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M01-22

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M01-22  [#permalink]

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New post 16 Sep 2014, 00:15
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Re M01-22  [#permalink]

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New post 16 Sep 2014, 00:15
12
5
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, 121, 221, 321, ...

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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Re M01-22  [#permalink]

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New post 21 Aug 2017, 16:26
I think this is a high-quality question and I agree with explanation.
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Re: M01-22  [#permalink]

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New post 09 Mar 2018, 11:33
3
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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Re: M01-22  [#permalink]

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New post 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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Re M01-22  [#permalink]

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New post 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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Re: M01-22  [#permalink]

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New post 15 Oct 2018, 07:33
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chetan2u, Bunuel

I was trying to solve this question by 324721 to 458521 .. where last 2 digits are fixed.
From 324721 to 399999 -- we can have only 8 (2 to 9) *6(4 to 9) * 3 (7 to 9) = 144 numbers ( as 3 and last 2 digits are fixed).
Similarly from 400000 to 458521 .. we can have 6(0 to 5 for ten thousands digit) * 9 * 6=324 ways.
so total 144+324 =468 ways.
Although i understood the answer for the question given... i did not understand through my above approach where am i goiing wrong. could you please help with this
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Re: M01-22  [#permalink]

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New post 15 Oct 2018, 09:19
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R999 wrote:
chetan2u, Bunuel

I was trying to solve this question by 324721 to 458521 .. where last 2 digits are fixed.
From 324721 to 399999 -- we can have only 8 (2 to 9) *6(4 to 9) * 3 (7 to 9) = 144 numbers ( as 3 and last 2 digits are fixed).
Similarly from 400000 to 458521 .. we can have 6(0 to 5 for ten thousands digit) * 9 * 6=324 ways.
so total 144+324 =468 ways.
Although i understood the answer for the question given... i did not understand through my above approach where am i goiing wrong. could you please help with this



Hi...
You miss out on a lot of combinations this way..
Just one example..
From 324721 to 399999 -- we can have only 8 (2 to 9) *6(4 to 9) * 3 (7 to 9) = 144 numbers ( as 3 and last 2 digits are fixed).
So you have taken hundreds place as only 7 to 9..
But if you have a number 325121.. this is a valid number but in your approach is not valid because of you taken hundreds as only 7, 8 or 9..
You could this by taking..
324700 to 324999....1*1*1*3*1*1=3
325000 to 329999....1*1*5*10*1*1=50
330000 to 399999..1* 7*10*10*1*1=700
400000 to 449999...1*5*10*10*1*1=500
450000 to 457999....1*1*8*10*1*1=80
458000 to 458600.. 1*1*1*6*1*1=6

Total ... 3+50+700+500+80+6=1339

A
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Re: M01-22  [#permalink]

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New post 15 Oct 2018, 10:30
chetan2u,

Thank you for such quick and clear reply
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M01-22  [#permalink]

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New post 22 Mar 2019, 17:34
Bunuel wrote:
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, 121, 221, 321, ...

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


Bunuel chetan2u

Can you explain why in the second explanation above range is considered till 4585 and not till 4586?

"count the numbers in the range of 3,247 to 4,585"

Thanks.

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Re: M01-22  [#permalink]

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New post 22 Mar 2019, 19:01
1
Helium wrote:
Bunuel wrote:
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, 121, 221, 321, ...

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


Bunuel chetan2u

Can you explain why in the second explanation above range is considered till 4585 and not till 4586?

"count the numbers in the range of 3,247 to 4,585"

Thanks.

Posted from my mobile device


Hi..

We are looking for numbers till 458600 and ending with 21..
The numbers are abcd21, abcd is the four digits sucht that abcd21 remains in the range 324700<abcd21<458600.
The moment you go above 4585__, it becomes 458621, which is>458600.
So you go uptil 4585 only.
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Re: M01-22   [#permalink] 22 Mar 2019, 19:01
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