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For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Question Stats: 48% (02:17) correct 52% (02:18) wrong based on 326 sessions

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For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

Solution -

Its given that, (m – r) > (s – n), which is equivalent to (m + n) > (r + s). So we need to determine the inequality holds true or false.

Stmt1 - 250 > r + s --> From the question m + n = 250, so the inequality m + n > r + s is true. Sufficient.

Stmt2 - m + r + s = 375 --> We know that m + n = 250 and m > n, m must be greater than 125. Subtracting 125 from 375 yields 250, so if m is greater than 125, then r + s must be smaller than 250. So the inequality m + n > r + s is true. Sufficient.
ANS D
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

Given, m>n, m+n=250

Question: is (m – r) > (s – n)? or is (m+n)>(s+r) ---> s+r<250 ?

Statement 1 , sufficient to say yes fr s+r<250.

Statement 2, m+r+s=375 ---> assuming (m – r) > (s – n) ---> m+n>375-m ---> 2m+n>375 or n >375-2m and n = 250-m

Thus, 250-m > 375-2m --- m >125

Thus, if m+n = 250 and m >125 ---> m>n which is a give. Thus our assumption above of (m – r) > (s – n) holds TRUE. Thus this statement is sufficient as well.

D is the correct answer.
Intern  Joined: 08 Jul 2012
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

given: m+n = 250 & m>n, hence n<125 & m>125

Asked: is (m-r)>(s-n), rearranging we get, is (m+n)>(s+r) ?

stmt1 : 250>r+s -->250 = m+n, hence m+n > r+s , sufficient

stmt2 : m+r+s = 375 -->r+s = 375-m, but m>125...so smallest possible value of m is 126 (m is given as integer)

so, r+s = 375-126 = 249 (this is max value of r+s) hence r+s is always equal to or less than 249 & m+n is 250 -->m+n > r+s....sufficient

Ans. D
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Math Expert V
Joined: 02 Sep 2009
Posts: 58450
Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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1
Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

800score Official Solution:

Statement (1) tells us that 250 > r + s. Since the question statement tells us that m + n = 250, we can determine that m + n > r + s.

Now, let us manipulate this inequality to see whether it is equivalent to the inequality in the question:
(m + n) > (r + s)
m > (r + s) – n
(m – r) > (s – n)

This is exactly what we were looking for. We can answer the question using Statement (1), hence it is sufficient.

Statement (2) tells us that m + r + s = 375.
Because we know that m + n = 250 and m > n, m must be greater than 125. Subtracting 125 from 375 yields 250, so if m is greater than 125, then r + s must be smaller than 250. We are now left with the same inequality that we were given in Statement (1), which can be manipulated to show that (m – r) > (s – n). So Statement (2) is also sufficient.

Since both statements are sufficient alone, the correct answer is choice (D).
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

Target question: Is (m - r) > (s - n)?

This is a great candidate for rephrasing the target question. We have a video on this at the bottom of this post

If we take the inequality in the target question and add r and n to both sides, we get . . .
REPHRASED target question: Is (m + n) > (s + r)?

Since m + n = 250, we can also rephrase it this way . . .
REPHRASED target question: Is 250 > (s + r)?

Given Information: m + n = 250 and m > n
If m and n were EQUAL, then m and n would both equal 125
Since m is GREATER THAN n, we can conclude that m > 125

Statement 1: 250 > r + s
Perfect!
One of our REPHRASED target questions is Is 250 > (s + r)?
Since statement 1 allows us to answer the REPHRASED target question with certainty, it is SUFFICIENT

Statement 2: m + r + s = 375
Earlier (in the Given Information part of the solution), we determined that m > 125
So, we can reword statement 2 as: (a number bigger than 125) + (r + s) = 375
This means that (r + s) must be LESS THAN 250
In other words, 250 > (s + r)
One of our REPHRASED target questions is Is 250 > (s + r)?
Since statement 2 allows us to answer the REPHRASED target question with certainty, it is SUFFICIENT

RELATED VIDEOS

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GMAT 1: 460 Q32 V22 Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

