Level 4-5 math questions for math experts.

Test Preparation SAT Preparation
1

<p>1- Each student in a group of 30 students studies German, Italian or both. The total number of students studying German is three more than the total number of students studying Italian. If the number of students that study both subjects is the same as the number of students that study exactly one subject, how many students in the group study only Italian?</p>

<p>A) 6
B) 9
C) 15
D) 21
E) 24</p>

<p>2- 90n + 23p = 4523.
If n and p are positive integers in the equation above, which is one possible value of n+p?
(Grid-in)</p>

<p>3- At a certain hospital, 89 children were born in the month of June. If more children were born on the fifteenth of June than on any other day in June, what is the least number of children that could have been born on the fifteenth June?</p>

<p>4- When x is expressed as a decimal, the hundredths digit is 8. What is the greatest possible value of 1/x?</p>

<p>A) 0.00125
B) 0.010101 ...
C) 0.98999
D) 11.11111111 ...
E) 12.5</p>

<p>5- 2/x > 3. In which of the following intervals does every value of x satisfy the inequality above?</p>

<p>A) -1 < x < -1/2
B) -1/2 < x < -1/4
C) 1/4 < x < 1/2
D) 1/2 < x < 3/4
E) 3/4 < x < 1</p>

<p>6- If (root of a)^n = 5, what is the value of 1/a^n? (Grid-in)</p>

<p>7- If 10a + 10b = 32, what is the value of a+b? (Grid-in)
PS. I got this one by luck while I was trying out numbers, what is the most advisable approach to this kind of problem though?</p>

<p>8- In a volleyball league with 4 teams, each team plays exactly 2 games with each of the other 3 teams in the league. What is the total number of games played in this league?</p>

<p>9- If (x-y)^x = 1 and y^x = 1, where x and y are positive integers, what is the value of x?</p>

<p>A) 1
B) 2
C) 3
D) 4
E) 5</p>

<p>10- In the decimal representation of 1/k, where 0<1/k<1, the tenths digit is 1, the hundredths digit is 3, and at least one other digit is nonzero. What is the tenths digit in the decimal representation of k-1/k? (One of the hardest I've ever come across)</p>

<p>A)5
B)6
C)7
D)8
E)9</p>

<p>11- 5, a, b, 5...
In the sequence above, the first term is 5 and the second term is a, Each term after the second is the product of the two immediately preceding terms. If a <0, what is the 10th term of the sequence?</p>

<p>A) -5^21
B) -5^10
C) 5
D) 5^10
E) 5^21</p>

<p>12- If (a+b)^1/2 = (a-b)^1/2, which of the following must be true?</p>

<p>A) b = 0
B) a+b = 1
C) a-b = 1
D) a^2 + b^2 = 1
E) a^2 - b^2 = 1</p>

<p>Happy solving! :-)</p>

2

<h1>1. Is it 6?</h1>

<p>Info from the question:
G + It + B = 30
G = 3+ It
G + It = B</p>

<p>So:
G + It + B = 30 and G + It = B, so G + It + G + It = 30
2G + 2It = 30</p>

<p>And:
G = 3 + It and 2G + 2It = 30, so 2(3 + It) + 2It = 30</p>

<p>Therefore:
2(3 + It) + 2It = 30
6 + 2It + 2It = 30
6 + 4It = 30
4It = 24
It = 6</p>

3

<p>You’re right, but do you have a non-algebraic method though?</p>

<p>For other people who are perplexed with the same questions, I found this explanation more elaborative. Here:</p>

<p>"Let ‘G’ stand for those studying German,
‘I’ for those studying Italian,
and ‘B’ for those studying both</p>

<p>So,</p>

<p>G + I + B = 30</p>

<p>If The total number of students studying German is three more than the total number of students studying Italian, then:</p>

<p>G = I + 3</p>

<p>So, now the equation looks like this:</p>

<p>(I+3) + I + B = 30</p>

<p>If If the number of students that study both subjects is the same as the number of students that study exactly one subject, then</p>

<p>B = I + G and since we already figured out ‘G’ is I + 3,</p>

<p>B = I + (I + 3)</p>

<p>so now the equation looks like:</p>

<p>(I + 3) + I + [ I + (I + 3)] = 30</p>

<p>Because it’s all addition, the parentheses don’t matter, now that we are down to just one variable.</p>

<p>So take all the I’s and put them together, and then add up all the numbers to the left of the equal sign. There are 4 I’s, and the numbers add up to 6, so now the equasion looks like this:</p>

