Hi all, This is my first time posting. I recently finished a class with MGMAT and I've been focusing on my MBA.com tests. I recently completed two and was happily surprised with both scores but I have a few questions on some math problems I got wrong. I will list them out below:


1) If two of the four expressions x + y, x + 5y, x-y, 5x-y are chosen at random, what is the probability that their product will be the form x^2 - (by) ^2, where b is an integer

a. 1/2
b. 1/3
c. 1/4
d. 1/5
e. 1/6

Correct Answer is e.

2) If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd

Correct answer is E (1) and (2) are not sufficient

3) The next question I think I may have misinterpreted

A store purchased 20 coats that each cost an equal amount and then sold each of the 20 coats at an equal price. What was the store's gross profit on the 20 coats?

(1) If the selling price had been twice as much the store's gross profit on the 20 coats would have been $2,400
(2) If the selling price had been $2 more, the store's gross profit on the 20 coats would have been $440

Answer is B (2) is sufficient

4) Tanya prepared 4 different letters. For each letter, she prepared an envelope with the correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address.

a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

Answer is d. Thank you all for any insight.

Best,
Sohail

For 3. I was thinking something along the lines of

P = Price per sweater C= Cost per sweater

Gross profit = 20P - 20C?

1) Statement 1 tells me 20(2P) - 20C = 2400

2) Statement 2 tells me 20(P+2)-20C = 440

Maybe I'm missing something obvious?

1) If two of the four expressions x + y, x + 5y, x-y, 5x-y are chosen at random, what is the probability that their product will be the form x^2 - (by) ^2, where b is an integer

a. 1/2
b. 1/3
c. 1/4
d. 1/5
e. 1/6



Sample Space = Total number of outcomes for selecting 2 expression out of 4 are 4C2
Number of ways we can select so that product are in form of x^2 - (by) ^2 is 1.
Since it is possible only when we select (1) x + y and x -y.

So the probability would be 1/4C2 = 1/6



(E)

2) If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd


The Sum w/x + y/z can be written as (wz + xy)/xz

Stmt (1) wx + yz is odd, Clearly this is insufficient as it doesn't relate to oue question.
Stmt(2) wz + xy is odd, This is also insufficient as we dont know the value of denominator xz.

Answere (E)



3) The next question I think I may have misinterpreted

A store purchased 20 coats that each cost an equal amount and then sold each of the 20 coats at an equal price. What was the store's gross profit on the 20 coats?

(1) If the selling price had been twice as much the store's gross profit on the 20 coats would have been $2,400
(2) If the selling price had been $2 more, the store's gross profit on the 20 coats would have been $440


Lets C.P of each coat be x and S.P be y, so profit is 20(y-x), so we have to find the values of 20(y-x)

(1) S.P = 2y, profit = 20(2y -x) = 2400
=> Using this we cannot get the value of 20(y-x) . Hence Insufficient

(2) S.P = y+2 , profit => 20(y+2 - x) = 440
=> 20y + 40 -20x = 440
=> 20(y-x)=400 .

We are only interested in this expression 20(y-x) which is the profit , not the individual values of x and y.
Hence this is sufficient.


4) Tanya prepared 4 different letters. For each letter, she prepared an envelope with the correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address.

a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

Take letters l1,l2,l3,l4 with their respective address as e1,e2,e3,e4.

We are asked to find out the probabilty of only 1 letter in the right envelope and others in different envelope.

There are 4 cases that needs to be considered:-
(1)When l1 goes in right envelope all other goes in wrong envelopes
(2)When l2 goes in right envelope all other goes in wrong envelopes
(3)When l3 goes in right envelope all other goes in wrong envelopes
(4)When l4 goes in right envelope all other goes in wrong envelopes


CASE(1) :- When l1 goes in right envelope all other goes in wrong envelopes
= (1/4)(as there is only 1 correct envelope out of 4 envelopes) *
(2/3)(as there are 2 wrong envelopes out of 3 envelopes left without letters) *
(1/2)(as there are 1 wrong envelopes out of 2 envelopes left without letters) *
(1/1) ( as there is 1 envelop left)
= 1/12



Similarly for all other cases , probability would be 1/12

So, Final probability is P(case1) + p(case2) + P(case3) + p(case4) = 1/12 + 1/12 + 1/12 + 12 = 1/3


1) If two of the four expressions x + y, x + 5y, x-y, 5x-y are chosen at random, what is the probability that their product will be the form x^2 - (by) ^2, where b is an integer

a. 1/2
b. 1/3
c. 1/4
d. 1/5
e. 1/6



Sample Space = Total number of outcomes for selecting 2 expression out of 4 are 4C2
Number of ways we can select so that product are in form of x^2 - (by) ^2 is 1.
Since it is possible only when we select (1) x + y and x -y.

