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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
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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?
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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 :)
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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.

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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

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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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