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M32-15

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M32-15 [#permalink]

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New post 18 Jul 2017, 04:32
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If \(x\) is the tenths digit in the decimal \(9.x5\), what is the value of \(x\)?



(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is \(5.4\).

(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is \(4.7\).

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Re M32-15 [#permalink]

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New post 18 Jul 2017, 04:32
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1
Official Solution:


If \(x\) is the tenths digit in the decimal \(9.x5\), what is the value of \(x\)?

(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is \(5.4\).

This implies that \(15 - 9.x5\) must be between \(5.35\) (inclusive) and \(5.45\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(5.4\). So, we can write the following inequality:

\(5.35 \leq (15 - 9.x5) < 5.45\);

Subtract 15 from all parts: \(-9.65 \leq -9.x5 < -9.55\);

Multiply by -1 and flip the signs: \(9.65 \geq 9.x5 > 9.55\), which is the same as \(9.55 < 9.x5 \leq 9.65\). From this we can deduce that \(x\) can be only \(6\). Sufficient.

(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is \(4.7\).

This implies that \(9.x5 – 5\) must be between \(4.65\) (inclusive) and \(4.75\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(4.7\). So, we can write the following inequality:

\(4.65 \leq (9.x5 – 5) < 4.75\);

Add 5 to all parts: \(9.65 \leq 9.x5 < 9.75\). From this we can deduce that \(x\) can be only \(6\). Sufficient.


Answer: D
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M32-15 [#permalink]

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New post 03 Aug 2017, 19:07
Hi Bunuel,

Please can you tell me if my below approach is correct.

15−9.x5 -->
15.00
- 9.x5
______
5.y5
where y is (9-x).
Given-When 15−9.x515−9.x5 is rounded to the nearest tenth, the result is 5.4-->5.x5=5.35

Thanks,
CP
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M32-15 [#permalink]

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New post 26 Oct 2017, 12:57
Bunuel wrote:
Official Solution:


If \(x\) is the tenths digit in the decimal \(9.x5\), what is the value of \(x\)?

(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is \(5.4\).

This implies that \(15 - 9.x5\) must be between \(5.35\) (inclusive) and \(5.45\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(5.4\). So, we can write the following inequality:

\(5.35 \leq (15 - 9.x5) < 5.45\);

Subtract 15 from all parts: \(-9.65 \leq -9.x5 < -9.55\);

Multiply by -1 and flip the signs: \(9.65 \geq 9.x5 > 9.55\), which is the same as \(9.55 < 9.x5 \leq 9.65\). From this we can deduce that \(x\) can be only \(6\). Sufficient.

(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is \(4.7\).

This implies that \(9.x5 – 5\) must be between \(4.65\) (inclusive) and \(4.75\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(4.7\). So, we can write the following inequality:

\(4.65 \leq (9.x5 – 5) < 4.75\);

Add 5 to all parts: \(9.65 \leq 9.x5 < 9.75\). From this we can deduce that \(x\) can be only \(6\). Sufficient.


Answer: D



Hello Bunuel
Can you please explain/elaborate as to how you arrived at highlighted statements?
Thanks

This is what i did -
1)
15.00
- 9.x5
5.y5 = 5.4
y = [9 - (x+1)] ; (x+1) because we have 5.x5 ~ 5.4. So whatever value we get would be +1 because of 0.05 - i hope i am clarifying myself....

9 - x - 1 = 4
8 - x = 4
x = 4

2)
9.x5
- 5.00
4.y5 ~ 4.7
y = x+1 = 7
x = 6

Thus D
Please let me know how did you arrived at highlighted portion and what i did was wrong?
Thanks
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Re: M32-15 [#permalink]

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New post 26 Oct 2017, 21:43
manishtank1988 wrote:
Bunuel wrote:
Official Solution:


If \(x\) is the tenths digit in the decimal \(9.x5\), what is the value of \(x\)?

(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is \(5.4\).

This implies that \(15 - 9.x5\) must be between \(5.35\) (inclusive) and \(5.45\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(5.4\). So, we can write the following inequality:

\(5.35 \leq (15 - 9.x5) < 5.45\);

Subtract 15 from all parts: \(-9.65 \leq -9.x5 < -9.55\);

Multiply by -1 and flip the signs: \(9.65 \geq 9.x5 > 9.55\), which is the same as \(9.55 < 9.x5 \leq 9.65\). From this we can deduce that \(x\) can be only \(6\). Sufficient.

(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is \(4.7\).

This implies that \(9.x5 – 5\) must be between \(4.65\) (inclusive) and \(4.75\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(4.7\). So, we can write the following inequality:

\(4.65 \leq (9.x5 – 5) < 4.75\);

Add 5 to all parts: \(9.65 \leq 9.x5 < 9.75\). From this we can deduce that \(x\) can be only \(6\). Sufficient.


Answer: D



Hello Bunuel
Can you please explain/elaborate as to how you arrived at highlighted statements?
Thanks

This is what i did -
1)
15.00
- 9.x5
5.y5 = 5.4
y = [9 - (x+1)] ; (x+1) because we have 5.x5 ~ 5.4. So whatever value we get would be +1 because of 0.05 - i hope i am clarifying myself....

9 - x - 1 = 4
8 - x = 4
x = 4

2)
9.x5
- 5.00
4.y5 ~ 4.7
y = x+1 = 7
x = 6

Thus D
Please let me know how did you arrived at highlighted portion and what i did was wrong?
Thanks


Take any number from that range, round it to the nearest tenth and you will be \(5.4\).
_________________

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M32-15 [#permalink]

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New post 06 Jun 2018, 17:20
Hi Bunuel

I am a bit confused by this explanation. Is there any easier way to deduce this?

I just assumed D because if we know that the hundredths digit ends in a five, then we know we must round the tenth digit up. Thus, we should be able to determine the tenths digit as a result. Is my logic flawed?

Thanks in advance for your help.
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Re: M32-15 [#permalink]

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New post 08 Jun 2018, 03:23
1
norovers wrote:
Hi Bunuel

I am a bit confused by this explanation. Is there any easier way to deduce this?

I just assumed D because if we know that the hundredths digit ends in a five, then we know we must round the tenth digit up. Thus, we should be able to determine the tenths digit as a result. Is my logic flawed?

Thanks in advance for your help.


This is a tough question and does not have a silver bullet solution. You can check alternative solutions here: https://gmatclub.com/forum/if-x-represe ... 43252.html

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: M32-15 [#permalink]

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New post 08 Jun 2018, 07:02
Thanks Bunuel! I think the first response is a very helpful way to go about it.
Re: M32-15   [#permalink] 08 Jun 2018, 07:02
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