Let's start with the official scoring table from the horse's mouth:
The Quantitative Score range is 0 - 60, and it says that scores over 50 are extremely rare. I consequently was shooting for a 60 when I took the test on June 6th 2012. I was dismayed to receive this score report:
Although the pursuit of a perfect score may be a fools errand, I wanted a 60, and was very disappointing that I might have made a mistake on a question, so I took to the internet to find out how rare scores higher than 51 really are. What I found was a lot of confusion and misinformation, but almost no one even claimed to score higher than a 51. Curious.
To get to the bottom of it, I spoke to an authoritative source within GMAC who was very helpful. Let me set the record straight and say that for at least 14 years there has not been a score higher than 51 on the Quantitative GMAT. In the past there were some scores higher than 51, but when they migrated to a computer-based test, 51 became the maximum. (it's also mentioned here.)
This lead me to believe that, unlike the old version of the GRE (which ran out of room in its scoring range), the GMAT could differentiate the top couple percent without having to goto a new scoring range. They could use the scores 52-60 to differentiate the top two percentile. In fact they were going to do this recently, but the schools said it would be too confusing if some 51s were maximum scores, and others weren't. So that was scrapped, and they will probably only differentiate the top better when/if they change to an entirely new scoring range/system (like the ETS did with the revised GRE in 2011).
Speaking of the top of the Quantitative Scorers, 2% getting the max score seems a bit high [not nearly as bad as the old GRE (~6%!), but not as good as the new GRE (<1%)]... let's examine what's going on: (the following distribution graphs come from this great tool from GMAC.com)
Here's the Verbal Scores:
A nice normal curve, max and min's are rare, and the mean is also the mode.
Now for the Quantitative:
Which doesn't look nearly as normal, the most frequent score is not the maximum, but it's close. And the max extreme of 51 is more frequent than each scores less than 23, this is not an ideal distribution.
Now lets look at Quantitative Scores from Americans (maybe we are just getting smarter over the years?):
Here we have a relatively normal distribution, the mean is the mode, and the extremes are rare... So it's not that we are getting much smarter.... Let's look at GMAT Quantitative scores from international test takers:
This is rather shocking, the Chinese and other Asian test takers score really well... And since the gmat is getting more popular internationally, this explains why the total curve is getting top-heavy today.
In 2010, 30,264 GMAT tests were taken by Chinese citizens (out of china's population of 1.3 billion), and 127,061 Americans took the test in that period (out of our population of 0.31 billion). So it might be that only an elite class of students takes the test in china, and that is why they all score so well... Or, maybe they are all just smarter in math than us Americans : )
Either way, this is why so many test-takers (2%) got a perfect 51 on the Quantitative section of the GMAT this year.
Laura is 3 times as old as Maria was when Laura was as old as Maria is now.
In 2 years Laura will be twice as old as Maria was 2 years ago.
How old are they now?
[Spoiler Below]
Translating english to algebra:
"Laura is 3 times as old as Maria was ":
`1_a`: `Laura_(now) = 3 * Maria_(then)`
"Laura was as old as Maria is now":
`2_a`: `Laura_(then) = Maria_(now)`
"In 2 years Laura will be twice as old as Maria was 2 years ago":
`3_a`: `Laura_(now) + 2 = 2 * (Maria_(now) - 2)`
With four unknowns and three equations, it's not yet solvable, and it seems like we've used all of the information from the problem.... except:
"Maria was when Laura was", this tells us this:
`4_a`: `Maria_(now) - Maria_(then) = Laura_(now) - Laura_(then)`
Now that we have four independent equations, let's solve with substitution:
Substituting (`1_a`) into (`3_a`) and (`4_a`) to get:
`3_b`: `3*Maria_(then) + 2 = 2 * (Maria_(now) - 2)`
and:
`4_b`: `Maria_(now) - Maria_(then) = 3*Maria_(then) - Laura_(then)`
The prime factorization of `40320` yields:
`40320 = 2^7*3^2*5^1*7^1`
so any factor of of 40320 can be written in this form:
`2^a*3^b*5^c*7^d`
where `a`,`b`,`c`, and `d` are all non-negative integers and:
`a<=7`
`b<=2`
`c<=1`
`d<=1`
so the number of options for `a` is 8 (including 0), and for `b`: 3, `c`: 2, and `d`: 2.
so to count all the permutations of factors, we take the product:
`8*3*2*2 = 96`
If the two-digit integers `M` and `N` are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of `N` and `M`?
A) 181
B) 165
C) 121
D) 99
E) 44
[Spoiler Below]
Without loss of generality, le's say `M` has `a` as the tens digit, and `b` as the units digit.
Consequently, `N` would have `b` as the tens, and `a` as the units digit.
remembering how the decimal system works:
`M = 10*a + b`
`N = 10*b + a`
(adding these two equations)
`M + N = 11*a + 11*b = 11*(a + b)`
since `a` and `b` are both integers, `a + b` must also be an integer, so `M + N = 11* an_integer`
so M + N is a multiple of 11.
A) `181` is prime
B) `165 = 11 * 5 * 3`
C) `121 = 11 * 11`
D) `99 = 11 * 3^2`
E) `44 = 11 * 4`
(A) is the only choice that isn't a multiple of 11, and therefore cannot be the the sum of `M` and `N`.
How many nonzero digits does `1/(2^13 * 5^19)` have when written as a decimal?
