sherxon wrote:
ab and ba are two-digit numbers. Find these numbers.
(1) ab/ba = 1.75
(2) ab*a = 3.5*ba
There's a notation problem in the question (it's not a mathematical convention that a and b are the two digits of a number ab when ab is underlined). It would be reasonable for a test taker to guess that ab is a product of a and b here, which is not what is intended. What is the source?
For Statement 1, we can write the two digit number AB as 10A + B, and similarly BA = 10B + A. We then learn
(10A + B)/(10B + A) = 7/4
40A + 4B = 70B + 7A
33A = 66B
A = 2B
and AB can be 21, 42, 63 or 84.
For Statement 2, we have the equation
A*(AB) = (7/2)*(BA)
where A and B are the digits of the two-digit numbers AB and BA. The left side of this equation must be an integer, so the right side must be too. So that '2' needs to cancel out, which can only happen if BA is even, and that means the units digit of BA, which is A, is even. We also have a factor of '7' on the right side, so we must have a factor of 7 on the left side. If A is even, and is a single digit number, A can't be divisible by 7. So AB must be divisible by 7.
Finally, you could notice in the equation A*(AB) = 3.5*(BA) that if A < 3.5, then AB must be larger than BA (otherwise both numbers we're multiplying on the left are smaller than both numbers we're multiplying on the right). If AB > BA, then A > B. If A < 3.5, and A is even, A must be 2, and AB can only be a multiple of 7 with a units digit less than its tens digit, so can only be 21.
Conversely, if A > 3.5, then BA must be larger than AB, and B > A. So AB needs to start with 4, 6 or 8, and AB needs to be a multiple of 7 with a units digit larger than its tens digit. There aren't any numbers like that, so 21 is the only possible value of AB, and Statement 2 is sufficient.