Please help me solve these Math problems

Test Preparation SAT Preparation
1

<p>What are the shortest ways of solving the following problems:</p>

<ol>
<li><p>When a coin is tossed in an experiment, the result is
either a head or a tail. A head is given a point value
of 1 and a tail is given a point value of 1. If the sum
of the point values after 50 tosses is 14, how many of
the tosses must have resulted in heads?
(A) 14
(B) 18
(C) 32
(D) 36
(E) 39
The answer is "C". How do you go about it?</p></li>
<li><p>If a triangle has exactly one of its vertices on a circle,
which of the following CANNOT be the number of
points that the triangle and the circle have in common?
(A) Two
(B) Three
(C) Four
(D) Five
(E) Six</p></li>
</ol>

<p>The answer is "E"</p>

<ol>
<li>List I List II
2 3
4 5
7 6</li>
</ol>

<p>One number is to be selected at random from each
of the lists above. What is the probability that both of
the numbers selected will be less than 5 ?
(A) 1/9
(B) 2/9
(C) 1/3
(D) 4/9
(E) 5/9
The answer is "B".</p>

<ol>
<li>How many positive integers less than 1,001 are
divisible by either 2 or 5 or both?
(A) 400
(B) 500
(C) 540
(D) 600
(E) 700</li>
</ol>

<p>The answer is 600. How do arrive at that?</p>

<ol>
<li>A rectangle measuring 9 centimeters by 12 centimeters
is completely divided into t nonoverlapping triangles,
each with sides of length 3 centimeters, 4 centimeters,
and 5 centimeters. What is the value of t ?</li>
</ol>

<p>The answer is 18.</p>

<ol>
<li>A group of 110 people is divided into 4 committees. If
each committee contains at least 2 people, which of the
following statements must be true?</li>
</ol>

<p>(A) Each committee has at least 4 people.
(B) No 2 committees have the same number of
people.
(C) No committee has more than 100 people.
(D) At least 1 committee has more than 25 people.
(E) The largest committee has 3 more people than
the smallest committee.</p>

<p>The answer is D. Why is it so?</p>

<ol>
<li>n = 1234567891011 . . . 787980</li>
</ol>

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

<p>The answer is "E". I had to write all the numbers down and count up to the 90th digit. Is there a shorter way of going about it?</p>

<ol>
<li>If 2x+3y=1, what is x/2 + y/3 in terms of y ?</li>
</ol>

<p>(A) y/5
(B) (1-3y)/2
(C) (1-3y)/4
(D) (3y+4)/15
(E) (3-5y)/12</p>

<p>The answer is "E".</p>

2

<ol>
<li>When a coin is tossed in an experiment, the result is
either a head or a tail. A head is given a point value
of 1 and a tail is given a point value of 1. If the sum
of the point values after 50 tosses is 14, how many of
the tosses must have resulted in heads?
(A) 14
(B) 18
(C) 32
(D) 36
(E) 39</li>
</ol>

<p>Now I’m assuming you meant -1 for tails.</p>

<p>so H+T=50 (number of heads + number of Tails)
and H-T=14</p>

<p>go back to the first formula and solve for T in terms of H
H+T=50
T=50-H</p>

<p>Now plug that value into the second equation
H-T=14
H-(50-H)=14
2H=64
H=32</p>

3

<p>**4. How many positive integers less than 1,001 are
divisible by either 2 or 5 or both? **</p>

<p>Positive integers are integers >0 </p>

<p>So you are dealing with numbers 1-1001. I would consider 1000 as the upper value in the range instead of 1001 for ease of calculation, and we already know that 1001 is neither divisible by 2 nor 5. Then split the range into something easier to work with- in a 1000 there are 10 100’s.</p>

<p>100/2= 50, so there are 50 numbers from 1-100 that are divisible by 2
100/5= 20, so there are 20 numbers from 1-100 that are divisible by 2</p>

<p>Now we have 20+50=70 multiples of 5 & 2 within 1-100</p>

<p>But, there is an overlapping of multiples of 2 & 5 here, meaning the multiples of 10 (like 10,20,30,40) so to find out how many multiples of 10 are there within 100… 100/10= 10.</p>

<p>Get rid of the overlappers, 70-10= 60. Then simply multiply 60 by 10 since the question asks for multiples in the range 1-1000. </p>

