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Intern  Joined: 15 Nov 2010
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If d represents the hundredths digit and e represents the  [#permalink]

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Question Stats: 50% (02:27) correct 50% (02:25) wrong based on 526 sessions

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If d represents the hundredths digit and e represents the thousandths digit in the decimal 0.4de, what is the value of this decimal rounded to the nearest tenth?

(1) d – e is equal to a positive perfect square.
(2) √d > e^2

Originally posted by ghostdude on 10 Jul 2011, 08:44.
Last edited by Bunuel on 03 Aug 2012, 00:01, edited 1 time in total.
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Re: If d represents the hundredths digit and e represents the  [#permalink]

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pbull78 wrote:
can anyone help me with detailed explanation ?

If d represents the hundredths digit and e represents the thousandths digit in the decimal 0.4de, what is the value of this decimal rounded to the nearest tenth?

To answer the question we should know whether $$d\geq{5}$$.

(1) d – e is equal to a positive perfect square --> easy to get two different result: $$0.4de=0.451$$ (5-1=4=2^2), then 0.4de rounded to the nearest tenth will be $$0.5$$ but if $$0.4de=0.421$$ (2-1=1=1^2), then 0.4de rounded to the nearest nearest tenth will be $$0.4$$. Not sufficient.

(2) $$\sqrt{d}>e^2$$ --> also easy to get two different result: if $$\sqrt{d}=\sqrt{5}>1^2=e^2$$ or $$\sqrt{d}=\sqrt{2}>1^2=e^2$$. Not sufficient.

(1)+(2) 0.451 and 0.421 satisfy both statements and give different values of 0.4de when rounded to the nearest tenth: 0.5 and 0.4. Not sufficient.

Rounding rules
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Hope it helps.
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ghostdude wrote:
If d represents the hundredths digit and e represents the thousandths digit in the decimal 0.4de, what is the value of this decimal rounded to the nearest tenth?

(1) d – e is equal to a positive perfect square.
(2) √d > e^2

E
0.451. Decimal rounded tenth=0.5
0.421. Decimal rounded tenth=0.4
Intern  Joined: 16 Dec 2011
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can anyone help me with detailed explanation ?
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1
0.4de . question -0.x?

st1-d-e =perfect square.
let d-e=4 (d=4; e=0) or d-e=9 (d=9 e =0) insuff

st2- sqrootd>e^2
sqroot4>0 or sqroot9>0. again insuff. as everyone can understand, together these stmnts are also insuff
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Re: If d represents the hundredths digit and e represents the  [#permalink]

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ghostdude wrote:
If d represents the hundredths digit and e represents the thousandths digit in the decimal 0.4de, what is the value of this decimal rounded to the nearest tenth?

(1) d – e is equal to a positive perfect square.
(2) √d > e^2

we are allowed to use numbers < 10.

1. there are only 3 perfect squares in the available range. 1; 4; 9.

Pick 4. If d-e=4 then d might be 4 and e 0 or d might be 7 and e 3. If d=4 then our decimal remains 0.4; If d=7 our decimal becomes 0.5

NS.

2. since we know for sure that we are dealing with positive integers+0 we can safely say that if the square root of a number is greater than another number to the power of n, then the number under square root must be greater than the number to the power of n. This conveys us that d>e.

NS

1+2) Both cases on statement 1) hold. (E)
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GRE 1: Q169 V154 Re: If d represents the hundredths digit and e represents the  [#permalink]

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ghostdude wrote:
If d represents the hundredths digit and e represents the thousandths digit in the decimal 0.4de, what is the value of this decimal rounded to the nearest tenth?

(1) d – e is equal to a positive perfect square.
(2) √d > e^2

Amazing Question
Let n=0.4de
Here when we round n to nearest thousandths we will get either=>
0.4 for d≤4
OR
0.5 for d≥5
hence in this question we are concerned about the value of d.
statement 1
d-e=positive perfect square
numerous values are possible
e.g=>
n=0.454
n=0.487
n=0.421
n=9.451
Hence insufficient as we can have d<5 so ≥5.
Statement 2
hmm since d and e are both positive and we can square on both sides of any inequality for which both of its sides are positive
we get=> d>e^4
hmm numerous values are possible
e.g=>
n=0.421
n=0.491
n=0.451
etc
hence insufficient
combining the two statements we can still get two-bound cases
n=0.421
n=0.451
hence E
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