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SAT I Practice Test 2002-2003 (from Taking the SAT I: Reasoning Test booklet)

OK, I took the practice test in there, and I don't know if I was really tired or what because I seriously made some really stupid mistakes. However, in the last math section, there are three questions I am absolutely stumped on how to do. If some of you geniuses out there would help me, I would be much obliged...

The integer n is formed by writing the positive integers in a row, starting with 1 and ending with 80, as shown above. Counting from the left, what is the 90th digit of n?
n = 1234567891011...787980
answer: 5
how on earth do you figure that out besides actually physically counting digits which is a little time consuming to say the least?

If 2x + 3y = 1, what is x/2 + y/3 in terms of y?
answer: (3 - 5y)/12

When I did out the problem I ended up getting an answer that wasn't even an option.

For positive integers x, let symbol x (double underline) represent the sum of the digits of x. For example 74 (double underline) = 7 + 4 = 11. If n is a positive integer and n (double underline) = 33 (double underline) + 17 (double underline), which of the following could be the value of n?

So if you follow their procedure and add up the digits, you get 33 (double u) = 6 and 17 (double u) = 8, and 6 + 8 = 14. So it either has to be that or 5 (1 + 4), right? How on earth do you get 86 (the answer)? Does the + sign in between the two double underlined numbers mean that they are just adjacent to each other as opposed to actually being added? I don't know, I just feel like this was a poorly phrased question...

Thanks everyone...

By Curiousone (Curiousone) on Saturday, March 22, 2003 - 11:18 am: Edit

You can find that test btw right here...
http://www.collegeboard.com/prod_downloads/sat/satguide/SAT_Math.pdf

By Dferraro (Dferraro) on Saturday, March 22, 2003 - 11:51 am: Edit

For the first one, you know the 1st 9 digits are going to be 1-9 and the next 20 digits are 10-19 and the next 20 are 20-29, and the next 20 are 30-39, and the next 20 are 40-49. I'll stop here because I've already counted 89 digits. I need the 90th--if the 88th and 89th are 4 and 9, respectively, the 90th is going to be 5 adn the 91st is going to be 0. there you go, FIVE is your answer.

This didnt take very long.

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Number two is a little tricky---you could always just use reasoning, and take the MC answers, and see if they "work"--But i'm assuming you want the algebraic way.

I don't know have the time to solve this algebraically..but on the test if i got something like this--i'd just plug in.

-------------------
Number three is real easy, you just made a small erorr. You were right in your reasoning up to when you got 14. The question is not asking what is 14 double underline...but what value of *N* double underline would you give 14. One possible value of N is 77. However, it seems they want a different possible value (there are a few) for N, so another would be 86..and that is one of the answer choices.

By Incognito (Incognito) on Saturday, March 22, 2003 - 12:17 pm: Edit

#2
Here is one of the algebraic methods that you may use:
If 2x + 3y = 1, what is x/2 + y/3 in terms of y?

basically, just figure out what x is in terms of y:
2x + 3y = 1
2x = 1-3y
x = (1-3y)/2

Now that we know what x is in terms of y, we just need to replace the "y term" for x in the second equation. If we do this, then we can figure out what x/2 + y/3 is in terms of y:

x/2 + y/3
= [(1-3y)/2]/2 + y/3
= (1-3y)/4 + y/3
= (3-9y)/12 + 4y/12
= (3-9y+4y)/12
= (3-5y)/12

By Pat57575 (Pat57575) on Saturday, March 22, 2003 - 12:18 pm: Edit

2) 2x+3y=1, x/2 + y/3 = ?

Okay, you need to move some numbers around in the first equation so that when you divide the left side of the equation by some number, you get x/2 + y/3 = something. You may notice that if you divide the equation, as it is now, by 4, you would get the desired x term (2x/4= x/2), but not the desired y term. So we need to find out what number divided by 4 equals y/3.

c/4=y/3
c= 4y/3

So if we can get 2x + 4y/3 isolated on the left side, we can divide by 4 and get our answer. To do this you just have to subtract 5y/3 from both sides of the original equation. Then divide by 4, rearrange the terms, and you'll have your answer.

2x + 3y = 1
2x + 4y/3 = 1- 5y/3
x/2 + y/3 = (1- 5y/3)/4
x/2 + y/3 = ([3-5y]/3)/4
x/2 + y/3 = (3-5y)/12

By Curiousone (Curiousone) on Saturday, March 22, 2003 - 01:01 pm: Edit

Thanks everyone for the help. I made a really stupid error in the algebra for the second problem and at the point where I plugged in ((1-3y)/2)/2 + y/3, instead of multiplying the 2 by the denominator, I multiplied by the numerator...yeah. So, I feel really stupid now... ;)

By K22 (K22) on Monday, March 24, 2003 - 11:03 am: Edit

where can i get answers to the math questions on collegeboard site?

By Incognito (Incognito) on Monday, March 24, 2003 - 04:31 pm: Edit

If nobody posts a link, and you still cant find it, I'd recommend that you email them to find out. But are you sure that there are no links to an answer sheen on the site?

By Arthur (Arthur) on Monday, March 24, 2003 - 04:48 pm: Edit

i just took this practice a few days ago. i got a 770 (two wrong). both of my mistakes were on the last 10-question section. one of them was one of the three problems you listed too. it was a tough section.

however, i looked over it after and figured out where i went wrong. dferraro's logic is spot on in all cases.

definitely plug in for that question, it's the fastest way imo.

By K22 (K22) on Tuesday, March 25, 2003 - 11:18 am: Edit

hey i found the link to the answers. I need help w/ the 1 prob i missed: Diana goes 700 meters in 2 laps one 7m/s the other 5m/s wuz her average speed for entire race? do any of you know how to do that problem?

By Incognito (Incognito) on Tuesday, March 25, 2003 - 03:07 pm: Edit

To find the average speed for the entire race:

Determine how long she ran for all together:
Well, she ran at 7 m/s in the first, so the first lap took a total of 50 seconds (700/2 divided by 7 = 50).

She ran @ 5m/s in the second one, so she spent a total of 70 seconds running the second lap.

The total time of the race was thus 120 seconds (70+50=120).
The average speed would be the total distance divided by the total amount of time...

700 meters/120 seconds = 5.83 m/s

By Curiousone (Curiousone) on Tuesday, March 25, 2003 - 03:08 pm: Edit

700 m / 2 laps = 350 m
350 m / 7 (m/s) = 50 seconds
350 m / 5 (m/s) = 70 seconds
50 + 70 = 120 seconds
700m/120s = 5.83 m/s

It's a trick question...do NOT just avg 7 m/s and 5 m/s...I got it wrong too but I figured it out.

By Incognito (Incognito) on Tuesday, March 25, 2003 - 03:16 pm: Edit

Here's a formula that may help for determining average rate:

If something travels @ a certain distance at A mph (or whatever speed unit you want to use) and travels the same distance at B mph a second time, then the average rate is:

(2ab)/(a+b)


So for the preceding problem...
(2*7*5)/(7+5) = 70/12 = 5.8

HTH

By K22 (K22) on Tuesday, March 25, 2003 - 05:08 pm: Edit

cool, thanks a lot incognito and curiousone, i appreciate the help.

By Mac (Mac) on Wednesday, March 26, 2003 - 07:42 pm: Edit

What's dferraro's logic?

By Curiousone (Curiousone) on Wednesday, March 26, 2003 - 07:49 pm: Edit

dferraro posted the third message in the thread that explains a quick way of solving several of the problems I had asked about.