SOLUTION: If p and q are positive integers such that 7/10 < p/q < 11/15. then the smallest possible value of q is (A) 25 (B) 60 (C) 30 (D) 7 (E) 6 Help i have no idea how to do it. but i

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Question 974901: If p and q are positive integers such that 7/10 < p/q < 11/15. then the smallest possible value of q is
(A) 25 (B) 60 (C) 30 (D) 7 (E) 6
Help i have no idea how to do it. but i know that 7/10 is 0.7 and 11/15 is 0.733.. so it would have to be in between. How can i find the answer? thanks
Found 4 solutions by josgarithmetic, jim_thompson5910, solver91311, MathTherapy:
Answer by josgarithmetic(39614)   (Show Source):
You can put this solution on YOUR website!
Here is something to try.

Check the factorizations of the denominators carefully and find lowest common denominator is 30. Raise the known numbers to terms of 30th's.

, and the unknown ratio need no such treatment.
You know that you cannot use q=30 because p would not be an integer. You could multiply p/q by 1 so that you force both p and q to be integers.

Starting at denominator q=30, ratio is
. ---begin fixing mistake.

You have exactly one choice for factor of 1 which turns the ratio into that of integers. Use factor .

~~but~~ this is obviously NOT reducible.

ANSWER: q is 4.
Obviously this is wrong and a mistake must be in my solution method...
3/4 is NOT within the bound requirements.
ANSWER: q is 60

Choice B, for q is 60

Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Let's pick the smallest answer choice and test our way up.

Let q = 6 (from choice E)

Now cycle through all of the smaller values of p (p = 1 through p = 5). The value of p must be smaller so that p/q < 1. Make sure q is fixed at 6.

If p = 1, then p/q = 1/6 = 0.166666666666667
this is NOT between 0.7 and 0.733 so we move on

If p = 2, then p/q = 2/6 = 0.333333333333333
this is NOT between 0.7 and 0.733 so we move on

If p = 3, then p/q = 3/6 = 0.5
this is NOT between 0.7 and 0.733 so we move on

If p = 4, then p/q = 4/6 = 0.666666666666667
this is NOT between 0.7 and 0.733 so we move on

If p = 5, then p/q = 5/6 = 0.833333333333333
this is NOT between 0.7 and 0.733 so we move on

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Let q = 7 (from choice D)

Again we will cycle through all of the positive integers smaller than 7 testing various values of p. The value of q will stick to 7.

If p = 1, then p/q = 1/7 = 0.142857142857143
this is NOT between 0.7 and 0.733 so we move on

If p = 2, then p/q = 2/7 = 0.285714285714286
this is NOT between 0.7 and 0.733 so we move on

If p = 3, then p/q = 3/7 = 0.428571428571429
this is NOT between 0.7 and 0.733 so we move on

If p = 4, then p/q = 4/7 = 0.571428571428571
this is NOT between 0.7 and 0.733 so we move on

If p = 5, then p/q = 5/7 = 0.714285714285714
this is between 0.7 and 0.733 so we have a winner

We don't need to test any more answer choices because we have found the final answer.

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So if p = 5 and q = 7, then p/q = 5/7 = 0.714285714285714 which is between 0.7 and 0.733

Therefore, 7/10 < 5/7 < 11/15 is true

So D) 7 is the correct answer

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Jim
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Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

The question asks for the smallest possible value of selecting from the given answers. So let's try 6.

Since must be a positive integer, then our choices, given that we know must be smaller than 1 based on the fact that , are . But since , we can eliminate 1, 2, and 3. Hence, .

Let's compare to . Finding the common denominator, we find that and , so is outside the given interval and is not a possible value for the desired fraction.

and , so is outside the range on the other side. Therefore no number where is a positive integer exists inside of the given interval.

On to the next smallest choice, namely 7. Let's examine numerators 4 and up.

but so 4 is eliminated as a candidate.

However, and that IS larger than . Things is lookin' up. Let's check against the other end of the interval.

and . is smaller than the high end of the interval.

Since we can say that , and there was no fraction with a denominator of 6 that fit in the interval, for positive integers and such that , 7 is the smallest of the given possible values for q.

Indeed, 7 is the smallest possible integer value for q without qualifiying the value as belonging to a particular list. Proof of this assertion is left as an exercise for the student.

John

My calculator said it, I believe it, that settles it

Answer by MathTherapy(10549)   (Show Source):
You can put this solution on YOUR website!
If p and q are positive integers such that 7/10 < p/q < 11/15. then the smallest possible value of q is
(A) 25 (B) 60 (C) 30 (D) 7 (E) 6
Help i have no idea how to do it. but i know that 7/10 is 0.7 and 11/15 is 0.733.. so it would have to be in between. How can i find the answer? thanks

(CHOICE D).