If k is a 3 digit positive number, what is the hundred digit

If k is a positive three-digit integer, what is the hundreds digit of k?

(1) The hundreds digit of K +150 is 4 --> \(400\leq{k+150}\leq{499}\) --> \(250\leq{k}\leq{349}\), not sufficient, as hundreds digit of k can be 2 or 3.

(2) The tens digit of K + 25 is 7 --> \(x70\leq{k+25}\leq{x79}\) --> \(x45\leq{k}\leq{x54}\). Clearly insufficient to get \(x\).

(1)+(2) Hundreds digit still can be 2 (for example if k=250) or 3 (for example if k=345). Not sufficient.

Basically we have that: \(250\leq{k}\leq{254}\) or \(345\leq{k}\leq{349}\).

Answer: E.
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In general, with problems like this one, just consider the EXTREMES of any possible range they throw at you.

If the hundreds digit of (k + 150) is 4, then take that to mean that
400 <= k + 150 <= 499
or
250 <= k <= 349.

Therefore, the hundreds digit of k can be either 2 or 3, so this statement is insufficient.

Note that you don't have to set up the inequalities to solve the problem: you can just notice that the biggest possible value for (k + 150) is 499, and that the smallest possible value for that expression is 400.

Hence E.

Let us assume that the k="ABC" => K= 100 (A) +10 (B) + C

According to statement 1: The hundred digit of k+150 is 4
K+150= 100 (A) +10 (B) + C + 150
k+150= 100 (A) + 100 + 10 (B) + 50 + C
K+150= 100 (A+1) +10 (B+5) + C

A+1=4 => A=3

Statement 1 is sufficient.

According to statement 2: The tens digit of k+25 is 7

K+25= 100 (A) +10 (B) + C + 25
= 100 (A) + 10 (B+2) + (C+5)

B+2=7 => B=5

Statement 2 is insufficient because it tell us nothing about the hundreds place



Answer is A

Red part is not correct. OA for this question is E, not A. Refer for the solution to my post above.
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Thank You for your response, I will go through the solution provided by you.

But I am interested to know why the red part highlighted by you is not correct?

Note that when you represent 3-digit number as \(100A+10B+C\): A, B and C must be digits from 0 to 9 (for A from 1 to 9).

Now, when you get \(K+150=100(A+1)+10(B+5)+C\) you can not say that it's a representation of 3-digit number where A+1 is hundreds digit, B+5 tens digit and C units digit as we don't know whether B+5 is in the range from 0 to 9, because if it's more than 9 then you'll have A+1 plus 1 hundreds: A+1+1=4 --> A=2.

Hope it's clear.
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Perfect! Thank You

I thought about possible "carry overs", i.e. adding an imaginary digit when you do the written method for addition and a sum of digits is >10.

Ex. 18
+ 23

= 41

Here 8+3=11-> 11 is 1 bigger than 10, so you carry the 1 over to the next "column".

Using this approach let us assess the 2 statements:

(1) No info about hundreds digit, thus insufficient.

(2) Assuming no carry over from the tens to the hundreds digit, the hundreds digit could be 3.
However, assuming a carry over of 1 from the tens to the hundreds digit, the hundreds digit could also be 2. Since more than 1 value for the hundreds digtit is possible, statement (2) is insufficient.

(1) & (2) Using the same reasoning from above, combining the 2 statements does not provide any new information. The hundreds digit could still be 3 or 2.

Hence, the answer is E.

If you liked this post, feel free to hit Kudos!

Hi,

here are my two cents for this question.


We need to determine the hundred's digit of a given number.

K= _ _ _

We are told that K+150 , the hundreds digit is 4, this means if there is no carry bought in by tens digit then the value is 3 else the value is 2

from statement 2 we have k+25 , the tens digit is 7 . Similarly if there is if there is no carry bought in by ones digit then the value is 5 else the value is 4

Combining 1 and 2 we have
if tens digit of k is 4 and k+150 has hundreds digit as 4 then hundreds digit of k is 3.

if tens digit of k is 5 and k+150 has hundreds digit as 4 then hundreds digit of k is 2.

So we cannot conclusively say if hundred's digit is 2 or 3

Hence E

Probus
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Hello, Bunuel!

We know that the tenths digit is a 4 or a 5 and that 250 <= K <= 349. Hence, K could equal anything from 250-259 or 340-349 and the hundreds digit could still be 2 or 3.

Why just 250≤k≤254 or 345≤k≤349 ?

Thank you!

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