Lesson Classic entertainment problems
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<H2>Classic entertainment problems</H2> Well known old, FAMOUS and legendary entertainment problems are collected in this lesson. Each generation re-opens them again and again. <H3>Problem 1</H3>If 3 hens lay 3 eggs in 3 days so 6 hens lay how many eggs in 6 days? <B>Solution</B> I think that the best service I can make for you regarding this problem is to leave it to you solve it on your own . . . <H3>Problem 2</H3>If a grandfather clock takes 30 seconds to strike six, how long does it take to strike 12? <B>Solution</B> Wikipedia article https://en.wikipedia.org/wiki/Striking_clock says: <pre> A striking clock (also known as chiming clock) is a clock that sounds the hours audibly on a bell or gong. In 12-hour striking, used most commonly in striking clocks today, the clock strikes once at one a.m., twice at two a.m., continuing in this way up to twelve times at 12 noon, then starts again, striking once at one p.m., twice at two p.m., up to twelve times at 12 midnight. </pre> Armed with this knowledge, we are ready to solve the problem. The striking clock, when striking six, has 5 (five) time intervals between the strikes. It means that each such time interval is {{{30/5}}} = 6 seconds. 12 strikes have 11 time intervals between the very first and very last strikes. Hence, it will take (6 seconds * 11 intervals) = 66 seconds from the first strike to the last one. <H3>Problem 3</H3>A roll of news paper of length 7000 m and thickness 0.022 cm is rolled into a solid cylinder. Find its radius. <B>Solution</B> <A HREF=https://www.algebra.com/algebra/homework/Bodies-in-space/Bodies-in-space.faq.question.1100341.html>https://www.algebra.com/algebra/homework/Bodies-in-space/Bodies-in-space.faq.question.1100341.html</A> <B>Solution</B> <pre> Let x be the width in meters of the roll of newspaper. We need to use consistent units; so call the thickness of the newspaper 0.00022 m. Spread out into a rectangular sheet, the newspaper is a very thin rectangular solid with length 7000 m, width x m, and thickness 0.00022 m. The volume is then 1.54x cubic meters. When rolled into a cylinder, the volume will be the same. The volume of a cylinder is (pi)*(radius squared)*(height); the height of the cylinder is the width of the newspaper. So {{{1.54x}}} = {{{pi*r^2*x}}} {{{r^2}}} = {{{1.54/pi}}} = {{{0.49}}} (approximately... 0.490197....) {{{r}}} = {{{sqrt(0.49)}}} = {{{0.7}}} The radius of the roll of newspaper is approximately 0.7 m. </pre> <H3>Problem 4</H3>A cake shaped like a rectangular prism with sides of 9cm by 14 cm by 15cm is completely dipped in chocolate and then cut into small 1 cm x 1 cm x 1 cm cubes. How many of the cubes have chocolate on at least one side? How many of the cubes have chocolate on exactly one side? How many of the cubes have no chocolate on their sides? <B>Solution</B> <pre> According to the condition, the entire/(the whole) cake is cut into small (1 cm X 1 cm x 1 cm) cubes. The total number of these small cubes is 9*14*15 = 1890. Of them, have chocolate on at least one side those and only those cubes that are in one slice adjacent to some (to any) face of the prism. The rest of the small cubes HAVE NO chocolate on any small faces. Those small cubes that have no chocolate on any small face are ALL INTERNAL cubes. They form smaller prism whose dimensions are 2 cm less than dimensions of the whole prism. Namely, these dimensions are (9-2) = 7 cm, (14-2) = 12 cm and (15-2) = 13 cm. So, the number of small INTERNAL cubes is 7*12*13 = 1092. Thus the number of cubes that have chocolate <U>at least on one side</U> is 1890 - 1092 = 798. The number of small cubes that have chocolate <U>exactly on one side</U> is 2*(7*12 + 7*13 + 12*13) = 662. <U>Answer</U>. The number of small cubes that have chocolate <U>at least on one side</U> is 798. The number of small cubes that have chocolate <U>exactly on one side</U> is 662. </pre> <H3>Problem 5</H3>Among the children in a family, each boy has as many sisters as brothers, but each girl has only half as many sisters as brothers. Find the number of children in the family. <B>Answer</B>. 4 boys and 3 girls. 7 children, in all. <B>Solution</B> <pre> Let b be the number of boys and g be the number of girls. Then b - 1 = g (1) ("each boy has as many sisters as brothers") 2*(g-1) = b (2) ("each girl has only half as many sisters as brothers") First equation says "each boy has b-1 brothers, and this number is equal to the number of sisters". Second equation says "each girl has g-1 sisters, and .. . and so on . . . 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