[Spoiler Below]
`1/(2^13*5^19) = 1/(2^13*5^13*5^6) = 1/((2*5)^13*5^6) = 1/(10^13*5^6) = 10^-13 * 1/(5^6)`
note I can multiple by 1 whenever I want without changing the value, and `1 = (2^6/2^6)`, continuing: ` = 10^-13 * (2^6/2^6)*1/5^6` ` = 10^-13 * 2^6/(2^6*5^6)` ` = 10^-13 * 2^6/(2*5)^6` ` = 10^-13*2^6/10^6 = 10^-13 * 10^-6 * 2^6 = 2^6 * 10^-19` ` = (2^3)^2 * 10^-19 = 8^2 *10^-19 = 64 * 10^-19`
the `10^-19` is only going to change the place of the decimal, not the number of non-zero digits, so the answer is simply 2, the number of digits in `64`.
The average of 8 numbers is 20. If after one number is removed, the average of the remaining 7 numbers is 16. What number was removed?
[Spoiler Below]
Remembering that the mean is the sum divided by the number of numbers:
1: `"Average"_8 = 20 = "Sum"_8/8`
2: `"Average"_7 = 16 = "Sum"_7/7`
so:
3: `160 = "Sum"_8`
and:
4: `112 = "Sum"_7`
What is the smallest non-prime integer that isn't a factor of `20!`?
[Spoiler Below]
Since the solution to this puzzle (let's call it `n`) isn't a factor of `20!`, it must either:
1. Have more of a specific prime in its prime factorization than `20!` does.
2. Have a prime factor that isn't in the prime factorization of `20!`
Let's examine what #1 yields:
since `20! = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1`,
we can see 10 even numbers there, so at the very least we would need 11 `2`s in `n`, which is `2^11` which is `2048`, (actually there will be far more `2`s needed ).
if we focused on `3`s, I see 6 multiples of `3`, meaning we'd at least need 7 `3`s in `n`, `3^7` which is `2187`
I see 4 multiples of `5`, so we would need `5^5` which is `3125`
2 multiples of `7`, which would require `7^3` which is `343`
and the smallest prime that only occurs once in `20!` is `11`, which would mean we'd need `11^2`, which is `121`.
so the smallest potential n value we get from this approach is `121`, let's think about #2:
let's choose the smallest prime that isn't a factor of `20!`, to find it we can start with `20` and increment until we get a prime:
`21=3*7`
`22=11*2`
`23` is prime
since `23` is prime, and larger than `20`, it can't be in the prime factorization of `20!`, if that's not clear then look at how we wrote out `20!` above into a product of decreasing integers... each of those integers is less than `20`, therefore none of those integers has `23` as a factor, and therefore the product of all those integers doesn't have `23` in its prime factorization.
but the problem is the find the smallest non-prime non-factor of `20!`, so we need to make it not-prime.
Since `23` is not a factor of `20!` any multiple of 23 will also not be a factor of `20!`, the smallest multiple of `23` which is greater than `23` is `2*23=46`, this is no longer prime, but definitely isn't a factor of `20!`.
this is much better (smaller) than `121`, and therefore must be the solution.
If `x+y=a` and `x-y=b`, then `2*x*y=`
(A) `(a^2-b^2)/2`
(B) `(b^2-a^2)/2`
(C) `(a-b)/2`
(D) `(a*b)/2`
(E) `(a^2+b^2)/2`
A classic problem that I see all the time on both the GMAT and GRE, take a look at one way to solve it:
[spoiler below]
we start out with two equations:
1: `x+y=a`
2: `x-y=b`
after squaring both sides of (1):
3: `(x+y)^2=x^2 + 2*x*y + y^2 = a^2`
after squaring both sides of (2):
4: `(x-y)^2=x^2 -2*x*y+y^2=b^2`
now lets subtract (4) from (3) to get:
5: `(x^2 + 2*x*y + y^2) - (x^2 -2*x*y+y^2) = 4*x*y = a^2 - b^2`
now if we divide both sides of (5) by 2 we get:
6: `2*x*y=(a^2-b^2)/2`
What's the second smallest integer with a units digit of 5, which is the square of an integer, and the cube of an integer?
[Spoiler Below]
Video:
Text:
Let's call the number we are looking for `n`.
we know it's the square of an integer, so:
`n = "int"_a^2`
and we know it's the cube of an integer, so:
`n = "int"_b^3`
since any integer can be prime-factorized, let's consider the prime factorization of `"int"_1` and `"int"_2`:
`"int"_a = "prime"_1^("int"_(a1)) * "prime"_2 ^("int"_(a2)) ...`
so `"prime"_1^(2*"int"_(a1))*"prime"_2^(2*"int"_(a2))... = "prime"_1^(3*"int"_(b1))*"prime"_2^(3*"int"_(b2))...`
and so we can say:
`"prime"_x^(2*"int"_(ax)) = "prime"_x^(3*"int"_(bx))` for all x.
and therefore: `2*"int"_ax = 3*"int"_bx` this is the number of `"prime"_x` in the prime factorization of `n`.
since it has a factor of `2` and `3`, then the number of `"prime"_x` in `n` must be a multiple of 6.
to summarize, for any prime factor of `n`,there are a multiple of 6 of them in `n`.
we also know that the units digit of `n` is `5`, which means it has a factor of `5`, so:
`n = 5 * "int"_3`
so `n` has at least one `5` in it's prime factorization, and consequently it must have at least 6.
so `5^6` is the least integer that fits the description, but we are looking for the 2nd smallest.
well `5^12` would fit the description, and it's not the smallest, but maybe it's not the 2nd smallest,
`5^6 * 2^6` is worth considering, but `5^6 * 2^6 = 10^6`, which is one million, which doesn't have a `5` as the units digit. This does not fit the description in the problem.
`5^6 * 3^6` fits the description and is the second smallest integer that does.
`n = 5^6 * 3^6 = 11390625`
