The number x is two-digit positive integer, x^2 is a four-digit intege

Correction
Conditions not satisfy for 94.

Posted from my mobile device

10^2 = 100
90^2 = 8100
95^2 = 9025
100^2 = 10000
Hence option D is right answer.

The condition in the question is true for any integer greater than square root of 9000 and less than 100. This condition is satisfied for 95, 96, 97 98 and 99. Hence D is the right answer,

Correction
I thought sub option lll is x^2 has unit digit atleast 5
But it is asked x has unit digit at least 5
Hence answer is option D.

Posted from my mobile device

When we have such questions, we need to check if the question asks which statements can be true or which must be true.

If it asks for must be true as in the case above, then under no condition can the statement be false.

If it asks for can be true, we need to test for a single condition where it would be true (even if other conditions do not satisfy the statements)



We have the following data

(a) A 2 digit number (say x), who's square is a 4 digit number. So 32 \(\leq\) x < 100

(b) The thousands place of \(x^2\) is 9. So, x can be either 95, 96, 97, 98 or 99



Statement I: Tens digit of x is 9. This is always true for the numbers listed.



Statement II: This requires calculation and an area where time can be lost.

If we glance through the options, since statement I is true, Options A, B and E are eliminated.

If statement III is true, then the answer is D, else Option C.



Statement III: The units digit of x is at least 5 (i.e the last digit can be 5, 6, 7, 8 or 9).

This is always true.




Option D

Arun Kumar

x is a two digit number and x^2 is a 4 digit number something like 9abc i.e. x^2 is 9000 something.
Now note that 100^2 = 10,000 and 90^2 = 8100. So x will lie between 90 and 100.
Let's narrow down the range further. We can easily calculate the squares of numbers ending in 5.
95^2 = 9025 (We write 25 at the end and multiply the previous leftover number with its successor i.e. 9*10 = 90. So we get 9025)

Certainly, this would be the smallest perfect square greater than 9000. So x has 5 possible values from 95 to 99.

So we see that statements I and III are certainly true.

Answer (D)


Statement II is a bit unwieldy.
It is possible that the sum of digits of the square may not be greater than the sum of digits of x. We will need to check the squares of all numbers from 95 to 99. But since the options do not have I, II and III as one of the options, we don't need to do this extra work.
Though there are some tips that we can use to make it easier if we were required to check that too.
96^2 would have 9 in the beginning and 6 at the end. In the middle it can't be two 0s (because 9006 will not be the square of 96 since 95^2 is 9025) so sum of digits of the square will be greater.
97^2 will have 9 in the beginning and 9 at the end (because 7^2 is 49) so sum of square will certainly be greater.
98^2 will have 9 in the beginning and 4 at the end but will be above 9500 so it will have at least another 5 in the middle so sum of digits of the square will be greater.
99^2 will have 9 in the beginning and 1 at the end. It will need to be found out but we can use (100 - 1)^2 = 10,000 + 1 - 200 = 9801 to get its value.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.