<p>I don’t think I can explain the other problems all that well. :frowning: I can’t do 1, 7 & 8. And I think #6 is from the old SATs, I haven’t seen questions like that on tests.</p>

4

<ol>
<li>If a triangle has exactly one of its vertices on a circle,
which of the following CANNOT be the number of
points that the triangle and the circle have in common?
(A) Two
(B) Three
(C) Four
(D) Five
(E) Six</li>
</ol>

<p>You’re going to have to draw this one, it’s really easy. just put the one vertex on the edge and have the other two outside the circle and you can see that it crosses the circle 4 times, plus the vertex so the max is 5</p>

<ol>
<li>List I List II
2 3
4 5
7 6</li>
</ol>

<p>One number is to be selected at random from each
of the lists above. What is the probability that both of
the numbers selected will be less than 5 ?
(A) 1/9
(B) 2/9
(C) 1/3
(D) 4/9
(E) 5/9</p>

<p>2/3 chance of picking number less then 5 in first list. 1/3 chance in second list. (2/3)(1/3)=2/9</p>

<ol>
<li>A rectangle measuring 9 centimeters by 12 centimeters
is completely divided into t nonoverlapping triangles,
each with sides of length 3 centimeters, 4 centimeters,
and 5 centimeters. What is the value of t ?</li>
</ol>

<p>well the area of the rectangle is 9 x 12 = 108. So the area of the rectangle divided by the area of one triangle will give you t. 108/(.5 x 3 x 4) = 18</p>

<ol>
<li>A group of 110 people is divided into 4 committees. If
each committee contains at least 2 people, which of the
following statements must be true?</li>
</ol>

<p>(A) Each committee has at least 4 people.
(B) No 2 committees have the same number of
people.
(C) No committee has more than 100 people.
(D) At least 1 committee has more than 25 people.
(E) The largest committee has 3 more people than
the smallest committee.</p>

<p>Because if each committee had 25 people in it and there were 4 committees, 25 x 4= 100 people, but there are 110
The answer is D so one must be more then 25. </p>

<ol>
<li>n = 1234567891011 . . . 787980</li>
</ol>

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

<p>well first thing you have to do is subtract the first 9 digits because the only have one digit space so 90-9=81. So now we are starting with 10111213… we can look at it and realize that all of the even digits will be tens values and all the odds will be ones values.(so we will be finding the tens value because 90 is even) then you count how many digits it takes to get from 10 to 20 (2 x 20-10 = 20). Then you divide 81/20 and you get 4.something, but all you need to know is that the tens value will increase four times, and since you started at 10 it will be 50 something so the answer is 5)</p>

<ol>
<li>If 2x+3y=1, what is x/2 + y/3 in terms of y ?</li>
</ol>

<p>(A) y/5
(B) (1-3y)/2
(C) (1-3y)/4
(D) (3y+4)/15
(E) (3-5y)/12</p>

<p>So first solve for x/2
2x+3y=1
2x=1-3y
x=(1/2)-(3/2)y
x/2=(1/4)-(3/4)y</p>

<p>then plug it in to x/2 + y/3 so
(1/4)-(3/4)y + y/3 add -(3/4)y + y/3<br>
(1/4)- (5y/12) 1/4 = 3/12
(3-5y)/12</p>

5

<p>Thanks Josh for your solution for number 1. But I don’t understand how you came about with “H-T = 14”. Please explain that to me.</p>

6

<p>ok well heads counts as +1 and tails counts as -1. and we know that when the points have been netted(summed up the 1s and -1s), there were 14 more heads than tails because we ended up at +14. we can write this as 1(H)+ -1(T) = 14 or more simply H-T=14</p>

7

<p>Thanks Josh. I understand that one now. But I don’t understand your solution for question 7. “we can look at it and realize that all of the even digits will be tens values and all the odds will be ones values”. I don’t understand that because taking 12 for instance, 2 is even but it is in the tens.</p>

8

<p>when I say even i am not referring the the individual number, but to the number digit in the sequence. 123456789101112131415. so if the nth digit is an even umber it must be the tens place and if it is odd it is the ones place. If we look at the 16th digit in the sequence, we can see that it is 1(tens value 13) If we look at the 19th digit we see that it is 4 (ones value of 14)</p>