If two of the four expressions x+y, x+5y, x-y, and 5x-y are

[chester]

Hi I don't know how to apporach these problems, thanks for your help!

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

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

# 25 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 = 3x5x5. How many two-digit positive integers have length 6?

a. none b. one c. two d. three e. four


#27 If m and r are two nmbers on a number line, what is the value of r?

1. the distance between r and 0 is 3 times the distance between m and 0

2. 12 is halfway between m and r

I chose C because I thought m=6, and r=18, which would satisfy both statemtn 1 and 2.

Is there another possibility that makes r to be a different value?

[StaceyKoprince]

#14

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.


#25
They give us the example that 75 = length 3 b/c 75 = 3*5*5. 3*5*5 represents the prime factorization of 75. Then they want a 2-digit integer whose length is 6 - the key shortcut here is to recognize that 6 is a lot of digits but the product can be only a 2-digit integer, so there can't be that many options. Start with the smallest possibility.

The smallest one is going to be based on prime number 2 (b/c this is the smallest prime number). Try 6 2's: 2*2*2*2*2*2 = 64. So there's one possibliity. The next smallest possibility would be to replace one of those 2's with the next largest prime number (3): 2*2*2*2*2*3 = 96. Possibility #2. Next, we could replace another 2 with a 3... but if you think about it, you may be able to tell that this won't give you a 2-digit integer. (If you're not sure, actually check the math.)

#27
Tricky one. Here they ask us for the actual value (of r), not just the distance between the two points, which means we have to be able to find one (and ONLY one) number for r.

I assume you dealt with statements 1 and 2 individually okay, because you got yourself down to C and E. Don't forget about negative numbers. Statement 1 only says that r is 3 times the distance - it doesn't say the actual number is 3x. It could be -3x. So, m could be 6 and r 18. But m could also be -12 and r 36. That pair still satisfies both requirements.

[rtfact]

Q14.

x^2-(by)^2=(x+by)(x-by)

so we are looking for a structure similar to x+by or x-by
we can eliminate 5x-y, x+5y.
the only possibility is using x+y, x-y. this gives us only one option.
2/4 for choosing any one of the four numbers. 1/3 is for choosing the other number.
(2/4)*(1/3)=1/6.

[RonPurewal]

Q14.

x^2-(by)^2=(x+by)(x-by)

so we are looking for a structure similar to x+by or x-by
we can eliminate 5x-y, x+5y.
the only possibility is using x+y, x-y. this gives us only one option.
2/4 for choosing any one of the four numbers. 1/3 is for choosing the other number.
(2/4)*(1/3)=1/6.

correct.

notice that you eliminate (5x - y) because it has the wrong form: you can't have a coefficient in front of x. (i.e., it needs to be (x + by), not (ax + by))

however, you're eliminating (x + 5y) not because it has the wrong form but, rather, because its "partner", (x - 5y), is absent.

[imanemekouar]

Hi Ron,
Can you please explain the last part of the problem
My understanding stop after you kept X-Y ,X+y and eliminate the others
how did you get to those number.
2/4 for choosing any one of the four numbers. 1/3 is for choosing the other number.
(2/4)*(1/3)=1/6.

[agha79]

If m and r are two nmbers on a number line, what is the value of r?

1. the distance between r and 0 is 3 times the distance between m and 0

2. 12 is halfway between m and r

I am still lost what is the correct answer for above question and how to go about doing it

[esledge]

If m and r are two nmbers on a number line, what is the value of r?

1. the distance between r and 0 is 3 times the distance between m and 0

2. 12 is halfway between m and r

I am still lost what is the correct answer for above question and how to go about doing it

The correct answer is E.

Method 1: Visual/Number Line approach.
(1) r is 3 times farther away from 0 than m is. But we have no "distances" given, nor any info about sign (i.e. is m left or right of 0?)

(2) On a number line, put a dot at 12. Put two dots on either side of it for m and r. What can vary? The distance between m and r--they can be very close to 12, or both very far away. Also, we don't know whether m is the dot to the left or to the right of 12.

(1)&(2) together: We still don't know distances (from 12 or 0), or whether m is left or right of r.
We can either have (case A) r = 18 and m = 6 or (case B) r = 36 and m = -12.

Method 2: Algebra approach
(1) r = +/-3m
(2) r-12 = 12-m, or r+m = 24.
(1)&(2) together: r+m= (+/-3m)+m = 24. Either 4m = 24 (i.e. m=6) or -2m = 24 (i.e. m = -12).

[agha79]

Thanks Emily:)

Makes sense now

[RonPurewal]

Hi Ron,
Can you please explain the last part of the problem
My understanding stop after you kept X-Y ,X+y and eliminate the others
how did you get to those number.
2/4 for choosing any one of the four numbers. 1/3 is for choosing the other number.
(2/4)*(1/3)=1/6.

this is basic sequential probability. you can choose either of these items first, out of four, giving you a probability of 2/4. after you make that choice, only one desired item remains, out of three, giving you a probability of 1/3.
as with all sequential probability, once you find these different probabilities, you must multiply them together.

[zaarathelab]

[quote="chester"]

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

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

quote]

Hi Ron, sorry for opening this thread once again, but could you pls tell me if my approach is incorrect?

The probability of choosing (x+y) =1/4, the probability of choosing (x-y) after the selection of (x+y)= 1/3

Now either of them can be selected first. Hence,

(1/4*1/3) + (1/4*1/3) = 1/6

[RonPurewal]

Hi Ron, sorry for opening this thread once again, but could you pls tell me if my approach is incorrect?

The probability of choosing (x+y) =1/4, the probability of choosing (x-y) after the selection of (x+y)= 1/3

Now either of them can be selected first. Hence,

(1/4*1/3) + (1/4*1/3) = 1/6

that works.

[visitdhiraj]

Hi Ron & Stacey & Emity Sledge

You guys have made my life very easy with your magical answers

I have one small query with regards to this question

if you consider that the coefficient of y (B=1) is 1, which is b.

In the same way can't we consider that there is a similar a =1, which is the coefficient of x.

In that case even this choice would not form the correct answer.

[tim]

can you demonstrate what you mean by your example not giving the correct answer?

[crusade]

Hi,

Shouldnt the answer be:

2/4*1/3*2! as these two expressions could be selected in two ways?

For eg. in a question that asks whats the probability of 2 selected balls out of 10 to be blue and red,

probability=b/10*r/9*2! (as this could happen in 2 ways). -- b and r are the no. of blue and red balls respectively.

Why dont we use the same principal here?

[RonPurewal]

Hi,

Shouldnt the answer be:

2/4*1/3*2! as these two expressions could be selected in two ways?

no, because the numerator "2" (in the fraction 2/4 on the left) already accounts for both possibilities.

if you want to account for each possibility separately, then note that the probability of each individual possibility is (1/4)(1/3). then you can take this approach and write (1/4)(1/3) x 2, which will give you the correct answer.

by the way, if you have any doubts about a problem like this one, just write out all the possibilities (= here, just write out a list of all the different pairings of factors that you could pick). then, you can just count the total number of "winners", and then there's no question as to what the correct probability should be.