude there are 5 such integers.

### 43% out of 222,000 concluded the same.

### Other techniques involve solving the inequality – I suggest the graphical approach since it is faster.

Aug 04

## SAT Question of the Day July 18th 2013

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

### A.

### B.

### C.

### D.

### E.

### 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.

Aug 04

## SAT Question of the Day July 15th 2013

### A number cube, with faces numbered through , 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.

### B.

### C.

### D.

### E.

### 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.

Jul 13

## SAT Question of The Day July 12th 2013

### Triangle is isosceles, and the measure of angle is . What must be the measure of another angle of this triangle?

### A.

### B.

### C.

### D.

### 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.

Jul 13

## SAT Question of The Day July 9th 2013

### If , what is the value of ?

### A.

### B.

### C.

### D.

### E.

### 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.

Jul 13

## SAT Question of The Day 6th July 2013

### , , ,, …

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

### A.

### B.

### C.

### D.

### E.

### 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.

Jul 13

## 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.

### B.

### C.

### D.

### E.

### 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

Jul 13

## SAT Question of The Day June 30th 2013

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

Jul 13

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