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Amusements in Mathematics

SOLUTIONS 101-200
101.—THE TRUSSES OF HAY.—solution
Add together the ten weights and divide by 4, and we get 289 lbs. as the weight of the
five trusses together. If we call the five trusses in the order of weight A, B, C, D, and E,
the lightest being A and the heaviest E, then the lightest, no lbs., must be the weight of A
and B; and the next lightest, 112 lbs., must be the weight of A and C. Then the two
heaviest, D and E, must weigh 121 lbs., and C and E must weigh 120 lbs. We thus know
that A, B, D, and E weigh together 231 lbs., which, deducted from 289 lbs. (the weight of
the five trusses), gives us the weight of C as 58 lbs. Now, by mere subtraction, we find
the weight of each of the five trusses—54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62 lbs.
respectively.
102.—MR. GUBBINS IN A FOG.—solution
The candles must have burnt for three hours and three-quarters. One candle had one-
sixteenth of its total length left and the other four-sixteenths.
103.—PAINTING THE LAMP-POSTS.—solution
Pat must have painted six more posts than Tim, no matter how many lamp-posts there
were. For example, suppose twelve on each side; then Pat painted fifteen and Tim nine. If
a hundred on each side, Pat painted one hundred and three, and Tim only ninety-seven
104.—CATCHING THE THIEF.—solution
The constable took thirty steps. In the same time the thief would take forty-eight, which,
added to his start of twenty-seven, carried him seventy-five steps. This distance would be
exactly equal to thirty steps of the constable.
105.—THE PARISH COUNCIL ELECTION,—solution
The voter can vote for one candidate in 23 ways, for two in 253 ways, for three in 1,771,
for four in 8,855, for five in 33,649, for six in 100,947, for seven in 245,157, for eight in
490,314, and for nine candidates in 817,190 different ways. Add these together, and we
get the total of 1,698,159 ways of voting.
 
 
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