the 1st question,

lets call big s as s'
and Height =H, hight of bigger triangle as H'

Now since both triangles have same angles, we can say that s'=sX, and H'=HX, where X is a number that tells us how big one is to another

so to find X, just use the triangle formula.
the smaller triangle:
(1/2)*SH,
Bigger triangle
(1/2)*S'H'

we are told that bigger triangle has twice the area of smaller one, so

SH=(1/2)S'H'

replace S' and H' with SX and HX we get

SH=(1/2) SX*HX
SH=(1/2) SH*X^2
SHs cancel out,
1=(1/2)*X^2
2=X^2
X=Sqrt(2)

plug X in S'=SX, we get S'=sqrt(2)s

Question 2:Working alone at its constant rate, a machine seals K cartons in 8 hours, and working alone at its own constant rate, a second machine seals K cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

I have no idea, I think it's 66.67 too

Question 3:For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example the length of 75 is 3, since 75=3*5*5. How many two digit positive integers have length 6?

Answer is 2.
the smallest 2 digit number with length of 6 is 2*2*2*2*2*2 =4*4*4=16*4=64, change one of 2 to 3, we get, 16*2*3=96,
if we change 2 of 2s, we get 16*9=144

Question 4: see attached

Pi*(r^2)-Pi*(r-s)^2
Pi*((r^2)-((r^2)-2rs+(s^2))
pi*(2rs-(s^2))
pi*s*(2r-s)

Question 3: answer has to be 2.
Question 4: tricky wording. "medallion made of a circular piece of colored glass surrounded by a metal frame" means that frame is included in medallion and the radius of 'empty' circle is (r-s)
Question 7: https://en.wikipedia.org/wiki/Binomial_coefficient

#7 says that order doesn't matter, so it should be 12P4... byt the answer is 12C4

does someone knows the other questions?

zenithexe, i didn't really get the part: 'big s as s'
and Height =H, hight of bigger triangle as H'

Now since both triangles have same angles, we can say that s'=sX, and H'=HX, where X is a number that tells us how big one is to another '

could u pls give more details??

shibal wrote:#7 says that order doesn't matter, so it should be 12P4... byt the answer is 12C4

does someone knows the other questions?

There is no such thing as 12 pick 4 in Combinatorics. 12 choose 4 represents the nummber of possible combinations of 12 objects taken 4 at a time when order is irrelevant. If order was relevant, the answer would be 12 * 11 * 10 * 9. Since order is irrelevant, we need to divide that number by 4! (combinations of 4 stores when order is relevant).

12 * 11 * 10 * 9 / 4! = 12! / (4! * 8!) which is 12 choose 4

P.S.: what program/site are these screenshots from?

shibal wrote:zenithexe, i didn't really get the part: 'big s as s'
and Height =H, hight of bigger triangle as H'

Now since both triangles have same angles, we can say that s'=sX, and H'=HX, where X is a number that tells us how big one is to another '

could u pls give more details??

Yes, I certainly can. First please see the picture attached. Those are what I mean by S, S',H,H' . I had to rename the sides because it will be confusing to explain with s and S.


>Now since both triangles have same angles, we can say that s'=sX, and H'=HX, where X is a number that tells us how big one is to another '

To understand this part, I need you just forget about the question, for now.
Imagine a square with sides 1cm and 1cm, when you enlarge that square so that the one side becomes 2cm, what is the length of the the other side? 2cm.
Even diagonal would be twice as long. Original square had diagonal of sqrt(1^2+1^2)=sqrt(2)
and modified square has
sqrt(2^2+2^2)=sqrt(8)=2*sqrt(2)

The dimension of any polygon/shape will sustain their ratio when simply enlarged. No matter how much you enlarge a equilateral triangle, each side will have same length.

Now, back to the question. I have defined X as ratio of how much this triangle is enlarged by. So if we refer back to the square with 1cm sides, X would be 2, because 1cm*2=2cm.

Since the question is asking what is S' in terms of s? It is asking "how much is large triangle is bigger than the smaller triangle?" So we just need to solve for X.
With the answer being sqrt(2), we can say that
The larger triangle is sqrt(2) times bigger than the smaller triangle.


hmm... I hope this explains. Please let me not if its not clear.=D

Question 5:In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average (arithmetic mean) number of books borrowed per students was 2, what is the maximum number of books that any single student could have borrowed?
A)3 B)5 c)8 D)13 E)15
OA D)

So total of 60 books were borrowed, so we need to figure out how many books were borrowed by people who borrowed more than 3 books.
60-12-20=28 books
The number of people who borrowed at least 3 books is
30-2-12-10=6
Subtract the least possible no of books that can be borrowed by 5 people who borrowed 3 books
28-3*5=13
The ans is D)

Question 6: See attached

Choice I,
if X and Y are both 0.000000000000000000000000.......00000001,
than it can be smaller, therefore I is out

Choice II,

lets modify 1/sqrt(X+Y) to sqrt(x+Y)/(X+Y) by multiplying sqrt(X+Y)/sqrt(x+Y).

compare choice II with it
since bottom part is same, compare only top part.
is sqrt(X)+sqrt(Y)>sqrt(X+Y) for any positive X,Y? yes

choice III

same thing as II,
is sqrt(X)-sqrt(Y)>sqrt(X+Y) for any positive X,Y? no


so C is correct.

I just want to clarify Question 2:
Working alone at its constant rate, a machine seals K cartons in 8 hours, and working alone at its own constant rate, a second machine seals K cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

machine A seals (k/8) cartons in 1hour, machine B seals (k/4) cartons in 1 hour, therefore together they seal (k/8)+(k/4) = (12k/32) cartons in 1hour or k cartons in (16/6)=2.666... hours.

if machine A and B work (16/6)hours, A seals (16/6)*(k/8) = (1/3)k
and B seals (16/6)*(k/4)=(2/3)k

Because machine A works at the faster rate (it only needs 4h instead of 8h for k cartons), machine A seals 66.667% of the cartons.

Hope that this will help,
greetz