<p>4(I) + 6 = 30</p>

<p>Now get rid of the six by subtracting it from both sides.</p>

<p>(You can always get rid of a number on one side by doing the opposite with that number on both sides. If it’s x + 6 on one side, SUBTRACT 6 on both sides. If it’s x/6 on one side, MULTIPLY 6 on both sides, etc…):</p>

<p>4(I) + 6 - 6 = 30 - 6 (or)
4(I) = 24</p>

<p>now, get rid of the four by dividing both sides by four:</p>

<p>4(I)/4 = 24/4 (or)
I = 6</p>

<p>There you have it. ‘I’ equals six!"</p>

4

<h1>2. Is 51 a possibility?</h1>

<p>90n + 23p = 4523</p>

<p>If p = 1, then 23p = 23</p>

<p>90n + 23(1) = 4523
90n = 4500
n =50</p>

<p>n+p = 50 + 1 = 51</p>

5

<p>Correct, you are right. Looks like it’s a logic-based question, with no systematic way of solving. Thanks. :)</p>

6

<h1>3. Is it 45?</h1>

<p>For the least number of kids to be born on the 15th, imagine the remainder being born on another day, say the 1st. If 44 are born on the 1st, then 45 have to be born on the 15th. </p>

<p>If 1 had been born on the 1st, and 1 on the 2nd, 1 on the 3rd, etc., then only 29 babies would have been born on days other than the 15th. So 89-29=60, not the least. </p>

<p>Even if you scatter the 44 births out among 2 or 3 days, 45 still have to be born on the 15th.</p>

7

<p>Wrong, the answer is 4.</p>

<p>EDIT: I took the liberty to use the search function, and came up with the solution to #3.</p>

<p>“In this case, since there are 30 days in June, you have 29 days and the fifteenth day of June. To solve this question, you have to find the max value of kids that are born on 29 days in order to get the least number of kids born on 15th of June. So, the max value is 28<em>3 + 1.
In words, 3 kids are born every day for 28 days, and 1 kid is born on the 29th day. 28</em>3+1=85. So, the min value of kids born on the 15th is 4.”</p>

8

<h1>4. E?</h1>

<p>I tried two extremes:</p>

<p>x = 0.08 and x = 0.98</p>

<p>1/0.08 = 12.5</p>

<p>1/0.98 = 1.02</p>

<p>To confirm, I tried x = 0.081 and noticed that 1/0.081 was only 12.3</p>

9

<p>Seems logical enough to incorporate substitution tactics in number 4, indeed, the answer is E.</p>

10

<p>Hmm, I see why that’s the answer for #3. The other 29 days, there are 2 or 3 babies born. So 4 is the max on the 15th. But not sure the mathematical solution for that one. Hopefully someone here knows. T</p>

11

<h1>9 - obviously, x needs to be a larger number than y. y needs to be 1 so that (y^x) is true (the only other solution would be x=0 but x is a positive integer). Now that y is 1, x needs to be one larger than 1 which is 2. (2 - 1)^2=1 TRUE.</h1>

<p>The answer is B) 2</p>

12

<p>For 5, I just plugged in the values for x, selectively starting with C. Choices (A) and (B) would make the quantity on the left side of the greater than sign negative, thus making them less than 3. </p>

<p>When I plugged the values in C in, they worked:
2/x > 3
2/(.25) > 3
8 > 3 check</p>

<p>2/x > 3
2/(.5) > 3
4 > 3 check</p>

<p>Just to be sure, I tried 3/4 since it was in (D) and (E):</p>

<p>2/x > 3
2/(3/4) > 3
2.66 > 3 FALSE</p>

13

<p>@ Courts, EssayTees, both your answers are correct…and appreciated! :-)</p>

14

<h1>6 is evil, so time to take a break.</h1>

15

<h1>11 E ?</h1>

<p>replace b with 5a then find the value of a … then it is solved !</p>

16

<p>Sorry, but I’m not really getting it, Omar. How is it solved when you replace b with 5a? You still have to find out the value for the tenth term in the sequence.</p>

17

<p>Anyone want to make a further contribution to solve questions 6, 7, 8, 10, 11, and 12 please? :-)</p>

18

<p>I’d say the answer to 8 is 12 and that to 12 it is A.</p>

19

<p>

</p>

<p>Why would you try numbers? You need to find a + b, and not find a value for a or b.</p>

<p>10a + 10b = 10 (a+b) = 32
(a+b) = 32/10 or 3.2</p>

20

<p>^ Because I’m a ■■■■■■ who thinks ‘substitution’ whenever confronted with a weird looking problem.</p>

<p>Thanks lol.</p>