<p>Hi! So I just took the 2010 practice test for the PSAT and I had some questions about the math.</p>
<p>First:
Question 19) How many positive integers less than 1000 do NOT have 7 as a digit?
a) 700
b) 728
c) 736
d) 770
c) 819</p>
<p>Collegeboard's answer is (B) - 728. My question is...how are we supposed to figure this out. I realize you need to determine the number of numbers under 1000 that do have 7 as a digit, then subtract that from 1000, but that's really hard to do! Can any of you math whiz folks help me out?</p>
<p>Second:
Question 26) If there are (m) minutes in (h) hours, what is (m) in terms of (h)?
a) 60/h
b) h/60
c) 60+h
d) 60h
e)3600h</p>
<p>My answer that I was pretty confident in was (B) - h/60. Collegeboard says the answer is (D) - 60h. Can someone explain what I did wrong and what CB did right?</p>
<p>THANK YOU SO MUCH, and good luck on Wednesday (or Saturday!)</p>
<p>For the first question:</p>
<p>How many positive integers less than 1000 do NOT have 7 as a digit?</p>
<p>The numbers range between 1 and 999. We’ll count the number of valid possibilities. It suffices to look at 3 digit, 2 digit and 1 digit numbers.</p>
<p>For the leftmost digit there are 9 possible values (excluding “7”). For the middle digit there are again 9 possible values (since we exclude “7”), and the same for the unit digit. So the total is 9x9x9 = 729. From this we subtract 1 since “0” is excluded – i.e. it’s not a positive number. So: 728.</p>
<p>For 2, you may just want to plug in some numbers. Take the case of 60 minutes per hour. Then take 5 hours. How many minutes in 5 hours? Well 300. Now go through the choices. Only d works. This is a good technique to use to check your answers! In this case you can use it to work around a possible misunderstanding and to quickly arrive at the right answer.</p>
<p>This is not to difficult to solve “generally”. You’re probably getting a bit confused. Try a few examples with real numbers and think about how you’re manipulating the numbers. I’m sure you’ll get the right equation in a minute or so.</p>
<p>Wow, thanks, this helps a lot!</p>
<p>Q16)The sum of all even integers from 2 to 202, inclusive, is 10.302.what is the sum of all even intergers from 4 to 200, inclusive?
a) 10,098
B) 10,100
c) 10,200
D) 10,300
E ) 10,506</p>
<p>CB says (E), I did not get it how a detailed explation to help me understand is appreciated.</p>
<p>THANK YOU , and good luck on Wednesday (or Saturday!)</p>
<p>
</p>
<p>Are you sure? I’m pretty sure the answer is A.</p>
<p>sorry yes rechecked the answer it is A, can anyone pls. explain in detail.
thanks much</p>
<p>The sum of all even integers from 2 to 202, inclusive, is 10,302.what is the sum of all even intergers from 4 to 200, inclusive?</p>
<p>2+4+6+8+10…+202 = 10,302
4+6+8+10+12…+200 = 10,302 - 202 - 2 = 10,098</p>
<p>The second set of numbers does not include 202 and 2 from the first set. Therefore, subtract 202 and 2 from the sum of the first set, 10,302, to get the answer.</p>
<p>^ Lol, I missed that part… should’ve known there was an easier way to do it than doing finding the # of terms for 4 to 200 inclusive (99), and then averaging the terms (204/2 = 102).</p>
<p>
</p>
<p>I like your method also. In any method, this problem shouldn’t take much time…</p>
<p>thank you very much this clears my doubt.</p>
<p>Grid in question:
Q37) If x^2=25, y^2=16, z^2=9, what is the greates possible value of (x+y-z)^2?</p>
<p>I have used the valuses of x as +/- 5, y as +/- 4, z as +/- 3, sub’ed in the values and getting 36, but as per CB it is 144, can any one please provide a detailed explation to help me understand .
thanks</p>
<p>x=±5
y=±4
z=±3
[(5+4-(-3)]² = (12)² = 144
[(-5)+(-4)-3]² = (-12)² = 144</p>
<p>Think logically. Since all the numbers can be positive or negative, x+y-z doesn’t really play any role. Therefore, 5+4+3 or -5-4-3 to the second power equals the greatest possible value.</p>