A technician is installing two bundles of cables for an online system. Bundle A is 138 inches in diameter, and Bundle B is 234 inches in diameter. The installation requires a hole to be drilled with a diameter 25% larger than the diameter of the larger bundle. To the nearest tenth of a centimeter, what is the diameter of the hole that the technician must drill? (1 inch = 2.54 centimeters)A) 6.5 centimeters B) 7.2 centimeters C) 7.8 centimeters D) 8.7 centimeters

Accepted Solution

Answer:Answer D: 8.7 cmStep-by-step explanation:First compare the actual diameters of both cables to find out which one is larger. If we write both diameters in fraction form (as they are given), we need to have them expressed with the same denominator for a straight forward comparison: Bundle A: [tex]1\frac{3}{8} =1+\frac{3}{8} = \frac{8}{8} +\frac{3}{8} =\frac{11}{8}[/tex]Bundle B: [tex]2\frac{3}{4} =2+\frac{3}{4}=\frac{8}{4} +\frac{3}{4}=\frac{11}{4}[/tex]So we multiply this last fraction by 2 in numerator and denominator to obtain the same denominator as for Bundle A without actually changing its numerical value: [tex]\frac{11}{4} = \frac{22}{8}[/tex]So now we compare Bundle A with bundle B, and see that bundle B has the larger diameter: 22/8 inchesThe hole has to be 25% larger than this larger diameter, so we estimate 25% of 22/8 in is:[tex]\frac{22}{8} * \frac{25}{100} = 0.6875[/tex] inchesTherefore, the hole must be of diameter 22/8 in plus 0.6875 in = 3.4375 inNow we convert this value into centimeters by multiplying it by 2.54 (since one inch is 2.54 cm):3.4375 * 2.54 = 8.73125 cmwhich rounded to the nearest tenth as requested in the problem is: 8.7 cm