From the question stem we have to prove that s+r< 250
1. SUFFICIENT
2. m+r+s =375, m+n= 250 i.e. r+s=125+n, now it is given that m > n. So maximum value of n can be 124. Hence we use n=124 r+s would be 249 <250.
Hence SUFFICIENT
Correct answer is D.
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Hello Bunuel,

I have a doubt. Please help me see what am I missing...

m, n, r, and s are integers-> nowhere it's mentioned that it is positive integer, it can as well be negative integer... in that case how can we reduce the equation (m – r) > (s – n) into (m + n) > (r + s)

thanks

Bunuel wrote:
Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

800score Official Solution:

Statement (1) tells us that 250 > r + s. Since the question statement tells us that m + n = 250, we can determine that m + n > r + s.

Now, let us manipulate this inequality to see whether it is equivalent to the inequality in the question:
(m + n) > (r + s)
m > (r + s) – n
(m – r) > (s – n)

This is exactly what we were looking for. We can answer the question using Statement (1), hence it is sufficient.

Statement (2) tells us that m + r + s = 375.
Because we know that m + n = 250 and m > n, m must be greater than 125. Subtracting 125 from 375 yields 250, so if m is greater than 125, then r + s must be smaller than 250. We are now left with the same inequality that we were given in Statement (1), which can be manipulated to show that (m – r) > (s – n). So Statement (2) is also sufficient.

Since both statements are sufficient alone, the correct answer is choice (D).

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Math Expert V
Joined: 02 Sep 2009
Posts: 58450
Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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1
DaenerysStormborn wrote:
Hello Bunuel,

I have a doubt. Please help me see what am I missing...

m, n, r, and s are integers-> nowhere it's mentioned that it is positive integer, it can as well be negative integer... in that case how can we reduce the equation (m – r) > (s – n) into (m + n) > (r + s)

thanks

Bunuel wrote:
Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

800score Official Solution:

Statement (1) tells us that 250 > r + s. Since the question statement tells us that m + n = 250, we can determine that m + n > r + s.

Now, let us manipulate this inequality to see whether it is equivalent to the inequality in the question:
(m + n) > (r + s)
m > (r + s) – n
(m – r) > (s – n)

This is exactly what we were looking for. We can answer the question using Statement (1), hence it is sufficient.

Statement (2) tells us that m + r + s = 375.
Because we know that m + n = 250 and m > n, m must be greater than 125. Subtracting 125 from 375 yields 250, so if m is greater than 125, then r + s must be smaller than 250. We are now left with the same inequality that we were given in Statement (1), which can be manipulated to show that (m – r) > (s – n). So Statement (2) is also sufficient.

Since both statements are sufficient alone, the correct answer is choice (D).

We are concerned about the sign of a number when multiplying/dividing an inequality by that number. However we can safely add/subtract a number from both sides of an inequality, which is done in that example: add n+r to both sides of (m – r) > (s – n) to get (m + n) > (r + s).

Hope it's clear.
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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Bunuel wrote:
For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375

Kudos for a correct solution.

To prove : m-r > s-n => m-r+n > s => m+n > s+r => 250 > s+r or s+r < 250
Also from the question stem m+n = 250 , m > n
so maximum value for n can be 124 and minimum value of m can be 126

1. 250 > r+s ...... sufficient
2. m + r + s = 375
r + s = 375 - m

lets take minimum value of m ie 126 as this will give us maximum value of r+s

r + s = 375 - 126 => r + s = 249 => r + s < 250
This is also sufficient

So answer is D
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Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)  [#permalink]

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For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s

Rewriting $$m-r > s-n$$

$$m+n > s+r$$

We know that m + n = 250

Thus, $$250 > s+r$$

SUFFICIENT

(2) m + r + s = 375

$$m+n=250$$ and $$m>n$$, we know that m is greater than 125.

$$375-125=250[/b] As m is greater than 125, [m]r+s$$ has to be less than 250

Inequality stands true

SUFFICIENT.

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Kudos are appreciated. Re: For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r)   [#permalink] 16 Oct 2018, 05:32
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