So the probability would be 1/4C2 = 1/6



(E)

2) If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd


The Sum w/x + y/z can be written as (wz + xy)/xz

Stmt (1) wx + yz is odd, Clearly this is insufficient as it doesn't relate to oue question.
Stmt(2) wz + xy is odd, This is also insufficient as we dont know the value of denominator xz.

Answere (E)



3) The next question I think I may have misinterpreted

A store purchased 20 coats that each cost an equal amount and then sold each of the 20 coats at an equal price. What was the store's gross profit on the 20 coats?

(1) If the selling price had been twice as much the store's gross profit on the 20 coats would have been $2,400
(2) If the selling price had been $2 more, the store's gross profit on the 20 coats would have been $440


Lets C.P of each coat be x and S.P be y, so profit is 20(y-x), so we have to find the values of 20(y-x)

(1) S.P = 2y, profit = 20(2y -x) = 2400
=> Using this we cannot get the value of 20(y-x) . Hence Insufficient

(2) S.P = y+2 , profit => 20(y+2 - x) = 440
=> 20y + 40 -20x = 440
=> 20(y-x)=400 .

We are only interested in this expression 20(y-x) which is the profit , not the individual values of x and y.
Hence this is sufficient.


4) Tanya prepared 4 different letters. For each letter, she prepared an envelope with the correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address.

a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

Take letters l1,l2,l3,l4 with their respective address as e1,e2,e3,e4.

We are asked to find out the probabilty of only 1 letter in the right envelope and others in different envelope.

There are 4 cases that needs to be considered:-
(1)When l1 goes in right envelope all other goes in wrong envelopes
(2)When l2 goes in right envelope all other goes in wrong envelopes
(3)When l3 goes in right envelope all other goes in wrong envelopes
(4)When l4 goes in right envelope all other goes in wrong envelopes


CASE(1) :- When l1 goes in right envelope all other goes in wrong envelopes
= (1/4)(as there is only 1 correct envelope out of 4 envelopes) *
(2/3)(as there are 2 wrong envelopes out of 3 envelopes left without letters) *
(1/2)(as there are 1 wrong envelopes out of 2 envelopes left without letters) *
(1/1) ( as there is 1 envelop left)
= 1/12



Similarly for all other cases , probability would be 1/12

So, Final probability is P(case1) + p(case2) + P(case3) + p(case4) = 1/12 + 1/12 + 1/12 + 12 = 1/3

2) If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd

i see question 2 differently

we know tht y/z and w/x are integers.

So the sum w/x and y/x is an integer.

w/x + y/z = wz+xy/xz
From 2, numerator is odd.


If the sum of the two ratios is an integer, denominator has to be odd since an odd numerator/ even denominator cannot be an integer. Now we know that denominator and num are odd and the ratio is even.

Thus sum is even.

Am I doing somethng wrong here.

I meant ratio is odd and sum is odd.

hi -

welcome to the forums.
could you please do the following 3 things:
(1) SEARCH using the search box at the upper right, to make sure that these questions aren't already in the forum;
(2) read the 'sticky' posts at the top of each folder, containing the posting guidelines;
(3) follow those guidelines, most notably the following:
---- post EACH QUESTION in a SEPARATE THREAD
---- TITLE the thread using the first 8(ish) words of the question prompt (this is easy - you can just copy and paste from the question you've already typed into the main post body)

once you create these separate threads, we'll answer your questions.

thanks.

actually i'm feeling altruistic today, so i'll take care of these. by the way, i found the solutions to #1, #2, and #4 in a combined total of 26 seconds using the google search box at the upper right. that's a ridiculously small fraction of the time it must have taken you to type up all this stuff, so use the search box next time and everybody wins!

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#1 - i just cut and pasted this from http://www.manhattangmat.com/forums/if-two-of-the-four-expressions-x-y-x-5y-x-y-and-5x-y-are-t497.html
Start to write out the possibilities for the products (but keep an eye out for shortcuts):
1) (x+y)(x-y) = x^2 - y^2 (you should have this memorized, as it is one of the 3 common quadratics) This matches the form, with b = 1
2) (x+y)(x+5y) = x^2 +5xy + xy + 5y^2 = x^2 + 6xy + 5y^2 This does not match the form because we've got a 6xy term.
3) (x+y)(5x-y) = don't do this one - because of the previous one, you should see this is not going to give you the right form
4) (x+5y)(x-y) = ditto
5) (x+5y)(5x-y) = 5x^2 stop here - this is not the right form
6) (x-y)(5x-y) = ditto

Six possibilities and only one gives you the right form, so the answer is 1/6.

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#2
http://www.manhattangmat.com/forums/if-w-x-y-and-z-are-integers-t2821.html

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#3

i don't think we have a thread on this problem yet. i'll create one.

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#4
there is a ridiculously long thread on this problem (one of the longest in the history of these forums, actually). you'll find it here.

done
http://www.manhattangmat.com/forums/post14336.html