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Math Expert
Joined: 02 Sep 2009
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D
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Question Stats:
72% (01:12) correct 28% (01:39) wrong based on 145 sessions
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How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit? A. 1,339 B. 1,352 C. 1,353 D. 10,030 E. 10,300
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Math Expert
Joined: 02 Sep 2009
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Re M01-22 [#permalink]
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 16 Sep 2014, 00:15
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Official Solution:How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit? A. 1,339 B. 1,352 C. 1,353 D. 10,030 E. 10,300 Approach #1: There is one number in each hundred with 2 in the tens digit and 1 in the units digit: 21, 1 21, 2 21, 3 21, ... The difference between 324,700 and 458,600 is \(458,600-324,700=133,900\) - one number per each hundred gives \(\frac{133,900}{100}=1,339\) numbers. Approach #2: Look at the series of all numbers ending with 21 within the given range. - 324,721
- 324,821
- .....
- 458,421
- 458,521
Following the series, we can see that only the first four numbers are changing. Instead of counting the numbers ending in 21 in the given range (324,700 to 458,600), simplify and only count the numbers in the range of 3,247 to 4,585. Therefore, the number of integers ending in 21 within the range is \(4,585-3,247+1 = 1,339\). Answer: A
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Collection of Questions:
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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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Intern
Joined: 21 May 2015
Posts: 18
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Re M01-22 [#permalink]
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21 Aug 2017, 16:26
I think this is a high-quality question and I agree with explanation.
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Intern
Joined: 10 Mar 2015
Posts: 13
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Re: M01-22 [#permalink]
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09 Mar 2018, 11:33
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Check out the series below:
324,721
324,821
324,921
.....
458521
So, we could see that above series is in AP.
so, apply the formula and find out the number of terms.
first term=324,721
Common diff= 100
last term= 458521
458521=324721 + (n-1)100
this will give n=1339.
The right choice.
Please give kudos, if you really like my explanation.
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Manager
Joined: 26 Feb 2018
Posts: 67
Location: United Arab Emirates
GMAT 1: 710 Q47 V41 GMAT 2: 770 Q49 V47
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Re: M01-22 [#permalink]
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12 Mar 2018, 13:39
I think the answer choices make this question a bit too easy.
The question asks "How many numbers that are xxx,x21" - obviously there would be one number for every 100, (and 10 for every complete 1000). So the real question is how many hundreds between the two numbers? You can immediately see that the difference between the hundreds digit is 9, and there is only one answer choice that ends in a 9. That means you can go for this one immediately without any calculation.
You can also simply subtract the two numbers which gives you 133900. This is 1339 * 100 and you have 1 number per 100
I feel a real GMAT question would throw in some "trip up" answers into the options, like 139 or 13390 or 1329, etc
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Intern
Joined: 26 Feb 2018
Posts: 7
Location: India
WE: Web Development (Computer Software)
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Re M01-22 [#permalink]
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18 Apr 2018, 14:09
I think this is a high-quality question and I agree with explanation. This is a good quality question. The explanation provided is appropriate.
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