If m, p, s and v are positive, and m/p < s/v, which of the following

Responding to a pm:

You can work on this question using some number line and averaging concepts.
Let's look at statement II and III first since they are very easy.

We know \(\frac{m}{p} <\frac{s}{v}\)

On the number line: .............0....................m/p ........................s/v (since m, p, s and v are all positive so m/p and s/v are to the right of 0)

II. \(\frac{ms}{pv}\)
Think of the case when m/p and s/v are both less than 1. When you multiply them, they will become even smaller. Say .2*.3 = .06. So the product may not lie between them.

III. \(\frac{s}{v} - \frac{m}{p}\)
Think of a case such as this: .............0..............................m/p .......s/v
\(\frac{s}{v} - \frac{m}{p}\) will be much smaller than both m/p and s/v and will lie somewhere here:
.............0.......Here...........................m/p .......s/v
So it needn't be between them.

Now only issue is (I). You can check some numbers for it including fractions and non fractions. Or try to understand it using number line.
Think of 4 numbers as N1, N2, D1, D2 for ease and given fractions as N1/D1 and N2/D2.

\(\frac{m+s}{p+v} = \frac{m+s/2}{p+v/2}\) = (Avg of N1 and N2)/(Avg of D1 and D2)

Now numerator of avg will lie between N1 and N2 and denominator of avg will lie between D1 and D2. So Avg N/Avg D will lie between N1/D1 and N2/D2. Try to think this through.

If N1/D1 < N2/D2, it could be because N1 < N2 and D1 = D2. So AvgN will lie between N1 and N2 and AvgD = D1 = D2. It could also be because N1 < N2 and D1 > D2. AvgN will be larger than N1 but smaller than N2. AvgD will be smaller than D1 but greater than D2 so AvgN/AvgD will be greater than N1/D1 but smaller than N2/D2. It could also be because N1 << N2 and D1 < D2 i.e. N1 is much smaller than N2 as compared to D1 to D2.
It could be because N1=N2 but D1>D2. Again, AvgD will lie between D1 and D2 and AvgN = N1 = N2.
It could also be because N1 > N2 but D1 >> D2.
Take some numbers to understand why this makes sense.

This is one of those very few problems were picking numbers is almost your only weapon

Let's say that m/p<s/v,

Let's say m=4, p=2, s=3 and v=1

So let's begin with statements

I. 4+3/3=7/3=2.3 is it between 2<x<3?

Yes, so this one is true

2. Clearly not true
3. Clearly not true

Therefore only I is our best answer choice

Hope this helps
Cheers
J

Let’s analyze each statement using specific values for the variables.

We can let m = 2, s = 3, p = 4, and v = 5. Thus:

m/p = 2/4 = 0.5 and s/v = 3/5 = 0.6.

Notice that m/p = 0.5 is less than s/v = 0.6. Now let’s analyze each statement.

I. (m+s)/(p+v)

(2 + 3)/(4 + 5) = 5/9 = 0.555… is between m/p = 0.5 and s/v = 0.6.

II. (ms)/(pv)

(2 x 3)/(4 x 5) = 6/20 = 0.3 is NOT between 0.5 and 0.6.

III. s/v - m/p

3/5 - 2/4 = 0.6 - 0.5 = 0.1 is NOT between 0.5 and 0.6.

From the above, we see that only statement I is true. However, this was illustrated by using one set of numbers (m = 2, s = 3, p = 4, and v = 5). It’s possible that it could be false when we use another set of values for m, s, p, and m.

However, we can prove that (m+s)/(p+v) is between m/p and s/v; that is, we can prove that m/p < (m+s)/(p+v) < s/v regardless of the values we use for m, s, p, and m, as long as the values are positive.

Notice that m/p < (m+s)/(p+v) < s/v means m/p < (m+s)/(p+v) and (m+s)/(p+v) < s/v. Also, keep in mind that we are given that m/p < s/v, which is equivalent to mv < ps.

Let’s prove that m/p < (m+s)/(p+v):

m/p < (m+s)/(p+v) ?

m(p+v) < p(m + s) ?

mp + mv < mp + ps?

mv < ps ? (YES)

Since mv < ps is true, m/p < (m+s)/(p+v) is true. Finally, let’s prove that (m+s)/(p+v) < s/v:

(m+s)/(p+v) < s/v ?

v(m+s) < s(p+v)?

mv + sv < sp + sv ?

mv < ps ? (YES)

Again, since mv < ps is true, (m+s)/(p+v) < s/v is true. Thus we have shown that m/p < (m+s)/(p+v) < s/v is always true.

Answer: B

for these kind of problems, I feel it is better to keep as close as possible so that we can be confident with every option

I chose 3,4,4,5 so that 3/4<4/5 (i.e. 0.75<0.8)

I=> (3+4)/(4+5) = 7/9 = 0.77 satisfies
II=>12/20=>3/5=0.6 incorrect
III=>4/5 - 3/4 = 0.8 - 0.75= 0.05

only I is correct.

