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P = n+1-n, and Q = m+1-m for positive integers m and n. Which one is greater than the other?

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P = n+1-n, and Q = m+1-m for positive integers m and n. Which one is greater than the other?

by Max@Math Revolution » Thu May 21, 2020 12:13 am

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[GMAT math practice question]

P = \(\sqrt{n+1}-\sqrt{n}\) , and Q = \(\sqrt{m+1}-\sqrt{m}\) for positive integers m and n. Which one is greater than the other?

1) n > m.
2) n and m are consecutive integers.
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Source: — Data Sufficiency |
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Re: P = n+1-n, and Q = m+1-m for positive integers m and n. Which one is greater than the other?

by deloitte247 » Sat May 23, 2020 2:35 am

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Given that m and n are positive integers

Target question => Which one is greater than the other (between P and Q)

Statement 1 => n > m
$$If\ n=8\ and\ m=5$$
$$P=\sqrt{8+1}-\sqrt{8}=\sqrt{9}-\sqrt{8}$$
$$=3-2.83=0.17$$
$$Q=\sqrt{5+1}-\sqrt{5}=\sqrt{6}-\sqrt{5}$$
$$=2.45\ -\ 2.24=0.21$$
Q>P when n and m is not consecutive

$$If\ n=8\ and\ m=7$$
$$P=\sqrt{8+1}-\sqrt{8}=\sqrt{9}-\sqrt{8}$$
$$=3-2.83\ =\ 0.17$$
$$Q=\sqrt{7+1}-\sqrt{7}=\sqrt{8}-\sqrt{7}$$
$$=2.83-2.65=0.18$$
Q>P when is consecutive
Hence, statement 1 is SUFFICIENT as Q>P

Statement 2 => n and m are consecutive integers
$$If\ \ n=4\ and\ m\ =3$$
$$P=\sqrt{4+1}-\sqrt{4}=\sqrt{5}-\sqrt{4}$$
$$=2.24-2=0.24$$
$$Q=\sqrt{3+1}-\sqrt{3}=\sqrt{4}-\sqrt{3}$$
$$=2-1.73=0.27$$
Q>P when n>m

$$If\ n=7\ and\ m=8$$
$$P=\sqrt{7+1}-\sqrt{7}=\sqrt{8}-\sqrt{7}$$
$$=2.83-2.65=0.18$$
$$Q=\sqrt{8+1}-\sqrt{8}=\sqrt{9}-\sqrt{8}$$
$$=3-2.83=0.17$$
P>Q when n<m

Possible scenarios in this statement yield difficult result as there is no definite answer for the target question, statement 2 is NOT SUFFICIENT

Since statement 1 alone is SUFFICIENT,
Answer = A
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Re: P = n+1-n, and Q = m+1-m for positive integers m and n. Which one is greater than the other?

by Max@Math Revolution » Sun May 24, 2020 5:04 pm

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

1/P = 1 / ( \(\sqrt{n+1}-\sqrt{n}\) ) = \(\sqrt{n+1}+\sqrt{n}\)
1/Q = 1 / ( \(\sqrt{m+1}-\sqrt{m}\)) = \(\sqrt{m+1}+\sqrt{m}\)
Since n > m from condition 1), we have 1/P – 1/Q = ( \(\sqrt{n+1}+\sqrt{n}\)) – ( \(\sqrt{m+1}+\sqrt{m}\)) > 0 or 1/P > 1/Q.
Since P and Q are positive, we have P < Q.
Thus, condition 1) is sufficient.
Condition 2) is not sufficient, since we don’t know which one of m or n is greater.
Therefore, A is the answer.
Answer: A
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