Friday Quiz

I dropped by Fairfax High after school yesterday to visit with the Varsity Math League. It was not a lively gathering of math enthusiasts, more like a library. A few dozen students were hunched over their desks, chewing on pencil erasers and puzzling over a timed test. Every couple of weeks, these students take a quiz. Their answers are sent to the county math department and compared with results from other high schools. The Varsity league recruits from students in Honors Algebra II or higher

The questions on yesterday's quiz stumped many of them.

See how you fare. Here is a sample:

1. Of the 1 million integers from 1 through 1 million, how many are neither squares nor cubes nor fourth powers of integers?

2. If sin(A + B) = 0.3 and sin(A - B) = 0.2, what is the value of the product (tan A)(cot B)?

3. The yearly salary increments received by a man equal the yearly salary increments received by his son. The man earned (yearly) twice as much as his son earned when the man earned what his son earns now. When the son earns what the man now earns, their combined salary will be $126,000. How many dollars does the man now earn?

***The third question is a candidate for the "most poorly worded word problem that I have ever seen" prize. If you can figure out what they are asking here, I will be doubly impressed.

By Michael Alison Chandler  |  February 20, 2009; 11:28 AM ET  | Category:  Friday Quiz
Previous: Life After Algebra II | Next: Who is a Successful Honors Math Student?

Comments



1) The square root of 1,000,000 is 1,000. So for x <= 1,000, x^2 <= 1,000,000.

The cube root of 1,000,000 is 100. So for x <= 100, x^3 <= 1,000,000.

The fourth root of 1,000,000 is 31.6. So for x <= 32, x^4 < 1,000,000.

That means we can pull out 1,000 + 100 + 32 numbers of the original million. But how many are duplicates? (I.e. 2^4 = 4^2 = 16, and we can't pull out 16 twice.)

We know x^4 = (x^2)^2, so all of the fourth powers are also squares...so all 32 of them are being double counted.

Also, any number of the form (x^2)^3 = (x^3)^2. So when you cube a square number, it's being double counted. We're cubing 1-100, and there are 10 squares in those numbers, so we need to take those 10 out.

So, the total is 1,000,000 - 1,000 - (100-10) - (32-32) = 998,910 integers between 1 and 1,000,000 that aren't squares, cubes, or 4th powers.

2) Unfortunately, I don't remember my trig identities, but I know where to find them! (Don't know if the kids had to have them memorized, I may be cheating.)

Anyway, sin(a+b) = sin a cos b + cos a sin b. Since sin (-x) = -sin x (sin is antisymmetric about 0°), and cos (-x) = cos x (cos is symmetric about 0°), we also have sin(a-b) = sin a cos b - cos a sin b.

From the definition of tan and cot, tan x = sin x/cos x, and cot x = cos x/sin x. (These I remember!)

So,

sin(a+b) = sin a cos b + cos a sin b = 0.3
sin(a-b) = sin a cos b - cos a sin b = 0.2

Adding the two eqns gives

2 sin a cos b = 0.5

Subtracting gives

2 cos a sin b = 0.1

Dividing those two gives

(sin a/cos a) * (cos b/sin b) = 0.5/0.1

Simplifying,

tan a cot b = 5

3) The way I read the question, it translates to this:

M_past = S_now = 2*S_past
S_future = M_now
S_future + M_future = $126,000

The confusion to me is, what yearly "increment" is the same? Percentage? Dollars? I'll assume dollars.

M_now - M_past = S_now - S_past
M_future - M_now = S_future - S_now

That's it for now, I'll have to take a stab at solving it later...

Posted by: tomsing | February 20, 2009 1:00 PM | Report abuse

Okay, here's my stab at #3. We've got 6 variables (S_past, S_now, S_future, M_past, M_now, M_future) and 6 eqns (there are 2 eqns on the first line). Assuming they're all independent, we can solve for the variables.

But rather than try to solve the system of equations directly, I'm going to take a guess. Let's say that S_now = $2.

That directly leads to S_past = $1, and M_past = $2.

Since S_now - S_past = M_now - M_past, $2 - $1 = M_now - $2, so M_now = $3.

Since S_future = M_now, S_future is also $3.

Finally, since M_future - M_now = S_future - S_now, M_future - $3 = $3 - $2, so M_future = $4.

Now, let's check with the one remaining eqn. S_future + M_future = $3 + $4 = $126,000. Well, not quite. But since we have a linear system, we can scale up.

$126,000/($3+$4) = 18,000

So, multiply everything by 18,000.

S_past = $18,000
S_now = $36,000
S_future = $54,000
M_past = $36,000
M_now = $54,000
M_future = $72,000

So, the answer to the question is, M_now = $54,000.

Note that you don't need to know anything about how far in the past or in the future is being considered.

By the way, with a similar method, if we were talking about the same percentage increase for each instead of the same dollar increase, I use

(M_now - M_past)/M_past = (S_now - S_past)/S_past

(M_future - M_now)/M_now = (S_future - S_now)/S_now

Those simplify to

M_now/M_past - 1 = S_now/S_past - 1
M_now/M_past = S_now/S_past
M_now*S_past = S_now*M_past

and

M_future/M_now - 1 = S_future/S_now - 1
M_future/M_now = S_future/S_now
M_future*S_now = S_future*M_now

Even though these aren't linear eqns anymore, the same technique works, giving:

S_past = $10,500
S_now = $21,000
S_future = $42,000
M_past = $21,000
M_now = $42,000
M_future = $84,000

Posted by: tomsing | February 20, 2009 3:03 PM | Report abuse

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