The key word here is MUST.
So, if we can show that a certain statement is NOT TRUE we can eliminate some answer choices.

GIVEN: m/p < s/v

So, let's see what happens when m = 1, p = 3, s = 1 and v = 2
We get the inequality: 1/3 < 1/2, which works.

Now test the 3 statements:

I) (m+s)/(p+v) = (1+1)/(3+2) = 2/5
Since it IS the case that 1/3 < 2/5 < 1/2, statement I COULD be true

II) ms/pv = (1)(1)/(3)(2) = 1/6
Here, 1/6 < 1/3 < 1/2
In other words, 1/6 is NOT between 1/3 and 1/6
So, statement II need not be true.
ELIMINATE II

III) s/v - m/p = 1/2 - 1/3 = 1/6
Here, 1/6 < 1/3 < 1/2
In other words, 1/6 is NOT between 1/3 and 1/6
So, statement III need not be true.
ELIMINATE III

At this point, only answer choices A and B remain.
We COULD try testing more values in the hopes that statement I may not be true.
Or we can try to convince ourselves that statement I IS true.
Let's try the latter.

I) (m+s)/(p+v)
Is it the case that (m+s)/(p+v) is BETWEEN m/p and s/v?
Let's first see whether (m+s)/(p+v) < s/v
Since v is POSITIVE, we can safely multiply both sides by v to get: v(m+s)/(p+v) < s
Since (p+v) is POSITIVE, we can safely multiply both sides by (p+v) to get: v(m+s) < s(p+v)
Expand to get: vm + sv < sp + sv
Subtract sv to get: vm < sp
Divide both sides by p to get: vm/p < s
Divide both sides by v to get: m/p < s/v
So, our (m+s)/(p+v) < s/v turns into the inequality m/p < s/v, which is GIVEN information.
Since we know that the inequality m/p < s/v is true, it must also be the case that the inequality (m+s)/(p+v) < s/v is also true

Using the same technique to show that m/p < (m+s)/(p+v) [I'll leave it to you to do that ]

Since we can show that m/p < (m+s)/(p+v) < s/v, we can conclude that statement I is true.

Answers: B

Cheers,
Brent

Testing numbers for "must be true" doesn't always work. One set of values could help you establish that statements II and III do not hold. But how can you PROVE that statement I will always hold just because it holds for one set of values?
You do need to establish algebraically or logically that it will hold.

Check for algebraic solution: if-m-p-s-and-v-are-positive-and-m-p-s-v-which-of-the-fol-160298.html#p1269854
Check for logical solution: if-m-p-s-and-v-are-positive-and-m-p-s-v-which-of-the-fol-160298.html#p1270389

Statement 3 is (s/v) - (m/p)
When you take values as 1, 2, 3 and 4, it becomes (3/4) - (1/2) = 1/4 (This is not between 1/2 and 3/4 and hence you know that statement 3 may not hold always)

Plugging in numbers is not the best strategy for 'must be true' questions. You know that statement 1 holds for these particular values of m , p, s and v (1, 2, 3 and 4) but how do you know that it will be true for every set of valid values of m, p, s and v? You cannot try every set. You can certainly ignore statements II and III since you have already got values for which they are not satisfied. But you must focus more on statement I and try to figure out using logic whether it must always hold.

"Must be" means for any (possible) values of m, p, s and v. So, \(\frac{m}{p}< (option) <\frac{s}{v}\), must hold true fo any positive values of m, p, s and v.

Must or Could be True Questions to practice: search.php?search_id=tag&tag_id=193

Hope it helps.

We need "MUST BE" between, not could be between. In the case I mentioned above, s/v - m/p will not be between them. Even if in some cases it will be between them, it doesn't matter. We need those in which it will ALWAYS be between them.
This is how we deal with "must be true" questions. If we can find one case in which it is not true, that is enough to say that the statement does not hold.

Also, check this: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/0 ... -question/

You can take numbers such 1,2,3 and 4 as m,p,s and v
find the value between m/p and s/v
try to enter the numbers in the answer choices.
only option B will satisfy .

Thanks
Rahul

I. if \(\frac{m+s}{p+v}\) is to be more than \(\frac{m}{p}\),

==> \(\frac{m}{p}\)<\(\frac{m+s}{p+v}\)
==> pm+mv<pm+ps and since all numbers are positive,
==> \(\frac{m}{p}\)<\(\frac{s}{v}\) which is true as per question stem.

if \(\frac{m+s}{p+v}\) is to be less than \(\frac{s}{v}\),
==> \(\frac{s}{v}\)>\(\frac{m+s}{p+v}\)
==> sp+sv>mv+sv
==> \(\frac{m}{p}\)<\(\frac{s}{v}\) which is true as per question stem.

II. if \(\frac{ms}{pv}\) is to be more than \(\frac{m}{p}\),

==> \(\frac{s}{v}\) must be greater than 1, which is not conclusive as per stem of question.

