How many integers are there between 25000 and 33000..

Question

How many integers are there between 25000 and 33000 that have 2 in the unit’s place and 3 in the ten’s place?

A. 79 
B. 80 
C. 81 
D. 78 
E. 76

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Ranasaymon · Answer

# of items between 25000 and 33000 that have 2 in the unit’s place and 3 in the ten’s place
Inclusive set: 25032 to 32932

The set is, 25032, 25132, 25232,....,32832,32932.
Here, the difference between terms is 100

So, the number of terms = (last-first)/difference +1
 =(32932-25032)/100 +1= 79+1=80

Ans. B

Gauravji21 · Answer

Numbers will be 25032, 25132,25232,..., 25932..Total 10 such numbers between 25000 to 26000
i.e. for every 1000 increase there are 10 such numbers.

Difference between 25000 to 33000 is 8000 and for every 1000 we have 10 such numbers.So, in total we will have 10 x 8 = 80 such numbers.

IanStewart · Answer

There are 8000 numbers between 25,001 and 33,000, and exactly 1% of them will have a specific tens and units digit (a number in this range is as likely to end in '32' as in any other two digits), so the answer is 80.

AliciaSierra · Answer

Hello Mr. Ian  IanStewart

Can you please explain  exactly 1% of them will have a specific tens and units digit  little bit more.

IanStewart · Answer

Among the numbers from 25,001 through 33,000, each tens and units digit combination occurs equally often. A number in that range is just as likely to end in "03" as it is in "55" or "97" or "32". Since there are 100 different tens and units digit combinations in total, 1/100 of the numbers in that range will end specifically in "32" (or in any other two specific digits). Hope that answers your question!

rsrighosh · Answer

25000 - 33000

So let us see number of digits

_ _ _ 3 2

Hundreds digit can have 10 digits (0 to 9)

Thousands digit has dependency with Ten Thousands digit

Case 1:-

if Ten Thousands digit is 2, then Thousands digit will have 5, 6, 7, 8, 9 = 5 digits

So, we have

2   (5 digits)   (10 digits)   3   2  which makes 5 x 10 = 50

Case 2:-

if Ten Thousands digit is 3, then Thousands digit will have 0, 1, 2 = 3 digits

So, we have

3   (3 digits)   (10 digits)   3   2  which makes 3 x 10 = 30

Adding both the cases..  50 + 30 = 80

B

Fdambro294 · Answer

Scenario 1: 3 in the 1st Place (Ten Thousands Place)

10,000s Place ---- 3 Anchored ---- 1 Option

and

1,000s Place ---- Options 0 or 1 or 2 (33, 000 is the MAX) ----- 3 Options

and

100s Place ----- can take any Digit ----- 10 Options

and

Tens Place ---- Anchored 3 Digit ---- 1 Option

and

Units Place ---- Anchored 2 Digit ---- 1 Option

1 * 3 * 10 * 1 * 1 = 30 Numbers

OR

Scenario 2: 2 in the Ten Thousands Place Anchored

10,000s Place ---- 2 Anchored ---- 1 Option

and

1,000s Place ---- since Lower Boundary is 25, 000: Available Options are 5 or 6 or 7 or 8 or 9 ---- 5 Options

and

100s Place --- any Digit is Possible ---- 10 Options

and

Tens Place --- 3 Digit Anchored --- 1 Option

and

Units Place ---- 2 Digit Anchored ---- 1 Options

1 * 5 * 10 * 1 * 1 = 50 Numbers

Scenario 1 + Scenario 2 =

30 + 50 = 80 Possible Numbers

-B-

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How many integers are there between 25000 and 33000..

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How many integers are there between 25000 and 33000..

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