Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
The Competition Continues - Game of Timers is a team-based competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th
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 \(4,585-3,247+1 = 1,339\).
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.
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
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
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
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 \(4,585-3,247+1 = 1,339\).
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 _________________
Everything will fall into place…
There is perfect timing for everything and everyone. Never doubt, But Work on improving yourself, Keep the faith and Stay ready. When it’s finally your turn, It will all make sense.
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 \(4,585-3,247+1 = 1,339\).
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.
_________________
close
78% (01:48) correct 
