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Re M3215
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18 Jul 2017, 03:32
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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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 (9x). GivenWhen 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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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 BunuelCan 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: M3215
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26 Oct 2017, 20:43
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 BunuelCan 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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Re: M3215
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06 Jun 2018, 16:20
Hi BunuelI 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: M3215
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08 Jun 2018, 02:23



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Re: M3215
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08 Jun 2018, 06:02
Thanks Bunuel! I think the first response is a very helpful way to go about it.



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Re: M3215
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11 Sep 2018, 04:34
Bunuel wrote: manishtank1988 wrote: Bunuel wrote: Official Solution: 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:
Hi Bunuel, can you please help me with the reason for inclusion and exclusion. thanks.



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Re: M3215
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11 Sep 2018, 04:41
harsh8686 wrote: Bunuel wrote: Official Solution: 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:
Hi Bunuel, can you please help me with the reason for inclusion and exclusion. thanks. 5.35 rounded to the nearest tenth is 5.4 but 5.45 rounded to the nearest tenth is 5.5. 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. So, according to the above 8.35y rounded to the nearest tenth will be 8.4 irrespective of the value of y. For mote on this check the following posts: Math: Number TheoryRounding Rules on the GMAT: Slip to the Side and Look for a Five!3. Fractions, Decimals, Ratios and Proportions For more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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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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