vomhorizon wrote:

What two-digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5?

(A) 95

(B) 99

(C) 26

(D) 73

(E) None of the Above

Use the answer choices.

Arithmetically, the operations on the given numbers are easy.

And there are two conditions. It's likely that the first one will eliminate at least one choice.

Start with just the first condition: the number is "less than the sum of the square of its digits by 11."

Translate: Square each digit. Add the results. Subtract 11. That should be the number.

(A) 95

\(9^2 = 81\)

\(5^2 = 25\)

Sum is 106. (196 - 11 ) = 95 KEEP

(B) 99

\(9^2 = 81\)

\(9^2 = 81\)

Sum is 162. STOP. Not even close to a difference of 11. REJECT

(C) 26

\(2^2 = 4\)

\(6^2 = 36\)

Sum is 40. (40 - 11) = 29 REJECT

(D) 73

\(7^2 = 49\)

\(3^2 = 6\)

Sum is 55. STOP. 55 < 73, let alone (55 - 11). REJECT

(E) None of the Above.

KEEP FOR NOW

One number, 95, satisfies first condition.

Second condition? It "must exceed [the digits'] doubled product by 5."

Muktiply digits, double, add 5:

2(9*5) = 90. Add 5.

90 + 5 = 95. KEEP

Answer A

_________________

The only thing more dangerous than ignorance is arrogance.

-- Albert Einstein