III. if \(\frac{s}{v} - \frac{m}{p}\) is to be more than \(\frac{m}{p}\),

==> \(\frac{s}{v} must be greater than 2 X [m]\frac{m}{p}\), which is not conclusive as per stem of question.

So only (I) can be derived with 100% confidence.
Hence Answer must be B.

Hi,

I make the plug in as follows but it does not work: 1/2 < 3/4

1. (1+3)/(2+4) = 4/6 = 2/3 so in between
2. (1*3)/(2*4) = 3/8 less than 1/2
3. (3-1)/(4-2) = 1 > 3/4

So this problem cannot solve by using plugin?? Please help to correct me if I went wrong.

Thanks!

I have an awesome trick for the the first "statement", that is also a handy tool for other questions. Here's what and why:

The trick: If you have a fraction (or an integer) and add specific values to the numerator and the denominator, the fraction moves closer on to a fraction consisting of those two values (on the number line)!

Example 1: If you have fraction \(\frac{2}{3}\) and you add 1 to both the numerator and the denominator ( \(\frac{2+1}{3+1}\) ) you get \(\frac{3}{4}\) which is closer to \(\frac{1}{1}=1\) than it was.

Example 2: If you have fraction \(\frac{1}{5}\) and you add 3 to the numerator and 4 to the denominator ( \(\frac{1+3}{5+4}\) ) you get \(\frac{4}{9}\) which is closer to \(\frac{1}{1}=1\) than it was.

Note that it will never be more than the fraction consisting of what you add, but closer to. Try it out yourself with different fractions and integers!

So for this question: \(\frac{m+s}{p+v}\) gives a fraction that starts as \(\frac{m}{p}\) and gets closer to \(\frac{s}{v}\), but it can never become or surpass \(\frac{s}{v}\), as long as the fractions don't have the same value. Hence, \(\frac{m+s}{p+v}\) must lie somewhere in between the two fractions.


Other uses: You can use this knowledge when comparing two fractions.
Example: Compare \(\frac{147}{200}\) and \(\frac{150}{203}\). Which is bigger?
Answer: Clearly, both are below 1. And if you add 3 to the numerator and denominator of \(\frac{147}{200}\) like this: \(\frac{147+3}{200+3}\), you get \(\frac{150}{203}\). So when you add 3 to both the numerator and the denominator, the number gets closer to \(\frac{3}{3}=1\) and it actually becomes the fraction you are comparing it to. This means that \(\frac{147}{200}\) must become bigger to become \(\frac{150}{203}\), hence \(\frac{147}{200}\) is smaller than \(\frac{150}{203}\).

i would try with simple values through brute force which will take us faster to the options
let m=s=1 and p=8 and v=4

then the condition is met

in 2) 1/32 is recieved when we plug into the equation which is not between 1/8 and 1/4

in 3) 1/4 -1/8 =1/8 which is not in between so we safely eleminate

in 1) however we get 2/12 = 1/6 which is between 1/8 and 1/4 therefore we take this value

Therefore IMO B

If m, p, s and v are positive, and \(\frac{m}{p} <\frac{s}{v}\), which of the following must be between \(\frac{m}{p}\) and \(\frac{s}{v}\)

I. \(\frac{m+s}{p+v}\)
II. \(\frac{ms}{pv}\)
III. \(\frac{s}{v} - \frac{m}{p}\)

A. None
B. I only
C. II only
D. III only
E. I and II both

I.
\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\)

\(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)

I is always true.

II.
we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{m}{p} \)
=\((\frac{m}{p})^2<\frac{ms}{pv}\)

we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{s}{v} \)
=\(\frac{ms}{pv}<(\frac{s}{p})^2\)

combined
\((\frac{m}{p})^2<\frac{ms}{pv}<(\frac{s}{p})^2\)

squareroot
\(\frac{m}{p}<\sqrt{\frac{ms}{pv}}<\frac{s}{p}\)

it's not always true that
\(\sqrt{\frac{ms}{pv}} = \frac{ms}{pv}\)

hence II is not always true.

III.
let's start with the stem;
\(\frac{m}{p} <\frac{s}{v}\)
= \(0<\frac{s}{v} - \frac{m}{p} \)
= \(\frac{m}{p}-\frac{s}{v} <0 \)
= \(\frac{m}{p}-\frac{s}{v} < \frac{s}{v} - \frac{m}{p} \)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)

now lets evaluate III
if its between then that means
\(\frac{m}{p} < \frac{s}{v} - \frac{m}{p} <\frac{s}{v}\)
= add \(\frac{m}{p}\) and subtract \(\frac{s}{v}\)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)

therefore III is not inbetween. its less than.

so we know that its not always the case where
\(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)
based on our analysis of the stem it can be = \(\frac{m}{p} < \frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)

How can I solve this using numbers?

which of the following 'must be' between m/p and s/v?

I didn't understand the question. What does the question mean here by 'must be'?