Olympiad Maths Prep

Track / Stage 3 / 77 of 260 #77 of 2000

Problem 77

AMC 10/12, early questions
Algebra Difficulty 3.2 Find the answer

Which of the following is closest to 6563\sqrt{65}-\sqrt{63}?
(A) .12(B) .13(C) .14(D) .15(E) .16\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16

Official solution

We have 65>8>7.5\sqrt{65} > 8 > 7.5. Also 7.52=(7+0.5)2=72+270.5+0.52=49+7+0.25=56.257.57.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 7.5. Thus 65+63>7.5+7.5=15\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15. Now notice that 6563=(6563)(65+63)65+63=265+63\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}, so 65630.125    6526563+63>0.015625    1280.015625>24095    4095<640.0078125    4095<40961280.0078125+0.00781252=40961+0.00781252\sqrt{65} - \sqrt{63} 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2 which is true. Hence as the expression is greater than 0.1250.125, and less than or equal to 0.130.13 (since we showed it is certainly less than 0.1333333...0.1333333...), it is closest to 0.130.13, which is answer B\boxed{\text{B}}.
(6563\sqrt{65} - \sqrt{63} is approximately equal to 0.1250038150.125003815)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.