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.
_________________

Given that \(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)Adding 1 on both sides we have \(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)

Again,\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\) Adding 1 on both sides, we have \(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\) . Thus, I is always true. We just have to check for Option II now.

Assming it to be true, we should have\(\frac{ms}{pv}<\frac{s}{v} \to \frac{m}{p}<1\) Which is not always true. Thus the answer is B.
_________________

How can I solve this using numbers?

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

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!

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.
_________________

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

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.

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'?

"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.
_________________

You have the best approach in this thread, but I would never have thought of your solution to #1. I think you made it more complicated than it needed to be.

This is actually a very easy problem once you spend a few moments understanding what you need to do.

You're told the following:
1) m, p, s and v are positive
2) \(\frac{m}{p} <\frac{s}{v}\)

So, you need to know whether each option is both less than \(\frac{s}{v}\) and greater than \(\frac{m}{p}\). That's it. Once that clicks, it's very easy.

I. \(\frac{m+s}{p+v}\)

So part A:

Is \(\frac{m+s}{p+v} < \frac{s}{v}\) ?
Is \(mv+sv < sp + sv\) ?
Is \(mv < sp\) ?
Hmmm, that looks familar. In fact, it's the second thing they gave us: \(\frac{m}{p} <\frac{s}{v}\).
So yes!

Part B:

Is \(\frac{m+s}{p+v} > \frac{m}{p}\) ?
Is \(mp+sp > mp + mv\) ?
Is \(sp > mv\) ?
Hmmm, that looks familar too. In fact, it's also the second thing they gave us: \(\frac{m}{p} <\frac{s}{v}\).
So yes!

Then you perform the same (quicker) process for II.

Hi guys,

I got this problem on an exam pack 1 CAT and failed to solve it correctly due to a technical error (DRIVING ME CRAZY). Otherwise I picked numbers and was 99% on the right path. One question I have for you is:

Do you think testing one pari of numbers is enough? I picked 2 pairs:
- m=1, p=3, s=3 and v=4
- m=3, p=1, s=4 and v=1

These 2 options are more conservative as I wanted to test these equations for fractions and integers. Do you think this was overkill (question stem says we're talking about positive numbers and not integers)? As far as I can see, all of you tested just one pair of numbers, which of course almost certainly decreases the time needed to solve the problem.

Thanks!

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
_________________

Nowhere it says m,p,s and v can't be same numbers. Therefore I plugged in numbers to disprove the statements.
Shall they not mention that the numbers are distict.

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
_________________

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.

Dear Amanrohra,

I'm happy to respond here,
the question doesn't explicitly mentions that m,p,s and v are distinct, but the question gives us a condition:
\(\frac{m}{p}< \frac{s}{v}\)
which can not be true with map,s and v being the same numbers.
Hence they need to be distinct, at least in a manner to fulfil the condition.
for e.g. \(\frac{2}{5} < \frac{5}{2}\)
I hope this helps.

Thanks VeritasKarishma. Wouldn't this logic also hold true for "could be true" questions.

The given statement may not be true for the particular values we consider, but how do we know that the given statement will not be true for any values?

Absolutely! The logic will hold for could be true.

Must be true for all - Usually, the values you will try would be true. You might be hard pressed to find a value for which it doesn't hold. Here lies the challenge.

Could be true for some value of x - Usually, the values you will try would not be true. You will need to find a value for which it will hold. Doing that is often a bit easier by putting in 0, 1 etc. But yes, there could be a challenge here too.
_________________