Some practice questions

Test Preparation SAT Preparation
1

<p>I didnt get these. Help please :)</p>

<li>Let # be defined as a#b = ab + a + b for all numbers a and b
For what value of x is the statement x#y = x always true?</li>
</ol>

<p>A) -2
B) -1
C) 0
D) 1
E) 2</p>

<p>17) If the length of Segment AB is 5 and the length of Segment BC is 6, which of the following could be the length of Segment AC?
a 10
b 12
c 13
d 15
e 16</p>

<p>15) Molly is 64 inches tall. At 10 AM one day, her shadow is 16 inches long, and the shadow of a nearby tree is s inches long. In terms of s, what is the height, in inches, of the tree?
a) s+48
b) 2*sqrt(s)
c) s/4
d) 4s
e) (s/2)^2</p>

<p>6) Rita’s dog weighed 5 pounds when she bought i. Over the next several years, the dog’s weight increased by 10 % per year. Which of the following function gives the weight, w, in pounds, of the dog after n years of weight gain at this rate?
a) w(n) = 5 +0.1n
b) w(n) = 5(0.1)^n
c) w(n) = 5(0.9)^n
d) w(n) = 5(1.1)^n
e) w(n) = 5(n)^1.1</p>

<p>Grid In**/
11) The sum of the positive odd integers less than 100 is subtracted from the sum of the positive integers less than or equal to 100. What is the resulting difference?</p>

<p>Thanks</p>

2

<p>sry forgot to include answers:</p>

<p>9) B
17) A
15) D
6) D
11) 50</p>

<p>Can anyone explain them to me please?</p>

3

<p>9) Just plug in numbers till you get the right one. Say x is -1 and y is 6.</p>

<p>Then -1#6 = (-1 * 6) -1 + 6</p>

<p>After a couple tries you'll notice that the first and third terms always cancel each other out when x = -1.</p>

<p>17) Not sure.</p>

<p>15) An object's height is always proportional to its shadow, if the time of day is consistent.</p>

<p>16/64 = s/h</p>

<p>1/4 = s/h</p>

<p>1/4h = s</p>

<p>h = 4s</p>

<p>6) Don't know how to explain it, but weight increases by 10% = 110% of the weight each year = weight(1.1), raised by the number of years.</p>

<p>11) Count the difference for 10. </p>

<p>1 + 3 + 5 + 7 + 9 = 30
2 + 4 + 8 + 10 = 25</p>

<p>The difference is 5 for one set, and for each set of those numbers (10s, 20s, 30s) the difference grows by 5 each time. There are 10 sets, so 10 x 5 = 50.</p>

4

<ol>
<li>The longest that AC could possibly be is 11 (5 + 6). Thus, you are looking for an answer of 11 or less.</li>
</ol>

5

<p>Another way to think of 11: think in terms of odd and even pairs...</p>

<p>(2 + 4 + 6 ... ) - (1 + 3 + 5 ... ) = (2 - 1) + (4 - 3) + (6 - 5) ...</p>

<p>Each pair equals 1. There are 50 pairs (100 / 2). 50 x 1 = 50.</p>

6

<p>ALSO, for 17, whoever said that it was a straight line? Let's say that it's a TRIANGLE, so the two sides added together must be less than 11. If you couldn't think about that, then I guess you could logically assume that it could never get to 12 or above if it was only 11. </p>

<p>To skatj, I do'nt get your explanation for 11. Did you think that it said evens for the second one? Well I believe the sum of integers from 1 to N is N*N+1 divided by 2, so you get 5050. Now I might get a little careless and subtract the odd #s 1-99 from 5050, which would take a while, and I'm sure there's some algebra expression for sum of odd integers from 1-100. But did I misread the question? Sum of integers from 1-100 minus the sum of odd integers from 1-100?</p>

7

<p>
[quote]
But did I misread the question?

[/quote]

The OP appears to have mistyped it.</p>

8

<p>wow thanks guys, they seem so easy now that you explained it :P</p>

<p>thank you skatj!</p>

<p>"An object's height is always proportional to its shadow" was the key concept i was missing on all the shadow problems come across.
-I think i will remember that for the rest of my life!</p>

9

<p>Nevertheless, what is the expression for the sum of odd integers from 1 to 100? But if it's even integers, why did he leave out 6 in that list?</p>

10

<p>Oh, I mistyped that by a lot. The sums were actually the other way around, and I forgot to type in the 6.</p>

<p>And TBH I did it wrong the first time and modified my explanation when I saw the correct answer, so someone tell me if I did it the wrong way.</p>

<p>btw, thanks for the line segment explanation, I didn't consider that it could be a triangle.</p>

<p>The real explanation would be sort of like this:</p>

<p>11) Count the difference for 10.</p>

<p>1 + 3 + 5 + 7 + 9 = 25
2 + 4 + 6 + 8 + 10 = 30</p>

<p>The difference is 5 for one set, and for each set of those numbers (10s, 20s, 30s) the difference grows by 5 each time. There are 10 sets, so 10 x 5 = 50.</p>

<p>Kind of like alpha2018's way, except on a slightly larger scale.</p>

<p>edit: Yea I assumed that he meant evens for the second one and judging from the answer I guess that assumption was true.</p>

11

<p>oooh i get it now :)
thanks</p>

12

<p>Dear classmate in distress, </p>

<p>These problems are so easy I could almost crap myself.</p>

<p>9)</p>

<p>x#y= xy+x+y</p>

<p>we want to find x, such that x=xy+x+y</p>

<p>the 2 x's cancel and we're left with</p>

<p>xy+y=0</p>

<p>xy=-y</p>

<p>x=-1</p>

<p>17)</p>

<p>AB+BC=11</p>

<p>What # is there smaller than 11????????????????????</p>

<p>10, duh</p>

<p>15)</p>

<p>Since Molly's shadow is 16, and her "heighth" is 64, her shadow is 1/4 her height. This translates to the tree's shadow being 1/4 its height, or the tree being 4 times its shadow.</p>

<p>Hence, we get 4s.</p>

<p>6)</p>

<p>The dog increases by 10%, which means 110% or 1.1 times. Her original weight is 5. So, you multiply by 1.1 each time</p>

<p>We must write 1.1^x because if x years elapse, let's say 3, it's the same thing as writing (1.1x1.1x1.1)5</p>

<p>Generalizing, we get</p>

<p>5(1.1)^x</p>

<p>11)</p>

<p>The odd integers from 1-100 are represented by 2n-1 [1,50] and the evens are 2n [1,50]</p>

<p>The Gaussian summation formula is n(n+1)/2</p>

<p>Let s(o) represent odd sum and s(e) equal even sum</p>

<p>We then seek the difference 2n(2n+1)/2-[(n(2n-1+1)/2] for n of 50</p>

<p>Clean this mess up to get</p>

<p>n(n+1)-n^2</p>

<p>n^2+n-n^2</p>

<p>We're left with n, which we said earlier was 50</p>

<p>Thus n(50) = 50 which is the same as 2pi times the volume of a line of length 9 million, all divided by 6000 Coloumbs</p>

<p>That is all</p>